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

Optimal. Leaf size=63 \[ -\frac{2 b^5}{9 f (b \sec (e+f x))^{9/2}}+\frac{4 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac{2 b}{f \sqrt{b \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0497374, 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} \[ -\frac{2 b^5}{9 f (b \sec (e+f x))^{9/2}}+\frac{4 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac{2 b}{f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.205191, size = 48, normalized size = 0.76 \[ -\frac{(554 \cos (e+f x)-47 \cos (3 (e+f x))+5 \cos (5 (e+f x))) \sqrt{b \sec (e+f x)}}{360 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((554*Cos[e + f*x] - 47*Cos[3*(e + f*x)] + 5*Cos[5*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(360*f)

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Maple [B]  time = 0.137, size = 507, normalized size = 8.1 \begin{align*} \text{result too large to display} \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]

-1/90/f*(-1+cos(f*x+e))^2*(20*cos(f*x+e)^5-72*cos(f*x+e)^3+45*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)-45*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)
+45*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)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-45*(-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)+180*cos(f*x+e))*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(1/2)/sin(f*x+e)^4

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Maxima [A]  time = 1.03814, size = 68, normalized size = 1.08 \begin{align*} -\frac{2 \,{\left (5 \, b^{4} - \frac{18 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac{45 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{45 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{9}{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]

-2/45*(5*b^4 - 18*b^4/cos(f*x + e)^2 + 45*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(9/2))

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Fricas [A]  time = 2.21484, size = 117, normalized size = 1.86 \begin{align*} -\frac{2 \,{\left (5 \, \cos \left (f x + e\right )^{5} - 18 \, \cos \left (f x + e\right )^{3} + 45 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{45 \, 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]

-2/45*(5*cos(f*x + e)^5 - 18*cos(f*x + e)^3 + 45*cos(f*x + e))*sqrt(b/cos(f*x + e))/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 [A]  time = 1.14488, size = 112, normalized size = 1.78 \begin{align*} -\frac{2 \,{\left (5 \, \sqrt{b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{4} - 18 \, \sqrt{b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{2} + 45 \, \sqrt{b \cos \left (f x + e\right )} b^{4}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{45 \, b^{4} 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="giac")

[Out]

-2/45*(5*sqrt(b*cos(f*x + e))*b^4*cos(f*x + e)^4 - 18*sqrt(b*cos(f*x + e))*b^4*cos(f*x + e)^2 + 45*sqrt(b*cos(
f*x + e))*b^4)*sgn(cos(f*x + e))/(b^4*f)