3.2.24 \(\int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx\) [124]

Optimal. Leaf size=90 \[ -\frac {2 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 90, 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 = {2673, 2681, 2719} \begin {gather*} \frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}-\frac {2 b^2 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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2673

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*(n - 1))), x] - Dist[b^2*((m + 2)/(a^2*(n - 1))), Int[(a*Sin[e
+ f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ
[m, -1] && EqQ[n, 3/2])) && 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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin {align*} \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx &=\frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}-\frac {b^2 \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{a^2}\\ &=\frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}-\frac {\left (b^2 \sqrt {a \sin (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 b^2 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)}}+\frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.30, size = 92, normalized size = 1.02 \begin {gather*} \frac {\left (2 \cos (e+f x) \cos ^2(e+f x)^{3/4}-\cos ^3(e+f x) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right )\right ) (b \tan (e+f x))^{3/2}}{a f \cos ^2(e+f x)^{3/4} \sqrt {a \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.41, size = 316, normalized size = 3.51

method	result	size
default	\(-\frac {2 \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-i \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )-1\right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right )}{f \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sin \left (f x +e \right )}\)	\(316\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((b*tan(f*x + e))^(3/2)/(a*sin(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 = 0.11, size = 110, normalized size = 1.22 \begin {gather*} \frac {\sqrt {2} \sqrt {-a b} b {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {2} \sqrt {-a b} b {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, \sqrt {a \sin \left (f x + e\right )} b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(2)*sqrt(-a*b)*b*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + sqrt
(2)*sqrt(-a*b)*b*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + 2*sqrt(a*
sin(f*x + e))*b*sqrt(b*sin(f*x + e)/cos(f*x + e)))/(a^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((b*tan(f*x+e))**(3/2)/(a*sin(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((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(a*sin(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 (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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