3.2.39 \(\int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx\) [139]

Optimal. Leaf size=167 \[ -\frac {4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt {b \tan (e+f x)}}-\frac {2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt {b \tan (e+f x)}}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}+\frac {8 a^6 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{77 b^2 f \sqrt {a \sin (e+f x)}} \]

[Out]

-4/77*a^4*(a*sin(f*x+e))^(3/2)/b/f/(b*tan(f*x+e))^(1/2)-2/77*a^2*(a*sin(f*x+e))^(7/2)/b/f/(b*tan(f*x+e))^(1/2)
+2/11*(a*sin(f*x+e))^(11/2)/b/f/(b*tan(f*x+e))^(1/2)+8/77*a^6*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*
EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/b^2/f/(a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2676, 2678, 2681, 2720} \begin {gather*} \frac {8 a^6 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{77 b^2 f \sqrt {a \sin (e+f x)}}-\frac {4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt {b \tan (e+f x)}}-\frac {2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt {b \tan (e+f x)}}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(11/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-4*a^4*(a*Sin[e + f*x])^(3/2))/(77*b*f*Sqrt[b*Tan[e + f*x]]) - (2*a^2*(a*Sin[e + f*x])^(7/2))/(77*b*f*Sqrt[b*
Tan[e + f*x]]) + (2*(a*Sin[e + f*x])^(11/2))/(11*b*f*Sqrt[b*Tan[e + f*x]]) + (8*a^6*Sqrt[Cos[e + f*x]]*Ellipti
cF[(e + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(77*b^2*f*Sqrt[a*Sin[e + f*x]])

Rule 2676

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 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 2720

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

Rubi steps

\begin {align*} \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx &=\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}+\frac {a^2 \int (a \sin (e+f x))^{7/2} \sqrt {b \tan (e+f x)} \, dx}{11 b^2}\\ &=-\frac {2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt {b \tan (e+f x)}}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}+\frac {\left (6 a^4\right ) \int (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)} \, dx}{77 b^2}\\ &=-\frac {4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt {b \tan (e+f x)}}-\frac {2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt {b \tan (e+f x)}}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}+\frac {\left (4 a^6\right ) \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}} \, dx}{77 b^2}\\ &=-\frac {4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt {b \tan (e+f x)}}-\frac {2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt {b \tan (e+f x)}}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}+\frac {\left (4 a^6 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{77 b^2 \sqrt {a \sin (e+f x)}}\\ &=-\frac {4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt {b \tan (e+f x)}}-\frac {2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt {b \tan (e+f x)}}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}+\frac {8 a^6 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{77 b^2 f \sqrt {a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 118, normalized size = 0.71 \begin {gather*} \frac {a^5 \left (\sqrt [4]{\cos ^2(e+f x)} (-22 \cos (e+f x)-17 \cos (3 (e+f x))+7 \cos (5 (e+f x)))+64 \cot (e+f x) F\left (\left .\frac {1}{2} \text {ArcSin}(\sin (e+f x))\right |2\right )\right ) \sqrt {a \sin (e+f x)} \tan ^2(e+f x)}{616 f \sqrt [4]{\cos ^2(e+f x)} (b \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(11/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(a^5*((Cos[e + f*x]^2)^(1/4)*(-22*Cos[e + f*x] - 17*Cos[3*(e + f*x)] + 7*Cos[5*(e + f*x)]) + 64*Cot[e + f*x]*E
llipticF[ArcSin[Sin[e + f*x]]/2, 2])*Sqrt[a*Sin[e + f*x]]*Tan[e + f*x]^2)/(616*f*(Cos[e + f*x]^2)^(1/4)*(b*Tan
[e + f*x])^(3/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.84, size = 181, normalized size = 1.08

method	result	size
default	\(-\frac {2 \left (-7 \left (\cos ^{6}\left (f x +e \right )\right )+4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )+7 \left (\cos ^{5}\left (f x +e \right )\right )+13 \left (\cos ^{4}\left (f x +e \right )\right )-13 \left (\cos ^{3}\left (f x +e \right )\right )-4 \left (\cos ^{2}\left (f x +e \right )\right )+4 \cos \left (f x +e \right )\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {11}{2}}}{77 f \left (\cos \left (f x +e \right )-1\right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{3} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/77/f*(-7*cos(f*x+e)^6+4*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)+7*cos(f*x+e)^5+13*cos(f*x+e)^4-13*cos(f*x+e)^3-4*cos(f*x+e)^2+4*cos(f*x+e))*(a*s
in(f*x+e))^(11/2)/(cos(f*x+e)-1)/cos(f*x+e)^2/sin(f*x+e)^3/(b*sin(f*x+e)/cos(f*x+e))^(3/2)

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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))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(11/2)/(b*tan(f*x + e))^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 150, normalized size = 0.90 \begin {gather*} \frac {2 \, {\left (2 \, \sqrt {2} \sqrt {-a b} a^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 2 \, \sqrt {2} \sqrt {-a b} a^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + {\left (7 \, a^{5} \cos \left (f x + e\right )^{5} - 13 \, a^{5} \cos \left (f x + e\right )^{3} + 4 \, a^{5} \cos \left (f x + e\right )\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 )}}{77 \, b^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

2/77*(2*sqrt(2)*sqrt(-a*b)*a^5*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 2*sqrt(2)*sqrt(-a*b
)*a^5*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + (7*a^5*cos(f*x + e)^5 - 13*a^5*cos(f*x + e)^
3 + 4*a^5*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e)))/(b^2*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))**(11/2)/(b*tan(f*x+e))**(3/2),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))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^(11/2)/(b*tan(f*x + e))^(3/2), 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 )}^{11/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(e + f*x))^(11/2)/(b*tan(e + f*x))^(3/2),x)

[Out]

int((a*sin(e + f*x))^(11/2)/(b*tan(e + f*x))^(3/2), x)

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