3.183 \(\int \frac{\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=273 \[ -\frac{2 a \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{a^{2/3}-b^{2/3}}}+\frac{2 a \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{5/3} d \sqrt{b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac{2 a \tanh ^{-1}\left (\frac{(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac{\sin (c+d x) \cos (c+d x)}{2 b d}+\frac{x}{2 b} \]

[Out]

x/(2*b) - (2*a*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^(5/3)*d) + (2*a*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/
3)]])/(3*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*b^(5/3)*d) + (2*a*ArcTanh[(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c
+ d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]*b^(5/3)*d) - (Cos[c + d*
x]*Sin[c + d*x])/(2*b*d)

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Rubi [A]  time = 0.566971, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3220, 2635, 8, 2660, 618, 204, 206} \[ -\frac{2 a \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{a^{2/3}-b^{2/3}}}+\frac{2 a \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{5/3} d \sqrt{b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac{2 a \tanh ^{-1}\left (\frac{(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac{\sin (c+d x) \cos (c+d x)}{2 b d}+\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/(a + b*Sin[c + d*x]^3),x]

[Out]

x/(2*b) - (2*a*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^(5/3)*d) + (2*a*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/
3)]])/(3*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*b^(5/3)*d) + (2*a*ArcTanh[(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c
+ d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]*b^(5/3)*d) - (Cos[c + d*
x]*Sin[c + d*x])/(2*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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx &=\int \left (\frac{\sin ^2(c+d x)}{b}-\frac{a \sin ^2(c+d x)}{b \left (a+b \sin ^3(c+d x)\right )}\right ) \, dx\\ &=\frac{\int \sin ^2(c+d x) \, dx}{b}-\frac{a \int \frac{\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx}{b}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int 1 \, dx}{2 b}-\frac{a \int \left (\frac{1}{3 b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}+\frac{1}{3 b^{2/3} \left (-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}+\frac{1}{3 b^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}\right ) \, dx}{b}\\ &=\frac{x}{2 b}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}-\frac{a \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{5/3}}-\frac{a \int \frac{1}{-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{5/3}}-\frac{a \int \frac{1}{(-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{5/3}}\\ &=\frac{x}{2 b}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}-\frac{(2 a) \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^{5/3} d}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{-\sqrt [3]{-1} \sqrt [3]{a}+2 \sqrt [3]{b} x-\sqrt [3]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 b^{5/3} d}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{(-1)^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x+(-1)^{2/3} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 b^{5/3} d}\\ &=\frac{x}{2 b}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{(4 a) \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^{5/3} d}+\frac{(4 a) \operatorname{Subst}\left (\int \frac{1}{-4 \left ((-1)^{2/3} a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{b}-2 \sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 b^{5/3} d}+\frac{(4 a) \operatorname{Subst}\left (\int \frac{1}{4 \left (\sqrt [3]{-1} a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{b}+2 (-1)^{2/3} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 b^{5/3} d}\\ &=\frac{x}{2 b}-\frac{2 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^{5/3} d}+\frac{2 a \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt{-(-1)^{2/3} a^{2/3}+b^{2/3}} b^{5/3} d}+\frac{2 a \tanh ^{-1}\left (\frac{\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}} b^{5/3} d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}

Mathematica [C]  time = 0.277522, size = 255, normalized size = 0.93 \[ \frac{-2 i a \text{RootSum}\left [8 \text{$\#$1}^3 a+i \text{$\#$1}^6 b-3 i \text{$\#$1}^4 b+3 i \text{$\#$1}^2 b-i b\& ,\frac{-i \text{$\#$1}^4 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 i \text{$\#$1}^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-i \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^4 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-4 \text{$\#$1}^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-4 i \text{$\#$1}^2 a+\text{$\#$1}^5 b-2 \text{$\#$1}^3 b+\text{$\#$1} b}\& \right ]+6 (c+d x)-3 \sin (2 (c+d x))}{12 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5/(a + b*Sin[c + d*x]^3),x]

[Out]

(6*(c + 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)] - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] - 4*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]
*#1^2 + (2*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*Log
[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)*a*#1^2 - 2*b*#1^3 + b*#1^5) & ] - 3*Sin[2*(c + d*x)])/(12*b
*d)

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Maple [C]  time = 0.161, size = 163, normalized size = 0.6 \begin{align*} -{\frac{4\,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}}^{2}}{{{\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 ) }}+{\frac{1}{bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-{\frac{1}{bd}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+{\frac{1}{bd}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a+b*sin(d*x+c)^3),x)

[Out]

-4/3/d*a/b*sum(_R^2/(_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))+1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2
*d*x+1/2*c)+1/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)^5/(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)^5/(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)**5/(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 )^{5}}{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)^5/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^5/(b*sin(d*x + c)^3 + a), x)