3.1358 \(\int \frac {\csc ^3(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=332 \[ \frac {b^2 \sec ^3(c+d x)}{3 a^3 d}+\frac {b^2 \sec (c+d x)}{a^3 d}-\frac {b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {b \tan ^3(c+d x)}{3 a^2 d}-\frac {2 b \tan (c+d x)}{a^2 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {2 b^7 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}+\frac {b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 d \left (a^2-b^2\right )}-\frac {b^3 \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a^3 d \left (a^2-b^2\right )^2}+\frac {5 \sec ^3(c+d x)}{6 a d}+\frac {5 \sec (c+d x)}{2 a d}-\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d} \]

[Out]

-2*b^7*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/(a^2-b^2)^(5/2)/d-5/2*arctanh(cos(d*x+c))/a/d-b^2*
arctanh(cos(d*x+c))/a^3/d+b*cot(d*x+c)/a^2/d+5/2*sec(d*x+c)/a/d+b^2*sec(d*x+c)/a^3/d+5/6*sec(d*x+c)^3/a/d+1/3*
b^2*sec(d*x+c)^3/a^3/d-1/2*csc(d*x+c)^2*sec(d*x+c)^3/a/d+1/3*b^3*sec(d*x+c)^3*(b-a*sin(d*x+c))/a^3/(a^2-b^2)/d
-1/3*b^3*sec(d*x+c)*(3*b^3+a*(2*a^2-5*b^2)*sin(d*x+c))/a^3/(a^2-b^2)^2/d-2*b*tan(d*x+c)/a^2/d-1/3*b*tan(d*x+c)
^3/a^2/d

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Rubi [A]  time = 0.55, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {2898, 2622, 302, 207, 2620, 270, 288, 2696, 2866, 12, 2660, 618, 204} \[ -\frac {2 b^7 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}+\frac {b^2 \sec ^3(c+d x)}{3 a^3 d}+\frac {b^2 \sec (c+d x)}{a^3 d}-\frac {b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 d \left (a^2-b^2\right )}-\frac {b^3 \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a^3 d \left (a^2-b^2\right )^2}-\frac {b \tan ^3(c+d x)}{3 a^2 d}-\frac {2 b \tan (c+d x)}{a^2 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {5 \sec ^3(c+d x)}{6 a d}+\frac {5 \sec (c+d x)}{2 a d}-\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^3*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]

[Out]

(-2*b^7*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/2)*d) - (5*ArcTanh[Cos[c + d*x]]
)/(2*a*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a^3*d) + (b*Cot[c + d*x])/(a^2*d) + (5*Sec[c + d*x])/(2*a*d) + (b^2*S
ec[c + d*x])/(a^3*d) + (5*Sec[c + d*x]^3)/(6*a*d) + (b^2*Sec[c + d*x]^3)/(3*a^3*d) - (Csc[c + d*x]^2*Sec[c + d
*x]^3)/(2*a*d) + (b^3*Sec[c + d*x]^3*(b - a*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)*d) - (b^3*Sec[c + d*x]*(3*b^3 +
a*(2*a^2 - 5*b^2)*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)^2*d) - (2*b*Tan[c + d*x])/(a^2*d) - (b*Tan[c + d*x]^3)/(3*
a^2*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 2898

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rubi steps

\begin {align*} \int \frac {\csc ^3(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \left (\frac {b^2 \csc (c+d x) \sec ^4(c+d x)}{a^3}-\frac {b \csc ^2(c+d x) \sec ^4(c+d x)}{a^2}+\frac {\csc ^3(c+d x) \sec ^4(c+d x)}{a}-\frac {b^3 \sec ^4(c+d x)}{a^3 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac {\int \csc ^3(c+d x) \sec ^4(c+d x) \, dx}{a}-\frac {b \int \csc ^2(c+d x) \sec ^4(c+d x) \, dx}{a^2}+\frac {b^2 \int \csc (c+d x) \sec ^4(c+d x) \, dx}{a^3}-\frac {b^3 \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx}{a^3}\\ &=\frac {b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 \left (a^2-b^2\right ) d}+\frac {b^3 \int \frac {\sec ^2(c+d x) \left (-2 a^2+3 b^2-2 a b \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 a^3 \left (a^2-b^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{a d}-\frac {b \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d}+\frac {b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 \left (a^2-b^2\right ) d}-\frac {b^3 \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 d}-\frac {b^3 \int \frac {3 b^4}{a+b \sin (c+d x)} \, dx}{3 a^3 \left (a^2-b^2\right )^2}+\frac {5 \operatorname {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 a d}-\frac {b \operatorname {Subst}\left (\int \left (2+\frac {1}{x^2}+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {b^2 \operatorname {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=\frac {b \cot (c+d x)}{a^2 d}+\frac {b^2 \sec (c+d x)}{a^3 d}+\frac {b^2 \sec ^3(c+d x)}{3 a^3 d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d}+\frac {b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 \left (a^2-b^2\right ) d}-\frac {b^3 \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 d}-\frac {2 b \tan (c+d x)}{a^2 d}-\frac {b \tan ^3(c+d x)}{3 a^2 d}-\frac {b^7 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )^2}+\frac {5 \operatorname {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{2 a d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=-\frac {b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {5 \sec (c+d x)}{2 a d}+\frac {b^2 \sec (c+d x)}{a^3 d}+\frac {5 \sec ^3(c+d x)}{6 a d}+\frac {b^2 \sec ^3(c+d x)}{3 a^3 d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d}+\frac {b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 \left (a^2-b^2\right ) d}-\frac {b^3 \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 d}-\frac {2 b \tan (c+d x)}{a^2 d}-\frac {b \tan ^3(c+d x)}{3 a^2 d}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 a d}-\frac {\left (2 b^7\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right )^2 d}\\ &=-\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {5 \sec (c+d x)}{2 a d}+\frac {b^2 \sec (c+d x)}{a^3 d}+\frac {5 \sec ^3(c+d x)}{6 a d}+\frac {b^2 \sec ^3(c+d x)}{3 a^3 d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d}+\frac {b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 \left (a^2-b^2\right ) d}-\frac {b^3 \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 d}-\frac {2 b \tan (c+d x)}{a^2 d}-\frac {b \tan ^3(c+d x)}{3 a^2 d}+\frac {\left (4 b^7\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right )^2 d}\\ &=-\frac {2 b^7 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2} d}-\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {5 \sec (c+d x)}{2 a d}+\frac {b^2 \sec (c+d x)}{a^3 d}+\frac {5 \sec ^3(c+d x)}{6 a d}+\frac {b^2 \sec ^3(c+d x)}{3 a^3 d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d}+\frac {b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 \left (a^2-b^2\right ) d}-\frac {b^3 \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 d}-\frac {2 b \tan (c+d x)}{a^2 d}-\frac {b \tan ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 6.27, size = 947, normalized size = 2.85 \[ 16 \left (-\frac {\tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right ) \csc (c+d x) (a+b \sin (c+d x)) b^7}{8 a^3 \left (a^2-b^2\right )^{5/2} d (b+a \csc (c+d x))}+\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc (c+d x) (a+b \sin (c+d x)) b}{32 a^2 d (b+a \csc (c+d x))}-\frac {\csc (c+d x) (a+b \sin (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right ) b}{32 a^2 d (b+a \csc (c+d x))}+\frac {\csc (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \sin (c+d x))}{128 a d (b+a \csc (c+d x))}+\frac {\left (-5 a^2-2 b^2\right ) \csc (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sin (c+d x))}{32 a^3 d (b+a \csc (c+d x))}+\frac {\left (5 a^2+2 b^2\right ) \csc (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sin (c+d x))}{32 a^3 d (b+a \csc (c+d x))}+\frac {\csc (c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \sin (c+d x))}{96 (a+b) d (b+a \csc (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\csc (c+d x) \left (16 b \sin \left (\frac {1}{2} (c+d x)\right )-13 a \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sin (c+d x))}{96 (a-b)^2 d (b+a \csc (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\csc (c+d x) \left (13 a \sin \left (\frac {1}{2} (c+d x)\right )+16 b \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sin (c+d x))}{96 (a+b)^2 d (b+a \csc (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right ) \csc (c+d x) (a+b \sin (c+d x))}{128 a d (b+a \csc (c+d x))}+\frac {a \left (13 a^2-19 b^2\right ) \csc (c+d x) (a+b \sin (c+d x))}{96 \left (a^2-b^2\right )^2 d (b+a \csc (c+d x))}+\frac {\csc (c+d x) (a+b \sin (c+d x))}{192 (a+b) d (b+a \csc (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\csc (c+d x) (a+b \sin (c+d x))}{192 (a-b) d (b+a \csc (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {\csc (c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \sin (c+d x))}{96 (a-b) d (b+a \csc (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^3*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]

[Out]

16*((a*(13*a^2 - 19*b^2)*Csc[c + d*x]*(a + b*Sin[c + d*x]))/(96*(a^2 - b^2)^2*d*(b + a*Csc[c + d*x])) - (b^7*A
rcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]]*Csc[c + d*x]*(a + b*Sin[c
+ d*x]))/(8*a^3*(a^2 - b^2)^(5/2)*d*(b + a*Csc[c + d*x])) + (b*Cot[(c + d*x)/2]*Csc[c + d*x]*(a + b*Sin[c + d*
x]))/(32*a^2*d*(b + a*Csc[c + d*x])) - (Csc[(c + d*x)/2]^2*Csc[c + d*x]*(a + b*Sin[c + d*x]))/(128*a*d*(b + a*
Csc[c + d*x])) + ((-5*a^2 - 2*b^2)*Csc[c + d*x]*Log[Cos[(c + d*x)/2]]*(a + b*Sin[c + d*x]))/(32*a^3*d*(b + a*C
sc[c + d*x])) + ((5*a^2 + 2*b^2)*Csc[c + d*x]*Log[Sin[(c + d*x)/2]]*(a + b*Sin[c + d*x]))/(32*a^3*d*(b + a*Csc
[c + d*x])) + (Csc[c + d*x]*Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x]))/(128*a*d*(b + a*Csc[c + d*x])) + (Csc[c +
d*x]*(a + b*Sin[c + d*x]))/(192*(a + b)*d*(b + a*Csc[c + d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (Cs
c[c + d*x]*Sin[(c + d*x)/2]*(a + b*Sin[c + d*x]))/(96*(a + b)*d*(b + a*Csc[c + d*x])*(Cos[(c + d*x)/2] - Sin[(
c + d*x)/2])^3) - (Csc[c + d*x]*Sin[(c + d*x)/2]*(a + b*Sin[c + d*x]))/(96*(a - b)*d*(b + a*Csc[c + d*x])*(Cos
[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + (Csc[c + d*x]*(a + b*Sin[c + d*x]))/(192*(a - b)*d*(b + a*Csc[c + d*x])
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (Csc[c + d*x]*(-13*a*Sin[(c + d*x)/2] + 16*b*Sin[(c + d*x)/2])*(a
+ b*Sin[c + d*x]))/(96*(a - b)^2*d*(b + a*Csc[c + d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (Csc[c + d*x]
*(13*a*Sin[(c + d*x)/2] + 16*b*Sin[(c + d*x)/2])*(a + b*Sin[c + d*x]))/(96*(a + b)^2*d*(b + a*Csc[c + d*x])*(C
os[(c + d*x)/2] - Sin[(c + d*x)/2])) - (b*Csc[c + d*x]*(a + b*Sin[c + d*x])*Tan[(c + d*x)/2])/(32*a^2*d*(b + a
*Csc[c + d*x])))

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fricas [A]  time = 3.10, size = 1182, normalized size = 3.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[-1/12*(4*a^8 - 8*a^6*b^2 + 4*a^4*b^4 - 6*(5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 - a^2*b^6)*cos(d*x + c)^4 + 4*(5*a^8
- 13*a^6*b^2 + 8*a^4*b^4)*cos(d*x + c)^2 + 6*(b^7*cos(d*x + c)^5 - b^7*cos(d*x + c)^3)*sqrt(-a^2 + b^2)*log(-
((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x +
c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 3*((5*a^8 - 13*a^6*b^2 + 9*a^4*
b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^5 - (5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^3)*log
(1/2*cos(d*x + c) + 1/2) - 3*((5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^5 - (5*a^8 - 13*
a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) - 4*(a^7*b - 2*a^5*b^3 + a
^3*b^5 - (8*a^7*b - 22*a^5*b^3 + 17*a^3*b^5 - 3*a*b^7)*cos(d*x + c)^4 + (4*a^7*b - 11*a^5*b^3 + 7*a^3*b^5)*cos
(d*x + c)^2)*sin(d*x + c))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)^5 - (a^9 - 3*a^7*b^2 + 3*a^
5*b^4 - a^3*b^6)*d*cos(d*x + c)^3), -1/12*(4*a^8 - 8*a^6*b^2 + 4*a^4*b^4 - 6*(5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 -
a^2*b^6)*cos(d*x + c)^4 + 4*(5*a^8 - 13*a^6*b^2 + 8*a^4*b^4)*cos(d*x + c)^2 - 12*(b^7*cos(d*x + c)^5 - b^7*co
s(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 3*((5*a^8 - 13*a^
6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^5 - (5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d
*x + c)^3)*log(1/2*cos(d*x + c) + 1/2) - 3*((5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^5
- (5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) - 4*(a^7*b -
2*a^5*b^3 + a^3*b^5 - (8*a^7*b - 22*a^5*b^3 + 17*a^3*b^5 - 3*a*b^7)*cos(d*x + c)^4 + (4*a^7*b - 11*a^5*b^3 +
7*a^3*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)^5 - (a^9 - 3*
a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)^3)]

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giac [A]  time = 0.24, size = 417, normalized size = 1.26 \[ -\frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{7}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{2}} - \frac {12 \, {\left (5 \, a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {16 \, {\left (6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 14 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 18 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, a^{3} + 10 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}} + \frac {3 \, {\left (30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/24*(48*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*b^7
/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(a^2 - b^2)) - 3*(a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 -
12*(5*a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 16*(6*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 9*b^3*tan(1/2*d*
x + 1/2*c)^5 - 9*a^3*tan(1/2*d*x + 1/2*c)^4 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 8*a^2*b*tan(1/2*d*x + 1/2*c)^3
+ 14*b^3*tan(1/2*d*x + 1/2*c)^3 + 12*a^3*tan(1/2*d*x + 1/2*c)^2 - 18*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 6*a^2*b*t
an(1/2*d*x + 1/2*c) - 9*b^3*tan(1/2*d*x + 1/2*c) - 7*a^3 + 10*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1
/2*c)^2 - 1)^3) + 3*(30*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 4*a*b*tan(1/2*d*x + 1/2*c
) + a^2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d

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maple [A]  time = 0.55, size = 376, normalized size = 1.13 \[ -\frac {5 a}{2 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 b}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 d \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 d \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{2 d \,a^{2}}-\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{3}}+\frac {b}{2 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {5 a}{2 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 b}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{3 d \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 d \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 b^{7} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{3} \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^4/(a+b*sin(d*x+c)),x)

[Out]

-5/2/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)*a-3/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)*b-1/3/d/(a+b)/(tan(1/2*d*x+1/2*c)-1
)^3-1/2/d/(a+b)/(tan(1/2*d*x+1/2*c)-1)^2+1/8/a/d*tan(1/2*d*x+1/2*c)^2-1/2/d/a^2*tan(1/2*d*x+1/2*c)*b-1/8/a/d/t
an(1/2*d*x+1/2*c)^2+5/2/a/d*ln(tan(1/2*d*x+1/2*c))+1/d/a^3*ln(tan(1/2*d*x+1/2*c))*b^2+1/2/d/a^2*b/tan(1/2*d*x+
1/2*c)+5/2/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*a-3/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*b+1/3/d/(a-b)/(tan(1/2*d*x+1/
2*c)+1)^3-1/2/d/(a-b)/(tan(1/2*d*x+1/2*c)+1)^2-2/d/a^3*b^7/(a-b)^2/(a+b)^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan
(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 18.01, size = 5035, normalized size = 15.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*sin(c + d*x)^3*(a + b*sin(c + d*x))),x)

[Out]

((b^9*sin(c + d*x))/24 + (41*b^9*sin(3*c + 3*d*x))/48 + (23*b^9*sin(5*c + 5*d*x))/48)/(d*sin(c + d*x)^2*((3*co
s(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (a*((19*b^8
)/48 - (5*b^8*cos(c + d*x))/12 + (17*b^8*cos(2*c + 2*d*x))/12 + (5*b^8*cos(3*c + 3*d*x))/24 + (11*b^8*cos(4*c
+ 4*d*x))/16 + (5*b^8*cos(5*c + 5*d*x))/24 - (5*b^8*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/1
6 + (5*b^8*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/32 + (5*b^8*log(sin(c/2 + (d*x)/2)/cos
(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/32))/(d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4
- 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (a^8*((b*sin(3*c + 3*d*x))/6 - (b*sin(c + d*x))/3 + (b*si
n(5*c + 5*d*x))/6))/(d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 -
b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + ((3*b^10)/16 + (b^10*cos(2*c + 2*d*x))/4 + (b^10*cos(4*c + 4*d*x))/16 + (5*b^
10*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/16 - (5*b^10*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x
)/2))*cos(3*c + 3*d*x))/32 - (5*b^10*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/32)/(a*d*sin
(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b
^2)) - ((b^11*sin(c + d*x))/8 + (3*b^11*sin(3*c + 3*d*x))/16 + (b^11*sin(5*c + 5*d*x))/16)/(a^2*d*sin(c + d*x)
^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (a
^6*((19*b^3*sin(3*c + 3*d*x))/24 - (4*b^3*sin(c + d*x))/3 + (19*b^3*sin(5*c + 5*d*x))/24))/(d*sin(c + d*x)^2*(
(3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (a^4*(
(25*b^5*sin(3*c + 3*d*x))/16 - (15*b^5*sin(c + d*x))/8 + (23*b^5*sin(5*c + 5*d*x))/16))/(d*sin(c + d*x)^2*((3*
cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (a^2*((77
*b^7*sin(3*c + 3*d*x))/48 - (23*b^7*sin(c + d*x))/24 + (59*b^7*sin(5*c + 5*d*x))/48))/(d*sin(c + d*x)^2*((3*co
s(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (a^9*((5*co
s(2*c + 2*d*x))/12 - (7*cos(c + d*x))/24 + (7*cos(3*c + 3*d*x))/48 + (5*cos(4*c + 4*d*x))/16 + (7*cos(5*c + 5*
d*x))/48 - (5*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/16 + (5*log(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2))*cos(3*c + 3*d*x))/32 + (5*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/32 - 11/48)
)/(d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 -
3*a^4*b^2)) - (a^5*((7*b^4*cos(2*c + 2*d*x))/2 - b^4 - (17*b^4*cos(c + d*x))/8 + (17*b^4*cos(3*c + 3*d*x))/16
+ (5*b^4*cos(4*c + 4*d*x))/2 + (17*b^4*cos(5*c + 5*d*x))/16 - (5*b^4*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(
c/2 + (d*x)/2)))/2 + (5*b^4*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + (5*b^4*log(sin(c/
2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/4))/(d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x
)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (a^3*((19*b^6*cos(2*c + 2*d*x))/6 - b^6/6
- (37*b^6*cos(c + d*x))/24 + (37*b^6*cos(3*c + 3*d*x))/48 + 2*b^6*cos(4*c + 4*d*x) + (37*b^6*cos(5*c + 5*d*x))
/48 - (15*b^6*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + (15*b^6*log(sin(c/2 + (d*x)/2)/cos(
c/2 + (d*x)/2))*cos(3*c + 3*d*x))/16 + (15*b^6*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/16
))/(d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4
- 3*a^4*b^2)) + (a^7*((23*b^2*cos(2*c + 2*d*x))/12 - (41*b^2)/48 - (31*b^2*cos(c + d*x))/24 + (31*b^2*cos(3*c
+ 3*d*x))/48 + (23*b^2*cos(4*c + 4*d*x))/16 + (31*b^2*cos(5*c + 5*d*x))/48 - (23*b^2*cos(c + d*x)*log(sin(c/2
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/16 + (23*b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/32 +
(23*b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/32))/(d*sin(c + d*x)^2*((3*cos(c + d*x))
/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (b^12*log(sin(c/2 + (d
*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/(16*a^3*d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4
)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (b^12*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/
2))*cos(5*c + 5*d*x))/(16*a^3*d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^
2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (b^12*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(8*a^
3*d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 -
3*a^4*b^2)) - (b^7*atan((8*b^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b
^2)^(1/2) - 5*a^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + 4
*a*b^7*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + 5*a^7*b*cos(
c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + a^3*b^5*cos(c/2 + (d*x)
/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) - 8*a^5*b^3*cos(c/2 + (d*x)/2)*(b^10
- a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) - 17*a^4*b^4*sin(c/2 + (d*x)/2)*(b^10 - a^10
- 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + 18*a^6*b^2*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*
b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2))/(a^13*cos(c/2 + (d*x)/2)*5i - b^13*sin(c/2 + (d*x)/2)*8i - a
*b^12*cos(c/2 + (d*x)/2)*4i + a^12*b*sin(c/2 + (d*x)/2)*10i + a^3*b^10*cos(c/2 + (d*x)/2)*9i + a^5*b^8*cos(c/2
+ (d*x)/2)*3i - a^7*b^6*cos(c/2 + (d*x)/2)*30i + a^9*b^4*cos(c/2 + (d*x)/2)*40i - a^11*b^2*cos(c/2 + (d*x)/2)
*23i + a^2*b^11*sin(c/2 + (d*x)/2)*20i + a^4*b^9*sin(c/2 + (d*x)/2)*2i - a^6*b^7*sin(c/2 + (d*x)/2)*58i + a^8*
b^5*sin(c/2 + (d*x)/2)*80i - a^10*b^3*sin(c/2 + (d*x)/2)*46i))*cos(c + d*x)*(-(a + b)^5*(a - b)^5)^(1/2)*1i)/(
4*a^3*d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^
4 - 3*a^4*b^2)) + (b^7*atan((8*b^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a
^8*b^2)^(1/2) - 5*a^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)
+ 4*a*b^7*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + 5*a^7*b*
cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + a^3*b^5*cos(c/2 + (
d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) - 8*a^5*b^3*cos(c/2 + (d*x)/2)*(
b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) - 17*a^4*b^4*sin(c/2 + (d*x)/2)*(b^10 - a
^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + 18*a^6*b^2*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*
a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2))/(a^13*cos(c/2 + (d*x)/2)*5i - b^13*sin(c/2 + (d*x)/2)*8i
- a*b^12*cos(c/2 + (d*x)/2)*4i + a^12*b*sin(c/2 + (d*x)/2)*10i + a^3*b^10*cos(c/2 + (d*x)/2)*9i + a^5*b^8*cos
(c/2 + (d*x)/2)*3i - a^7*b^6*cos(c/2 + (d*x)/2)*30i + a^9*b^4*cos(c/2 + (d*x)/2)*40i - a^11*b^2*cos(c/2 + (d*x
)/2)*23i + a^2*b^11*sin(c/2 + (d*x)/2)*20i + a^4*b^9*sin(c/2 + (d*x)/2)*2i - a^6*b^7*sin(c/2 + (d*x)/2)*58i +
a^8*b^5*sin(c/2 + (d*x)/2)*80i - a^10*b^3*sin(c/2 + (d*x)/2)*46i))*cos(3*c + 3*d*x)*(-(a + b)^5*(a - b)^5)^(1/
2)*1i)/(8*a^3*d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6 - b^6 +
3*a^2*b^4 - 3*a^4*b^2)) + (b^7*atan((8*b^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b
^4 + 5*a^8*b^2)^(1/2) - 5*a^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^
2)^(1/2) + 4*a*b^7*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) +
5*a^7*b*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + a^3*b^5*cos
(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) - 8*a^5*b^3*cos(c/2 + (d
*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) - 17*a^4*b^4*sin(c/2 + (d*x)/2)*(
b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) + 18*a^6*b^2*sin(c/2 + (d*x)/2)*(b^10 - a
^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2))/(a^13*cos(c/2 + (d*x)/2)*5i - b^13*sin(c/2 + (d*
x)/2)*8i - a*b^12*cos(c/2 + (d*x)/2)*4i + a^12*b*sin(c/2 + (d*x)/2)*10i + a^3*b^10*cos(c/2 + (d*x)/2)*9i + a^5
*b^8*cos(c/2 + (d*x)/2)*3i - a^7*b^6*cos(c/2 + (d*x)/2)*30i + a^9*b^4*cos(c/2 + (d*x)/2)*40i - a^11*b^2*cos(c/
2 + (d*x)/2)*23i + a^2*b^11*sin(c/2 + (d*x)/2)*20i + a^4*b^9*sin(c/2 + (d*x)/2)*2i - a^6*b^7*sin(c/2 + (d*x)/2
)*58i + a^8*b^5*sin(c/2 + (d*x)/2)*80i - a^10*b^3*sin(c/2 + (d*x)/2)*46i))*cos(5*c + 5*d*x)*(-(a + b)^5*(a - b
)^5)^(1/2)*1i)/(8*a^3*d*sin(c + d*x)^2*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)*(a^4 + b^4 - 2*a^2*b^2)*(a^6
- b^6 + 3*a^2*b^4 - 3*a^4*b^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3*sec(d*x+c)**4/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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