Integrand size = 21, antiderivative size = 132 \[ \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\csc (e+f x)}{12 b f \sqrt {b \sec (e+f x)}}+\frac {\csc ^3(e+f x)}{30 b f \sqrt {b \sec (e+f x)}}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}-\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{12 b^2 f} \] Output:
1/12*csc(f*x+e)/b/f/(b*sec(f*x+e))^(1/2)+1/30*csc(f*x+e)^3/b/f/(b*sec(f*x+ e))^(1/2)-1/5*csc(f*x+e)^5/b/f/(b*sec(f*x+e))^(1/2)-1/12*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
Time = 0.86 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.56 \[ \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {5 \csc (e+f x)+2 \csc ^3(e+f x)-12 \csc ^5(e+f x)-\frac {5 \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{\sqrt {\cos (e+f x)}}}{60 b f \sqrt {b \sec (e+f x)}} \] Input:
Integrate[Csc[e + f*x]^6/(b*Sec[e + f*x])^(3/2),x]
Output:
(5*Csc[e + f*x] + 2*Csc[e + f*x]^3 - 12*Csc[e + f*x]^5 - (5*EllipticF[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]])/(60*b*f*Sqrt[b*Sec[e + f*x]])
Time = 1.02 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3103, 3042, 3105, 3042, 3105, 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.
\(\displaystyle \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc (e+f x)^6}{(b \sec (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3103 |
\(\displaystyle -\frac {\int \csc ^4(e+f x) \sqrt {b \sec (e+f x)}dx}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \csc (e+f x)^4 \sqrt {b \sec (e+f x)}dx}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3105 |
\(\displaystyle -\frac {\frac {5}{6} \int \csc ^2(e+f x) \sqrt {b \sec (e+f x)}dx-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {5}{6} \int \csc (e+f x)^2 \sqrt {b \sec (e+f x)}dx-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3105 |
\(\displaystyle -\frac {\frac {5}{6} \left (\frac {1}{2} \int \sqrt {b \sec (e+f x)}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {5}{6} \left (\frac {1}{2} \int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle -\frac {\frac {5}{6} \left (\frac {1}{2} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {5}{6} \left (\frac {1}{2} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\frac {5}{6} \left (\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{f}-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}}\right )-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}-\frac {\csc ^5(e+f x)}{5 b f \sqrt {b \sec (e+f x)}}\) |
Input:
Int[Csc[e + f*x]^6/(b*Sec[e + f*x])^(3/2),x]
Output:
-1/5*Csc[e + f*x]^5/(b*f*Sqrt[b*Sec[e + f*x]]) - (-1/3*(b*Csc[e + f*x]^3)/ (f*Sqrt[b*Sec[e + f*x]]) + (5*(-((b*Csc[e + f*x])/(f*Sqrt[b*Sec[e + f*x]]) ) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/f) )/6)/(10*b^2)
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-a)*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n + 1)/(f*b*(m - 1))), x] + Simp[a^2*((n + 1)/(b^2*(m - 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[m, 1] && LtQ[n, -1] && IntegersQ[2*m, 2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Simp[(-a)*b*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(m - 1))), x] + Simp[a^2*((m + n - 2)/(m - 1)) Int[(a*Csc[e + f* x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[ m, 1] && IntegersQ[2*m, 2*n] && !GtQ[n, m]
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]
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]
Result contains complex when optimal does not.
Time = 2.89 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {\frac {i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (-5-5 \sec \left (f x +e \right )\right )}{60}+\frac {\left (5 \cos \left (f x +e \right )^{4}-12 \cos \left (f x +e \right )^{2}-5\right ) \csc \left (f x +e \right )^{5}}{60}}{f b \sqrt {b \sec \left (f x +e \right )}}\) | \(116\) |
Input:
int(csc(f*x+e)^6/(b*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/f*(1/60*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ell ipticF(I*(cot(f*x+e)-csc(f*x+e)),I)*(-5-5*sec(f*x+e))+1/60*(5*cos(f*x+e)^4 -12*cos(f*x+e)^2-5)*csc(f*x+e)^5)/b/(b*sec(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.49 \[ \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {5 \, \sqrt {2} {\left (-i \, \cos \left (f x + e\right )^{4} + 2 i \, \cos \left (f x + e\right )^{2} - i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 \, \sqrt {2} {\left (i \, \cos \left (f x + e\right )^{4} - 2 i \, \cos \left (f x + e\right )^{2} + i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (5 \, \cos \left (f x + e\right )^{5} - 12 \, \cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{120 \, {\left (b^{2} f \cos \left (f x + e\right )^{4} - 2 \, b^{2} f \cos \left (f x + e\right )^{2} + b^{2} f\right )} \sin \left (f x + e\right )} \] Input:
integrate(csc(f*x+e)^6/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")
Output:
-1/120*(5*sqrt(2)*(-I*cos(f*x + e)^4 + 2*I*cos(f*x + e)^2 - I)*sqrt(b)*sin (f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 5*sq rt(2)*(I*cos(f*x + e)^4 - 2*I*cos(f*x + e)^2 + I)*sqrt(b)*sin(f*x + e)*wei erstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*(5*cos(f*x + e) ^5 - 12*cos(f*x + e)^3 - 5*cos(f*x + e))*sqrt(b/cos(f*x + e)))/((b^2*f*cos (f*x + e)^4 - 2*b^2*f*cos(f*x + e)^2 + b^2*f)*sin(f*x + e))
\[ \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {\csc ^{6}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(csc(f*x+e)**6/(b*sec(f*x+e))**(3/2),x)
Output:
Integral(csc(e + f*x)**6/(b*sec(e + f*x))**(3/2), x)
\[ \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{6}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(f*x+e)^6/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")
Output:
integrate(csc(f*x + e)^6/(b*sec(f*x + e))^(3/2), x)
\[ \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{6}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(f*x+e)^6/(b*sec(f*x+e))^(3/2),x, algorithm="giac")
Output:
integrate(csc(f*x + e)^6/(b*sec(f*x + e))^(3/2), x)
Timed out. \[ \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^6\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(1/(sin(e + f*x)^6*(b/cos(e + f*x))^(3/2)),x)
Output:
int(1/(sin(e + f*x)^6*(b/cos(e + f*x))^(3/2)), x)
\[ \int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \csc \left (f x +e \right )^{6}}{\sec \left (f x +e \right )^{2}}d x \right )}{b^{2}} \] Input:
int(csc(f*x+e)^6/(b*sec(f*x+e))^(3/2),x)
Output:
(sqrt(b)*int((sqrt(sec(e + f*x))*csc(e + f*x)**6)/sec(e + f*x)**2,x))/b**2