3.4.88 \(\int (b \sec (e+f x))^{3/2} \sin (e+f x) \, dx\) [388]

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

[Out]

2*b*(b*sec(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.02, 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 b \sqrt {b \sec (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sqrt[b*Sec[e + f*x]])/f

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 (b \sec (e+f x))^{3/2} \sin (e+f x) \, dx &=\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {2 b \sqrt {b \sec (e+f x)}}{f}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 18, normalized size = 1.00 \begin {gather*} \frac {2 b \sqrt {b \sec (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sqrt[b*Sec[e + f*x]])/f

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Maple [A]
time = 0.05, size = 17, normalized size = 0.94

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*(b*sec(f*x+e))^(1/2)/f

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Maxima [A]
time = 0.28, size = 25, normalized size = 1.39 \begin {gather*} \frac {2 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} \cos \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 19, normalized size = 1.06 \begin {gather*} \frac {2 \, b \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*sqrt(b/cos(f*x + e))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**(3/2)*sin(f*x+e),x)

[Out]

Integral((b*sec(e + f*x))**(3/2)*sin(e + f*x), x)

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Giac [A]
time = 4.59, size = 27, normalized size = 1.50 \begin {gather*} \frac {2 \, b^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{\sqrt {b \cos \left (f x + e\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.48, size = 18, normalized size = 1.00 \begin {gather*} \frac {2\,b\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)*(b/cos(e + f*x))^(3/2),x)

[Out]

(2*b*(b/cos(e + f*x))^(1/2))/f

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