3.404 \(\int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx\)

Optimal. Leaf size=130 \[ \frac{20 b^3 \sin ^3(e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{40 b^3 \sin (e+f x)}{21 f \sqrt{b \sec (e+f x)}}-\frac{80 b^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{21 f}+\frac{2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

(-80*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(21*f) + (40*b^3*Sin[e + f*x])/(21
*f*Sqrt[b*Sec[e + f*x]]) + (20*b^3*Sin[e + f*x]^3)/(21*f*Sqrt[b*Sec[e + f*x]]) + (2*b*(b*Sec[e + f*x])^(3/2)*S
in[e + f*x]^5)/(3*f)

________________________________________________________________________________________

Rubi [A]  time = 0.146661, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2624, 2627, 3771, 2641} \[ \frac{20 b^3 \sin ^3(e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{40 b^3 \sin (e+f x)}{21 f \sqrt{b \sec (e+f x)}}-\frac{80 b^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{21 f}+\frac{2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^6,x]

[Out]

(-80*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(21*f) + (40*b^3*Sin[e + f*x])/(21
*f*Sqrt[b*Sec[e + f*x]]) + (20*b^3*Sin[e + f*x]^3)/(21*f*Sqrt[b*Sec[e + f*x]]) + (2*b*(b*Sec[e + f*x])^(3/2)*S
in[e + f*x]^5)/(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 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 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 ^6(e+f x) \, dx &=\frac{2 b (b \sec (e+f x))^{3/2} \sin ^5(e+f x)}{3 f}-\frac{1}{3} \left (10 b^2\right ) \int \sqrt{b \sec (e+f x)} \sin ^4(e+f x) \, dx\\ &=\frac{20 b^3 \sin ^3(e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2} \sin ^5(e+f x)}{3 f}-\frac{1}{7} \left (20 b^2\right ) \int \sqrt{b \sec (e+f x)} \sin ^2(e+f x) \, dx\\ &=\frac{40 b^3 \sin (e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{20 b^3 \sin ^3(e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2} \sin ^5(e+f x)}{3 f}-\frac{1}{21} \left (40 b^2\right ) \int \sqrt{b \sec (e+f x)} \, dx\\ &=\frac{40 b^3 \sin (e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{20 b^3 \sin ^3(e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2} \sin ^5(e+f x)}{3 f}-\frac{1}{21} \left (40 b^2 \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx\\ &=-\frac{80 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)}}{21 f}+\frac{40 b^3 \sin (e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{20 b^3 \sin ^3(e+f x)}{21 f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2} \sin ^5(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.205723, size = 74, normalized size = 0.57 \[ -\frac{b^2 \sqrt{b \sec (e+f x)} \left (-58 \sin (2 (e+f x))+3 \sin (4 (e+f x))-56 \tan (e+f x)+320 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )\right )}{84 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^6,x]

[Out]

-(b^2*Sqrt[b*Sec[e + f*x]]*(320*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] - 58*Sin[2*(e + f*x)] + 3*Sin[4*(
e + f*x)] - 56*Tan[e + f*x]))/(84*f)

________________________________________________________________________________________

Maple [C]  time = 0.177, size = 168, normalized size = 1.3 \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}}{21\,f \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( 40\,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 ) -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+16\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-16\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+7\,\cos \left ( fx+e \right ) -7 \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)^6,x)

[Out]

2/21/f*(-1+cos(f*x+e))*(40*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)*(co
s(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-3*cos(f*x+e)^5+3*cos(f*x+e)^4+16*cos(f*x+e)^3-16*cos(f*x+e)^2+7*cos(
f*x+e)-7)*cos(f*x+e)*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(5/2)/sin(f*x+e)^3

________________________________________________________________________________________

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 )^{6}\,{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)^6,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(5/2)*sin(f*x + e)^6, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{2} \cos \left (f x + e\right )^{6} - 3 \, b^{2} \cos \left (f x + e\right )^{4} + 3 \, 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)^6,x, algorithm="fricas")

[Out]

integral(-(b^2*cos(f*x + e)^6 - 3*b^2*cos(f*x + e)^4 + 3*b^2*cos(f*x + e)^2 - b^2)*sqrt(b*sec(f*x + e))*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((b*sec(f*x+e))**(5/2)*sin(f*x+e)**6,x)

[Out]

Timed out

________________________________________________________________________________________

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 )^{6}\,{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)^6,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(5/2)*sin(f*x + e)^6, x)