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

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

[Out]

-2/45*a^2*(a*sin(f*x+e))^(5/2)/b/f/(b*tan(f*x+e))^(1/2)+2/9*(a*sin(f*x+e))^(9/2)/b/f/(b*tan(f*x+e))^(1/2)-8/45
*a^4*(a*sin(f*x+e))^(1/2)/b/f/(b*tan(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 109, 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 = {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 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]

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]

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 = 0.22, 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]])

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 71, normalized size = 0.65 \[ \frac {2 \, {\left (5 \, a^{4} \cos \left (f x + e\right )^{5} - 9 \, a^{4} \cos \left (f x + e\right )^{3}\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{45 \, b^{2} f \sin \left (f x + e\right )} \]

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))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {9}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.46, size = 60, normalized size = 0.55 \[ \frac {2 \left (a \sin \left (f x +e \right )\right )^{\frac {9}{2}} \left (5 \left (\cos ^{2}\left (f x +e \right )\right )-9\right ) \cos \left (f x +e \right )}{45 f \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )^{3}} \]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {9}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [B]  time = 5.48, size = 94, normalized size = 0.86 \[ \frac {a^4\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (47\,\sin \left (2\,e+2\,f\,x\right )+16\,\sin \left (4\,e+4\,f\,x\right )-5\,\sin \left (6\,e+6\,f\,x\right )\right )}{360\,b^2\,f\,\left (\cos \left (2\,e+2\,f\,x\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^4*(a*sin(e + f*x))^(1/2)*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2)*(47*sin(2*e + 2*f*x) + 16*sin(
4*e + 4*f*x) - 5*sin(6*e + 6*f*x)))/(360*b^2*f*(cos(2*e + 2*f*x) - 1))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________