3.3.25 \(\int (d \sec (a+b x))^{3/2} \sin (a+b x) \, dx\) [225]

Optimal. Leaf size=18 \[ \frac {2 d \sqrt {d \sec (a+b x)}}{b} \]

[Out]

2*d*(d*sec(b*x+a))^(1/2)/b

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Rubi [A]
time = 0.03, antiderivative size = 18, 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 \sqrt {d \sec (a+b x)}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[a + b*x])^(3/2)*Sin[a + b*x],x]

[Out]

(2*d*Sqrt[d*Sec[a + b*x]])/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))^{3/2} \sin (a+b x) \, dx &=\frac {d \text {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,d \sec (a+b x)\right )}{b}\\ &=\frac {2 d \sqrt {d \sec (a+b x)}}{b}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 18, normalized size = 1.00 \begin {gather*} \frac {2 d \sqrt {d \sec (a+b x)}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[a + b*x])^(3/2)*Sin[a + b*x],x]

[Out]

(2*d*Sqrt[d*Sec[a + b*x]])/b

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Maple [A]
time = 0.40, size = 17, normalized size = 0.94

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(b*x+a))^(3/2)*sin(b*x+a),x,method=_RETURNVERBOSE)

[Out]

2*d*(d*sec(b*x+a))^(1/2)/b

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Maxima [A]
time = 0.30, size = 23, normalized size = 1.28 \begin {gather*} \frac {2 \, \left (\frac {d}{\cos \left (b x + a\right )}\right )^{\frac {3}{2}} \cos \left (b x + a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(3/2)*sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.35, size = 18, normalized size = 1.00 \begin {gather*} \frac {2 \, d \sqrt {\frac {d}{\cos \left (b x + a\right )}}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(3/2)*sin(b*x+a),x, algorithm="fricas")

[Out]

2*d*sqrt(d/cos(b*x + a))/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \sec {\left (a + b x \right )}\right )^{\frac {3}{2}} \sin {\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))**(3/2)*sin(b*x+a),x)

[Out]

Integral((d*sec(a + b*x))**(3/2)*sin(a + b*x), x)

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Giac [A]
time = 0.43, size = 25, normalized size = 1.39 \begin {gather*} \frac {2 \, d^{2} \mathrm {sgn}\left (\cos \left (b x + a\right )\right )}{\sqrt {d \cos \left (b x + a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(3/2)*sin(b*x+a),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.10, size = 18, normalized size = 1.00 \begin {gather*} \frac {2\,d\,\sqrt {\frac {d}{\cos \left (a+b\,x\right )}}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*(d/cos(a + b*x))^(3/2),x)

[Out]

(2*d*(d/cos(a + b*x))^(1/2))/b

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