3.2.26 \(\int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx\) [126]

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

[Out]

8/15*a^4*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*(a*sin(f*x+e))^
(1/2)/f/cos(f*x+e)^(1/2)/(b*tan(f*x+e))^(1/2)-4/15*a^2*b*(a*sin(f*x+e))^(5/2)/f/(b*tan(f*x+e))^(3/2)-2/9*b*(a*
sin(f*x+e))^(9/2)/f/(b*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 123, 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 = {2678, 2681, 2719} \begin {gather*} \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 {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}} \end {gather*}

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 2678

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*
Tan[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 2681

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 2719

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.54, size = 100, normalized size = 0.81 \begin {gather*} \frac {a^4 \left (\cos ^2(e+f x)^{3/4} (-17+5 \cos (2 (e+f x)))+12 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right )\right ) \sqrt {a \sin (e+f x)} \sin (2 (e+f x))}{90 f \cos ^2(e+f x)^{3/4} \sqrt {b \tan (e+f x)}} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.73, size = 349, normalized size = 2.84

method	result	size
default	\(-\frac {2 \left (5 \left (\cos ^{6}\left (f x +e \right )\right )+12 i \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )-12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-16 \left (\cos ^{4}\left (f x +e \right )\right )+23 \left (\cos ^{2}\left (f x +e \right )\right )-12 \cos \left (f x +e \right )\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {9}{2}}}{45 f \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5} \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}\)	\(349\)

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,method=_RETURNVERBOSE)

[Out]

-2/45/f*(5*cos(f*x+e)^6+12*I*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos
(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)-12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ellip
ticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)*cos(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)-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)-16*cos(f*x+e)^4+23*cos(f*x+e)^2-12*cos(f
*x+e))*(a*sin(f*x+e))^(9/2)/cos(f*x+e)/sin(f*x+e)^5/(b*sin(f*x+e)/cos(f*x+e))^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 145, normalized size = 1.18 \begin {gather*} -\frac {2 \, {\left (6 \, \sqrt {2} \sqrt {-a b} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 6 \, \sqrt {2} \sqrt {-a b} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - {\left (5 \, a^{4} \cos \left (f x + e\right )^{4} - 11 \, a^{4} \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}\right )}}{45 \, b f} \end {gather*}

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]

-2/45*(6*sqrt(2)*sqrt(-a*b)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e
))) + 6*sqrt(2)*sqrt(-a*b)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)
)) - (5*a^4*cos(f*x + e)^4 - 11*a^4*cos(f*x + e)^2)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e)))/(b
*f)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^{9/2}}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(e + f*x))^(9/2)/(b*tan(e + f*x))^(1/2),x)

[Out]

int((a*sin(e + f*x))^(9/2)/(b*tan(e + f*x))^(1/2), x)

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