3.477 \(\int \frac{(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=172 \[ \frac{a^4 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{24 b^2 f \sqrt{a \sin (e+f x)}}-\frac{a^3 \sqrt{a \sin (e+f x)}}{12 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{9/2}}{5 a b f \sqrt{b \sec (e+f x)}}-\frac{a (a \sin (e+f x))^{5/2}}{30 b f \sqrt{b \sec (e+f x)}} \]

[Out]

-(a^3*Sqrt[a*Sin[e + f*x]])/(12*b*f*Sqrt[b*Sec[e + f*x]]) - (a*(a*Sin[e + f*x])^(5/2))/(30*b*f*Sqrt[b*Sec[e +
f*x]]) + (a*Sin[e + f*x])^(9/2)/(5*a*b*f*Sqrt[b*Sec[e + f*x]]) + (a^4*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[
e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(24*b^2*f*Sqrt[a*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.274728, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{a^4 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{24 b^2 f \sqrt{a \sin (e+f x)}}-\frac{a^3 \sqrt{a \sin (e+f x)}}{12 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{9/2}}{5 a b f \sqrt{b \sec (e+f x)}}-\frac{a (a \sin (e+f x))^{5/2}}{30 b f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(7/2)/(b*Sec[e + f*x])^(3/2),x]

[Out]

-(a^3*Sqrt[a*Sin[e + f*x]])/(12*b*f*Sqrt[b*Sec[e + f*x]]) - (a*(a*Sin[e + f*x])^(5/2))/(30*b*f*Sqrt[b*Sec[e +
f*x]]) + (a*Sin[e + f*x])^(9/2)/(5*a*b*f*Sqrt[b*Sec[e + f*x]]) + (a^4*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[
e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(24*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))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx &=\frac{(a \sin (e+f x))^{9/2}}{5 a b f \sqrt{b \sec (e+f x)}}+\frac{\int \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2} \, dx}{10 b^2}\\ &=-\frac{a (a \sin (e+f x))^{5/2}}{30 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{9/2}}{5 a b f \sqrt{b \sec (e+f x)}}+\frac{a^2 \int \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx}{12 b^2}\\ &=-\frac{a^3 \sqrt{a \sin (e+f x)}}{12 b f \sqrt{b \sec (e+f x)}}-\frac{a (a \sin (e+f x))^{5/2}}{30 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{9/2}}{5 a b f \sqrt{b \sec (e+f x)}}+\frac{a^4 \int \frac{\sqrt{b \sec (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx}{24 b^2}\\ &=-\frac{a^3 \sqrt{a \sin (e+f x)}}{12 b f \sqrt{b \sec (e+f x)}}-\frac{a (a \sin (e+f x))^{5/2}}{30 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{9/2}}{5 a b f \sqrt{b \sec (e+f x)}}+\frac{\left (a^4 \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}{24 b^2}\\ &=-\frac{a^3 \sqrt{a \sin (e+f x)}}{12 b f \sqrt{b \sec (e+f x)}}-\frac{a (a \sin (e+f x))^{5/2}}{30 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{9/2}}{5 a b f \sqrt{b \sec (e+f x)}}+\frac{\left (a^4 \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}{24 b^2 \sqrt{a \sin (e+f x)}}\\ &=-\frac{a^3 \sqrt{a \sin (e+f x)}}{12 b f \sqrt{b \sec (e+f x)}}-\frac{a (a \sin (e+f x))^{5/2}}{30 b f \sqrt{b \sec (e+f x)}}+\frac{(a \sin (e+f x))^{9/2}}{5 a b f \sqrt{b \sec (e+f x)}}+\frac{a^4 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)}}{24 b^2 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.861685, size = 103, normalized size = 0.6 \[ -\frac{a^5 \left (-20 \left (-\tan ^2(e+f x)\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\sec ^2(e+f x)\right )+17 \cos (2 (e+f x))-16 \cos (4 (e+f x))+3 \cos (6 (e+f x))-4\right )}{480 b f (a \sin (e+f x))^{3/2} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(7/2)/(b*Sec[e + f*x])^(3/2),x]

[Out]

-(a^5*(-4 + 17*Cos[2*(e + f*x)] - 16*Cos[4*(e + f*x)] + 3*Cos[6*(e + f*x)] - 20*Hypergeometric2F1[1/2, 3/4, 3/
2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(3/4)))/(480*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.138, size = 246, normalized size = 1.4 \begin{align*} -{\frac{\sqrt{2}}{120\,f \left ( -1+\cos \left ( fx+e \right ) \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( -12\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}\sqrt{2}+12\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}\sqrt{2}+5\,\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 ) }}},1/2\,\sqrt{2} \right ) +22\,\sqrt{2} \left ( \cos \left ( fx+e \right ) \right ) ^{4}-22\,\sqrt{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}-5\,\sqrt{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}+5\,\sqrt{2}\cos \left ( fx+e \right ) \right ) \left ( a\sin \left ( fx+e \right ) \right ) ^{{\frac{7}{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))^(7/2)/(b*sec(f*x+e))^(3/2),x)

[Out]

-1/120/f*2^(1/2)*(-12*cos(f*x+e)^6*2^(1/2)+12*cos(f*x+e)^5*2^(1/2)+5*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))+22*2^(1/2)*cos(f*x+e)^4-22*2^(1/2)*cos(f*x+e)^3-5*2^(1/2
)*cos(f*x+e)^2+5*2^(1/2)*cos(f*x+e))*(a*sin(f*x+e))^(7/2)/(-1+cos(f*x+e))/sin(f*x+e)^3/cos(f*x+e)^2/(b/cos(f*x
+e))^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \sin \left (f x + e\right )\right )^{\frac{7}{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))^(7/2)/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(7/2)/(b*sec(f*x + e))^(3/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a^{3} \cos \left (f x + e\right )^{2} - a^{3}\right )} \sqrt{b \sec \left (f x + e\right )} \sqrt{a \sin \left (f x + e\right )} \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))^(7/2)/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-(a^3*cos(f*x + e)^2 - a^3)*sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))*sin(f*x + e)/(b^2*sec(f*x + e)^
2), x)

________________________________________________________________________________________

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))**(7/2)/(b*sec(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \sin \left (f x + e\right )\right )^{\frac{7}{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))^(7/2)/(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^(7/2)/(b*sec(f*x + e))^(3/2), x)