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

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {b (7+\cos (2 (e+f x))) \sqrt {b \sec (e+f x)}}{3 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3102, 25, 27, 244, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(b \sec (e+f x))^{3/2}}{\csc (e+f x)^3}dx\)

\(\Big \downarrow \) 3102

\(\displaystyle \frac {b^3 \int -\frac {b^2-b^2 \sec ^2(e+f x)}{b^2 (b \sec (e+f x))^{5/2}}d(b \sec (e+f x))}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b^3 \int \frac {b^2-b^2 \sec ^2(e+f x)}{b^2 (b \sec (e+f x))^{5/2}}d(b \sec (e+f x))}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {b \int \frac {b^2-b^2 \sec ^2(e+f x)}{(b \sec (e+f x))^{5/2}}d(b \sec (e+f x))}{f}\)

\(\Big \downarrow \) 244

\(\displaystyle -\frac {b \int \left (\frac {b^2}{(b \sec (e+f x))^{5/2}}-\frac {1}{\sqrt {b \sec (e+f x)}}\right )d(b \sec (e+f x))}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b \left (-\frac {2 b^2}{3 (b \sec (e+f x))^{3/2}}-2 \sqrt {b \sec (e+f x)}\right )}{f}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3102

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S 
ymbol] :> Simp[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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(235\) vs. \(2(35)=70\).

Time = 1.30 (sec) , antiderivative size = 236, normalized size of antiderivative = 5.76

method	result	size
default	\(\frac {\left (\left (3 \cos \left (f x +e \right )+3\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )+\left (-3 \cos \left (f x +e \right )-3\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )+4 \cos \left (f x +e \right )^{2}+12\right ) b \sqrt {b \sec \left (f x +e \right )}}{6 f}\)	\(236\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{3 \, f} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {2 \, b {\left (\frac {b^{2}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}} + 3 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )}}{3 \, f} \] Input:

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

Output:

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)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\frac {2 \, {\left (\sqrt {b \cos \left (f x + e\right )} b \cos \left (f x + e\right ) + \frac {3 \, b^{2}}{\sqrt {b \cos \left (f x + e\right )}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{3 \, f} \] Input:

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

Output:

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

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

Output:

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

Reduce [F]

\[ \int (b \sec (e+f x))^{3/2} \sin ^3(e+f x) \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) \sin \left (f x +e \right )^{3}d x \right ) b \] Input:

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

Output:

sqrt(b)*int(sqrt(sec(e + f*x))*sec(e + f*x)*sin(e + f*x)**3,x)*b