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

Optimal. Leaf size=63 \[ -\frac{2 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac{4 b^3}{f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

(-2*b^5)/(5*f*(b*Sec[e + f*x])^(5/2)) + (4*b^3)/(f*Sqrt[b*Sec[e + f*x]]) + (2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

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Rubi [A]  time = 0.0557179, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2622, 270} \[ -\frac{2 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac{4 b^3}{f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^5)/(5*f*(b*Sec[e + f*x])^(5/2)) + (4*b^3)/(f*Sqrt[b*Sec[e + f*x]]) + (2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

Rule 2622

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int (b \sec (e+f x))^{5/2} \sin ^5(e+f x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{\left (-1+\frac{x^2}{b^2}\right )^2}{x^{7/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^5 \operatorname{Subst}\left (\int \left (\frac{1}{x^{7/2}}-\frac{2}{b^2 x^{3/2}}+\frac{\sqrt{x}}{b^4}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=-\frac{2 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac{4 b^3}{f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.340456, size = 42, normalized size = 0.67 \[ \frac{b (108 \cos (2 (e+f x))-3 \cos (4 (e+f x))+151) (b \sec (e+f x))^{3/2}}{60 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(151 + 108*Cos[2*(e + f*x)] - 3*Cos[4*(e + f*x)])*(b*Sec[e + f*x])^(3/2))/(60*f)

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Maple [B]  time = 0.148, size = 522, normalized size = 8.3 \begin{align*} \text{result too large to display} \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)^5,x)

[Out]

-1/15/f*(-1+cos(f*x+e))^2*(6*cos(f*x+e)^4-15*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-(2*cos(f*x+
e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/
sin(f*x+e)^2)+15*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x
+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-15*cos(f*x+e
)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2
*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+15*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^
2)^(1/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(
cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-60*cos(f*x+e)^2-10)*cos(f*x+e)*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(5/2)/s
in(f*x+e)^4

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Maxima [A]  time = 1.04202, size = 70, normalized size = 1.11 \begin{align*} \frac{2 \,{\left (5 \, \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{3}{2}} - \frac{3 \,{\left (b^{4} - \frac{10 \, b^{4}}{\cos \left (f x + e\right )^{2}}\right )}}{\left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{5}{2}}}\right )} b}{15 \, f} \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)^5,x, algorithm="maxima")

[Out]

2/15*(5*(b/cos(f*x + e))^(3/2) - 3*(b^4 - 10*b^4/cos(f*x + e)^2)/(b/cos(f*x + e))^(5/2))*b/f

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Fricas [A]  time = 2.23493, size = 135, normalized size = 2.14 \begin{align*} -\frac{2 \,{\left (3 \, b^{2} \cos \left (f x + e\right )^{4} - 30 \, b^{2} \cos \left (f x + e\right )^{2} - 5 \, b^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\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)^5,x, algorithm="fricas")

[Out]

-2/15*(3*b^2*cos(f*x + e)^4 - 30*b^2*cos(f*x + e)^2 - 5*b^2)*sqrt(b/cos(f*x + e))/(f*cos(f*x + e))

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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)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.17027, size = 108, normalized size = 1.71 \begin{align*} -\frac{2 \,{\left (3 \, \sqrt{b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - 30 \, \sqrt{b \cos \left (f x + e\right )} b^{2} - \frac{5 \, b^{3}}{\sqrt{b \cos \left (f x + e\right )} \cos \left (f x + e\right )}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{15 \, f} \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)^5,x, algorithm="giac")

[Out]

-2/15*(3*sqrt(b*cos(f*x + e))*b^2*cos(f*x + e)^2 - 30*sqrt(b*cos(f*x + e))*b^2 - 5*b^3/(sqrt(b*cos(f*x + e))*c
os(f*x + e)))*sgn(cos(f*x + e))/f