3.135 \(\int \frac{(a \sin (e+f x))^{9/2}}{(b \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=109 \[ -\frac{2 a^2 (a \sin (e+f x))^{5/2}}{45 b f \sqrt{b \tan (e+f x)}}-\frac{8 a^4 \sqrt{a \sin (e+f x)}}{45 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{9/2}}{9 b f \sqrt{b \tan (e+f x)}} \]

[Out]

(-8*a^4*Sqrt[a*Sin[e + f*x]])/(45*b*f*Sqrt[b*Tan[e + f*x]]) - (2*a^2*(a*Sin[e + f*x])^(5/2))/(45*b*f*Sqrt[b*Ta
n[e + f*x]]) + (2*(a*Sin[e + f*x])^(9/2))/(9*b*f*Sqrt[b*Tan[e + f*x]])

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Rubi [A]  time = 0.150827, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2596, 2598, 2589} \[ -\frac{2 a^2 (a \sin (e+f x))^{5/2}}{45 b f \sqrt{b \tan (e+f x)}}-\frac{8 a^4 \sqrt{a \sin (e+f x)}}{45 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{9/2}}{9 b f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(9/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-8*a^4*Sqrt[a*Sin[e + f*x]])/(45*b*f*Sqrt[b*Tan[e + f*x]]) - (2*a^2*(a*Sin[e + f*x])^(5/2))/(45*b*f*Sqrt[b*Ta
n[e + f*x]]) + (2*(a*Sin[e + f*x])^(9/2))/(9*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))^{9/2}}{(b \tan (e+f x))^{3/2}} \, dx &=\frac{2 (a \sin (e+f x))^{9/2}}{9 b f \sqrt{b \tan (e+f x)}}+\frac{a^2 \int (a \sin (e+f x))^{5/2} \sqrt{b \tan (e+f x)} \, dx}{9 b^2}\\ &=-\frac{2 a^2 (a \sin (e+f x))^{5/2}}{45 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{9/2}}{9 b f \sqrt{b \tan (e+f x)}}+\frac{\left (4 a^4\right ) \int \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)} \, dx}{45 b^2}\\ &=-\frac{8 a^4 \sqrt{a \sin (e+f x)}}{45 b f \sqrt{b \tan (e+f x)}}-\frac{2 a^2 (a \sin (e+f x))^{5/2}}{45 b f \sqrt{b \tan (e+f x)}}+\frac{2 (a \sin (e+f x))^{9/2}}{9 b f \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.212795, size = 57, normalized size = 0.52 \[ \frac{a^4 \cos ^2(e+f x) (5 \cos (2 (e+f x))-13) \sqrt{a \sin (e+f x)}}{45 b f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(9/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(a^4*Cos[e + f*x]^2*(-13 + 5*Cos[2*(e + f*x)])*Sqrt[a*Sin[e + f*x]])/(45*b*f*Sqrt[b*Tan[e + f*x]])

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Maple [A]  time = 0.115, size = 60, normalized size = 0.6 \begin{align*}{\frac{ \left ( 10\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-18 \right ) \cos \left ( fx+e \right ) }{45\,f \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( a\sin \left ( fx+e \right ) \right ) ^{{\frac{9}{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))^(9/2)/(b*tan(f*x+e))^(3/2),x)

[Out]

2/45/f*(a*sin(f*x+e))^(9/2)*(5*cos(f*x+e)^2-9)*cos(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(3/2)/sin(f*x+e)^3

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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{9}{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))^(9/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(9/2)/(b*tan(f*x + e))^(3/2), x)

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Fricas [A]  time = 1.70836, size = 173, normalized size = 1.59 \begin{align*} \frac{2 \,{\left (5 \, a^{4} \cos \left (f x + e\right )^{5} - 9 \, a^{4} \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 )}}}{45 \, 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))^(9/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

2/45*(5*a^4*cos(f*x + e)^5 - 9*a^4*cos(f*x + e)^3)*sqrt(a*sin(f*x + e))*sqrt(b*sin(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))**(9/2)/(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{9}{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))^(9/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^(9/2)/(b*tan(f*x + e))^(3/2), x)