3.139 \(\int \frac{(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=167 \[ \frac{8 a^6 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \tan (e+f x)}}{77 b^2 f \sqrt{a \sin (e+f x)}}-\frac{4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt{b \tan (e+f x)}}-\frac{2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt{b \tan (e+f x)}} \]

[Out]

(-4*a^4*(a*Sin[e + f*x])^(3/2))/(77*b*f*Sqrt[b*Tan[e + f*x]]) - (2*a^2*(a*Sin[e + f*x])^(7/2))/(77*b*f*Sqrt[b*
Tan[e + f*x]]) + (2*(a*Sin[e + f*x])^(11/2))/(11*b*f*Sqrt[b*Tan[e + f*x]]) + (8*a^6*Sqrt[Cos[e + f*x]]*Ellipti
cF[(e + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(77*b^2*f*Sqrt[a*Sin[e + f*x]])

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Rubi [A]  time = 0.226937, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2596, 2598, 2601, 2641} \[ \frac{8 a^6 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \tan (e+f x)}}{77 b^2 f \sqrt{a \sin (e+f x)}}-\frac{4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt{b \tan (e+f x)}}-\frac{2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(11/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-4*a^4*(a*Sin[e + f*x])^(3/2))/(77*b*f*Sqrt[b*Tan[e + f*x]]) - (2*a^2*(a*Sin[e + f*x])^(7/2))/(77*b*f*Sqrt[b*
Tan[e + f*x]]) + (2*(a*Sin[e + f*x])^(11/2))/(11*b*f*Sqrt[b*Tan[e + f*x]]) + (8*a^6*Sqrt[Cos[e + f*x]]*Ellipti
cF[(e + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(77*b^2*f*Sqrt[a*Sin[e + f*x]])

Rule 2596

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sin[e + f
*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] - Dist[(a^2*(n + 1))/(b^2*m), Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan
[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && GtQ[m, 1] && IntegersQ[2*m, 2*n]

Rule 2598

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[(b*(a*Sin[
e + f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] + Dist[(a^2*(m + n - 1))/m, Int[(a*Sin[e + f*x])^(m - 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && IntegersQ[2
*m, 2*n]

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx &=\frac{2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt{b \tan (e+f x)}}+\frac{a^2 \int (a \sin (e+f x))^{7/2} \sqrt{b \tan (e+f x)} \, dx}{11 b^2}\\ &=-\frac{2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt{b \tan (e+f x)}}+\frac{\left (6 a^4\right ) \int (a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)} \, dx}{77 b^2}\\ &=-\frac{4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt{b \tan (e+f x)}}-\frac{2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt{b \tan (e+f x)}}+\frac{\left (4 a^6\right ) \int \frac{\sqrt{b \tan (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx}{77 b^2}\\ &=-\frac{4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt{b \tan (e+f x)}}-\frac{2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt{b \tan (e+f x)}}+\frac{\left (4 a^6 \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx}{77 b^2 \sqrt{a \sin (e+f x)}}\\ &=-\frac{4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt{b \tan (e+f x)}}-\frac{2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt{b \tan (e+f x)}}+\frac{8 a^6 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \tan (e+f x)}}{77 b^2 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.749951, size = 118, normalized size = 0.71 \[ \frac{a^5 \tan ^2(e+f x) \sqrt{a \sin (e+f x)} \left (\sqrt [4]{\cos ^2(e+f x)} (-22 \cos (e+f x)-17 \cos (3 (e+f x))+7 \cos (5 (e+f x)))+64 \cot (e+f x) F\left (\left .\frac{1}{2} \sin ^{-1}(\sin (e+f x))\right |2\right )\right )}{616 f \sqrt [4]{\cos ^2(e+f x)} (b \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(11/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(a^5*((Cos[e + f*x]^2)^(1/4)*(-22*Cos[e + f*x] - 17*Cos[3*(e + f*x)] + 7*Cos[5*(e + f*x)]) + 64*Cot[e + f*x]*E
llipticF[ArcSin[Sin[e + f*x]]/2, 2])*Sqrt[a*Sin[e + f*x]]*Tan[e + f*x]^2)/(616*f*(Cos[e + f*x]^2)^(1/4)*(b*Tan
[e + f*x])^(3/2))

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Maple [C]  time = 0.221, size = 181, normalized size = 1.1 \begin{align*} -{\frac{2}{77\,f \left ( \cos \left ( fx+e \right ) -1 \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( -7\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+4\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sin \left ( fx+e \right ) +7\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}+13\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-13\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4\,\cos \left ( fx+e \right ) \right ) \left ( a\sin \left ( fx+e \right ) \right ) ^{{\frac{11}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x)

[Out]

-2/77/f*(-7*cos(f*x+e)^6+4*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e
)-1)/sin(f*x+e),I)*sin(f*x+e)+7*cos(f*x+e)^5+13*cos(f*x+e)^4-13*cos(f*x+e)^3-4*cos(f*x+e)^2+4*cos(f*x+e))*(a*s
in(f*x+e))^(11/2)/(cos(f*x+e)-1)/sin(f*x+e)^3/(b*sin(f*x+e)/cos(f*x+e))^(3/2)/cos(f*x+e)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \sin \left (f x + e\right )\right )^{\frac{11}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(11/2)/(b*tan(f*x + e))^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{5} \cos \left (f x + e\right )^{4} - 2 \, a^{5} \cos \left (f x + e\right )^{2} + a^{5}\right )} \sqrt{a \sin \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )} \sin \left (f x + e\right )}{b^{2} \tan \left (f x + e\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral((a^5*cos(f*x + e)^4 - 2*a^5*cos(f*x + e)^2 + a^5)*sqrt(a*sin(f*x + e))*sqrt(b*tan(f*x + e))*sin(f*x +
e)/(b^2*tan(f*x + e)^2), x)

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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((a*sin(f*x+e))**(11/2)/(b*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \sin \left (f x + e\right )\right )^{\frac{11}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^(11/2)/(b*tan(f*x + e))^(3/2), x)