3.134 \(\int \frac{(a \sin (e+f x))^{13/2}}{(b \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=146 \[ -\frac{2 a^2 (a \sin (e+f x))^{9/2}}{117 b f \sqrt{b \tan (e+f x)}}-\frac{16 a^4 (a \sin (e+f x))^{5/2}}{585 b f \sqrt{b \tan (e+f x)}}-\frac{64 a^6 \sqrt{a \sin (e+f x)}}{585 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{13/2}}{13 b f \sqrt{b \tan (e+f x)}} \]

[Out]

(-64*a^6*Sqrt[a*Sin[e + f*x]])/(585*b*f*Sqrt[b*Tan[e + f*x]]) - (16*a^4*(a*Sin[e + f*x])^(5/2))/(585*b*f*Sqrt[
b*Tan[e + f*x]]) - (2*a^2*(a*Sin[e + f*x])^(9/2))/(117*b*f*Sqrt[b*Tan[e + f*x]]) + (2*(a*Sin[e + f*x])^(13/2))
/(13*b*f*Sqrt[b*Tan[e + f*x]])

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Rubi [A]  time = 0.20672, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2596, 2598, 2589} \[ -\frac{2 a^2 (a \sin (e+f x))^{9/2}}{117 b f \sqrt{b \tan (e+f x)}}-\frac{16 a^4 (a \sin (e+f x))^{5/2}}{585 b f \sqrt{b \tan (e+f x)}}-\frac{64 a^6 \sqrt{a \sin (e+f x)}}{585 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{13/2}}{13 b f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(13/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-64*a^6*Sqrt[a*Sin[e + f*x]])/(585*b*f*Sqrt[b*Tan[e + f*x]]) - (16*a^4*(a*Sin[e + f*x])^(5/2))/(585*b*f*Sqrt[
b*Tan[e + f*x]]) - (2*a^2*(a*Sin[e + f*x])^(9/2))/(117*b*f*Sqrt[b*Tan[e + f*x]]) + (2*(a*Sin[e + f*x])^(13/2))
/(13*b*f*Sqrt[b*Tan[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 2598

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

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))^{13/2}}{(b \tan (e+f x))^{3/2}} \, dx &=\frac{2 (a \sin (e+f x))^{13/2}}{13 b f \sqrt{b \tan (e+f x)}}+\frac{a^2 \int (a \sin (e+f x))^{9/2} \sqrt{b \tan (e+f x)} \, dx}{13 b^2}\\ &=-\frac{2 a^2 (a \sin (e+f x))^{9/2}}{117 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{13/2}}{13 b f \sqrt{b \tan (e+f x)}}+\frac{\left (8 a^4\right ) \int (a \sin (e+f x))^{5/2} \sqrt{b \tan (e+f x)} \, dx}{117 b^2}\\ &=-\frac{16 a^4 (a \sin (e+f x))^{5/2}}{585 b f \sqrt{b \tan (e+f x)}}-\frac{2 a^2 (a \sin (e+f x))^{9/2}}{117 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{13/2}}{13 b f \sqrt{b \tan (e+f x)}}+\frac{\left (32 a^6\right ) \int \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)} \, dx}{585 b^2}\\ &=-\frac{64 a^6 \sqrt{a \sin (e+f x)}}{585 b f \sqrt{b \tan (e+f x)}}-\frac{16 a^4 (a \sin (e+f x))^{5/2}}{585 b f \sqrt{b \tan (e+f x)}}-\frac{2 a^2 (a \sin (e+f x))^{9/2}}{117 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{13/2}}{13 b f \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.423783, size = 67, normalized size = 0.46 \[ \frac{a^6 \cos ^2(e+f x) (340 \cos (2 (e+f x))-45 \cos (4 (e+f x))-551) \sqrt{a \sin (e+f x)}}{2340 b f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(13/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(a^6*Cos[e + f*x]^2*(-551 + 340*Cos[2*(e + f*x)] - 45*Cos[4*(e + f*x)])*Sqrt[a*Sin[e + f*x]])/(2340*b*f*Sqrt[b
*Tan[e + f*x]])

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Maple [A]  time = 0.151, size = 70, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 90\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-260\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+234 \right ) \cos \left ( fx+e \right ) }{585\,f \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( a\sin \left ( fx+e \right ) \right ) ^{{\frac{13}{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))^(13/2)/(b*tan(f*x+e))^(3/2),x)

[Out]

-2/585/f*(45*cos(f*x+e)^4-130*cos(f*x+e)^2+117)*(a*sin(f*x+e))^(13/2)*cos(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(3/
2)/sin(f*x+e)^5

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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{13}{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))^(13/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(13/2)/(b*tan(f*x + e))^(3/2), x)

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Fricas [A]  time = 1.81169, size = 213, normalized size = 1.46 \begin{align*} -\frac{2 \,{\left (45 \, a^{6} \cos \left (f x + e\right )^{7} - 130 \, a^{6} \cos \left (f x + e\right )^{5} + 117 \, a^{6} \cos \left (f x + e\right )^{3}\right )} \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{585 \, 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))^(13/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-2/585*(45*a^6*cos(f*x + e)^7 - 130*a^6*cos(f*x + e)^5 + 117*a^6*cos(f*x + e)^3)*sqrt(a*sin(f*x + e))*sqrt(b*s
in(f*x + e)/cos(f*x + e))/(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))**(13/2)/(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{13}{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))^(13/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^(13/2)/(b*tan(f*x + e))^(3/2), x)