3.189 \(\int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=293 \[ \frac {2 a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {a x}{b^2}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cos (c+d x)}{b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3220, 2633, 3213, 2660, 618, 204} \[ \frac {2 a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {a x}{b^2}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^3),x]

[Out]

-((a*x)/b^2) + (2*a^(4/3)*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^2*d) + (2*a^(4/3)*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^2*d) - (2*a^(4/3)*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1)^(2
/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^2*
d) - Cos[c + d*x]/(b*d) + Cos[c + d*x]^3/(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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 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 ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx &=\int \left (-\frac {a}{b^2}+\frac {\sin ^3(c+d x)}{b}+\frac {a^2}{b^2 \left (a+b \sin ^3(c+d x)\right )}\right ) \, dx\\ &=-\frac {a x}{b^2}+\frac {a^2 \int \frac {1}{a+b \sin ^3(c+d x)} \, dx}{b^2}+\frac {\int \sin ^3(c+d x) \, dx}{b}\\ &=-\frac {a x}{b^2}+\frac {a^2 \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^2}-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{b d}\\ &=-\frac {a x}{b^2}-\frac {\cos (c+d x)}{b d}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {a^{4/3} \int \frac {1}{-\sqrt [3]{a}-\sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^2}-\frac {a^{4/3} \int \frac {1}{-\sqrt [3]{a}+\sqrt [3]{-1} \sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^2}-\frac {a^{4/3} \int \frac {1}{-\sqrt [3]{a}-(-1)^{2/3} \sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^2}\\ &=-\frac {a x}{b^2}-\frac {\cos (c+d x)}{b d}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {\left (2 a^{4/3}\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^2 d}-\frac {\left (2 a^{4/3}\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^2 d}-\frac {\left (2 a^{4/3}\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^2 d}\\ &=-\frac {a x}{b^2}-\frac {\cos (c+d x)}{b d}+\frac {\cos ^3(c+d x)}{3 b d}+\frac {\left (4 a^{4/3}\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^2 d}+\frac {\left (4 a^{4/3}\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^2 d}+\frac {\left (4 a^{4/3}\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^2 d}\\ &=-\frac {a x}{b^2}-\frac {2 a^{4/3} \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^2 d}+\frac {2 a^{4/3} \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^2 d}+\frac {2 a^{4/3} \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^2 d}-\frac {\cos (c+d x)}{b d}+\frac {\cos ^3(c+d x)}{3 b d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.28, size = 164, normalized size = 0.56 \[ -\frac {8 i a^2 \text {RootSum}\left [i \text {$\#$1}^6 b-3 i \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 i \text {$\#$1}^2 b-i b\& ,\frac {2 \text {$\#$1} \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \text {$\#$1} \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )}{\text {$\#$1}^4 b-2 \text {$\#$1}^2 b-4 i \text {$\#$1} a+b}\& \right ]+12 a c+12 a d x+9 b \cos (c+d x)-b \cos (3 (c+d x))}{12 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^3),x]

[Out]

-1/12*(12*a*c + 12*a*d*x + 9*b*Cos[c + d*x] - b*Cos[3*(c + d*x)] + (8*I)*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)]*#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) & ])/(b^2*d)

________________________________________________________________________________________

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)^6/(a+b*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{6}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^6/(b*sin(d*x + c)^3 + a), x)

________________________________________________________________________________________

maple [C]  time = 0.58, size = 166, normalized size = 0.57 \[ -\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {4}{3 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}+\frac {a^{2} \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a+b*sin(d*x+c)^3),x)

[Out]

-4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-4/3/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3-2/d/b^2*a*arctan(tan
(1/2*d*x+1/2*c))+1/3/d*a^2/b^2*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))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {-96 \, a^{2} b^{2} d \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^{4} \cos \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b^{4} \cos \left (4 \, d x + 4 \, c\right )^{2} + 64 \, a^{2} b^{2} \cos \left (3 \, d x + 3 \, c\right )^{2} + 9 \, b^{4} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{4} \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b^{4} \sin \left (4 \, d x + 4 \, c\right )^{2} + 64 \, a^{2} b^{2} \sin \left (3 \, d x + 3 \, c\right )^{2} - 48 \, a b^{3} \cos \left (3 \, d x + 3 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b^{4} \sin \left (2 \, d x + 2 \, c\right )^{2} - 6 \, b^{4} \cos \left (2 \, d x + 2 \, c\right ) + b^{4} - 2 \, {\left (3 \, b^{4} \cos \left (4 \, d x + 4 \, c\right ) - 3 \, b^{4} \cos \left (2 \, d x + 2 \, c\right ) - 8 \, a b^{3} \sin \left (3 \, d x + 3 \, c\right ) + b^{4}\right )} \cos \left (6 \, d x + 6 \, c\right ) - 6 \, {\left (3 \, b^{4} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, a b^{3} \sin \left (3 \, d x + 3 \, c\right ) - b^{4}\right )} \cos \left (4 \, d x + 4 \, c\right ) - 2 \, {\left (8 \, a b^{3} \cos \left (3 \, d x + 3 \, c\right ) + 3 \, b^{4} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, b^{4} \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + 6 \, {\left (8 \, a b^{3} \cos \left (3 \, d x + 3 \, c\right ) - 3 \, b^{4} \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (4 \, d x + 4 \, c\right ) + 16 \, {\left (3 \, a b^{3} \cos \left (2 \, d x + 2 \, c\right ) - a b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}\,{d x} + 12 \, a d x - b \cos \left (3 \, d x + 3 \, c\right ) + 9 \, b \cos \left (d x + c\right )}{12 \, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

-1/12*(96*a^2*b^2*d*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^4*co
s(6*d*x + 6*c)^2 + 9*b^4*cos(4*d*x + 4*c)^2 + 64*a^2*b^2*cos(3*d*x + 3*c)^2 + 9*b^4*cos(2*d*x + 2*c)^2 + b^4*s
in(6*d*x + 6*c)^2 + 9*b^4*sin(4*d*x + 4*c)^2 + 64*a^2*b^2*sin(3*d*x + 3*c)^2 - 48*a*b^3*cos(3*d*x + 3*c)*sin(2
*d*x + 2*c) + 9*b^4*sin(2*d*x + 2*c)^2 - 6*b^4*cos(2*d*x + 2*c) + b^4 - 2*(3*b^4*cos(4*d*x + 4*c) - 3*b^4*cos(
2*d*x + 2*c) - 8*a*b^3*sin(3*d*x + 3*c) + b^4)*cos(6*d*x + 6*c) - 6*(3*b^4*cos(2*d*x + 2*c) + 8*a*b^3*sin(3*d*
x + 3*c) - b^4)*cos(4*d*x + 4*c) - 2*(8*a*b^3*cos(3*d*x + 3*c) + 3*b^4*sin(4*d*x + 4*c) - 3*b^4*sin(2*d*x + 2*
c))*sin(6*d*x + 6*c) + 6*(8*a*b^3*cos(3*d*x + 3*c) - 3*b^4*sin(2*d*x + 2*c))*sin(4*d*x + 4*c) + 16*(3*a*b^3*co
s(2*d*x + 2*c) - a*b^3)*sin(3*d*x + 3*c)), x) + 12*a*d*x - b*cos(3*d*x + 3*c) + 9*b*cos(d*x + c))/(b^2*d)

________________________________________________________________________________________

mupad [B]  time = 14.58, size = 1800, normalized size = 6.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(a + b*sin(c + d*x)^3),x)

[Out]

symsum(log((1073741824*a^13*tan(c/2 + (d*x)/2) + 2013265920*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8
*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)*a^12*b^2 - 4831838208*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^
4 + 27*a^6*b^4*z^2 + a^8, z, k)^2*a^10*b^5 + 268435456*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)^2*a^12*b^3 + 33722204160*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)^3*a^10*b^6 - 15703474176*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)^4*a^8*b^9 + 4831838208*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 +
27*a^6*b^4*z^2 + a^8, z, k)^4*a^10*b^7 - 130459631616*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 +
27*a^6*b^4*z^2 + a^8, z, k)^5*a^6*b^12 + 154014842880*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)^5*a^8*b^10 - 35332816896*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)^6*a^6*b^13 + 21743271936*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)^6*a^8*b^11 - 130459631616*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)^7*a^4*b^16 + 122305904640*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^
4 + 27*a^6*b^4*z^2 + a^8, z, k)^7*a^6*b^14 - 3221225472*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)*a^11*b^3*tan(c/2 + (d*x)/2) + 18589155328*root(729*a^2*b^12*z^6 - 729*b^14*z^6
+ 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^2*a^11*b^4*tan(c/2 + (d*x)/2) - 17716740096*root(729*a^2*b^12*
z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^3*a^9*b^7*tan(c/2 + (d*x)/2) + 2818572288*r
oot(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^3*a^11*b^5*tan(c/2 + (d*x)
/2) - 86973087744*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^4*a^7*b
^10*tan(c/2 + (d*x)/2) + 88181047296*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 +
a^8, z, k)^4*a^9*b^8*tan(c/2 + (d*x)/2) - 30802968576*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)^5*a^7*b^11*tan(c/2 + (d*x)/2) + 18119393280*root(729*a^2*b^12*z^6 - 729*b^14*z^6
+ 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^5*a^9*b^9*tan(c/2 + (d*x)/2) - 86973087744*root(729*a^2*b^12*
z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^6*a^5*b^14*tan(c/2 + (d*x)/2) + 70665633792
*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^6*a^7*b^12*tan(c/2 + (d*
x)/2) - 40768634880*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^7*a^5
*b^15*tan(c/2 + (d*x)/2) + 32614907904*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2
+ a^8, z, k)^7*a^7*b^13*tan(c/2 + (d*x)/2) + 134217728*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4
+ 27*a^6*b^4*z^2 + a^8, z, k)*a^13*b*tan(c/2 + (d*x)/2))/b^7)*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*
b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k), k, 1, 6)/d - (4*tan(c/2 + (d*x)/2)^2)/(b*d + 3*b*d*tan(c/2 + (d*x)/2)^2
+ 3*b*d*tan(c/2 + (d*x)/2)^4 + b*d*tan(c/2 + (d*x)/2)^6) - 4/(3*(b*d + 3*b*d*tan(c/2 + (d*x)/2)^2 + 3*b*d*tan
(c/2 + (d*x)/2)^4 + b*d*tan(c/2 + (d*x)/2)^6)) + (a*log(tan(c/2 + (d*x)/2) - 1i)*1i)/(b^2*d) - (a*log(tan(c/2
+ (d*x)/2) + 1i)*1i)/(b^2*d)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{6}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**6/(a+b*sin(d*x+c)**3),x)

[Out]

Integral(sin(c + d*x)**6/(a + b*sin(c + d*x)**3), x)

________________________________________________________________________________________