3.3.24 \(\int (d \sec (a+b x))^{5/2} \sin (a+b x) \, dx\) [224]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, 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 = {2702, 30} \begin {gather*} \frac {2 d (d \sec (a+b x))^{3/2}}{3 b} \end {gather*}

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 30

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

Rule 2702

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])

Rubi steps

\begin {align*} \int (d \sec (a+b x))^{5/2} \sin (a+b x) \, dx &=\frac {d \text {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*}

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Mathematica [A]
time = 0.05, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 d (d \sec (a+b x))^{3/2}}{3 b} \end {gather*}

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 = 0.40, size = 17, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {2 d \left (d \sec \left (b x +a \right )\right )^{\frac {3}{2}}}{3 b}\)	\(17\)
default	\(\frac {2 d \left (d \sec \left (b x +a \right )\right )^{\frac {3}{2}}}{3 b}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 23, normalized size = 1.15 \begin {gather*} \frac {2 \, \left (\frac {d}{\cos \left (b x + a\right )}\right )^{\frac {5}{2}} \cos \left (b x + a\right )}{3 \, b} \end {gather*}

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 = 2.07, size = 28, normalized size = 1.40 \begin {gather*} \frac {2 \, d^{2} \sqrt {\frac {d}{\cos \left (b x + a\right )}}}{3 \, b \cos \left (b x + a\right )} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).
time = 0.43, size = 33, normalized size = 1.65 \begin {gather*} \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 {gather*}

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))

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Mupad [B]
time = 0.25, size = 39, normalized size = 1.95 \begin {gather*} \frac {4\,d^2\,\cos \left (a+b\,x\right )\,\sqrt {\frac {d}{\cos \left (a+b\,x\right )}}}{3\,b\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*d^2*cos(a + b*x)*(d/cos(a + b*x))^(1/2))/(3*b*(cos(2*a + 2*b*x) + 1))

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