\(\int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx\) [392]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=-\frac {24 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)} \sin ^3(e+f x)}{f} \] Output:

-24/5*b^2*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/f/cos(f*x+e)^(1/2)/(b*sec( 
f*x+e))^(1/2)+12/5*b^3*sin(f*x+e)/f/(b*sec(f*x+e))^(3/2)+2*b*(b*sec(f*x+e) 
)^(1/2)*sin(f*x+e)^3/f
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.61 \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\frac {b \sqrt {b \sec (e+f x)} \left (-48 \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+21 \sin (e+f x)+\sin (3 (e+f x))\right )}{10 f} \] Input:

Integrate[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^4,x]
 

Output:

(b*Sqrt[b*Sec[e + f*x]]*(-48*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] 
+ 21*Sin[e + f*x] + Sin[3*(e + f*x)]))/(10*f)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3104, 3042, 3107, 3042, 4258, 3042, 3119}

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.

\(\displaystyle \int \sin ^4(e+f x) (b \sec (e+f x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(b \sec (e+f x))^{3/2}}{\csc (e+f x)^4}dx\)

\(\Big \downarrow \) 3104

\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \int \frac {\sin ^2(e+f x)}{\sqrt {b \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \int \frac {1}{\csc (e+f x)^2 \sqrt {b \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {2}{5} \int \frac {1}{\sqrt {b \sec (e+f x)}}dx-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {2}{5} \int \frac {1}{\sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {2 \int \sqrt {\cos (e+f x)}dx}{5 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {2 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\)

Input:

Int[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^4,x]
 

Output:

(2*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3)/f - 6*b^2*((4*EllipticE[(e + f*x 
)/2, 2])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (2*b*Sin[e + f*x] 
)/(5*f*(b*Sec[e + f*x])^(3/2)))
 

Defintions of rubi rules used

rule 3042

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

rule 3104

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[b*(a*Csc[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/ 
(f*a*(n - 1))), x] + Simp[b^2*((m + 1)/(a^2*(n - 1)))   Int[(a*Csc[e + f*x] 
)^(m + 2)*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ 
[n, 1] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
 

rule 3107

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_.), x_Symbol] :> Simp[b*(a*Csc[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1) 
/(a*f*(m + n))), x] + Simp[(m + 1)/(a^2*(m + n))   Int[(a*Csc[e + f*x])^(m 
+ 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] 
&& NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.79 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.17

method	result	size
default	\(-\frac {2 b \left (\left (-\cos \left (f x +e \right )^{3}-\cos \left (f x +e \right )^{2}+7 \cos \left (f x +e \right )-5\right ) \sin \left (f x +e \right )+12 i \left (\cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )+12 i \left (-\cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )\right ) \sqrt {b \sec \left (f x +e \right )}}{5 f \left (\cos \left (f x +e \right )+1\right )}\)	\(213\)

Input:

int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x,method=_RETURNVERBOSE)
 

Output:

-2/5/f*b*((-cos(f*x+e)^3-cos(f*x+e)^2+7*cos(f*x+e)-5)*sin(f*x+e)+12*I*(cos 
(f*x+e)^2+2*cos(f*x+e)+1)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e) 
+1))^(1/2)*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I)+12*I*(-cos(f*x+e)^2-2*co 
s(f*x+e)-1)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*Ell 
ipticE(I*(cot(f*x+e)-csc(f*x+e)),I))*(b*sec(f*x+e))^(1/2)/(cos(f*x+e)+1)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=-\frac {2 \, {\left (6 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 6 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - {\left (b \cos \left (f x + e\right )^{2} + 5 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{5 \, f} \] Input:

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="fricas")
 

Output:

-2/5*(6*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0 
, cos(f*x + e) + I*sin(f*x + e))) - 6*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4 
, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - (b*cos(f 
*x + e)^2 + 5*b)*sqrt(b/cos(f*x + e))*sin(f*x + e))/f
 

Sympy [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\text {Timed out} \] Input:

integrate((b*sec(f*x+e))**(3/2)*sin(f*x+e)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \] Input:

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e))^(3/2)*sin(f*x + e)^4, x)
 

Giac [F]

\[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \] Input:

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e))^(3/2)*sin(f*x + e)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \] Input:

int(sin(e + f*x)^4*(b/cos(e + f*x))^(3/2),x)
 

Output:

int(sin(e + f*x)^4*(b/cos(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) \sin \left (f x +e \right )^{4}d x \right ) b \] Input:

int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x)
 

Output:

sqrt(b)*int(sqrt(sec(e + f*x))*sec(e + f*x)*sin(e + f*x)**4,x)*b