\(\int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx\) [386]

Optimal result

Integrand size = 21, antiderivative size = 63 \[ \int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx=-\frac {2 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {4 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

[Out]

-2/7*b^5/f/(b*sec(f*x+e))^(7/2)+4/3*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 = 63, 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, 276} \[ \int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx=-\frac {2 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {4 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

[In]

Int[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^5,x]

[Out]

(-2*b^5)/(7*f*(b*Sec[e + f*x])^(7/2)) + (4*b^3)/(3*f*(b*Sec[e + f*x])^(3/2)) + (2*b*Sqrt[b*Sec[e + f*x]])/f

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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^5 \text {Subst}\left (\int \frac {\left (-1+\frac {x^2}{b^2}\right )^2}{x^{9/2}} \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^5 \text {Subst}\left (\int \left (\frac {1}{x^{9/2}}-\frac {2}{b^2 x^{5/2}}+\frac {1}{b^4 \sqrt {x}}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = -\frac {2 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {4 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 = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.67 \[ \int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx=\frac {b (215+44 \cos (2 (e+f x))-3 \cos (4 (e+f x))) \sqrt {b \sec (e+f x)}}{84 f} \]

[In]

Integrate[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^5,x]

[Out]

(b*(215 + 44*Cos[2*(e + f*x)] - 3*Cos[4*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(84*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(834\) vs. \(2(53)=106\).

Time = 0.24 (sec) , antiderivative size = 835, normalized size of antiderivative = 13.25

method	result	size
default	\(\text {Expression too large to display}\)	\(835\)

[In]

int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^5,x,method=_RETURNVERBOSE)

[Out]

-1/42/f*b*(b*sec(f*x+e))^(1/2)*(21*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)-21*ln(2*(2*c
os(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)+63*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)*c
os(f*x+e)-63*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)-co
s(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)+63*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)-63*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)+12*cos(f*x+e)^4+21*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))*(-co
s(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*sec(f*x+e)-21*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)-
56*cos(f*x+e)^2-84)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.68 \[ \int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx=-\frac {2 \, {\left (3 \, b \cos \left (f x + e\right )^{4} - 14 \, b \cos \left (f x + e\right )^{2} - 21 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{21 \, f} \]

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^5,x, algorithm="fricas")

[Out]

-2/21*(3*b*cos(f*x + e)^4 - 14*b*cos(f*x + e)^2 - 21*b)*sqrt(b/cos(f*x + e))/f

Sympy [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx=\text {Timed out} \]

[In]

integrate((b*sec(f*x+e))**(3/2)*sin(f*x+e)**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx=-\frac {2 \, b {\left (\frac {3 \, b^{4}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {7}{2}}} - \frac {14 \, b^{2}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}} - 21 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )}}{21 \, f} \]

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^5,x, algorithm="maxima")

[Out]

-2/21*b*(3*b^4/(b/cos(f*x + e))^(7/2) - 14*b^2/(b/cos(f*x + e))^(3/2) - 21*sqrt(b/cos(f*x + e)))/f

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19 \[ \int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx=-\frac {2 \, {\left (3 \, \sqrt {b \cos \left (f x + e\right )} b^{3} \cos \left (f x + e\right )^{3} - 14 \, \sqrt {b \cos \left (f x + e\right )} b^{3} \cos \left (f x + e\right ) - \frac {21 \, b^{4}}{\sqrt {b \cos \left (f x + e\right )}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{21 \, b^{2} f} \]

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^5,x, algorithm="giac")

[Out]

-2/21*(3*sqrt(b*cos(f*x + e))*b^3*cos(f*x + e)^3 - 14*sqrt(b*cos(f*x + e))*b^3*cos(f*x + e) - 21*b^4/sqrt(b*co
s(f*x + e)))*sgn(cos(f*x + e))/(b^2*f)

Mupad [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^5\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \]

[In]

int(sin(e + f*x)^5*(b/cos(e + f*x))^(3/2),x)

[Out]

int(sin(e + f*x)^5*(b/cos(e + f*x))^(3/2), x)