\(\int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [444]

Optimal result

Integrand size = 21, antiderivative size = 126 \[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {8 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{65 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {4 b \sin (e+f x)}{39 f (b \sec (e+f x))^{7/2}}+\frac {8 \sin (e+f x)}{195 b f (b \sec (e+f x))^{3/2}}-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}} \] Output:

8/65*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)-4/39*b*sin(f*x+e)/f/(b*sec(f*x+e))^(7/2)+8/195*sin(f*x+e)/b/f 
/(b*sec(f*x+e))^(3/2)-2/13*b*sin(f*x+e)^3/f/(b*sec(f*x+e))^(7/2)
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66 \[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {192 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\cos ^{\frac {3}{2}}(e+f x) (-6 \sin (e+f x)-55 \sin (3 (e+f x))+15 \sin (5 (e+f x)))}{1560 f \cos ^{\frac {5}{2}}(e+f x) (b \sec (e+f x))^{5/2}} \] Input:

Integrate[Sin[e + f*x]^4/(b*Sec[e + f*x])^(5/2),x]
 

Output:

(192*EllipticE[(e + f*x)/2, 2] + Cos[e + f*x]^(3/2)*(-6*Sin[e + f*x] - 55* 
Sin[3*(e + f*x)] + 15*Sin[5*(e + f*x)]))/(1560*f*Cos[e + f*x]^(5/2)*(b*Sec 
[e + f*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3107, 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.

\(\displaystyle \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc (e+f x)^4 (b \sec (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {6}{13} \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{5/2}}dx-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{13} \int \frac {1}{\csc (e+f x)^2 (b \sec (e+f x))^{5/2}}dx-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {6}{13} \left (\frac {2}{9} \int \frac {1}{(b \sec (e+f x))^{5/2}}dx-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\right )-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{13} \left (\frac {2}{9} \int \frac {1}{\left (b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\right )-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {6}{13} \left (\frac {2}{9} \left (\frac {3 \int \frac {1}{\sqrt {b \sec (e+f x)}}dx}{5 b^2}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\right )-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{13} \left (\frac {2}{9} \left (\frac {3 \int \frac {1}{\sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\right )-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {6}{13} \left (\frac {2}{9} \left (\frac {3 \int \sqrt {\cos (e+f x)}dx}{5 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\right )-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{13} \left (\frac {2}{9} \left (\frac {3 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\right )-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {6}{13} \left (\frac {2}{9} \left (\frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin (e+f x)}{9 f (b \sec (e+f x))^{7/2}}\right )-\frac {2 b \sin ^3(e+f x)}{13 f (b \sec (e+f x))^{7/2}}\)

Input:

Int[Sin[e + f*x]^4/(b*Sec[e + f*x])^(5/2),x]
 

Output:

(-2*b*Sin[e + f*x]^3)/(13*f*(b*Sec[e + f*x])^(7/2)) + (6*((-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))/13
 

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 = 8.74 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.87

method	result	size
default	\(\frac {\frac {2 \sin \left (f x +e \right ) \left (15 \cos \left (f x +e \right )^{6}+15 \cos \left (f x +e \right )^{5}-25 \cos \left (f x +e \right )^{4}-25 \cos \left (f x +e \right )^{3}+4 \cos \left (f x +e \right )^{2}+4 \cos \left (f x +e \right )+12\right )}{195}+\frac {8 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{65}-\frac {8 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticE}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{65}}{f \left (\cos \left (f x +e \right )+1\right ) \sqrt {b \sec \left (f x +e \right )}\, b^{2}}\)	\(235\)

Input:

int(sin(f*x+e)^4/(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/195/f/(cos(f*x+e)+1)/(b*sec(f*x+e))^(1/2)/b^2*(sin(f*x+e)*(15*cos(f*x+e) 
^6+15*cos(f*x+e)^5-25*cos(f*x+e)^4-25*cos(f*x+e)^3+4*cos(f*x+e)^2+4*cos(f* 
x+e)+12)+12*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)-12*I*(1/(c 
os(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 = 0.10 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left ({\left (15 \, \cos \left (f x + e\right )^{6} - 25 \, \cos \left (f x + e\right )^{4} + 4 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + 6 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 6 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )\right )}}{195 \, b^{3} f} \] Input:

integrate(sin(f*x+e)^4/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

2/195*((15*cos(f*x + e)^6 - 25*cos(f*x + e)^4 + 4*cos(f*x + e)^2)*sqrt(b/c 
os(f*x + e))*sin(f*x + e) + 6*I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, wei 
erstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) - 6*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]

\[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {\sin ^{4}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sin(f*x+e)**4/(b*sec(f*x+e))**(5/2),x)
 

Output:

Integral(sin(e + f*x)**4/(b*sec(e + f*x))**(5/2), x)
 

Maxima [F]

\[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(sin(f*x+e)^4/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

integrate(sin(f*x + e)^4/(b*sec(f*x + e))^(5/2), x)
 

Giac [F]

\[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(sin(f*x+e)^4/(b*sec(f*x+e))^(5/2),x, algorithm="giac")
 

Output:

integrate(sin(f*x + e)^4/(b*sec(f*x + e))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^4}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \] Input:

int(sin(e + f*x)^4/(b/cos(e + f*x))^(5/2),x)
 

Output:

int(sin(e + f*x)^4/(b/cos(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sin \left (f x +e \right )^{4}}{\sec \left (f x +e \right )^{3}}d x \right )}{b^{3}} \] Input:

int(sin(f*x+e)^4/(b*sec(f*x+e))^(5/2),x)
 

Output:

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