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]

-1/30*a*(a*sin(f*x+e))^(5/2)/b/f/(b*sec(f*x+e))^(1/2)+1/5*(a*sin(f*x+e))^(9/2)/a/b/f/(b*sec(f*x+e))^(1/2)-1/12
*a^3*(a*sin(f*x+e))^(1/2)/b/f/(b*sec(f*x+e))^(1/2)-1/24*a^4*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*Elli
pticF(cos(e+1/4*Pi+f*x),2^(1/2))*(b*sec(f*x+e))^(1/2)*sin(2*f*x+2*e)^(1/2)/b^2/f/(a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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*}

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Mathematica [C]  time = 0.85, size = 103, normalized size = 0.60 \[ -\frac {a^5 \left (-20 \left (-\tan ^2(e+f x)\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\sec ^2(e+f x)\right )+17 \cos (2 (e+f x))-16 \cos (4 (e+f x))+3 \cos (6 (e+f x))-4\right )}{480 b f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/480*(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)))/(b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2))

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{3} \cos \left (f x + e\right )^{2} - a^{3}\right )} \sqrt {b \sec \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right )} \sin \left (f x + e\right )}{b^{2} \sec \left (f x + e\right )^{2}}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.21, size = 246, normalized size = 1.43 \[ -\frac {\left (-12 \left (\cos ^{6}\left (f x +e \right )\right ) \sqrt {2}+12 \left (\cos ^{5}\left (f x +e \right )\right ) \sqrt {2}+5 \sin \left (f x +e \right ) \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )+22 \left (\cos ^{4}\left (f x +e \right )\right ) \sqrt {2}-22 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}-5 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}+5 \cos \left (f x +e \right ) \sqrt {2}\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {7}{2}} \sqrt {2}}{120 f \left (-1+\cos \left (f x +e \right )\right ) \sin \left (f x +e \right )^{3} \cos \left (f x +e \right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

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*(-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*cos(f*x+e)^4*2^(1/2)-22*cos(f*x+e)^3*2^(1/2)-5*cos(f*x+e)^2*2
^(1/2)+5*cos(f*x+e)*2^(1/2))*(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)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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