\(\int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) [431]

Optimal result

Integrand size = 21, antiderivative size = 126 \[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {8 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{77 b^2 f}-\frac {12 b \sin (e+f x)}{77 f (b \sec (e+f x))^{5/2}}+\frac {8 \sin (e+f x)}{77 b f \sqrt {b \sec (e+f x)}}-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}} \] Output:

8/77*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))*(b*sec(f*x+e) 
)^(1/2)/b^2/f-12/77*b*sin(f*x+e)/f/(b*sec(f*x+e))^(5/2)+8/77*sin(f*x+e)/b/ 
f/(b*sec(f*x+e))^(1/2)-2/11*b*sin(f*x+e)^3/f/(b*sec(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64 \[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\sec ^2(e+f x) \left (128 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-5 \sin (2 (e+f x))-24 \sin (4 (e+f x))+7 \sin (6 (e+f x))\right )}{1232 f (b \sec (e+f x))^{3/2}} \] Input:

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

Output:

(Sec[e + f*x]^2*(128*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] - 5*Sin[ 
2*(e + f*x)] - 24*Sin[4*(e + f*x)] + 7*Sin[6*(e + f*x)]))/(1232*f*(b*Sec[e 
 + f*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.16 (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, 3120}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc (e+f x)^4 (b \sec (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {6}{11} \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{3/2}}dx-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{11} \int \frac {1}{\csc (e+f x)^2 (b \sec (e+f x))^{3/2}}dx-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {6}{11} \left (\frac {2}{7} \int \frac {1}{(b \sec (e+f x))^{3/2}}dx-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\right )-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{11} \left (\frac {2}{7} \int \frac {1}{\left (b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\right )-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {\int \sqrt {b \sec (e+f x)}dx}{3 b^2}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\right )-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {\int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\right )-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 b^2}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\right )-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\right )-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {6}{11} \left (\frac {2}{7} \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 b^2 f}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\right )-\frac {2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\)

Input:

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

Output:

(-2*b*Sin[e + f*x]^3)/(11*f*(b*Sec[e + f*x])^(5/2)) + (6*((-2*b*Sin[e + f* 
x])/(7*f*(b*Sec[e + f*x])^(5/2)) + (2*((2*Sqrt[Cos[e + f*x]]*EllipticF[(e 
+ f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*b^2*f) + (2*Sin[e + f*x])/(3*b*f*Sqr 
t[b*Sec[e + f*x]])))/7))/11
 

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 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 = 6.39 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {-\frac {2 \sin \left (f x +e \right ) \left (-7 \cos \left (f x +e \right )^{4}+13 \cos \left (f x +e \right )^{2}-4\right )}{77}-\frac {2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (-4-4 \sec \left (f x +e \right )\right )}{77}}{f \sqrt {b \sec \left (f x +e \right )}\, b}\)	\(114\)

Input:

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

Output:

1/f*(-2/77*sin(f*x+e)*(-7*cos(f*x+e)^4+13*cos(f*x+e)^2-4)-2/77*I*(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)*(-4-4*sec(f*x+e)))/(b*sec(f*x+e))^(1/2)/b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.87 \[ \int \frac {\sin ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {2 \, {\left ({\left (7 \, \cos \left (f x + e\right )^{5} - 13 \, \cos \left (f x + e\right )^{3} + 4 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 2 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 2 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{77 \, b^{2} f} \] Input:

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

Output:

2/77*((7*cos(f*x + e)^5 - 13*cos(f*x + e)^3 + 4*cos(f*x + e))*sqrt(b/cos(f 
*x + e))*sin(f*x + e) - 2*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos 
(f*x + e) + I*sin(f*x + e)) + 2*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 
0, cos(f*x + e) - I*sin(f*x + e)))/(b^2*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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