3.1357 \(\int \frac{\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=220 \[ \frac{2 b^6 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}+\frac{\left (-10 a^2 b^2+6 a^4+b^4\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right )^2}+\frac{b \left (2 b^2-a^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right )^2}+\frac{b \sec ^3(c+d x) (b \sin (c+d x)-a)}{3 a d \left (a^2-b^2\right )}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\tan ^3(c+d x)}{3 a d}-\frac{\cot (c+d x)}{a d} \]

[Out]

(2*b^6*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(5/2)*d) + (b*ArcTanh[Cos[c + d*x]])
/(a^2*d) - Cot[c + d*x]/(a*d) + (b*(-a^2 + 2*b^2)*Sec[c + d*x])/((a^2 - b^2)^2*d) + (b*Sec[c + d*x]^3*(-a + b*
Sin[c + d*x]))/(3*a*(a^2 - b^2)*d) + ((6*a^4 - 10*a^2*b^2 + b^4)*Tan[c + d*x])/(3*a*(a^2 - b^2)^2*d) + Tan[c +
d*x]^3/(3*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.473599, antiderivative size = 247, normalized size of antiderivative = 1.12, number of steps used = 15, number of rules used = 12, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.414, Rules used = {2898, 2622, 302, 207, 2620, 270, 2696, 2866, 12, 2660, 618, 204} \[ \frac{2 b^6 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}-\frac{b^2 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^2 d \left (a^2-b^2\right )}+\frac{b^2 \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a^2 d \left (a^2-b^2\right )^2}-\frac{b \sec ^3(c+d x)}{3 a^2 d}-\frac{b \sec (c+d x)}{a^2 d}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\tan ^3(c+d x)}{3 a d}+\frac{2 \tan (c+d x)}{a d}-\frac{\cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]

[Out]

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

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])

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 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 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 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 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 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 12

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

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 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 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])

Rubi steps

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

Mathematica [B]  time = 6.44284, size = 450, normalized size = 2.05 \[ \frac{2 b^6 \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}-\frac{b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 d}+\frac{b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 d}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{6 d (a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{10 a \sin \left (\frac{1}{2} (c+d x)\right )-13 b \sin \left (\frac{1}{2} (c+d x)\right )}{6 d (a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{10 a \sin \left (\frac{1}{2} (c+d x)\right )+13 b \sin \left (\frac{1}{2} (c+d x)\right )}{6 d (a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{12 d (a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{12 d (a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{6 d (a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{2 a d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]

[Out]

(2*b^6*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^
(5/2)*d) - Cot[(c + d*x)/2]/(2*a*d) + (b*Log[Cos[(c + d*x)/2]])/(a^2*d) - (b*Log[Sin[(c + d*x)/2]])/(a^2*d) +
1/(12*(a + b)*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + Sin[(c + d*x)/2]/(6*(a + b)*d*(Cos[(c + d*x)/2] - S
in[(c + d*x)/2])^3) + Sin[(c + d*x)/2]/(6*(a - b)*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 1/(12*(a - b)*d
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (10*a*Sin[(c + d*x)/2] - 13*b*Sin[(c + d*x)/2])/(6*(a - b)^2*d*(Co
s[(c + d*x)/2] + Sin[(c + d*x)/2])) + (10*a*Sin[(c + d*x)/2] + 13*b*Sin[(c + d*x)/2])/(6*(a + b)^2*d*(Cos[(c +
d*x)/2] - Sin[(c + d*x)/2])) + Tan[(c + d*x)/2]/(2*a*d)

________________________________________________________________________________________

Maple [A]  time = 0.123, size = 317, normalized size = 1.4 \begin{align*}{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{a}{d \left ( a+b \right ) ^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}-{\frac{5\,b}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{3\,d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+2\,{\frac{{b}^{6}}{d{a}^{2} \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{a}{d \left ( a-b \right ) ^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{5\,b}{2\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{3\,d \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,d \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x)

[Out]

1/2/d/a*tan(1/2*d*x+1/2*c)-2/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)*a-5/2/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^2/(a+b)^2/(a-b)^2*b^6/(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))-2/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*a+5/2/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)^2-1/2/d/a/tan(
1/2*d*x+1/2*c)-1/d/a^2*b*ln(tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 5.39312, size = 1891, normalized size = 8.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**4/(a+b*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.25729, size = 482, normalized size = 2.19 \begin{align*} \frac{\frac{12 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \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^{6}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sqrt{a^{2} - b^{2}}} - \frac{6 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} + \frac{3 \,{\left (2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{4 \,{\left (6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 9 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 14 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a^{2} b + 7 \, b^{3}\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}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(12*(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^6/(
(a^6 - 2*a^4*b^2 + a^2*b^4)*sqrt(a^2 - b^2)) - 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 3*tan(1/2*d*x + 1/2*c)
/a + 3*(2*b*tan(1/2*d*x + 1/2*c) - a)/(a^2*tan(1/2*d*x + 1/2*c)) - 4*(6*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*a*b^2*t
an(1/2*d*x + 1/2*c)^5 - 6*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 9*b^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^3*tan(1/2*d*x + 1/
2*c)^3 + 14*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 12*b^3*tan(1/2*d*x + 1/2*c)^2 + 6*
a^3*tan(1/2*d*x + 1/2*c) - 9*a*b^2*tan(1/2*d*x + 1/2*c) - 4*a^2*b + 7*b^3)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*d
*x + 1/2*c)^2 - 1)^3))/d