3.182 \(\int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=335 \[ \frac {2 (-1)^{2/3} a^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {2 a^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \sqrt [3]{-1} a^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac {3 \sin (c+d x) \cos (c+d x)}{8 b d}+\frac {3 x}{8 b} \]

[Out]

3/8*x/b+a*cos(d*x+c)/b^2/d-3/8*cos(d*x+c)*sin(d*x+c)/b/d-1/4*cos(d*x+c)*sin(d*x+c)^3/b/d-2/3*a^(5/3)*arctan((b
^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/b^(7/3)/d/(a^(2/3)-b^(2/3))^(1/2)+2/3*(-1)^(1/3)*a
^(5/3)*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^(7/3)/d/(a
^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2)+2/3*(-1)^(2/3)*a^(5/3)*arctan(((-1)^(1/3)*b^(1/3)-a^(1/3)*tan(1/2*d*x+1/2*c))
/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/b^(7/3)/d/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2)

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Rubi [A]  time = 0.70, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3220, 2638, 2635, 8, 2660, 618, 204} \[ \frac {2 (-1)^{2/3} a^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {2 a^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \sqrt [3]{-1} a^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac {3 \sin (c+d x) \cos (c+d x)}{8 b d}+\frac {3 x}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^7/(a + b*Sin[c + d*x]^3),x]

[Out]

(3*x)/(8*b) + (2*(-1)^(2/3)*a^(5/3)*ArcTan[((-1)^(1/3)*b^(1/3) - a^(1/3)*Tan[(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^(7/3)*d) - (2*a^(5/3)*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^(7/3)*d) + (2*(-1)^(1/3)*a^(5/3)*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^(7/3)*d) + (a*Cos[c + d*x])/(b^2*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(8*b*d) - (Cos[c + d*x]*Sin[
c + d*x]^3)/(4*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 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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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 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 ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx &=\int \left (-\frac {a \sin (c+d x)}{b^2}+\frac {\sin ^4(c+d x)}{b}+\frac {a^2 \sin (c+d x)}{b^2 \left (a+b \sin ^3(c+d x)\right )}\right ) \, dx\\ &=-\frac {a \int \sin (c+d x) \, dx}{b^2}+\frac {a^2 \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx}{b^2}+\frac {\int \sin ^4(c+d x) \, dx}{b}\\ &=\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {a^2 \int \left (-\frac {1}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}-\frac {(-1)^{2/3}}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} \sin (c+d x)\right )}+\frac {\sqrt [3]{-1}}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} \sin (c+d x)\right )}\right ) \, dx}{b^2}+\frac {3 \int \sin ^2(c+d x) \, dx}{4 b}\\ &=\frac {a \cos (c+d x)}{b^2 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{8 b d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac {a^{5/3} \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{7/3}}+\frac {\left (\sqrt [3]{-1} a^{5/3}\right ) \int \frac {1}{\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{7/3}}-\frac {\left ((-1)^{2/3} a^{5/3}\right ) \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{7/3}}+\frac {3 \int 1 \, dx}{8 b}\\ &=\frac {3 x}{8 b}+\frac {a \cos (c+d x)}{b^2 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{8 b d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac {\left (2 a^{5/3}\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^{7/3} d}+\frac {\left (2 \sqrt [3]{-1} a^{5/3}\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^{7/3} d}-\frac {\left (2 (-1)^{2/3} a^{5/3}\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^{7/3} d}\\ &=\frac {3 x}{8 b}+\frac {a \cos (c+d x)}{b^2 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{8 b d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {\left (4 a^{5/3}\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^{7/3} d}-\frac {\left (4 \sqrt [3]{-1} a^{5/3}\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^{7/3} d}+\frac {\left (4 (-1)^{2/3} a^{5/3}\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^{7/3} d}\\ &=\frac {3 x}{8 b}+\frac {2 (-1)^{2/3} a^{5/3} \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^{7/3} d}-\frac {2 a^{5/3} \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^{7/3} d}+\frac {2 \sqrt [3]{-1} a^{5/3} \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^{7/3} d}+\frac {a \cos (c+d x)}{b^2 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{8 b d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}\\ \end {align*}

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Mathematica [C]  time = 0.52, size = 219, normalized size = 0.65 \[ \frac {-32 a^2 \text {RootSum}\left [i \text {$\#$1}^6 b-3 i \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 i \text {$\#$1}^2 b-i b\& ,\frac {-i \text {$\#$1}^2 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )+i \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )+2 \text {$\#$1}^2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )}{\text {$\#$1}^4 b-2 \text {$\#$1}^2 b-4 i \text {$\#$1} a+b}\& \right ]+96 a \cos (c+d x)+3 b (12 (c+d x)-8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{96 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^7/(a + b*Sin[c + d*x]^3),x]

[Out]

(96*a*Cos[c + d*x] - 32*a^2*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] + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x]
- #1)]*#1^2 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4) & ] + 3*b*(12*(c
+ d*x) - 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(96*b^2*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)^7/(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 \[ \int \frac {\sin \left (d x + c\right )^{7}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^7/(b*sin(d*x + c)^3 + a), x)

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maple [C]  time = 0.56, size = 366, normalized size = 1.09 \[ \frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {11 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b}+\frac {2 a^{2} \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^7/(a+b*sin(d*x+c)^3),x)

[Out]

3/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^
6*a+11/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5+6/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/
2*c)^4*a-11/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+6/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d
*x+1/2*c)^2*a-3/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*a+3/4/d
/b*arctan(tan(1/2*d*x+1/2*c))+2/3/d*a^2/b^2*sum((_R^3+_R)/(_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(-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)^7/(a+b*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 15.24, size = 1978, normalized size = 5.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^7/(a + b*sin(c + d*x)^3),x)

[Out]

symsum(log((150994944*a^12*b^3*sin(c/2 + (d*x)/2) - 56623104*a^13*b^2*cos(c/2 + (d*x)/2) - 12582912*a^15*cos(c
/2 + (d*x)/2) + 679477248*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)*a^11*b^5*sin(c
/2 + (d*x)/2) + 679477248*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^2*a^9*b^8*cos(
c/2 + (d*x)/2) - 42467328*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^2*a^11*b^6*cos
(c/2 + (d*x)/2) - 402653184*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^2*a^13*b^4*c
os(c/2 + (d*x)/2) + 4586471424*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^3*a^8*b^1
0*cos(c/2 + (d*x)/2) - 503316480*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^3*a^12*
b^6*cos(c/2 + (d*x)/2) + 1911029760*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^4*a^
7*b^12*cos(c/2 + (d*x)/2) + 1774190592*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^4
*a^9*b^10*cos(c/2 + (d*x)/2) - 301989888*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)
^4*a^11*b^8*cos(c/2 + (d*x)/2) - 18345885696*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z
, k)^5*a^4*b^16*cos(c/2 + (d*x)/2) + 17199267840*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^1
0, z, k)^5*a^6*b^14*cos(c/2 + (d*x)/2) + 32614907904*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 +
a^10, z, k)^5*a^8*b^12*cos(c/2 + (d*x)/2) + 9172942848*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^
4 + a^10, z, k)^6*a^5*b^16*cos(c/2 + (d*x)/2) + 4416602112*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10
*z^4 + a^10, z, k)^6*a^7*b^14*cos(c/2 + (d*x)/2) - 130459631616*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4
*b^10*z^4 + a^10, z, k)^7*a^4*b^18*cos(c/2 + (d*x)/2) + 122305904640*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 24
3*a^4*b^10*z^4 + a^10, z, k)^7*a^6*b^16*cos(c/2 + (d*x)/2) + 1613758464*root(729*a^2*b^14*z^6 - 729*b^16*z^6 +
243*a^4*b^10*z^4 + a^10, z, k)^2*a^10*b^7*sin(c/2 + (d*x)/2) + 1073741824*root(729*a^2*b^14*z^6 - 729*b^16*z^
6 + 243*a^4*b^10*z^4 + a^10, z, k)^2*a^12*b^5*sin(c/2 + (d*x)/2) - 4076863488*root(729*a^2*b^14*z^6 - 729*b^16
*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^3*a^7*b^11*sin(c/2 + (d*x)/2) + 2420637696*root(729*a^2*b^14*z^6 - 729*b
^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^3*a^9*b^9*sin(c/2 + (d*x)/2) + 4831838208*root(729*a^2*b^14*z^6 - 729
*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^3*a^11*b^7*sin(c/2 + (d*x)/2) + 2293235712*root(729*a^2*b^14*z^6 -
729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^4*a^8*b^11*sin(c/2 + (d*x)/2) + 11475615744*root(729*a^2*b^14*z^
6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^4*a^10*b^9*sin(c/2 + (d*x)/2) - 2293235712*root(729*a^2*b^14
*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^5*a^5*b^15*sin(c/2 + (d*x)/2) - 27844411392*root(729*a^2*
b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^5*a^7*b^13*sin(c/2 + (d*x)/2) + 25367150592*root(729*
a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^5*a^9*b^11*sin(c/2 + (d*x)/2) + 16307453952*root(
729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^6*a^8*b^13*sin(c/2 + (d*x)/2) - 40768634880*r
oot(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^7*a^5*b^17*sin(c/2 + (d*x)/2) + 326149079
04*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^7*a^7*b^15*sin(c/2 + (d*x)/2) + 33554
432*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)*a^15*b*sin(c/2 + (d*x)/2) - 70778880
*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)*a^12*b^4*cos(c/2 + (d*x)/2))/(b^9*cos(c
/2 + (d*x)/2)))*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k), k, 1, 6)/d + (log((cos(
c/2 + (d*x)/2)*1i + sin(c/2 + (d*x)/2))/cos(c/2 + (d*x)/2))*3i)/(8*b*d) - (log((cos(c/2 + (d*x)/2)*1i - sin(c/
2 + (d*x)/2))/cos(c/2 + (d*x)/2))*3i)/(8*b*d) - sin(2*c + 2*d*x)/(4*b*d) + sin(4*c + 4*d*x)/(32*b*d) + (a*cos(
c + d*x))/(b^2*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)**7/(a+b*sin(d*x+c)**3),x)

[Out]

Timed out

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