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]

2/3*b*(b*sec(f*x+e))^(3/2)*sin(f*x+e)^5/f+40/21*b^3*sin(f*x+e)/f/(b*sec(f*x+e))^(1/2)+20/21*b^3*sin(f*x+e)^3/f
/(b*sec(f*x+e))^(1/2)-80/21*b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2
^(1/2))*cos(f*x+e)^(1/2)*(b*sec(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.15, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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 (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*}

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Mathematica [A]  time = 0.19, 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]

-1/84*(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*Si
n[4*(e + f*x)] - 56*Tan[e + f*x]))/f

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b^{2} \cos \left (f x + e\right )^{6} - 3 \, b^{2} \cos \left (f x + e\right )^{4} + 3 \, b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \sqrt {b \sec \left (f x + e\right )} \sec \left (f x + e\right )^{2}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{6}\,{d x} \]

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)

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maple [C]  time = 0.24, size = 168, normalized size = 1.29 \[ -\frac {2 \left (-1+\cos \left (f x +e \right )\right ) \left (-40 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+3 \left (\cos ^{5}\left (f x +e \right )\right )-3 \left (\cos ^{4}\left (f x +e \right )\right )-16 \left (\cos ^{3}\left (f x +e \right )\right )+16 \left (\cos ^{2}\left (f x +e \right )\right )-7 \cos \left (f x +e \right )+7\right ) \cos \left (f x +e \right ) \left (\cos \left (f x +e \right )+1\right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{21 f \sin \left (f x +e \right )^{3}} \]

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)*sin(f*x+e)*(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)+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*co
s(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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{6}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^6\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^6*(b/cos(e + f*x))^(5/2),x)

[Out]

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

[Out]

Timed out

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