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

Optimal. Leaf size=86 \[ -\frac{\sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \tan (e+f x)}}{b^2 f \sqrt{a \sin (e+f x)}}-\frac{1}{b f \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.104479, antiderivative size = 86, 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 = {2597, 2601, 2641} \[ -\frac{\sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \tan (e+f x)}}{b^2 f \sqrt{a \sin (e+f x)}}-\frac{1}{b f \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2597

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 + n + 1)), x] - Dist[(n + 1)/(b^2*(m + 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] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

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

Mathematica [A]  time = 0.184657, size = 79, normalized size = 0.92 \[ \frac{\sin (e+f x) \left (-F\left (\left .\frac{1}{2} \sin ^{-1}(\sin (e+f x))\right |2\right )\right )-\sqrt [4]{\cos ^2(e+f x)}}{b f \sqrt [4]{\cos ^2(e+f x)} \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.16, size = 185, normalized size = 2.2 \begin{align*} -{\frac{\sin \left ( fx+e \right ) }{f \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \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\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 ) \right ){\frac{1}{\sqrt{a\sin \left ( fx+e \right ) }}} \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(1/(a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2),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*(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)/si
n(f*x+e),I)*sin(f*x+e)+cos(f*x+e))*sin(f*x+e)/cos(f*x+e)^2/(a*sin(f*x+e))^(1/2)/(b*sin(f*x+e)/cos(f*x+e))^(3/2
)

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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