3.2.29 \(\int \frac {(a \sin (e+f x))^{3/2}}{\sqrt {b \tan (e+f x)}} \, dx\) [129]

Optimal. Leaf size=32 \[ -\frac {2 b (a \sin (e+f x))^{3/2}}{3 f (b \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2669} \begin {gather*} -\frac {2 b (a \sin (e+f x))^{3/2}}{3 f (b \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2669

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*(a*Sin[e
+ f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 32, normalized size = 1.00 \begin {gather*} -\frac {2 b (a \sin (e+f x))^{3/2}}{3 f (b \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.31, size = 48, normalized size = 1.50

method	result	size
default	\(-\frac {2 \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \cos \left (f x +e \right )}{3 f \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (28) = 56\).
time = 0.36, size = 58, normalized size = 1.81 \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 )^{2}}{3 \, b f \sin \left (f x + e\right )} \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))^(1/2),x, algorithm="fricas")

[Out]

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

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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))**(1/2),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))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.63, size = 69, normalized size = 2.16 \begin {gather*} -\frac {a\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\left (\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{6\,b\,f\,{\sin \left (e+f\,x\right )}^2} \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))^(1/2),x)

[Out]

-(a*(a*sin(e + f*x))^(1/2)*(sin(e + f*x) + sin(3*e + 3*f*x))*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/
2))/(6*b*f*sin(e + f*x)^2)

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