3.136 \(\int \frac{(a \sin (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=32 \[ -\frac{2 b (a \sin (e+f x))^{5/2}}{5 f (b \tan (e+f x))^{5/2}} \]

[Out]

(-2*b*(a*Sin[e + f*x])^(5/2))/(5*f*(b*Tan[e + f*x])^(5/2))

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Rubi [A]  time = 0.0547281, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2589} \[ -\frac{2 b (a \sin (e+f x))^{5/2}}{5 f (b \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(5/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-2*b*(a*Sin[e + f*x])^(5/2))/(5*f*(b*Tan[e + f*x])^(5/2))

Rule 2589

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))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx &=-\frac{2 b (a \sin (e+f x))^{5/2}}{5 f (b \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.131872, size = 45, normalized size = 1.41 \[ -\frac{2 a^2 \cos ^2(e+f x) \sqrt{a \sin (e+f x)}}{5 b f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(5/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-2*a^2*Cos[e + f*x]^2*Sqrt[a*Sin[e + f*x]])/(5*b*f*Sqrt[b*Tan[e + f*x]])

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Maple [A]  time = 0.112, size = 48, normalized size = 1.5 \begin{align*} -{\frac{2\,\cos \left ( fx+e \right ) }{5\,f\sin \left ( fx+e \right ) } \left ( a\sin \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x)

[Out]

-2/5/f*(a*sin(f*x+e))^(5/2)*cos(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(3/2)/sin(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(5/2)/(b*tan(f*x + e))^(3/2), x)

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Fricas [B]  time = 1.60006, size = 136, normalized size = 4.25 \begin{align*} -\frac{2 \, \sqrt{a \sin \left (f x + e\right )} a^{2} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{3}}{5 \, b^{2} f \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-2/5*sqrt(a*sin(f*x + e))*a^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)^3/(b^2*f*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))**(5/2)/(b*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^(5/2)/(b*tan(f*x + e))^(3/2), x)