3.387 \(\int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx\)

Optimal. Leaf size=41 \[ \frac{2 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac{2 b \sqrt{b \sec (e+f x)}}{f} \]

[Out]

(2*b^3)/(3*f*(b*Sec[e + f*x])^(3/2)) + (2*b*Sqrt[b*Sec[e + f*x]])/f

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Rubi [A]  time = 0.0501463, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2622, 14} \[ \frac{2 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac{2 b \sqrt{b \sec (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^3)/(3*f*(b*Sec[e + f*x])^(3/2)) + (2*b*Sqrt[b*Sec[e + f*x]])/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{-1+\frac{x^2}{b^2}}{x^{5/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^3 \operatorname{Subst}\left (\int \left (-\frac{1}{x^{5/2}}+\frac{1}{b^2 \sqrt{x}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{2 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac{2 b \sqrt{b \sec (e+f x)}}{f}\\ \end{align*}

Mathematica [A]  time = 0.0630676, size = 30, normalized size = 0.73 \[ \frac{b (\cos (2 (e+f x))+7) \sqrt{b \sec (e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(7 + Cos[2*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(3*f)

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Maple [B]  time = 0.139, size = 949, normalized size = 23.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/f*(cos(f*x+e)+1)^2*(-1+cos(f*x+e))^2*(3*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*ln(-2*(2*cos(f*x
+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)
/sin(f*x+e)^2)-3*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e
)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+9*cos(f*x+e)^2
*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+
2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-9*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1
)^2)^(3/2)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(
cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+9*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*ln(-2*(2*cos(f*x+e)^
2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin
(f*x+e)^2)-9*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2
)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+3*ln(-2*(2*cos(f*x+e
)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/s
in(f*x+e)^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)-3*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-c
os(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2
)^(3/2)+4*cos(f*x+e)^3+12*cos(f*x+e))*(b/cos(f*x+e))^(3/2)/sin(f*x+e)^4

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Maxima [A]  time = 0.992971, size = 50, normalized size = 1.22 \begin{align*} \frac{2 \, b{\left (\frac{b^{2}}{\left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{3}{2}}} + 3 \, \sqrt{\frac{b}{\cos \left (f x + e\right )}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.20181, size = 72, normalized size = 1.76 \begin{align*} \frac{2 \,{\left (b \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt{\frac{b}{\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))^(3/2)*sin(f*x+e)^3,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.18802, size = 68, normalized size = 1.66 \begin{align*} \frac{2 \,{\left (\sqrt{b \cos \left (f x + e\right )} b \cos \left (f x + e\right ) + \frac{3 \, b^{2}}{\sqrt{b \cos \left (f x + e\right )}}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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