3.119 \(\int \frac{\sqrt{b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.103392, 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 = {2599, 2601, 2641} \[ \frac{\sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}}-\frac{b}{a^2 f \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]]/(a*Sin[e + f*x])^(5/2),x]

[Out]

-(b/(a^2*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*
Tan[e + f*x]])/(a^2*f*Sqrt[a*Sin[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 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{\sqrt{b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx &=-\frac{b}{a^2 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 a^2}\\ &=-\frac{b}{a^2 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 a^2 \sqrt{a \sin (e+f x)}}\\ &=-\frac{b}{a^2 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)}}{a^2 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.243335, size = 79, normalized size = 0.92 \[ \frac{b \left (\sin (e+f x) F\left (\left .\frac{1}{2} \sin ^{-1}(\sin (e+f x))\right |2\right )-\sqrt [4]{\cos ^2(e+f x)}\right )}{a^2 f \sqrt [4]{\cos ^2(e+f x)} \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[e + f*x]]/(a*Sin[e + f*x])^(5/2),x]

[Out]

(b*(-(Cos[e + f*x]^2)^(1/4) + EllipticF[ArcSin[Sin[e + f*x]]/2, 2]*Sin[e + f*x]))/(a^2*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 = 0.167, size = 178, normalized size = 2.1 \begin{align*}{\frac{\sin \left ( fx+e \right ) }{f} \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 ) \sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}} \left ( a\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),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)*sin
(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)/sin
(f*x+e),I)*sin(f*x+e)-cos(f*x+e))*(b*sin(f*x+e)/cos(f*x+e))^(1/2)*sin(f*x+e)/(a*sin(f*x+e))^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \tan \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e))/(a*sin(f*x + e))^(5/2), 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 )}}{{\left (a^{3} \cos \left (f x + e\right )^{2} - a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e))*sqrt(b*tan(f*x + e))/((a^3*cos(f*x + e)^2 - a^3)*sin(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((b*tan(f*x+e))**(1/2)/(a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \tan \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*tan(f*x + e))/(a*sin(f*x + e))^(5/2), x)