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

Optimal. Leaf size=20 \[ \frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

(2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

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Rubi [A]  time = 0.0357261, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2622, 30} \[ \frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0422307, size = 20, normalized size = 1. \[ \frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

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Maple [A]  time = 0.013, size = 17, normalized size = 0.9 \begin{align*}{\frac{2\,b}{3\,f} \left ( b\sec \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}} \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),x)

[Out]

2/3*b*(b*sec(f*x+e))^(3/2)/f

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Maxima [A]  time = 1.04531, size = 31, normalized size = 1.55 \begin{align*} \frac{2 \, \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{5}{2}} \cos \left (f x + e\right )}{3 \, 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),x, algorithm="maxima")

[Out]

2/3*(b/cos(f*x + e))^(5/2)*cos(f*x + e)/f

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Fricas [A]  time = 2.16279, size = 63, normalized size = 3.15 \begin{align*} \frac{2 \, b^{2} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{3 \, 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),x, algorithm="fricas")

[Out]

2/3*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),x)

[Out]

Timed out

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Giac [B]  time = 1.15035, size = 49, normalized size = 2.45 \begin{align*} \frac{2 \, b^{3} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{3 \, \sqrt{b \cos \left (f x + e\right )} 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),x, algorithm="giac")

[Out]

2/3*b^3*sgn(cos(f*x + e))/(sqrt(b*cos(f*x + e))*f*cos(f*x + e))