3.1356 \(\int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=194 \[ \frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a d \left (a^2-b^2\right )}-\frac {2 b^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{5/2}}-\frac {b \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a d \left (a^2-b^2\right )^2}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {\sec (c+d x)}{a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {2898, 2622, 302, 207, 2696, 2866, 12, 2660, 618, 204} \[ -\frac {2 b^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{5/2}}+\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a d \left (a^2-b^2\right )}-\frac {b \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a d \left (a^2-b^2\right )^2}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {\sec (c+d x)}{a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)*d) - ArcTanh[Cos[c + d*x]]/(a*d
) + Sec[c + d*x]/(a*d) + Sec[c + d*x]^3/(3*a*d) + (b*Sec[c + d*x]^3*(b - a*Sin[c + d*x]))/(3*a*(a^2 - b^2)*d)
- (b*Sec[c + d*x]*(3*b^3 + a*(2*a^2 - 5*b^2)*Sin[c + d*x]))/(3*a*(a^2 - b^2)^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 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 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 (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \left (\frac {\csc (c+d x) \sec ^4(c+d x)}{a}-\frac {b \sec ^4(c+d x)}{a (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac {\int \csc (c+d x) \sec ^4(c+d x) \, dx}{a}-\frac {b \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}+\frac {b \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 \left (a^2-b^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}-\frac {b \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 d}-\frac {b \int \frac {3 b^4}{a+b \sin (c+d x)} \, dx}{3 a \left (a^2-b^2\right )^2}+\frac {\operatorname {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac {\sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}-\frac {b \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 d}-\frac {b^5 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a \left (a^2-b^2\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {\sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}-\frac {b \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 d}-\frac {\left (2 b^5\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 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {\sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}-\frac {b \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 d}+\frac {\left (4 b^5\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 \left (a^2-b^2\right )^2 d}\\ &=-\frac {2 b^5 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2} d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {\sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}-\frac {b \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 4.88, size = 334, normalized size = 1.72 \[ \frac {-\frac {24 b^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac {2 (7 a+10 b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 (10 b-7 a) \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {1}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {1}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {12 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a}-\frac {12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-24*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)) - (12*Log[Cos[(c + d*x)/2]])
/a + (12*Log[Sin[(c + d*x)/2]])/a + 1/((a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (2*Sin[(c + d*x)/2])
/((a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (2*(7*a + 10*b)*Sin[(c + d*x)/2])/((a + b)^2*(Cos[(c + d*
x)/2] - Sin[(c + d*x)/2])) - (2*Sin[(c + d*x)/2])/((a - b)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + 1/((a -
b)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (2*(-7*a + 10*b)*Sin[(c + d*x)/2])/((a - b)^2*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])))/(12*d)

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fricas [A]  time = 2.40, size = 680, normalized size = 3.51 \[ \left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} b^{5} \cos \left (d x + c\right )^{3} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{6} + 4 \, a^{4} b^{2} - 2 \, a^{2} b^{4} + 3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5} + {\left (2 \, a^{5} b - 7 \, a^{3} b^{3} + 5 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{3}}, \frac {6 \, \sqrt {a^{2} - b^{2}} b^{5} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + 2 \, a^{6} - 4 \, a^{4} b^{2} + 2 \, a^{2} b^{4} - 3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5} + {\left (2 \, a^{5} b - 7 \, a^{3} b^{3} + 5 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*sqrt(-a^2 + b^2)*b^5*cos(d*x + c)^3*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)) - 2*a^6 + 4*a^4*b^2 - 2*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(d*x + c)^3*log(1/
2*cos(d*x + c) + 1/2) - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(d*x + c)^3*log(-1/2*cos(d*x + c) + 1/2) - 6*
(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + (2*a^5*b - 7*a^3*b^3 + 5*a*b^5)*
cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^3), 1/6*(6*sqrt(a^2 - b^2)
*b^5*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c)^3 + 2*a^6 - 4*a^4*b^2 + 2*a^2*b
^4 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(d*x + c)^3*log(1/2*cos(d*x + c) + 1/2) + 3*(a^6 - 3*a^4*b^2 + 3
*a^2*b^4 - b^6)*cos(d*x + c)^3*log(-1/2*cos(d*x + c) + 1/2) + 6*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c)^2 -
2*(a^5*b - 2*a^3*b^3 + a*b^5 + (2*a^5*b - 7*a^3*b^3 + 5*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 3*a^5*b^
2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^3)]

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giac [A]  time = 0.24, size = 308, normalized size = 1.59 \[ -\frac {\frac {6 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{5}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{3} + 7 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*(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^5/(
(a^5 - 2*a^3*b^2 + a*b^4)*sqrt(a^2 - b^2)) - 3*log(abs(tan(1/2*d*x + 1/2*c)))/a - 2*(3*a^2*b*tan(1/2*d*x + 1/2
*c)^5 - 6*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*a^3*tan(1/2*d*x + 1/2*c)^4 + 9*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 2*a^2*b
*tan(1/2*d*x + 1/2*c)^3 + 8*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a*b^2*tan(1/2*d*x +
1/2*c)^2 + 3*a^2*b*tan(1/2*d*x + 1/2*c) - 6*b^3*tan(1/2*d*x + 1/2*c) - 4*a^3 + 7*a*b^2)/((a^4 - 2*a^2*b^2 + b
^4)*(tan(1/2*d*x + 1/2*c)^2 - 1)^3))/d

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maple [A]  time = 0.49, size = 279, normalized size = 1.44 \[ -\frac {1}{3 d \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 d \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a}{2 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {1}{2 d \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{3 d \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3 a}{2 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} a \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)^2-3/2/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1
)*a-2/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)*b+1/a/d*ln(tan(1/2*d*x+1/2*c))-1/2/d/(a-b)/(tan(1/2*d*x+1/2*c)+1)^2+1/3
/d/(a-b)/(tan(1/2*d*x+1/2*c)+1)^3+3/2/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*a-2/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*b-
2/d*b^5/(a-b)^2/(a+b)^2/a/(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)*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 = 16.20, size = 2162, normalized size = 11.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^5*((15*b^5*sin(c + d*x))/4 + (7*b^5*sin(3*c + 3*d*x))/4) - a^7*((9*b^3*sin(c + d*x))/4 + (11*b^3*sin(3*c +
3*d*x))/12) - a^3*((11*b^7*sin(c + d*x))/4 + (17*b^7*sin(3*c + 3*d*x))/12) + a^9*((b*sin(c + d*x))/2 + (b*sin
(3*c + 3*d*x))/6) - a^10*(cos(c + d*x) + cos(2*c + 2*d*x)/2 + cos(3*c + 3*d*x)/3 + (3*cos(c + d*x)*log(sin(c/2
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + 5/6) +
a*((3*b^9*sin(c + d*x))/4 + (5*b^9*sin(3*c + 3*d*x))/12) - a^2*((7*b^8*cos(c + d*x))/4 + (4*b^8)/3 + b^8*cos(2
*c + 2*d*x) + (7*b^8*cos(3*c + 3*d*x))/12 + (15*b^8*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4
+ (5*b^8*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4) - a^6*((33*b^4*cos(c + d*x))/4 + (13
*b^4)/2 + (9*b^4*cos(2*c + 2*d*x))/2 + (11*b^4*cos(3*c + 3*d*x))/4 + (15*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))/2) + a^8*((19*
b^2*cos(c + d*x))/4 + (23*b^2)/6 + (5*b^2*cos(2*c + 2*d*x))/2 + (19*b^2*cos(3*c + 3*d*x))/12 + (15*b^2*cos(c +
d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (5*b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*
c + 3*d*x))/4) + a^4*((25*b^6*cos(c + d*x))/4 + (29*b^6)/6 + (7*b^6*cos(2*c + 2*d*x))/2 + (25*b^6*cos(3*c + 3*
d*x))/12 + (15*b^6*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (5*b^6*log(sin(c/2 + (d*x)/2)/
cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2) + (3*b^10*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4
+ (b^10*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + (b^5*atan((4*b^6*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^6*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) + 2*a*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) + a^5*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) - 2*a^3*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) - 5*a^2*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) + 4*a^4*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^11*cos(c/2 + (d*x)/2)*1i - b^11*sin(c/2 + (d*x)/2)*4i - a*b^10*cos(c/2 + (d*x)/2)*2i + a^10*b*sin(c/2 +
(d*x)/2)*2i + a^3*b^8*cos(c/2 + (d*x)/2)*7i - a^5*b^6*cos(c/2 + (d*x)/2)*11i + a^7*b^4*cos(c/2 + (d*x)/2)*10i
- a^9*b^2*cos(c/2 + (d*x)/2)*5i + a^2*b^9*sin(c/2 + (d*x)/2)*15i - a^4*b^7*sin(c/2 + (d*x)/2)*24i + a^6*b^5*s
in(c/2 + (d*x)/2)*21i - a^8*b^3*sin(c/2 + (d*x)/2)*10i))*cos(3*c + 3*d*x)*(-(a + b)^5*(a - b)^5)^(1/2)*1i)/2 +
(b^5*atan((4*b^6*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
^6*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) + 2*a*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) + a^5*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) - 2*a^3*b^3*cos(c/2 + (d*x)/2)*(b^10 - a^1
0 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2) - 5*a^2*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) + 4*a^4*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^11*cos(c/2 + (d*x)/2)*1i - b^11*sin(c/2 + (d*x)/2)*4i - a*b^10*co
s(c/2 + (d*x)/2)*2i + a^10*b*sin(c/2 + (d*x)/2)*2i + a^3*b^8*cos(c/2 + (d*x)/2)*7i - a^5*b^6*cos(c/2 + (d*x)/2
)*11i + a^7*b^4*cos(c/2 + (d*x)/2)*10i - a^9*b^2*cos(c/2 + (d*x)/2)*5i + a^2*b^9*sin(c/2 + (d*x)/2)*15i - a^4*
b^7*sin(c/2 + (d*x)/2)*24i + a^6*b^5*sin(c/2 + (d*x)/2)*21i - a^8*b^3*sin(c/2 + (d*x)/2)*10i))*cos(c + d*x)*(-
(a + b)^5*(a - b)^5)^(1/2)*3i)/2)/(a*d*((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)*sec(d*x+c)**4/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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