\(\int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx\) [404]

Optimal result

Integrand size = 21, antiderivative size = 130 \[ \int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx=-\frac {80 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{21 f}+\frac {40 b^3 \sin (e+f x)}{21 f \sqrt {b \sec (e+f x)}}+\frac {20 b^3 \sin ^3(e+f x)}{21 f \sqrt {b \sec (e+f x)}}+\frac {2 b (b \sec (e+f x))^{3/2} \sin ^5(e+f x)}{3 f} \] Output:

-80/21*b^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))*(b*sec( 
f*x+e))^(1/2)/f+40/21*b^3*sin(f*x+e)/f/(b*sec(f*x+e))^(1/2)+20/21*b^3*sin( 
f*x+e)^3/f/(b*sec(f*x+e))^(1/2)+2/3*b*(b*sec(f*x+e))^(3/2)*sin(f*x+e)^5/f
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.57 \[ \int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx=-\frac {b^2 \sqrt {b \sec (e+f x)} \left (320 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-58 \sin (2 (e+f x))+3 \sin (4 (e+f x))-56 \tan (e+f x)\right )}{84 f} \] Input:

Integrate[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^6,x]
 

Output:

-1/84*(b^2*Sqrt[b*Sec[e + f*x]]*(320*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x 
)/2, 2] - 58*Sin[2*(e + f*x)] + 3*Sin[4*(e + f*x)] - 56*Tan[e + f*x]))/f
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3104, 3042, 3107, 3042, 3107, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3104

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \int \frac {\sqrt {b \sec (e+f x)}}{\csc (e+f x)^4}dx\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \left (\frac {6}{7} \int \sqrt {b \sec (e+f x)} \sin ^2(e+f x)dx-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \left (\frac {6}{7} \int \frac {\sqrt {b \sec (e+f x)}}{\csc (e+f x)^2}dx-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\right )\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \left (\frac {6}{7} \left (\frac {2}{3} \int \sqrt {b \sec (e+f x)}dx-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \left (\frac {6}{7} \left (\frac {2}{3} \int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \left (\frac {6}{7} \left (\frac {2}{3} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \left (\frac {6}{7} \left (\frac {2}{3} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 b \sin ^5(e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {10}{3} b^2 \left (\frac {6}{7} \left (\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\right )\)

Input:

Int[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^6,x]
 

Output:

(2*b*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^5)/(3*f) - (10*b^2*((-2*b*Sin[e + 
 f*x]^3)/(7*f*Sqrt[b*Sec[e + f*x]]) + (6*((4*Sqrt[Cos[e + f*x]]*EllipticF[ 
(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*f) - (2*b*Sin[e + f*x])/(3*f*Sqrt 
[b*Sec[e + f*x]])))/7))/3
 

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 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 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 = 1012.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {\left (-\frac {2 \left (3 \cos \left (f x +e \right )^{4}-16 \cos \left (f x +e \right )^{2}-7\right ) \tan \left (f x +e \right )}{21}-\frac {80 i \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{21}\right ) b^{2} \sqrt {b \sec \left (f x +e \right )}}{f}\)	\(112\)

Input:

int((b*sec(f*x+e))^(5/2)*sin(f*x+e)^6,x,method=_RETURNVERBOSE)
 

Output:

1/f*(-2/21*(3*cos(f*x+e)^4-16*cos(f*x+e)^2-7)*tan(f*x+e)-80/21*I*EllipticF 
(I*(cot(f*x+e)-csc(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)+1)*(cos 
(f*x+e)/(cos(f*x+e)+1))^(1/2))*b^2*(b*sec(f*x+e))^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.99 \[ \int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx=-\frac {2 \, {\left (-20 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 20 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + {\left (3 \, b^{2} \cos \left (f x + e\right )^{4} - 16 \, b^{2} \cos \left (f x + e\right )^{2} - 7 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{21 \, f \cos \left (f x + e\right )} \] Input:

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^6,x, algorithm="fricas")
 

Output:

-2/21*(-20*I*sqrt(2)*b^(5/2)*cos(f*x + e)*weierstrassPInverse(-4, 0, cos(f 
*x + e) + I*sin(f*x + e)) + 20*I*sqrt(2)*b^(5/2)*cos(f*x + e)*weierstrassP 
Inverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + (3*b^2*cos(f*x + e)^4 - 16 
*b^2*cos(f*x + e)^2 - 7*b^2)*sqrt(b/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x 
 + e))
 

Sympy [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx=\text {Timed out} \] Input:

integrate((b*sec(f*x+e))**(5/2)*sin(f*x+e)**6,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{6} \,d x } \] Input:

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^6,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e))^(5/2)*sin(f*x + e)^6, x)
 

Giac [F]

\[ \int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{6} \,d x } \] Input:

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^6,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e))^(5/2)*sin(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^6\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \] Input:

int(sin(e + f*x)^6*(b/cos(e + f*x))^(5/2),x)
 

Output:

int(sin(e + f*x)^6*(b/cos(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int (b \sec (e+f x))^{5/2} \sin ^6(e+f x) \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \sin \left (f x +e \right )^{6}d x \right ) b^{2} \] Input:

int((b*sec(f*x+e))^(5/2)*sin(f*x+e)^6,x)
 

Output:

sqrt(b)*int(sqrt(sec(e + f*x))*sec(e + f*x)**2*sin(e + f*x)**6,x)*b**2