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) + (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)

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Rubi [A]  time = 0.39395, antiderivative size = 293, normalized size of antiderivative = 1., 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 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]))

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 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 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{\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 = 0.272872, size = 164, normalized size = 0.56 \[ -\frac{8 i a^2 \text{RootSum}\left [8 \text{$\#$1}^3 a+i \text{$\#$1}^6 b-3 i \text{$\#$1}^4 b+3 i \text{$\#$1}^2 b-i b\& ,\frac{2 \text{$\#$1} \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-i \text{$\#$1} \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )}{\text{$\#$1}^4 b-2 \text{$\#$1}^2 b-4 i \text{$\#$1} a+b}\& \right ]+12 a c+12 a d x+9 b \cos (c+d x)-b \cos (3 (c+d x))}{12 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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) & ])/(12*b^2*d)

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Maple [C]  time = 0.161, size = 166, normalized size = 0.6 \begin{align*}{\frac{{a}^{2}}{3\,{b}^{2}d}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{6}+3\,a{{\it \_Z}}^{4}+8\,b{{\it \_Z}}^{3}+3\,a{{\it \_Z}}^{2}+a \right ) }{\frac{{{\it \_R}}^{4}+2\,{{\it \_R}}^{2}+1}{{{\it \_R}}^{5}a+2\,{{\it \_R}}^{3}a+4\,{{\it \_R}}^{2}b+{\it \_R}\,a}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \right ) }}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{4}{3\,bd} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-2\,{\frac{a\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{b}^{2}d}} \end{align*}

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]

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))-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))

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

[Out]

Timed out

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{6}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \end{align*}

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)