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*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)

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Rubi [A]  time = 0.457509, antiderivative size = 259, normalized size of antiderivative = 1., 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 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 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 ^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*}

Mathematica [C]  time = 0.179957, size = 140, normalized size = 0.54 \[ \frac{2 i a \text{RootSum}\left [8 \text{$\#$1}^3 a+i \text{$\#$1}^6 b-3 i \text{$\#$1}^4 b+3 i \text{$\#$1}^2 b-i b\& ,\frac{2 \text{$\#$1} \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-i \text{$\#$1} \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )}{\text{$\#$1}^4 b-2 \text{$\#$1}^2 b-4 i \text{$\#$1} a+b}\& \right ]+3 c+3 d x}{3 b d} \]

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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Maple [C]  time = 0.153, size = 106, normalized size = 0.4 \begin{align*} -{\frac{a}{3\,bd}\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 ) }}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{bd}} \end{align*}

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]

-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))+2/d/b*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)^3/(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)^3/(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)**3/(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 )^{3}}{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)^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)