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/(a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)+(cos(1/2*e+1/2*f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*EllipticF(s
in(1/2*e+1/2*f*x),2^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/a^2/f/(a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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*}

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Mathematica [A]  time = 0.25, 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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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {a \sin \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )}}{{\left (a^{3} \cos \left (f x + e\right )^{2} - a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [C]  time = 0.51, size = 178, normalized size = 2.07 \[ \frac {\left (i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-\cos \left (f x +e \right )\right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}{f \left (a \sin \left (f x +e \right )\right )^{\frac {5}{2}}} \]

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/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*si
n(f*x+e)*cos(f*x+e)+I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/s
in(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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