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

Optimal. Leaf size=259 \[ -\frac {2 \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {2 \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {x}{b} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3220, 3213, 2660, 618, 204} \[ -\frac {2 \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {2 \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/b - (2*a^(1/3)*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*d) - (2*a^(1/3)*ArcTan[((-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)
]])/(3*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b*d) + (2*a^(1/3)*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)*T
an[(c + d*x)/2]))/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b*d)

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 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 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 3213

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

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

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Mathematica [C]  time = 0.18, size = 140, normalized size = 0.54 \[ \frac {2 i a \text {RootSum}\left [i \text {$\#$1}^6 b-3 i \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 i \text {$\#$1}^2 b-i b\& ,\frac {2 \text {$\#$1} \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \text {$\#$1} \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )}{\text {$\#$1}^4 b-2 \text {$\#$1}^2 b-4 i \text {$\#$1} a+b}\& \right ]+3 c+3 d x}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*c + 3*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)]*#1 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4)
& ])/(3*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)^3/(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 \[ \int \frac {\sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.47, size = 106, normalized size = 0.41 \[ \frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b}-\frac {a \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {-8 \, a b \int \frac {8 \, a \cos \left (3 \, d x + 3 \, c\right )^{2} - b \cos \left (3 \, d x + 3 \, c\right ) \sin \left (6 \, d x + 6 \, c\right ) + 3 \, b \cos \left (3 \, d x + 3 \, c\right ) \sin \left (4 \, d x + 4 \, c\right ) + b \cos \left (6 \, d x + 6 \, c\right ) \sin \left (3 \, d x + 3 \, c\right ) - 3 \, b \cos \left (4 \, d x + 4 \, c\right ) \sin \left (3 \, d x + 3 \, c\right ) + 8 \, a \sin \left (3 \, d x + 3 \, c\right )^{2} - 3 \, b \cos \left (3 \, d x + 3 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + {\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) - b\right )} \sin \left (3 \, d x + 3 \, c\right )}{b^{3} \cos \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b^{3} \cos \left (4 \, d x + 4 \, c\right )^{2} + 64 \, a^{2} b \cos \left (3 \, d x + 3 \, c\right )^{2} + 9 \, b^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{3} \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b^{3} \sin \left (4 \, d x + 4 \, c\right )^{2} + 64 \, a^{2} b \sin \left (3 \, d x + 3 \, c\right )^{2} - 48 \, a b^{2} \cos \left (3 \, d x + 3 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b^{3} \sin \left (2 \, d x + 2 \, c\right )^{2} - 6 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) + b^{3} - 2 \, {\left (3 \, b^{3} \cos \left (4 \, d x + 4 \, c\right ) - 3 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) - 8 \, a b^{2} \sin \left (3 \, d x + 3 \, c\right ) + b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right ) - 6 \, {\left (3 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, a b^{2} \sin \left (3 \, d x + 3 \, c\right ) - b^{3}\right )} \cos \left (4 \, d x + 4 \, c\right ) - 2 \, {\left (8 \, a b^{2} \cos \left (3 \, d x + 3 \, c\right ) + 3 \, b^{3} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, b^{3} \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + 6 \, {\left (8 \, a b^{2} \cos \left (3 \, d x + 3 \, c\right ) - 3 \, b^{3} \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (4 \, d x + 4 \, c\right ) + 16 \, {\left (3 \, a b^{2} \cos \left (2 \, d x + 2 \, c\right ) - a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}\,{d x} + x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*a*b*integrate(-(8*a*cos(3*d*x + 3*c)^2 - b*cos(3*d*x + 3*c)*sin(6*d*x + 6*c) + 3*b*cos(3*d*x + 3*c)*sin(4*d
*x + 4*c) + b*cos(6*d*x + 6*c)*sin(3*d*x + 3*c) - 3*b*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) + 8*a*sin(3*d*x + 3*c)
^2 - 3*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + (3*b*cos(2*d*x + 2*c) - b)*sin(3*d*x + 3*c))/(b^3*cos(6*d*x + 6*c
)^2 + 9*b^3*cos(4*d*x + 4*c)^2 + 64*a^2*b*cos(3*d*x + 3*c)^2 + 9*b^3*cos(2*d*x + 2*c)^2 + b^3*sin(6*d*x + 6*c)
^2 + 9*b^3*sin(4*d*x + 4*c)^2 + 64*a^2*b*sin(3*d*x + 3*c)^2 - 48*a*b^2*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 9*b
^3*sin(2*d*x + 2*c)^2 - 6*b^3*cos(2*d*x + 2*c) + b^3 - 2*(3*b^3*cos(4*d*x + 4*c) - 3*b^3*cos(2*d*x + 2*c) - 8*
a*b^2*sin(3*d*x + 3*c) + b^3)*cos(6*d*x + 6*c) - 6*(3*b^3*cos(2*d*x + 2*c) + 8*a*b^2*sin(3*d*x + 3*c) - b^3)*c
os(4*d*x + 4*c) - 2*(8*a*b^2*cos(3*d*x + 3*c) + 3*b^3*sin(4*d*x + 4*c) - 3*b^3*sin(2*d*x + 2*c))*sin(6*d*x + 6
*c) + 6*(8*a*b^2*cos(3*d*x + 3*c) - 3*b^3*sin(2*d*x + 2*c))*sin(4*d*x + 4*c) + 16*(3*a*b^2*cos(2*d*x + 2*c) -
a*b^2)*sin(3*d*x + 3*c)), x) + x)/b

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mupad [B]  time = 14.90, size = 1672, normalized size = 6.46 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log(134217728*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)*a^7*ta
n(c/2 + (d*x)/2) - 268435456*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k
)^2*a^7*b - 1073741824*a^6*tan(c/2 + (d*x)/2) + 4831838208*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^
4 + 27*a^2*b^2*z^2 + a^2, z, k)^2*a^5*b^3 + 33722204160*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 +
27*a^2*b^2*z^2 + a^2, z, k)^3*a^6*b^3 + 15703474176*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27
*a^2*b^2*z^2 + a^2, z, k)^4*a^5*b^5 - 4831838208*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2
*b^2*z^2 + a^2, z, k)^4*a^7*b^3 - 130459631616*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b
^2*z^2 + a^2, z, k)^5*a^4*b^7 + 154014842880*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2
*z^2 + a^2, z, k)^5*a^6*b^5 + 35332816896*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^
2 + a^2, z, k)^6*a^5*b^7 - 21743271936*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 +
a^2, z, k)^6*a^7*b^5 - 130459631616*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a
^2, z, k)^7*a^4*b^9 + 122305904640*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2
, z, k)^7*a^6*b^7 + 2013265920*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z,
k)*a^6*b - 3221225472*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)*a^5*
b^2*tan(c/2 + (d*x)/2) - 18589155328*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a
^2, z, k)^2*a^6*b^2*tan(c/2 + (d*x)/2) - 17716740096*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27
*a^2*b^2*z^2 + a^2, z, k)^3*a^5*b^4*tan(c/2 + (d*x)/2) + 2818572288*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a
^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^3*a^7*b^2*tan(c/2 + (d*x)/2) + 86973087744*root(729*a^2*b^6*z^6 - 729
*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^4*a^4*b^6*tan(c/2 + (d*x)/2) - 88181047296*root(729*a
^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^4*a^6*b^4*tan(c/2 + (d*x)/2) - 308029
68576*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^5*a^5*b^6*tan(c/2 + (
d*x)/2) + 18119393280*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^5*a^7
*b^4*tan(c/2 + (d*x)/2) + 86973087744*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 +
a^2, z, k)^6*a^4*b^8*tan(c/2 + (d*x)/2) - 70665633792*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 2
7*a^2*b^2*z^2 + a^2, z, k)^6*a^6*b^6*tan(c/2 + (d*x)/2) - 40768634880*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243
*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^7*a^5*b^8*tan(c/2 + (d*x)/2) + 32614907904*root(729*a^2*b^6*z^6 - 7
29*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^7*a^7*b^6*tan(c/2 + (d*x)/2))*root(729*a^2*b^6*z^6
- 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k), k, 1, 6)/d - (log(tan(c/2 + (d*x)/2) - 1i)*1i)/
(b*d) + (log(tan(c/2 + (d*x)/2) + 1i)*1i)/(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)**3/(a+b*sin(d*x+c)**3),x)

[Out]

Timed out

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