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

Optimal. Leaf size=87 \[ -\frac {b \sqrt {a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac {E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 87, 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, 2639} \[ -\frac {b \sqrt {a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac {E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.35, size = 89, normalized size = 1.02 \[ -\frac {b \sqrt {a \sin (e+f x)} \left (\sin ^2(e+f x) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right )+2 \cos ^2(e+f x)^{3/4}\right )}{2 a^2 f \cos ^2(e+f x)^{3/4} (b \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {a \sin \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )}}{{\left (a^{2} b \cos \left (f x + e\right )^{2} - a^{2} b\right )} \tan \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {b \tan \left (f x + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.55, size = 315, normalized size = 3.62 \[ -\frac {\left (i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-i \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )+i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )\right ) \sin \left (f x +e \right )}{f \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {b \tan \left (f x + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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