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

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

[Out]

-16/585*a^4*(a*sin(f*x+e))^(5/2)/b/f/(b*tan(f*x+e))^(1/2)-2/117*a^2*(a*sin(f*x+e))^(9/2)/b/f/(b*tan(f*x+e))^(1
/2)+2/13*(a*sin(f*x+e))^(13/2)/b/f/(b*tan(f*x+e))^(1/2)-64/585*a^6*(a*sin(f*x+e))^(1/2)/b/f/(b*tan(f*x+e))^(1/
2)

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Rubi [A]  time = 0.21, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2596, 2598, 2589} \[ -\frac {2 a^2 (a \sin (e+f x))^{9/2}}{117 b f \sqrt {b \tan (e+f x)}}-\frac {16 a^4 (a \sin (e+f x))^{5/2}}{585 b f \sqrt {b \tan (e+f x)}}-\frac {64 a^6 \sqrt {a \sin (e+f x)}}{585 b f \sqrt {b \tan (e+f x)}}+\frac {2 (a \sin (e+f x))^{13/2}}{13 b f \sqrt {b \tan (e+f x)}} \]

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 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))^{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*}

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Mathematica [A]  time = 0.43, 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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fricas [A]  time = 0.49, size = 84, normalized size = 0.58 \[ -\frac {2 \, {\left (45 \, a^{6} \cos \left (f x + e\right )^{7} - 130 \, a^{6} \cos \left (f x + e\right )^{5} + 117 \, a^{6} \cos \left (f x + e\right )^{3}\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{585 \, b^{2} f \sin \left (f x + e\right )} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {13}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.49, size = 70, normalized size = 0.48 \[ -\frac {2 \left (45 \left (\cos ^{4}\left (f x +e \right )\right )-130 \left (\cos ^{2}\left (f x +e \right )\right )+117\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {13}{2}} \cos \left (f x +e \right )}{585 f \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )^{5}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {13}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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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mupad [B]  time = 8.60, size = 296, normalized size = 2.03 \[ \frac {\left (\cos \left (7\,e+7\,f\,x\right )-\sin \left (7\,e+7\,f\,x\right )\,1{}\mathrm {i}\right )\,\sqrt {\frac {b\,\left (\sin \left (2\,e+2\,f\,x\right )-\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}}}\,\left (\frac {a^6\,\cos \left (3\,e+3\,f\,x\right )\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\left (\cos \left (7\,e+7\,f\,x\right )+\sin \left (7\,e+7\,f\,x\right )\,1{}\mathrm {i}\right )\,217{}\mathrm {i}}{9360\,b^2\,f}-\frac {a^6\,\cos \left (5\,e+5\,f\,x\right )\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\left (\cos \left (7\,e+7\,f\,x\right )+\sin \left (7\,e+7\,f\,x\right )\,1{}\mathrm {i}\right )\,41{}\mathrm {i}}{1872\,b^2\,f}+\frac {a^6\,\cos \left (7\,e+7\,f\,x\right )\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\left (\cos \left (7\,e+7\,f\,x\right )+\sin \left (7\,e+7\,f\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{208\,b^2\,f}+\frac {a^6\,\cos \left (e+f\,x\right )\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\left (\cos \left (7\,e+7\,f\,x\right )+\sin \left (7\,e+7\,f\,x\right )\,1{}\mathrm {i}\right )\,1991{}\mathrm {i}}{9360\,b^2\,f}\right )\,1{}\mathrm {i}}{2\,\sin \left (e+f\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((cos(7*e + 7*f*x) - sin(7*e + 7*f*x)*1i)*((b*(sin(2*e + 2*f*x) - cos(2*e + 2*f*x)*1i + 1i))/(cos(2*e + 2*f*x)
+ sin(2*e + 2*f*x)*1i + 1))^(1/2)*((a^6*cos(3*e + 3*f*x)*(a*sin(e + f*x))^(1/2)*(cos(7*e + 7*f*x) + sin(7*e +
7*f*x)*1i)*217i)/(9360*b^2*f) - (a^6*cos(5*e + 5*f*x)*(a*sin(e + f*x))^(1/2)*(cos(7*e + 7*f*x) + sin(7*e + 7*
f*x)*1i)*41i)/(1872*b^2*f) + (a^6*cos(7*e + 7*f*x)*(a*sin(e + f*x))^(1/2)*(cos(7*e + 7*f*x) + sin(7*e + 7*f*x)
*1i)*1i)/(208*b^2*f) + (a^6*cos(e + f*x)*(a*sin(e + f*x))^(1/2)*(cos(7*e + 7*f*x) + sin(7*e + 7*f*x)*1i)*1991i
)/(9360*b^2*f))*1i)/(2*sin(e + f*x))

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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))**(13/2)/(b*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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