Integrand size = 16, antiderivative size = 143 \[ \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\frac {a x^6}{6}-\frac {i b x^4 \arctan \left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b x^2 \operatorname {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i b x^2 \operatorname {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {b \operatorname {PolyLog}\left (3,i e^{i \left (c+d x^2\right )}\right )}{d^3} \]
1/6*a*x^6-I*b*x^4*arctan(exp(I*(d*x^2+c)))/d+I*b*x^2*polylog(2,-I*exp(I*(d *x^2+c)))/d^2-I*b*x^2*polylog(2,I*exp(I*(d*x^2+c)))/d^2-b*polylog(3,-I*exp (I*(d*x^2+c)))/d^3+b*polylog(3,I*exp(I*(d*x^2+c)))/d^3
Time = 0.05 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.02 \[ \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\frac {a x^6}{6}-\frac {i b x^4 \arctan \left (e^{i c+i d x^2}\right )}{d}+\frac {i b x^2 \operatorname {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i b x^2 \operatorname {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {b \operatorname {PolyLog}\left (3,i e^{i \left (c+d x^2\right )}\right )}{d^3} \]
(a*x^6)/6 - (I*b*x^4*ArcTan[E^(I*c + I*d*x^2)])/d + (I*b*x^2*PolyLog[2, (- I)*E^(I*(c + d*x^2))])/d^2 - (I*b*x^2*PolyLog[2, I*E^(I*(c + d*x^2))])/d^2 - (b*PolyLog[3, (-I)*E^(I*(c + d*x^2))])/d^3 + (b*PolyLog[3, I*E^(I*(c + d*x^2))])/d^3
Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2010, 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.
\(\displaystyle \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (a x^5+b x^5 \sec \left (c+d x^2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a x^6}{6}-\frac {i b x^4 \arctan \left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac {b \operatorname {PolyLog}\left (3,-i e^{i \left (d x^2+c\right )}\right )}{d^3}+\frac {b \operatorname {PolyLog}\left (3,i e^{i \left (d x^2+c\right )}\right )}{d^3}+\frac {i b x^2 \operatorname {PolyLog}\left (2,-i e^{i \left (d x^2+c\right )}\right )}{d^2}-\frac {i b x^2 \operatorname {PolyLog}\left (2,i e^{i \left (d x^2+c\right )}\right )}{d^2}\) |
(a*x^6)/6 - (I*b*x^4*ArcTan[E^(I*(c + d*x^2))])/d + (I*b*x^2*PolyLog[2, (- I)*E^(I*(c + d*x^2))])/d^2 - (I*b*x^2*PolyLog[2, I*E^(I*(c + d*x^2))])/d^2 - (b*PolyLog[3, (-I)*E^(I*(c + d*x^2))])/d^3 + (b*PolyLog[3, I*E^(I*(c + d*x^2))])/d^3
3.1.1.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
\[\int x^{5} \left (a +b \sec \left (d \,x^{2}+c \right )\right )d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (115) = 230\).
Time = 0.31 (sec) , antiderivative size = 495, normalized size of antiderivative = 3.46 \[ \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\frac {2 \, a d^{3} x^{6} - 6 i \, b d x^{2} {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - 6 i \, b d x^{2} {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + 6 i \, b d x^{2} {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 i \, b d x^{2} {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + 3 \, b c^{2} \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) - 3 \, b c^{2} \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) + 3 \, b c^{2} \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) - 3 \, b c^{2} \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) + 3 \, {\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, {\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + 3 \, {\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, {\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) - 6 \, b {\rm polylog}\left (3, i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 \, b {\rm polylog}\left (3, i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) - 6 \, b {\rm polylog}\left (3, -i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 \, b {\rm polylog}\left (3, -i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )}{12 \, d^{3}} \]
1/12*(2*a*d^3*x^6 - 6*I*b*d*x^2*dilog(I*cos(d*x^2 + c) + sin(d*x^2 + c)) - 6*I*b*d*x^2*dilog(I*cos(d*x^2 + c) - sin(d*x^2 + c)) + 6*I*b*d*x^2*dilog( -I*cos(d*x^2 + c) + sin(d*x^2 + c)) + 6*I*b*d*x^2*dilog(-I*cos(d*x^2 + c) - sin(d*x^2 + c)) + 3*b*c^2*log(cos(d*x^2 + c) + I*sin(d*x^2 + c) + I) - 3 *b*c^2*log(cos(d*x^2 + c) - I*sin(d*x^2 + c) + I) + 3*b*c^2*log(-cos(d*x^2 + c) + I*sin(d*x^2 + c) + I) - 3*b*c^2*log(-cos(d*x^2 + c) - I*sin(d*x^2 + c) + I) + 3*(b*d^2*x^4 - b*c^2)*log(I*cos(d*x^2 + c) + sin(d*x^2 + c) + 1) - 3*(b*d^2*x^4 - b*c^2)*log(I*cos(d*x^2 + c) - sin(d*x^2 + c) + 1) + 3* (b*d^2*x^4 - b*c^2)*log(-I*cos(d*x^2 + c) + sin(d*x^2 + c) + 1) - 3*(b*d^2 *x^4 - b*c^2)*log(-I*cos(d*x^2 + c) - sin(d*x^2 + c) + 1) - 6*b*polylog(3, I*cos(d*x^2 + c) + sin(d*x^2 + c)) + 6*b*polylog(3, I*cos(d*x^2 + c) - si n(d*x^2 + c)) - 6*b*polylog(3, -I*cos(d*x^2 + c) + sin(d*x^2 + c)) + 6*b*p olylog(3, -I*cos(d*x^2 + c) - sin(d*x^2 + c)))/d^3
\[ \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\int x^{5} \left (a + b \sec {\left (c + d x^{2} \right )}\right )\, dx \]
\[ \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\int { {\left (b \sec \left (d x^{2} + c\right ) + a\right )} x^{5} \,d x } \]
1/6*a*x^6 + 2*b*integrate((x^5*cos(2*d*x^2 + 2*c)*cos(d*x^2 + c) + x^5*sin (2*d*x^2 + 2*c)*sin(d*x^2 + c) + x^5*cos(d*x^2 + c))/(cos(2*d*x^2 + 2*c)^2 + sin(2*d*x^2 + 2*c)^2 + 2*cos(2*d*x^2 + 2*c) + 1), x)
\[ \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\int { {\left (b \sec \left (d x^{2} + c\right ) + a\right )} x^{5} \,d x } \]
Timed out. \[ \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\int x^5\,\left (a+\frac {b}{\cos \left (d\,x^2+c\right )}\right ) \,d x \]