Integrand size = 21, antiderivative size = 68 \[ \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {\csc (e+f x)}{b f \sqrt {b \sec (e+f x)}}-\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{b^2 f} \] Output:
-csc(f*x+e)/b/f/(b*sec(f*x+e))^(1/2)-cos(f*x+e)^(1/2)*InverseJacobiAM(1/2* f*x+1/2*e,2^(1/2))*(b*sec(f*x+e))^(1/2)/b^2/f
Time = 0.44 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {-\sqrt {\cos (e+f x)} \csc (e+f x)-\operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{f \cos ^{\frac {3}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \] Input:
Integrate[Csc[e + f*x]^2/(b*Sec[e + f*x])^(3/2),x]
Output:
(-(Sqrt[Cos[e + f*x]]*Csc[e + f*x]) - EllipticF[(e + f*x)/2, 2])/(f*Cos[e + f*x]^(3/2)*(b*Sec[e + f*x])^(3/2))
Time = 0.59 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3103, 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.
\(\displaystyle \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc (e+f x)^2}{(b \sec (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3103 |
\(\displaystyle -\frac {\int \sqrt {b \sec (e+f x)}dx}{2 b^2}-\frac {\csc (e+f x)}{b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{2 b^2}-\frac {\csc (e+f x)}{b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle -\frac {\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{2 b^2}-\frac {\csc (e+f x)}{b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 b^2}-\frac {\csc (e+f x)}{b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{b^2 f}-\frac {\csc (e+f x)}{b f \sqrt {b \sec (e+f x)}}\) |
Input:
Int[Csc[e + f*x]^2/(b*Sec[e + f*x])^(3/2),x]
Output:
-(Csc[e + f*x]/(b*f*Sqrt[b*Sec[e + f*x]])) - (Sqrt[Cos[e + f*x]]*EllipticF [(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(b^2*f)
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[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 = 2.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.35
method | result | size |
default | \(\frac {i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (-1-\sec \left (f x +e \right )\right )-\csc \left (f x +e \right )}{f \sqrt {b \sec \left (f x +e \right )}\, b}\) | \(92\) |
Input:
int(csc(f*x+e)^2/(b*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/f/(b*sec(f*x+e))^(1/2)/b*(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)*(-1-sec(f*x+e))-csc( f*x+e))
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.49 \[ \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {i \, \sqrt {2} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - i \, \sqrt {2} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, b^{2} f \sin \left (f x + e\right )} \] Input:
integrate(csc(f*x+e)^2/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")
Output:
1/2*(I*sqrt(2)*sqrt(b)*sin(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e ) + I*sin(f*x + e)) - I*sqrt(2)*sqrt(b)*sin(f*x + e)*weierstrassPInverse(- 4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*sqrt(b/cos(f*x + e))*cos(f*x + e) )/(b^2*f*sin(f*x + e))
\[ \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {\csc ^{2}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(csc(f*x+e)**2/(b*sec(f*x+e))**(3/2),x)
Output:
Integral(csc(e + f*x)**2/(b*sec(e + f*x))**(3/2), x)
\[ \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{2}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(f*x+e)^2/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")
Output:
integrate(csc(f*x + e)^2/(b*sec(f*x + e))^(3/2), x)
\[ \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{2}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(f*x+e)^2/(b*sec(f*x+e))^(3/2),x, algorithm="giac")
Output:
integrate(csc(f*x + e)^2/(b*sec(f*x + e))^(3/2), x)
Timed out. \[ \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^2\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(1/(sin(e + f*x)^2*(b/cos(e + f*x))^(3/2)),x)
Output:
int(1/(sin(e + f*x)^2*(b/cos(e + f*x))^(3/2)), x)
\[ \int \frac {\csc ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \csc \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{2}}d x \right )}{b^{2}} \] Input:
int(csc(f*x+e)^2/(b*sec(f*x+e))^(3/2),x)
Output:
(sqrt(b)*int((sqrt(sec(e + f*x))*csc(e + f*x)**2)/sec(e + f*x)**2,x))/b**2