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

Optimal. Leaf size=93 \[ \frac {2 a^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{3 b^2 f \sqrt {a \sin (e+f x)}}+\frac {2 (a \sin (e+f x))^{3/2}}{3 b f \sqrt {b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 93, 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 = {2596, 2601, 2641} \[ \frac {2 a^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{3 b^2 f \sqrt {a \sin (e+f x)}}+\frac {2 (a \sin (e+f x))^{3/2}}{3 b f \sqrt {b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2596

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), x] - Dist[(a^2*(n + 1))/(b^2*m), Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan
[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && GtQ[m, 1] && 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 {(a \sin (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx &=\frac {2 (a \sin (e+f x))^{3/2}}{3 b f \sqrt {b \tan (e+f x)}}+\frac {a^2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}} \, dx}{3 b^2}\\ &=\frac {2 (a \sin (e+f x))^{3/2}}{3 b f \sqrt {b \tan (e+f x)}}+\frac {\left (a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{3 b^2 \sqrt {a \sin (e+f x)}}\\ &=\frac {2 (a \sin (e+f x))^{3/2}}{3 b f \sqrt {b \tan (e+f x)}}+\frac {2 a^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{3 b^2 f \sqrt {a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 80, normalized size = 0.86 \[ \frac {2 a \sqrt {a \sin (e+f x)} \left (\sin (e+f x) \sqrt [4]{\cos ^2(e+f x)}+F\left (\left .\frac {1}{2} \sin ^{-1}(\sin (e+f x))\right |2\right )\right )}{3 b f \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sin \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )} a \sin \left (f x + e\right )}{b^{2} \tan \left (f x + e\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.48, size = 137, normalized size = 1.47 \[ -\frac {2 \sin \left (f x +e \right ) \left (i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )+\cos \left (f x +e \right )\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f \left (-1+\cos \left (f x +e \right )\right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/f*sin(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))/si
n(f*x+e),I)*sin(f*x+e)-cos(f*x+e)^2+cos(f*x+e))*(a*sin(f*x+e))^(3/2)/(-1+cos(f*x+e))/(b*sin(f*x+e)/cos(f*x+e))
^(3/2)/cos(f*x+e)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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