3.425 \(\int \frac{\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=65 \[ -\frac{2 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac{4 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac{2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[Out]

(-2*b^5)/(13*f*(b*Sec[e + f*x])^(13/2)) + (4*b^3)/(9*f*(b*Sec[e + f*x])^(9/2)) - (2*b)/(5*f*(b*Sec[e + f*x])^(
5/2))

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Rubi [A]  time = 0.0581976, antiderivative size = 65, 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} \[ -\frac{2 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac{4 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac{2 b}{5 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5/(b*Sec[e + f*x])^(3/2),x]

[Out]

(-2*b^5)/(13*f*(b*Sec[e + f*x])^(13/2)) + (4*b^3)/(9*f*(b*Sec[e + f*x])^(9/2)) - (2*b)/(5*f*(b*Sec[e + f*x])^(
5/2))

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 \frac{\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{\left (-1+\frac{x^2}{b^2}\right )^2}{x^{15/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^5 \operatorname{Subst}\left (\int \left (\frac{1}{x^{15/2}}-\frac{2}{b^2 x^{11/2}}+\frac{1}{b^4 x^{7/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=-\frac{2 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac{4 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac{2 b}{5 f (b \sec (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.252138, size = 42, normalized size = 0.65 \[ \frac{b (340 \cos (2 (e+f x))-45 \cos (4 (e+f x))-551)}{2340 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5/(b*Sec[e + f*x])^(3/2),x]

[Out]

(b*(-551 + 340*Cos[2*(e + f*x)] - 45*Cos[4*(e + f*x)]))/(2340*f*(b*Sec[e + f*x])^(5/2))

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Maple [A]  time = 0.114, size = 46, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 90\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-260\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+234 \right ) \cos \left ( fx+e \right ) }{585\,f} \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5/(b*sec(f*x+e))^(3/2),x)

[Out]

-2/585/f*(45*cos(f*x+e)^4-130*cos(f*x+e)^2+117)*cos(f*x+e)/(b/cos(f*x+e))^(3/2)

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Maxima [A]  time = 1.00042, size = 68, normalized size = 1.05 \begin{align*} -\frac{2 \,{\left (45 \, b^{4} - \frac{130 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac{117 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{585 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{13}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-2/585*(45*b^4 - 130*b^4/cos(f*x + e)^2 + 117*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(13/2))

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Fricas [A]  time = 2.64023, size = 134, normalized size = 2.06 \begin{align*} -\frac{2 \,{\left (45 \, \cos \left (f x + e\right )^{7} - 130 \, \cos \left (f x + e\right )^{5} + 117 \, \cos \left (f x + e\right )^{3}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{585 \, b^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-2/585*(45*cos(f*x + e)^7 - 130*cos(f*x + e)^5 + 117*cos(f*x + e)^3)*sqrt(b/cos(f*x + e))/(b^2*f)

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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(sin(f*x+e)**5/(b*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{5}}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^5/(b*sec(f*x + e))^(3/2), x)