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

Optimal. Leaf size=70 \[ \frac{2 b \sin (e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac{4 b^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 f} \]

[Out]

(-4*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*f) + (2*b*(b*Sec[e + f*x])^(3/2)
*Sin[e + f*x])/(3*f)

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Rubi [A]  time = 0.0646299, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2624, 3771, 2641} \[ \frac{2 b \sin (e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac{4 b^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*f) + (2*b*(b*Sec[e + f*x])^(3/2)
*Sin[e + f*x])/(3*f)

Rule 2624

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))/(f*a*(n - 1)), x] + Dist[(b^2*(m + 1))/(a^2*(n - 1)), Int[(a*Csc[e +
f*x])^(m + 2)*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && LtQ[m, -1] && Integer
sQ[2*m, 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]

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

Mathematica [A]  time = 0.124455, size = 52, normalized size = 0.74 \[ \frac{2 b^2 \sqrt{b \sec (e+f x)} \left (\tan (e+f x)-2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*Sqrt[b*Sec[e + f*x]]*(-2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] + Tan[e + f*x]))/(3*f)

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Maple [C]  time = 0.129, size = 126, normalized size = 1.8 \begin{align*}{\frac{ \left ( -2+2\,\cos \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{3\,f \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( 2\,i\cos \left ( fx+e \right ){\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sin \left ( fx+e \right ) +\cos \left ( fx+e \right ) -1 \right ) \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/f*(-1+cos(f*x+e))*(2*I*cos(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(
f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)*cos(f*x+e)*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(5/2)/sin(f*x
+e)^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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