3.183 \(\int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=273 \[ -\frac {2 a \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 a \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{5/3} d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac {2 a \tanh ^{-1}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}+\frac {x}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.57, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3220, 2635, 8, 2660, 618, 204, 206} \[ -\frac {2 a \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 a \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{5/3} d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac {2 a \tanh ^{-1}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}+\frac {x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/(a + b*Sin[c + d*x]^3),x]

[Out]

x/(2*b) - (2*a*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)
]*b^(5/3)*d) + (2*a*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/
3)]])/(3*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*b^(5/3)*d) + (2*a*ArcTanh[(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c
+ d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]*b^(5/3)*d) - (Cos[c + d*
x]*Sin[c + d*x])/(2*b*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*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 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

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

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Mathematica [C]  time = 0.28, size = 255, normalized size = 0.93 \[ \frac {-2 i a \text {RootSum}\left [i \text {$\#$1}^6 b-3 i \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 i \text {$\#$1}^2 b-i b\& ,\frac {2 \text {$\#$1}^4 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+2 i \text {$\#$1}^2 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )-i \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )-4 \text {$\#$1}^2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \text {$\#$1}^4 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )+2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )}{\text {$\#$1}^5 b-2 \text {$\#$1}^3 b-4 i \text {$\#$1}^2 a+\text {$\#$1} b}\& \right ]+6 (c+d x)-3 \sin (2 (c+d x))}{12 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5/(a + b*Sin[c + d*x]^3),x]

[Out]

(6*(c + d*x) - (2*I)*a*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (2*ArcTan[Sin[c
+ d*x]/(Cos[c + d*x] - #1)] - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] - 4*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]
*#1^2 + (2*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*Log
[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)*a*#1^2 - 2*b*#1^3 + b*#1^5) & ] - 3*Sin[2*(c + d*x)])/(12*b
*d)

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fricas [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(sin(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{5}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^5/(b*sin(d*x + c)^3 + a), x)

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maple [C]  time = 0.57, size = 163, normalized size = 0.60 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b}-\frac {4 a \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a+b*sin(d*x+c)^3),x)

[Out]

1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+1/d/
b*arctan(tan(1/2*d*x+1/2*c))-4/3/d*a/b*sum(_R^2/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=R
ootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))

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maxima [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(sin(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 14.47, size = 1962, normalized size = 7.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^5/(a + b*sin(c + d*x)^3),x)

[Out]

tan(c/2 + (d*x)/2)^3/(b*d + 2*b*d*tan(c/2 + (d*x)/2)^2 + b*d*tan(c/2 + (d*x)/2)^4) - tan(c/2 + (d*x)/2)/(b*d +
2*b*d*tan(c/2 + (d*x)/2)^2 + b*d*tan(c/2 + (d*x)/2)^4) + symsum(log((134217728*a^9*b^2 - 16777216*a^11 - 4026
53184*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)*a^8*b^4 + 50331648*
a^10*b*tan(c/2 + (d*x)/2) - 2415919104*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2
+ a^6, z, k)^2*a^7*b^6 + 914358272*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 +
a^6, z, k)^2*a^9*b^4 + 7247757312*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^
6, z, k)^3*a^6*b^8 - 478150656*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6,
z, k)^3*a^8*b^6 + 10871635968*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z
, k)^4*a^5*b^10 - 21214789632*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z
, k)^4*a^7*b^8 - 301989888*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k
)^4*a^9*b^6 - 32614907904*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)
^5*a^4*b^12 + 59567505408*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)
^5*a^6*b^10 + 4529848320*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^
5*a^8*b^8 + 55717134336*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^6
*a^5*b^12 - 42127589376*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^6
*a^7*b^10 - 130459631616*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^
7*a^4*b^14 + 122305904640*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)
^7*a^6*b^12 - 452984832*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)*a
^9*b^3*tan(c/2 + (d*x)/2) + 1509949440*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2
+ a^6, z, k)^2*a^8*b^5*tan(c/2 + (d*x)/2) + 201326592*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4
- 27*a^4*b^4*z^2 + a^6, z, k)^2*a^10*b^3*tan(c/2 + (d*x)/2) - 2717908992*root(729*a^2*b^10*z^6 - 729*b^12*z^6
+ 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^3*a^7*b^7*tan(c/2 + (d*x)/2) - 2717908992*root(729*a^2*b^10*z^
6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^3*a^9*b^5*tan(c/2 + (d*x)/2) + 4076863488*roo
t(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^4*a^6*b^9*tan(c/2 + (d*x)/2)
+ 6039797760*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^4*a^8*b^7*t
an(c/2 + (d*x)/2) - 4076863488*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6,
z, k)^5*a^5*b^11*tan(c/2 + (d*x)/2) - 679477248*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^
4*b^4*z^2 + a^6, z, k)^5*a^7*b^9*tan(c/2 + (d*x)/2) + 16307453952*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a
^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^6*a^6*b^11*tan(c/2 + (d*x)/2) - 40768634880*root(729*a^2*b^10*z^6 - 7
29*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^7*a^5*b^13*tan(c/2 + (d*x)/2) + 32614907904*root(7
29*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^7*a^7*b^11*tan(c/2 + (d*x)/2) +
33554432*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)*a^11*b*tan(c/2
+ (d*x)/2))/b^5)*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k), k, 1, 6
)/d - (log(tan(c/2 + (d*x)/2) - 1i)*1i)/(2*b*d) + (log(tan(c/2 + (d*x)/2) + 1i)*1i)/(2*b*d)

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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(sin(d*x+c)**5/(a+b*sin(d*x+c)**3),x)

[Out]

Timed out

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