3.130 \(\int \frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=50 \[ \frac{2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

[Out]

(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]])

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Rubi [A]  time = 0.0511875, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2601, 2639} \[ \frac{2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*Sin[e + f*x]]/Sqrt[b*Tan[e + f*x]],x]

[Out]

(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]])

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{\sqrt{a \sin (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx &=\frac{\sqrt{a \sin (e+f x)} \int \sqrt{\cos (e+f x)} \, dx}{\sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=\frac{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)}}\\ \end{align*}

Mathematica [C]  time = 0.148261, size = 69, normalized size = 1.38 \[ \frac{\sin (2 (e+f x)) \sqrt{a \sin (e+f x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )}{2 f \cos ^2(e+f x)^{3/4} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*Sin[e + f*x]]/Sqrt[b*Tan[e + f*x]],x]

[Out]

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

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Maple [C]  time = 0.166, size = 327, normalized size = 6.5 \begin{align*} 2\,{\frac{\sqrt{a\sin \left ( fx+e \right ) }}{f\sin \left ( fx+e \right ) \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 ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}+\cos \left ( fx+e \right ) \right ){\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((a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x)

[Out]

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

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

[Out]

integrate(sqrt(a*sin(f*x + e))/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 )}}{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))^(1/2),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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \sin{\left (e + f x \right )}}}{\sqrt{b \tan{\left (e + f x \right )}}}\, dx \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))**(1/2),x)

[Out]

Integral(sqrt(a*sin(e + f*x))/sqrt(b*tan(e + f*x)), x)

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

[Out]

integrate(sqrt(a*sin(f*x + e))/sqrt(b*tan(f*x + e)), x)