\(\int \frac {\sin ^4(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) [418]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {8 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}-\frac {2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}} \] Output:

8/15*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)-4/15*b*sin(f*x+e)/f/(b*sec(f*x+e))^(3/2)-2/9*b*sin(f*x+e)^3/f/(b* 
sec(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.66 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {\frac {192 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)}}-68 \sin (2 (e+f x))+10 \sin (4 (e+f x))}{360 f \sqrt {b \sec (e+f x)}} \] Input:

Integrate[Sin[e + f*x]^4/Sqrt[b*Sec[e + f*x]],x]
 

Output:

((192*EllipticE[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] - 68*Sin[2*(e + f*x)] 
+ 10*Sin[4*(e + f*x)])/(360*f*Sqrt[b*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 100, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3107

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3107

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2}{3} \left (\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )-\frac {2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}\)

Input:

Int[Sin[e + f*x]^4/Sqrt[b*Sec[e + f*x]],x]
 

Output:

(-2*b*Sin[e + f*x]^3)/(9*f*(b*Sec[e + f*x])^(3/2)) + (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))))/3
 

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 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 = 5.10 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.23

method	result	size
default	\(\frac {\frac {2 \sin \left (f x +e \right ) \left (5 \cos \left (f x +e \right )^{4}+5 \cos \left (f x +e \right )^{3}-11 \cos \left (f x +e \right )^{2}-11 \cos \left (f x +e \right )+12\right )}{45}+\frac {8 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{15}-\frac {8 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticE}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{15}}{f \left (\cos \left (f x +e \right )+1\right ) \sqrt {b \sec \left (f x +e \right )}}\)	\(212\)

Input:

int(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/45/f/(cos(f*x+e)+1)/(b*sec(f*x+e))^(1/2)*(sin(f*x+e)*(5*cos(f*x+e)^4+5*c 
os(f*x+e)^3-11*cos(f*x+e)^2-11*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/(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 = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {2 \, {\left ({\left (5 \, \cos \left (f x + e\right )^{4} - 11 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + 6 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 6 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )\right )}}{45 \, b f} \] Input:

integrate(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

2/45*((5*cos(f*x + e)^4 - 11*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))*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))) - 6*I*sqrt(2)*sqrt(b)*weierstrassZeta(- 
4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))))/(b*f)
 

Sympy [F]

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

integrate(sin(f*x+e)**4/(b*sec(f*x+e))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(sin(e + f*x)^4/(b/cos(e + f*x))^(1/2),x)
 

Output:

int(sin(e + f*x)^4/(b/cos(e + f*x))^(1/2), x)
 

Reduce [F]

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

int(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

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