3.478 \(\int \frac{(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=135 \[ \frac{a^2 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{12 b^2 f \sqrt{a \sin (e+f x)}}+\frac{(a \sin (e+f x))^{5/2}}{3 a b f \sqrt{b \sec (e+f x)}}-\frac{a \sqrt{a \sin (e+f x)}}{6 b f \sqrt{b \sec (e+f x)}} \]

[Out]

-(a*Sqrt[a*Sin[e + f*x]])/(6*b*f*Sqrt[b*Sec[e + f*x]]) + (a*Sin[e + f*x])^(5/2)/(3*a*b*f*Sqrt[b*Sec[e + f*x]])
+ (a^2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(12*b^2*f*Sqrt[a*Sin[e + f*x
]])

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Rubi [A]  time = 0.215809, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2582, 2583, 2585, 2573, 2641} \[ \frac{a^2 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{12 b^2 f \sqrt{a \sin (e+f x)}}+\frac{(a \sin (e+f x))^{5/2}}{3 a b f \sqrt{b \sec (e+f x)}}-\frac{a \sqrt{a \sin (e+f x)}}{6 b f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Sqrt[a*Sin[e + f*x]])/(6*b*f*Sqrt[b*Sec[e + f*x]]) + (a*Sin[e + f*x])^(5/2)/(3*a*b*f*Sqrt[b*Sec[e + f*x]])
+ (a^2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(12*b^2*f*Sqrt[a*Sin[e + f*x
]])

Rule 2582

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((a*Sin[e + f
*x])^(m + 1)*(b*Sec[e + f*x])^(n + 1))/(a*b*f*(m - n)), x] - Dist[(n + 1)/(b^2*(m - n)), Int[(a*Sin[e + f*x])^
m*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m - n, 0] && IntegersQ[2*
m, 2*n]

Rule 2583

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*b*(a*Sin[
e + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(m - n)), x] + Dist[(a^2*(m - 1))/(m - n), Int[(a*Sin[e + f*x])
^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m - n, 0] && IntegersQ[2*
m, 2*n]

Rule 2585

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

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

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{(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx &=\frac{(a \sin (e+f x))^{5/2}}{3 a b f \sqrt{b \sec (e+f x)}}+\frac{\int \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx}{6 b^2}\\ &=-\frac{a \sqrt{a \sin (e+f x)}}{6 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{5/2}}{3 a b f \sqrt{b \sec (e+f x)}}+\frac{a^2 \int \frac{\sqrt{b \sec (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx}{12 b^2}\\ &=-\frac{a \sqrt{a \sin (e+f x)}}{6 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{5/2}}{3 a b f \sqrt{b \sec (e+f x)}}+\frac{\left (a^2 \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{b \cos (e+f x)} \sqrt{a \sin (e+f x)}} \, dx}{12 b^2}\\ &=-\frac{a \sqrt{a \sin (e+f x)}}{6 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{5/2}}{3 a b f \sqrt{b \sec (e+f x)}}+\frac{\left (a^2 \sqrt{b \sec (e+f x)} \sqrt{\sin (2 e+2 f x)}\right ) \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{12 b^2 \sqrt{a \sin (e+f x)}}\\ &=-\frac{a \sqrt{a \sin (e+f x)}}{6 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{5/2}}{3 a b f \sqrt{b \sec (e+f x)}}+\frac{a^2 F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{b \sec (e+f x)} \sqrt{\sin (2 e+2 f x)}}{12 b^2 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.432342, size = 87, normalized size = 0.64 \[ \frac{a \sqrt{a \sin (e+f x)} \left (\left (-\tan ^2(e+f x)\right )^{3/4} \csc ^2(e+f x) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\sec ^2(e+f x)\right )-2 \cos (2 (e+f x))\right )}{12 b f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.123, size = 218, normalized size = 1.6 \begin{align*} -{\frac{\sqrt{2}}{12\,f \left ( -1+\cos \left ( fx+e \right ) \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( \sin \left ( fx+e \right ) \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ) +2\,\sqrt{2} \left ( \cos \left ( fx+e \right ) \right ) ^{4}-2\,\sqrt{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}-\sqrt{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}+\sqrt{2}\cos \left ( fx+e \right ) \right ) \left ( a\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/f*2^(1/2)*(sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e
))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2)
)+2*2^(1/2)*cos(f*x+e)^4-2*2^(1/2)*cos(f*x+e)^3-2^(1/2)*cos(f*x+e)^2+2^(1/2)*cos(f*x+e))*(a*sin(f*x+e))^(3/2)/
(-1+cos(f*x+e))/sin(f*x+e)/(b/cos(f*x+e))^(3/2)/cos(f*x+e)^2

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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{3}{2}}}{\left (b \sec \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((a*sin(f*x+e))^(3/2)/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(3/2)/(b*sec(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{b \sec \left (f x + e\right )} \sqrt{a \sin \left (f x + e\right )} a \sin \left (f x + e\right )}{b^{2} \sec \left (f x + e\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))*a*sin(f*x + e)/(b^2*sec(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((a*sin(f*x+e))**(3/2)/(b*sec(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{\left (a \sin \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (b \sec \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((a*sin(f*x+e))^(3/2)/(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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