3.132 \(\int \frac{1}{(a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=87 \[ -\frac{b \sqrt{a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac{E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{a^2 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.105374, antiderivative size = 87, 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 = {2599, 2601, 2639} \[ -\frac{b \sqrt{a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac{E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{a^2 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2599

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*(m + n + 1)), x] + Dist[(m + 2)/(a^2*(m + n + 1)), Int[(a*Sin
[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n + 1, 0]
&& 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{1}{(a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}} \, dx &=-\frac{b \sqrt{a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac{\int \frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx}{2 a^2}\\ &=-\frac{b \sqrt{a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac{\sqrt{a \sin (e+f x)} \int \sqrt{\cos (e+f x)} \, dx}{2 a^2 \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{b \sqrt{a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac{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)}}\\ \end{align*}

Mathematica [C]  time = 0.337826, size = 89, normalized size = 1.02 \[ -\frac{b \sqrt{a \sin (e+f x)} \left (\sin ^2(e+f x) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )+2 \cos ^2(e+f x)^{3/4}\right )}{2 a^2 f \cos ^2(e+f x)^{3/4} (b \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.171, size = 315, normalized size = 3.6 \begin{align*} -{\frac{\sin \left ( fx+e \right ) }{f\cos \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 ) \right ) \left ( a\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x)

[Out]

-1/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))*sin(f*x+e)/(a*sin(f*x+e))^(3/2)/(b*sin(f*x+e)/cos(f*x+e))^
(1/2)/cos(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sin \left (f x + e\right )\right )^{\frac{3}{2}} \sqrt{b \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((a*sin(f*x + e))^(3/2)*sqrt(b*tan(f*x + e))), 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 )}}{{\left (a^{2} b \cos \left (f x + e\right )^{2} - a^{2} b\right )} \tan \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e))*sqrt(b*tan(f*x + e))/((a^2*b*cos(f*x + e)^2 - a^2*b)*tan(f*x + e)), 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(1/(a*sin(f*x+e))**(3/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sin \left (f x + e\right )\right )^{\frac{3}{2}} \sqrt{b \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((a*sin(f*x + e))^(3/2)*sqrt(b*tan(f*x + e))), x)