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

Optimal. Leaf size=90 \[ \frac{2 b \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}}{a^2 f}-\frac{2 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{a^2 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

[Out]

(-2*b^2*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(a^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) + (2*b
*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(a^2*f)

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Rubi [A]  time = 0.111241, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2593, 2601, 2639} \[ \frac{2 b \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}}{a^2 f}-\frac{2 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{a^2 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(a^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) + (2*b
*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(a^2*f)

Rule 2593

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sin[e +
f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 1))/(a^2*f*(n - 1)), x] - Dist[(b^2*(m + 2))/(a^2*(n - 1)), Int[(a*Sin[e
+ f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ
[m, -1] && EqQ[n, 3/2])) && 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 2639

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

Rubi steps

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

Mathematica [C]  time = 0.264575, size = 92, normalized size = 1.02 \[ \frac{(b \tan (e+f x))^{3/2} \left (2 \cos (e+f x) \cos ^2(e+f x)^{3/4}-\cos ^3(e+f x) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )\right )}{a f \cos ^2(e+f x)^{3/4} \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Cos[e + f*x]*(Cos[e + f*x]^2)^(3/4) - Cos[e + f*x]^3*Hypergeometric2F1[1/4, 1/2, 3/2, Sin[e + f*x]^2])*(b*
Tan[e + f*x])^(3/2))/(a*f*(Cos[e + f*x]^2)^(3/4)*Sqrt[a*Sin[e + f*x]])

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Maple [C]  time = 0.163, size = 316, normalized size = 3.5 \begin{align*} -2\,{\frac{\cos \left ( fx+e \right ) }{f \left ( a\sin \left ( fx+e \right ) \right ) ^{3/2}\sin \left ( fx+e \right ) } \left ( 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 ) \cos \left ( fx+e \right ) -i\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) +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 ) -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 EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sin \left ( fx+e \right ) +\cos \left ( fx+e \right ) -1 \right ) \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{3/2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*(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)*si
n(f*x+e)*cos(f*x+e)-I*sin(f*x+e)*cos(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ellipti
cE(I*(cos(f*x+e)-1)/sin(f*x+e),I)+I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(co
s(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)-I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*
(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)+cos(f*x+e)-1)*cos(f*x+e)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)/(a*sin(f*x+e)
)^(3/2)/sin(f*x+e)

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

[Out]

integrate((b*tan(f*x + e))^(3/2)/(a*sin(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{\sqrt{a \sin \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )} b \tan \left (f x + e\right )}{a^{2} \cos \left (f x + e\right )^{2} - a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(a*sin(f*x + e))*sqrt(b*tan(f*x + e))*b*tan(f*x + e)/(a^2*cos(f*x + e)^2 - a^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((b*tan(f*x+e))**(3/2)/(a*sin(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 (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (a \sin \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((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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