3.445 \(\int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=98 \[ \frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {4 \sin (e+f x)}{45 b f (b \sec (e+f x))^{3/2}}-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}} \]

[Out]

-2/9*b*sin(f*x+e)/f/(b*sec(f*x+e))^(7/2)+4/45*sin(f*x+e)/b/f/(b*sec(f*x+e))^(3/2)+4/15*(cos(1/2*f*x+1/2*e)^2)^
(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/b^2/f/cos(f*x+e)^(1/2)/(b*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2627, 3769, 3771, 2639} \[ \frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {4 \sin (e+f x)}{45 b f (b \sec (e+f x))^{3/2}}-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2/(b*Sec[e + f*x])^(5/2),x]

[Out]

(4*EllipticE[(e + f*x)/2, 2])/(15*b^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (2*b*Sin[e + f*x])/(9*f*(b*
Sec[e + f*x])^(7/2)) + (4*Sin[e + f*x])/(45*b*f*(b*Sec[e + f*x])^(3/2))

Rule 2627

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

Rule 2639

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

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx &=-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}+\frac {2}{9} \int \frac {1}{(b \sec (e+f x))^{5/2}} \, dx\\ &=-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}+\frac {4 \sin (e+f x)}{45 b f (b \sec (e+f x))^{3/2}}+\frac {2 \int \frac {1}{\sqrt {b \sec (e+f x)}} \, dx}{15 b^2}\\ &=-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}+\frac {4 \sin (e+f x)}{45 b f (b \sec (e+f x))^{3/2}}+\frac {2 \int \sqrt {\cos (e+f x)} \, dx}{15 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\\ &=\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}+\frac {4 \sin (e+f x)}{45 b f (b \sec (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 66, normalized size = 0.67 \[ \frac {-4 \sin (2 (e+f x))-10 \sin (4 (e+f x))+\frac {96 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)}}}{360 b^2 f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^2/(b*Sec[e + f*x])^(5/2),x]

[Out]

((96*EllipticE[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] - 4*Sin[2*(e + f*x)] - 10*Sin[4*(e + f*x)])/(360*b^2*f*Sqrt
[b*Sec[e + f*x]])

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b \sec \left (f x + e\right )}}{b^{3} \sec \left (f x + e\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-(cos(f*x + e)^2 - 1)*sqrt(b*sec(f*x + e))/(b^3*sec(f*x + e)^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{2}}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^2/(b*sec(f*x + e))^(5/2), x)

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maple [C]  time = 0.18, size = 333, normalized size = 3.40 \[ \frac {\frac {4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )}{15}-\frac {4 i \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )}{15}+\frac {2 \left (\cos ^{6}\left (f x +e \right )\right )}{9}+\frac {4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )}{15}-\frac {4 i \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )}{15}-\frac {14 \left (\cos ^{4}\left (f x +e \right )\right )}{45}-\frac {8 \left (\cos ^{2}\left (f x +e \right )\right )}{45}+\frac {4 \cos \left (f x +e \right )}{15}}{f \cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2/(b*sec(f*x+e))^(5/2),x)

[Out]

2/45/f*(6*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),
I)*sin(f*x+e)*cos(f*x+e)-6*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*(1/(cos(f*x+e)+1)
)^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+5*cos(f*x+e)^6+6*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x
+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-6*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*s
in(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-7*cos(f*x+e)^4-4*cos(f*x+e)^2+6*cos(f*x+e
))/cos(f*x+e)^3/sin(f*x+e)/(b/cos(f*x+e))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{2}}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^2/(b*sec(f*x + e))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (e+f\,x\right )}^2}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^2/(b/cos(e + f*x))^(5/2),x)

[Out]

int(sin(e + f*x)^2/(b/cos(e + f*x))^(5/2), 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(sin(f*x+e)**2/(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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