Integrand size = 44, antiderivative size = 41 \[ \int \left (-\left (\left (1+b^2 n^2\right ) \sec \left (a+b \log \left (c x^n\right )\right )\right )+2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-x \sec \left (a+b \log \left (c x^n\right )\right )+b n x \sec \left (a+b \log \left (c x^n\right )\right ) \tan \left (a+b \log \left (c x^n\right )\right ) \]
Time = 1.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int \left (-\left (\left (1+b^2 n^2\right ) \sec \left (a+b \log \left (c x^n\right )\right )\right )+2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx=x \sec \left (a+b \log \left (c x^n\right )\right ) \left (-1+b n \tan \left (a+b \log \left (c x^n\right )\right )\right ) \]
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.32 (sec) , antiderivative size = 175, normalized size of antiderivative = 4.27, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {2009}
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.
\(\displaystyle \int \left (2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )-\left (b^2 n^2+1\right ) \sec \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {16 e^{3 i a} b^2 n^2 x \left (c x^n\right )^{3 i b} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {i}{b n}\right ),\frac {1}{2} \left (5-\frac {i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+3 i b n}-2 e^{i a} x (1-i b n) \left (c x^n\right )^{i b} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (1-\frac {i}{b n}\right ),\frac {1}{2} \left (3-\frac {i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )\) |
-2*E^(I*a)*(1 - I*b*n)*x*(c*x^n)^(I*b)*Hypergeometric2F1[1, (1 - I/(b*n))/ 2, (3 - I/(b*n))/2, -(E^((2*I)*a)*(c*x^n)^((2*I)*b))] + (16*b^2*E^((3*I)*a )*n^2*x*(c*x^n)^((3*I)*b)*Hypergeometric2F1[3, (3 - I/(b*n))/2, (5 - I/(b* n))/2, -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/(1 + (3*I)*b*n)
3.3.59.3.1 Defintions of rubi rules used
Time = 29.53 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \(-\frac {2 x \left (-\sin \left (a +b \ln \left (c \,x^{n}\right )\right ) b n +\cos \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{\cos \left (4 b \ln \left (\sqrt {c \,x^{n}}\right )+2 a \right )+1}\) | \(51\) |
risch | \(-\frac {2 i c^{i b} \left (x^{n}\right )^{i b} x \left (n b \,c^{2 i b} \left (x^{n}\right )^{2 i b} {\mathrm e}^{-\frac {3 b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}} {\mathrm e}^{\frac {3 b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}} {\mathrm e}^{\frac {3 b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}} {\mathrm e}^{-\frac {3 b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}} {\mathrm e}^{3 i a}-b n \,{\mathrm e}^{\frac {b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}} {\mathrm e}^{-\frac {b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}} {\mathrm e}^{-\frac {b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}} {\mathrm e}^{\frac {b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}} {\mathrm e}^{i a}-i c^{2 i b} \left (x^{n}\right )^{2 i b} {\mathrm e}^{-\frac {3 b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}} {\mathrm e}^{\frac {3 b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}} {\mathrm e}^{\frac {3 b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}} {\mathrm e}^{-\frac {3 b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}} {\mathrm e}^{3 i a}-i {\mathrm e}^{\frac {b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}} {\mathrm e}^{-\frac {b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}} {\mathrm e}^{-\frac {b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}} {\mathrm e}^{\frac {b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}} {\mathrm e}^{i a}\right )}{{\left (\left (x^{n}\right )^{2 i b} c^{2 i b} {\mathrm e}^{-b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}} {\mathrm e}^{b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )} {\mathrm e}^{b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}} {\mathrm e}^{-b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )} {\mathrm e}^{2 i a}+1\right )}^{2}}\) | \(525\) |
Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.15 \[ \int \left (-\left (\left (1+b^2 n^2\right ) \sec \left (a+b \log \left (c x^n\right )\right )\right )+2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {b n x \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - x \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{\cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2}} \]
integrate(-(b^2*n^2+1)*sec(a+b*log(c*x^n))+2*b^2*n^2*sec(a+b*log(c*x^n))^3 ,x, algorithm="fricas")
(b*n*x*sin(b*n*log(x) + b*log(c) + a) - x*cos(b*n*log(x) + b*log(c) + a))/ cos(b*n*log(x) + b*log(c) + a)^2
\[ \int \left (-\left (\left (1+b^2 n^2\right ) \sec \left (a+b \log \left (c x^n\right )\right )\right )+2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (2 b^{2} n^{2} \sec ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} - b^{2} n^{2} - 1\right ) \sec {\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 1696 vs. \(2 (41) = 82\).
Time = 0.62 (sec) , antiderivative size = 1696, normalized size of antiderivative = 41.37 \[ \int \left (-\left (\left (1+b^2 n^2\right ) \sec \left (a+b \log \left (c x^n\right )\right )\right )+2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Too large to display} \]
integrate(-(b^2*n^2+1)*sec(a+b*log(c*x^n))+2*b^2*n^2*sec(a+b*log(c*x^n))^3 ,x, algorithm="maxima")
-2*((b*n*sin(b*log(c)) + cos(b*log(c)))*x*cos(b*log(x^n) + a) + (b*n*cos(b *log(c)) - sin(b*log(c)))*x*sin(b*log(x^n) + a) + (((b*cos(3*b*log(c))*sin (4*b*log(c)) - b*cos(4*b*log(c))*sin(3*b*log(c)))*n + cos(4*b*log(c))*cos( 3*b*log(c)) + sin(4*b*log(c))*sin(3*b*log(c)))*x*cos(3*b*log(x^n) + 3*a) - ((b*cos(b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(b*log(c)))*n - cos(4*b*log(c))*cos(b*log(c)) - sin(4*b*log(c))*sin(b*log(c)))*x*cos(b*log (x^n) + a) - ((b*cos(4*b*log(c))*cos(3*b*log(c)) + b*sin(4*b*log(c))*sin(3 *b*log(c)))*n - cos(3*b*log(c))*sin(4*b*log(c)) + cos(4*b*log(c))*sin(3*b* log(c)))*x*sin(3*b*log(x^n) + 3*a) + ((b*cos(4*b*log(c))*cos(b*log(c)) + b *sin(4*b*log(c))*sin(b*log(c)))*n + cos(b*log(c))*sin(4*b*log(c)) - cos(4* b*log(c))*sin(b*log(c)))*x*sin(b*log(x^n) + a))*cos(4*b*log(x^n) + 4*a) - (2*((b*cos(2*b*log(c))*sin(3*b*log(c)) - b*cos(3*b*log(c))*sin(2*b*log(c)) )*n - cos(3*b*log(c))*cos(2*b*log(c)) - sin(3*b*log(c))*sin(2*b*log(c)))*x *cos(2*b*log(x^n) + 2*a) - 2*((b*cos(3*b*log(c))*cos(2*b*log(c)) + b*sin(3 *b*log(c))*sin(2*b*log(c)))*n + cos(2*b*log(c))*sin(3*b*log(c)) - cos(3*b* log(c))*sin(2*b*log(c)))*x*sin(2*b*log(x^n) + 2*a) + (b*n*sin(3*b*log(c)) - cos(3*b*log(c)))*x)*cos(3*b*log(x^n) + 3*a) - 2*(((b*cos(b*log(c))*sin(2 *b*log(c)) - b*cos(2*b*log(c))*sin(b*log(c)))*n - cos(2*b*log(c))*cos(b*lo g(c)) - sin(2*b*log(c))*sin(b*log(c)))*x*cos(b*log(x^n) + a) - ((b*cos(2*b *log(c))*cos(b*log(c)) + b*sin(2*b*log(c))*sin(b*log(c)))*n + cos(b*log...
\[ \int \left (-\left (\left (1+b^2 n^2\right ) \sec \left (a+b \log \left (c x^n\right )\right )\right )+2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { 2 \, b^{2} n^{2} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{3} - {\left (b^{2} n^{2} + 1\right )} \sec \left (b \log \left (c x^{n}\right ) + a\right ) \,d x } \]
integrate(-(b^2*n^2+1)*sec(a+b*log(c*x^n))+2*b^2*n^2*sec(a+b*log(c*x^n))^3 ,x, algorithm="giac")
Time = 27.40 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.12 \[ \int \left (-\left (\left (1+b^2 n^2\right ) \sec \left (a+b \log \left (c x^n\right )\right )\right )+2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {2\,x\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}\,\left (-1+b\,n\,1{}\mathrm {i}\right )-2\,x\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}\,\left (1+b\,n\,1{}\mathrm {i}\right )}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}+1\right )}^2} \]