\(\int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx\) [387]

Optimal result

Integrand size = 21, antiderivative size = 41 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {2 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

[Out]

2/3*b^3/f/(b*sec(f*x+e))^(3/2)+2*b*(b*sec(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2702, 14} \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {2 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

[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 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]]

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*} \text {integral}& = \frac {b^3 \text {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 \text {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] (verified)

Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {b (7+\cos (2 (e+f x))) \sqrt {b \sec (e+f x)}}{3 f} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(824\) vs. \(2(35)=70\).

Time = 0.23 (sec) , antiderivative size = 825, normalized size of antiderivative = 20.12

method	result	size
default	\(\text {Expression too large to display}\)	\(825\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{3 \, f} \]

[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

Sympy [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {2 \, b {\left (\frac {b^{2}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}} + 3 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )}}{3 \, f} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {2 \, {\left (\sqrt {b \cos \left (f x + e\right )} b \cos \left (f x + e\right ) + \frac {3 \, b^{2}}{\sqrt {b \cos \left (f x + e\right )}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{3 \, f} \]

[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

Mupad [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \]

[In]

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

[Out]

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