Integrand size = 12, antiderivative size = 66 \[ \int (b \sec (e+f x))^{3/2} \, dx=-\frac {2 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 b \sqrt {b \sec (e+f x)} \sin (e+f x)}{f} \]
-2*b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*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)+2*b*sin(f*x+e)* (b*sec(f*x+e))^(1/2)/f
Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73 \[ \int (b \sec (e+f x))^{3/2} \, dx=\frac {2 b \sqrt {b \sec (e+f x)} \left (-\sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\sin (e+f x)\right )}{f} \]
(2*b*Sqrt[b*Sec[e + f*x]]*(-(Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2]) + Sin[e + f*x]))/f
Time = 0.33 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4255, 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.
\(\displaystyle \int (b \sec (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-b^2 \int \frac {1}{\sqrt {b \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-b^2 \int \frac {1}{\sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-\frac {b^2 \int \sqrt {\cos (e+f x)}dx}{\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-\frac {b^2 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-\frac {2 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\) |
(-2*b^2*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f* x]]) + (2*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x])/f
3.4.94.3.1 Defintions of rubi rules used
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]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
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 = 0.51 (sec) , antiderivative size = 392, normalized size of antiderivative = 5.94
method | result | size |
default | \(\frac {2 \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \left (\cos ^{2}\left (f x +e \right )\right )-i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )-2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )+i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right )-i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right )+\sin \left (f x +e \right )\right ) \sqrt {b \sec \left (f x +e \right )}\, b}{f \left (\cos \left (f x +e \right )+1\right )}\) | \(392\) |
2/f*(I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Elliptic F(I*(-cot(f*x+e)+csc(f*x+e)),I)*cos(f*x+e)^2-I*(1/(cos(f*x+e)+1))^(1/2)*(c os(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-cot(f*x+e)+csc(f*x+e)),I)*co s(f*x+e)^2+2*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)*cos(f*x+e)-2*I*(1/(cos(f*x+e)+1))^ (1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-cot(f*x+e)+csc(f*x+e )),I)*cos(f*x+e)+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)-I*(1/(cos(f*x+e)+1))^(1/2)*(co s(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-cot(f*x+e)+csc(f*x+e)),I)+sin (f*x+e))*(b*sec(f*x+e))^(1/2)*b/(cos(f*x+e)+1)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27 \[ \int (b \sec (e+f x))^{3/2} \, dx=\frac {-i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, b \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{f} \]
(-I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos( f*x + e) + I*sin(f*x + e))) + I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, wei erstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + 2*b*sqrt(b/cos(f *x + e))*sin(f*x + e))/f
\[ \int (b \sec (e+f x))^{3/2} \, dx=\int \left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \]
\[ \int (b \sec (e+f x))^{3/2} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \,d x } \]
\[ \int (b \sec (e+f x))^{3/2} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (b \sec (e+f x))^{3/2} \, dx=\int {\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \]