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

Optimal. Leaf size=123 \[ -\frac{4 a^2 b (a \sin (e+f x))^{5/2}}{15 f (b \tan (e+f x))^{3/2}}+\frac{8 a^4 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{15 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}} \]

[Out]

(-4*a^2*b*(a*Sin[e + f*x])^(5/2))/(15*f*(b*Tan[e + f*x])^(3/2)) - (2*b*(a*Sin[e + f*x])^(9/2))/(9*f*(b*Tan[e +
f*x])^(3/2)) + (8*a^4*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(15*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e +
f*x]])

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Rubi [A]  time = 0.161566, antiderivative size = 123, 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 = {2598, 2601, 2639} \[ -\frac{4 a^2 b (a \sin (e+f x))^{5/2}}{15 f (b \tan (e+f x))^{3/2}}+\frac{8 a^4 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{15 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*b*(a*Sin[e + f*x])^(5/2))/(15*f*(b*Tan[e + f*x])^(3/2)) - (2*b*(a*Sin[e + f*x])^(9/2))/(9*f*(b*Tan[e +
f*x])^(3/2)) + (8*a^4*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(15*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e +
f*x]])

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 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a \sin (e+f x))^{9/2}}{\sqrt{b \tan (e+f x)}} \, dx &=-\frac{2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}+\frac{1}{3} \left (2 a^2\right ) \int \frac{(a \sin (e+f x))^{5/2}}{\sqrt{b \tan (e+f x)}} \, dx\\ &=-\frac{4 a^2 b (a \sin (e+f x))^{5/2}}{15 f (b \tan (e+f x))^{3/2}}-\frac{2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}+\frac{1}{15} \left (4 a^4\right ) \int \frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx\\ &=-\frac{4 a^2 b (a \sin (e+f x))^{5/2}}{15 f (b \tan (e+f x))^{3/2}}-\frac{2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}+\frac{\left (4 a^4 \sqrt{a \sin (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \, dx}{15 \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{4 a^2 b (a \sin (e+f x))^{5/2}}{15 f (b \tan (e+f x))^{3/2}}-\frac{2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}+\frac{8 a^4 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{a \sin (e+f x)}}{15 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.515641, size = 100, normalized size = 0.81 \[ \frac{a^4 \sin (2 (e+f x)) \sqrt{a \sin (e+f x)} \left (12 \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )+\cos ^2(e+f x)^{3/4} (5 \cos (2 (e+f x))-17)\right )}{90 f \cos ^2(e+f x)^{3/4} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*((Cos[e + f*x]^2)^(3/4)*(-17 + 5*Cos[2*(e + f*x)]) + 12*Hypergeometric2F1[1/4, 1/2, 3/2, Sin[e + f*x]^2])
*Sqrt[a*Sin[e + f*x]]*Sin[2*(e + f*x)])/(90*f*(Cos[e + f*x]^2)^(3/4)*Sqrt[b*Tan[e + f*x]])

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Maple [C]  time = 0.203, size = 349, normalized size = 2.8 \begin{align*}{\frac{2}{45\,f \left ( \sin \left ( fx+e \right ) \right ) ^{5}\cos \left ( fx+e \right ) } \left ( 12\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -12\,i\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) -5\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+12\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sin \left ( fx+e \right ) -12\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sin \left ( fx+e \right ) +16\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-23\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+12\,\cos \left ( fx+e \right ) \right ) \left ( a\sin \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \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))^(1/2),x)

[Out]

2/45/f*(12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),
I)*sin(f*x+e)*cos(f*x+e)-12*I*sin(f*x+e)*cos(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)
*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)-5*cos(f*x+e)^6+12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)
+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)-12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(
f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)+16*cos(f*x+e)^4-23*cos(f*x+e)^2+12*cos(f*
x+e))*(a*sin(f*x+e))^(9/2)/sin(f*x+e)^5/(b*sin(f*x+e)/cos(f*x+e))^(1/2)/cos(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{9}{2}}}{\sqrt{b \tan \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{4} \cos \left (f x + e\right )^{4} - 2 \, a^{4} \cos \left (f x + e\right )^{2} + a^{4}\right )} \sqrt{a \sin \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}{b \tan \left (f x + e\right )}, x\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))^(1/2),x, algorithm="fricas")

[Out]

integral((a^4*cos(f*x + e)^4 - 2*a^4*cos(f*x + e)^2 + a^4)*sqrt(a*sin(f*x + e))*sqrt(b*tan(f*x + e))/(b*tan(f*
x + e)), x)

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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))**(1/2),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}}}{\sqrt{b \tan \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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