3.224 \(\int (d \sec (a+b x))^{5/2} \sin (a+b x) \, dx\)

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

[Out]

(2*d*(d*Sec[a + b*x])^(3/2))/(3*b)

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Rubi [A]  time = 0.0359804, 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} \[ \frac{2 d (d \sec (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(d*Sec[a + b*x])^(3/2))/(3*b)

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

Mathematica [A]  time = 0.0412481, size = 20, normalized size = 1. \[ \frac{2 d (d \sec (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(d*Sec[a + b*x])^(3/2))/(3*b)

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Maple [A]  time = 0.017, size = 17, normalized size = 0.9 \begin{align*}{\frac{2\,d}{3\,b} \left ( d\sec \left ( bx+a \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(b*x+a))^(5/2)*sin(b*x+a),x)

[Out]

2/3*d*(d*sec(b*x+a))^(3/2)/b

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Maxima [A]  time = 1.14477, size = 31, normalized size = 1.55 \begin{align*} \frac{2 \, \left (\frac{d}{\cos \left (b x + a\right )}\right )^{\frac{5}{2}} \cos \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(5/2)*sin(b*x+a),x, algorithm="maxima")

[Out]

2/3*(d/cos(b*x + a))^(5/2)*cos(b*x + a)/b

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Fricas [A]  time = 1.62174, size = 63, normalized size = 3.15 \begin{align*} \frac{2 \, d^{2} \sqrt{\frac{d}{\cos \left (b x + a\right )}}}{3 \, b \cos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(5/2)*sin(b*x+a),x, algorithm="fricas")

[Out]

2/3*d^2*sqrt(d/cos(b*x + a))/(b*cos(b*x + a))

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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((d*sec(b*x+a))**(5/2)*sin(b*x+a),x)

[Out]

Timed out

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Giac [B]  time = 1.2794, size = 45, normalized size = 2.25 \begin{align*} \frac{2 \, d^{3} \mathrm{sgn}\left (\cos \left (b x + a\right )\right )}{3 \, \sqrt{d \cos \left (b x + a\right )} b \cos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(5/2)*sin(b*x+a),x, algorithm="giac")

[Out]

2/3*d^3*sgn(cos(b*x + a))/(sqrt(d*cos(b*x + a))*b*cos(b*x + a))