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

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

[Out]

-1/(3*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]])

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Rubi [A]  time = 0.168532, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2597, 2599, 2601, 2641} \[ -\frac{\sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \tan (e+f x)}}{6 a^2 b^2 f \sqrt{a \sin (e+f x)}}+\frac{1}{6 a^2 b f \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{1}{3 b f (a \sin (e+f x))^{5/2} \sqrt{b \tan (e+f x)}} \]

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*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 2597

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 2599

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 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 2641

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

Mathematica [A]  time = 0.342652, size = 96, normalized size = 0.74 \[ \frac{\sqrt [4]{\cos ^2(e+f x)} \left (1-2 \csc ^2(e+f x)\right )-\sin (e+f x) F\left (\left .\frac{1}{2} \sin ^{-1}(\sin (e+f x))\right |2\right )}{6 a^2 b f \sqrt [4]{\cos ^2(e+f x)} \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \]

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]  time = 0.172, size = 337, normalized size = 2.6 \begin{align*}{\frac{\sin \left ( fx+e \right ) }{6\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( i\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}\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 ) +i\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}\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 ) -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 ) -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 ) - \left ( \cos \left ( fx+e \right ) \right ) ^{3}-\cos \left ( fx+e \right ) \right ) \left ( a\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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)

[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)/cos(f*x+e)^2/(a*si
n(f*x+e))^(5/2)/(b*sin(f*x+e)/cos(f*x+e))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{a \sin \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}{{\left (a^{3} b^{2} \cos \left (f x + e\right )^{2} - a^{3} b^{2}\right )} \sin \left (f x + e\right ) \tan \left (f x + e\right )^{2}}, x\right ) \end{align*}

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]

integral(-sqrt(a*sin(f*x + e))*sqrt(b*tan(f*x + e))/((a^3*b^2*cos(f*x + e)^2 - a^3*b^2)*sin(f*x + e)*tan(f*x +
e)^2), 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(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)