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

Optimal. Leaf size=84 \[ \frac{2 b \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}}{f}-\frac{4 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

[Out]

(-4*b^2*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) + (2*b*Sqr
t[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])/f

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Rubi [A]  time = 0.101287, antiderivative size = 84, 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 = {2594, 2601, 2639} \[ \frac{2 b \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}}{f}-\frac{4 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b^2*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) + (2*b*Sqr
t[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])/f

Rule 2594

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*(n - 1)), x] - Dist[(b^2*(m + n - 1))/(n - 1), Int[(a*Sin[e + f*x])^m*(
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n] &&  !(GtQ[m,
1] &&  !IntegerQ[(m - 1)/2])

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

Mathematica [C]  time = 0.194348, size = 83, normalized size = 0.99 \[ \frac{2 b \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)} \left (\cos ^2(e+f x)^{3/4}-\cos ^2(e+f x) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )\right )}{f \cos ^2(e+f x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.172, size = 328, normalized size = 3.9 \begin{align*} -2\,{\frac{\cos \left ( fx+e \right ) \sqrt{a\sin \left ( fx+e \right ) }}{f \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( 2\,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 ) -2\,i\sin \left ( fx+e \right ) \cos \left ( fx+e \right ){\it EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}+2\,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 ) -2\,i\sin \left ( fx+e \right ){\it EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\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((a*sin(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x)

[Out]

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

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

[Out]

integrate(sqrt(a*sin(f*x + e))*(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 (\sqrt{a \sin \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )} b \tan \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [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))^(1/2)*(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out