3.2.43 \(\int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx\) [143]

Optimal. Leaf size=130 \[ -\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}+\frac {1}{6 a^2 b f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{6 a^2 b^2 f \sqrt {a \sin (e+f x)}} \]

[Out]

-1/3/b/f/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2)+1/6/a^2/b/f/(a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)-1/6*(
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)/a^2/b^2/f/(a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2677, 2679, 2681, 2720} \begin {gather*} -\frac {\sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{6 a^2 b^2 f \sqrt {a \sin (e+f x)}}+\frac {1}{6 a^2 b f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a*Sin[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2)),x]

[Out]

-1/3*1/(b*f*(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]]) + 1/(6*a^2*b*f*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*
x]]) - (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(6*a^2*b^2*f*Sqrt[a*Sin[e + f*x]])

Rule 2677

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 + n + 1))), x] - Dist[(n + 1)/(b^2*(m + n + 1)), Int[(a*Sin[e + f*x])
^m*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

Rule 2679

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[b*(a*Sin[e +
f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1)/(a^2*f*(m + n + 1))), x] + Dist[(m + 2)/(a^2*(m + n + 1)), Int[(a*Sin
[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n + 1, 0]
&& 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 {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx &=-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}-\frac {\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx}{6 b^2}\\ &=-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}+\frac {1}{6 a^2 b f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}} \, dx}{12 a^2 b^2}\\ &=-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}+\frac {1}{6 a^2 b f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\left (\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{12 a^2 b^2 \sqrt {a \sin (e+f x)}}\\ &=-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}+\frac {1}{6 a^2 b f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \tan (e+f x)}}{6 a^2 b^2 f \sqrt {a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 96, normalized size = 0.74 \begin {gather*} \frac {\sqrt [4]{\cos ^2(e+f x)} \left (1-2 \csc ^2(e+f x)\right )-F\left (\left .\frac {1}{2} \text {ArcSin}(\sin (e+f x))\right |2\right ) \sin (e+f x)}{6 a^2 b f \sqrt [4]{\cos ^2(e+f x)} \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a*Sin[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2)),x]

[Out]

((Cos[e + f*x]^2)^(1/4)*(1 - 2*Csc[e + f*x]^2) - EllipticF[ArcSin[Sin[e + f*x]]/2, 2]*Sin[e + f*x])/(6*a^2*b*f
*(Cos[e + f*x]^2)^(1/4)*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])

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Maple [C] Result contains complex when optimal does not.
time = 0.36, size = 337, normalized size = 2.59

method	result	size
default	\(\frac {\left (i \sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right )+i \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right )-i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-\left (\cos ^{3}\left (f x +e \right )\right )-\cos \left (f x +e \right )\right ) \sin \left (f x +e \right )}{6 f \left (a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right )^{2}}\)	\(337\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/6/f*(I*sin(f*x+e)*cos(f*x+e)^3*(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)+I*sin(f*x+e)*cos(f*x+e)^2*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*El
lipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)-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)-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)^3-cos(f*x+e))*sin(f*x+e)/(a*sin(f*x+e))^(5/
2)/(b*sin(f*x+e)/cos(f*x+e))^(3/2)/cos(f*x+e)^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(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((a*sin(f*x + e))^(5/2)*(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 = 0.12, size = 215, normalized size = 1.65 \begin {gather*} -\frac {{\left (\sqrt {2} \cos \left (f x + e\right )^{4} - 2 \, \sqrt {2} \cos \left (f x + e\right )^{2} + \sqrt {2}\right )} \sqrt {-a b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + {\left (\sqrt {2} \cos \left (f x + e\right )^{4} - 2 \, \sqrt {2} \cos \left (f x + e\right )^{2} + \sqrt {2}\right )} \sqrt {-a b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (\cos \left (f x + e\right )^{3} + \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 )}}}{12 \, {\left (a^{3} b^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} b^{2} f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/12*((sqrt(2)*cos(f*x + e)^4 - 2*sqrt(2)*cos(f*x + e)^2 + sqrt(2))*sqrt(-a*b)*weierstrassPInverse(-4, 0, cos
(f*x + e) + I*sin(f*x + e)) + (sqrt(2)*cos(f*x + e)^4 - 2*sqrt(2)*cos(f*x + e)^2 + sqrt(2))*sqrt(-a*b)*weierst
rassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(cos(f*x + e)^3 + cos(f*x + e))*sqrt(a*sin(f*x + e))*sq
rt(b*sin(f*x + e)/cos(f*x + e)))/(a^3*b^2*f*cos(f*x + e)^4 - 2*a^3*b^2*f*cos(f*x + e)^2 + a^3*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(1/(a*sin(f*x+e))**(5/2)/(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(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*sin(f*x + e))^(5/2)*(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 {1}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\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(1/((a*sin(e + f*x))^(5/2)*(b*tan(e + f*x))^(3/2)),x)

[Out]

int(1/((a*sin(e + f*x))^(5/2)*(b*tan(e + f*x))^(3/2)), x)

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