\(\int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [445]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}+\frac {4 \sin (e+f x)}{45 b f (b \sec (e+f x))^{3/2}} \] Output:

4/15*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/b^2/f/cos(f*x+e)^(1/2)/(b*sec(f 
*x+e))^(1/2)-2/9*b*sin(f*x+e)/f/(b*sec(f*x+e))^(7/2)+4/45*sin(f*x+e)/b/f/( 
b*sec(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.67 \[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\frac {96 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)}}-4 \sin (2 (e+f x))-10 \sin (4 (e+f x))}{360 b^2 f \sqrt {b \sec (e+f x)}} \] Input:

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

Output:

((96*EllipticE[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] - 4*Sin[2*(e + f*x)] - 
10*Sin[4*(e + f*x)])/(360*b^2*f*Sqrt[b*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3107, 3042, 4256, 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.

\(\displaystyle \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc (e+f x)^2 (b \sec (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {2}{9} \int \frac {1}{(b \sec (e+f x))^{5/2}}dx-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{9} \int \frac {1}{\left (b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {2}{9} \left (\frac {3 \int \frac {1}{\sqrt {b \sec (e+f x)}}dx}{5 b^2}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{9} \left (\frac {3 \int \frac {1}{\sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2}{9} \left (\frac {3 \int \sqrt {\cos (e+f x)}dx}{5 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{9} \left (\frac {3 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2}{9} \left (\frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\)

Input:

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

Output:

(-2*b*Sin[e + f*x])/(9*f*(b*Sec[e + f*x])^(7/2)) + (2*((6*EllipticE[(e + f 
*x)/2, 2])/(5*b^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) + (2*Sin[e + 
f*x])/(5*b*f*(b*Sec[e + f*x])^(3/2))))/9
 

Defintions of rubi rules used

rule 3042

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

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 4256

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

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 = 6.79 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.19

method	result	size
default	\(\frac {\frac {2 \sin \left (f x +e \right ) \left (-5 \cos \left (f x +e \right )^{4}-5 \cos \left (f x +e \right )^{3}+2 \cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+6\right )}{45}+\frac {4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{15}-\frac {4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticE}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{15}}{f \left (\cos \left (f x +e \right )+1\right ) \sqrt {b \sec \left (f x +e \right )}\, b^{2}}\)	\(215\)

Input:

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

Output:

2/45/f/(cos(f*x+e)+1)/(b*sec(f*x+e))^(1/2)/b^2*(sin(f*x+e)*(-5*cos(f*x+e)^ 
4-5*cos(f*x+e)^3+2*cos(f*x+e)^2+2*cos(f*x+e)+6)+6*I*(1/(cos(f*x+e)+1))^(1/ 
2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(2+cos(f*x+e)+sec(f*x+e))*EllipticF(I 
*(cot(f*x+e)-csc(f*x+e)),I)-6*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos( 
f*x+e)+1))^(1/2)*(2+cos(f*x+e)+sec(f*x+e))*EllipticE(I*(cot(f*x+e)-csc(f*x 
+e)),I))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09 \[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left ({\left (5 \, \cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 3 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )\right )}}{45 \, b^{3} f} \] Input:

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

Output:

-2/45*((5*cos(f*x + e)^4 - 2*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))*sin(f*x 
+ e) - 3*I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(f*x + e) + I*sin(f*x + e))) + 3*I*sqrt(2)*sqrt(b)*weierstrassZeta(- 
4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))))/(b^3*f)
 

Sympy [F]

\[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {\sin ^{2}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sin \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{3}}d x \right )}{b^{3}} \] Input:

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

Output:

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