3.411 \(\int \frac{\sin ^5(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx\)

Optimal. Leaf size=65 \[ -\frac{2 b^5}{11 f (b \sec (e+f x))^{11/2}}+\frac{4 b^3}{7 f (b \sec (e+f x))^{7/2}}-\frac{2 b}{3 f (b \sec (e+f x))^{3/2}} \]

[Out]

(-2*b^5)/(11*f*(b*Sec[e + f*x])^(11/2)) + (4*b^3)/(7*f*(b*Sec[e + f*x])^(7/2)) - (2*b)/(3*f*(b*Sec[e + f*x])^(
3/2))

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Rubi [A]  time = 0.0496222, 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} \[ -\frac{2 b^5}{11 f (b \sec (e+f x))^{11/2}}+\frac{4 b^3}{7 f (b \sec (e+f x))^{7/2}}-\frac{2 b}{3 f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5/Sqrt[b*Sec[e + f*x]],x]

[Out]

(-2*b^5)/(11*f*(b*Sec[e + f*x])^(11/2)) + (4*b^3)/(7*f*(b*Sec[e + f*x])^(7/2)) - (2*b)/(3*f*(b*Sec[e + f*x])^(
3/2))

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

Mathematica [A]  time = 0.162789, size = 42, normalized size = 0.65 \[ \frac{b (180 \cos (2 (e+f x))-21 \cos (4 (e+f x))-415)}{924 f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5/Sqrt[b*Sec[e + f*x]],x]

[Out]

(b*(-415 + 180*Cos[2*(e + f*x)] - 21*Cos[4*(e + f*x)]))/(924*f*(b*Sec[e + f*x])^(3/2))

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Maple [A]  time = 0.128, size = 46, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 42\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-132\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+154 \right ) \cos \left ( fx+e \right ) }{231\,f}{\frac{1}{\sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5/(b*sec(f*x+e))^(1/2),x)

[Out]

-2/231/f*(21*cos(f*x+e)^4-66*cos(f*x+e)^2+77)*cos(f*x+e)/(b/cos(f*x+e))^(1/2)

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Maxima [A]  time = 1.03342, size = 68, normalized size = 1.05 \begin{align*} -\frac{2 \,{\left (21 \, b^{4} - \frac{66 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac{77 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{231 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{11}{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))^(1/2),x, algorithm="maxima")

[Out]

-2/231*(21*b^4 - 66*b^4/cos(f*x + e)^2 + 77*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(11/2))

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Fricas [A]  time = 2.58573, size = 128, normalized size = 1.97 \begin{align*} -\frac{2 \,{\left (21 \, \cos \left (f x + e\right )^{6} - 66 \, \cos \left (f x + e\right )^{4} + 77 \, \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{231 \, b 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))^(1/2),x, algorithm="fricas")

[Out]

-2/231*(21*cos(f*x + e)^6 - 66*cos(f*x + e)^4 + 77*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))/(b*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))**(1/2),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}}{\sqrt{b \sec \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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