3.2.41 \(\int \frac {(a \sin (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx\) [141]

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

[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*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Elli
pticF(sin(1/2*f*x+1/2*e),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 = 0.08, 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 = {2676, 2681, 2720} \begin {gather*} \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)}} \end {gather*}

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 2676

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 2681

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], 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 = 0.25, size = 80, normalized size = 0.86 \begin {gather*} \frac {2 a \sqrt {a \sin (e+f x)} \left (F\left (\left .\frac {1}{2} \text {ArcSin}(\sin (e+f x))\right |2\right )+\sqrt [4]{\cos ^2(e+f x)} \sin (e+f x)\right )}{3 b f \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

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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Maple [C] Result contains complex when optimal does not.
time = 0.33, size = 137, normalized size = 1.47

method	result	size
default	\(-\frac {2 \sin \left (f x +e \right ) \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )+\cos \left (f x +e \right )\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f \left (\cos \left (f x +e \right )-1\right ) \cos \left (f x +e \right )^{2} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}\)	\(137\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 112, normalized size = 1.20 \begin {gather*} \frac {2 \, \sqrt {a \sin \left (f x + e\right )} a \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) + \sqrt {2} \sqrt {-a b} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} \sqrt {-a b} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}{3 \, b^{2} f} \end {gather*}

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]

1/3*(2*sqrt(a*sin(f*x + e))*a*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e) + sqrt(2)*sqrt(-a*b)*a*weierstras
sPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2)*sqrt(-a*b)*a*weierstrassPInverse(-4, 0, cos(f*x + e)
- I*sin(f*x + e)))/(b^2*f)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3007 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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