3.373 \(\int \sqrt{b \sec (e+f x)} \sin ^3(e+f x) \, dx\)

Optimal. Leaf size=41 \[ \frac{2 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac{2 b}{f \sqrt{b \sec (e+f x)}} \]

[Out]

(2*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 = 0.0428639, antiderivative size = 41, 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, 14} \[ \frac{2 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac{2 b}{f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3,x]

[Out]

(2*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \sqrt{b \sec (e+f x)} \sin ^3(e+f x) \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{-1+\frac{x^2}{b^2}}{x^{7/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^3 \operatorname{Subst}\left (\int \left (-\frac{1}{x^{7/2}}+\frac{1}{b^2 x^{3/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{2 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 = 0.1598, size = 36, normalized size = 0.88 \[ \frac{(\cos (3 (e+f x))-17 \cos (e+f x)) \sqrt{b \sec (e+f x)}}{10 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3,x]

[Out]

((-17*Cos[e + f*x] + Cos[3*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(10*f)

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Maple [B]  time = 0.148, size = 497, normalized size = 12.1 \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)^3*(b*sec(f*x+e))^(1/2),x)

[Out]

1/10/f*(-1+cos(f*x+e))^2*(4*cos(f*x+e)^3+5*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)/si
n(f*x+e)^2)-5*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)+5*(-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*(-c
os(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-5*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)-20*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 = 1.00714, size = 47, normalized size = 1.15 \begin{align*} \frac{2 \,{\left (b^{2} - \frac{5 \, b^{2}}{\cos \left (f x + e\right )^{2}}\right )} b}{5 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/5*(b^2 - 5*b^2/cos(f*x + e)^2)*b/(f*(b/cos(f*x + e))^(5/2))

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Fricas [A]  time = 2.16809, size = 84, normalized size = 2.05 \begin{align*} \frac{2 \,{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{5 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/5*(cos(f*x + e)^3 - 5*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)**3*(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.1112, size = 77, normalized size = 1.88 \begin{align*} \frac{2 \,{\left (\sqrt{b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - 5 \, \sqrt{b \cos \left (f x + e\right )} b^{2}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{5 \, b^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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