Integrand size = 22, antiderivative size = 562 \[ \int x (a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right ) \, dx=-\frac {b (d-e) x}{2 c}+\frac {b e x}{c}+\frac {b (d-e) \arctan (c x)}{2 c^2}+\frac {1}{2} d x^2 (a+b \arctan (c x))-\frac {1}{2} e x^2 (a+b \arctan (c x))-\frac {b e \sqrt {f} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{c \sqrt {g}}-\frac {b e \left (c^2 f-g\right ) \arctan (c x) \log \left (\frac {2}{1-i c x}\right )}{c^2 g}+\frac {b e \left (c^2 f-g\right ) \arctan (c x) \log \left (\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-i \sqrt {g}\right ) (1-i c x)}\right )}{2 c^2 g}+\frac {b e \left (c^2 f-g\right ) \arctan (c x) \log \left (\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{\left (c \sqrt {-f}+i \sqrt {g}\right ) (1-i c x)}\right )}{2 c^2 g}-\frac {b e x \log \left (f+g x^2\right )}{2 c}-\frac {b e \left (c^2 f-g\right ) \arctan (c x) \log \left (f+g x^2\right )}{2 c^2 g}+\frac {e \left (f+g x^2\right ) (a+b \arctan (c x)) \log \left (f+g x^2\right )}{2 g}+\frac {i b e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 c^2 g}-\frac {i b e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-i \sqrt {g}\right ) (1-i c x)}\right )}{4 c^2 g}-\frac {i b e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{\left (c \sqrt {-f}+i \sqrt {g}\right ) (1-i c x)}\right )}{4 c^2 g} \]
-1/2*b*(d-e)*x/c+b*e*x/c+1/2*b*(d-e)*arctan(c*x)/c^2+1/2*d*x^2*(a+b*arctan (c*x))-1/2*e*x^2*(a+b*arctan(c*x))-b*e*(c^2*f-g)*arctan(c*x)*ln(2/(1-I*c*x ))/c^2/g-1/2*b*e*x*ln(g*x^2+f)/c-1/2*b*e*(c^2*f-g)*arctan(c*x)*ln(g*x^2+f) /c^2/g+1/2*e*(g*x^2+f)*(a+b*arctan(c*x))*ln(g*x^2+f)/g+1/2*b*e*(c^2*f-g)*a rctan(c*x)*ln(2*c*((-f)^(1/2)-x*g^(1/2))/(1-I*c*x)/(c*(-f)^(1/2)-I*g^(1/2) ))/c^2/g+1/2*b*e*(c^2*f-g)*arctan(c*x)*ln(2*c*((-f)^(1/2)+x*g^(1/2))/(1-I* c*x)/(c*(-f)^(1/2)+I*g^(1/2)))/c^2/g+1/2*I*b*e*(c^2*f-g)*polylog(2,1-2/(1- I*c*x))/c^2/g-1/4*I*b*e*(c^2*f-g)*polylog(2,1-2*c*((-f)^(1/2)-x*g^(1/2))/( 1-I*c*x)/(c*(-f)^(1/2)-I*g^(1/2)))/c^2/g-1/4*I*b*e*(c^2*f-g)*polylog(2,1-2 *c*((-f)^(1/2)+x*g^(1/2))/(1-I*c*x)/(c*(-f)^(1/2)+I*g^(1/2)))/c^2/g-b*e*ar ctan(x*g^(1/2)/f^(1/2))*f^(1/2)/c/g^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1140\) vs. \(2(562)=1124\).
Time = 6.75 (sec) , antiderivative size = 1140, normalized size of antiderivative = 2.03 \[ \int x (a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right ) \, dx=\frac {-2 b c d g x+6 b c e g x+2 a c^2 d g x^2-2 a c^2 e g x^2+2 b d g \arctan (c x)-2 b e g \arctan (c x)+2 b c^2 d g x^2 \arctan (c x)-2 b c^2 e g x^2 \arctan (c x)-4 b c e \sqrt {f} \sqrt {g} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )+4 i b c^2 e f \arcsin \left (\sqrt {\frac {c^2 f}{c^2 f-g}}\right ) \arctan \left (\frac {c g x}{\sqrt {c^2 f g}}\right )-4 i b e g \arcsin \left (\sqrt {\frac {c^2 f}{c^2 f-g}}\right ) \arctan \left (\frac {c g x}{\sqrt {c^2 f g}}\right )-4 b c^2 e f \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+4 b e g \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+2 b c^2 e f \arcsin \left (\sqrt {\frac {c^2 f}{c^2 f-g}}\right ) \log \left (\frac {c^2 \left (1+e^{2 i \arctan (c x)}\right ) f+\left (-1+e^{2 i \arctan (c x)}\right ) g-2 e^{2 i \arctan (c x)} \sqrt {c^2 f g}}{c^2 f-g}\right )-2 b e g \arcsin \left (\sqrt {\frac {c^2 f}{c^2 f-g}}\right ) \log \left (\frac {c^2 \left (1+e^{2 i \arctan (c x)}\right ) f+\left (-1+e^{2 i \arctan (c x)}\right ) g-2 e^{2 i \arctan (c x)} \sqrt {c^2 f g}}{c^2 f-g}\right )+2 b c^2 e f \arctan (c x) \log \left (\frac {c^2 \left (1+e^{2 i \arctan (c x)}\right ) f+\left (-1+e^{2 i \arctan (c x)}\right ) g-2 e^{2 i \arctan (c x)} \sqrt {c^2 f g}}{c^2 f-g}\right )-2 b e g \arctan (c x) \log \left (\frac {c^2 \left (1+e^{2 i \arctan (c x)}\right ) f+\left (-1+e^{2 i \arctan (c x)}\right ) g-2 e^{2 i \arctan (c x)} \sqrt {c^2 f g}}{c^2 f-g}\right )-2 b c^2 e f \arcsin \left (\sqrt {\frac {c^2 f}{c^2 f-g}}\right ) \log \left (1+\frac {e^{2 i \arctan (c x)} \left (c^2 f+g+2 \sqrt {c^2 f g}\right )}{c^2 f-g}\right )+2 b e g \arcsin \left (\sqrt {\frac {c^2 f}{c^2 f-g}}\right ) \log \left (1+\frac {e^{2 i \arctan (c x)} \left (c^2 f+g+2 \sqrt {c^2 f g}\right )}{c^2 f-g}\right )+2 b c^2 e f \arctan (c x) \log \left (1+\frac {e^{2 i \arctan (c x)} \left (c^2 f+g+2 \sqrt {c^2 f g}\right )}{c^2 f-g}\right )-2 b e g \arctan (c x) \log \left (1+\frac {e^{2 i \arctan (c x)} \left (c^2 f+g+2 \sqrt {c^2 f g}\right )}{c^2 f-g}\right )+2 a c^2 e f \log \left (f+g x^2\right )-2 b c e g x \log \left (f+g x^2\right )+2 a c^2 e g x^2 \log \left (f+g x^2\right )+2 b e g \arctan (c x) \log \left (f+g x^2\right )+2 b c^2 e g x^2 \arctan (c x) \log \left (f+g x^2\right )+2 i b e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-i b e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,\frac {e^{2 i \arctan (c x)} \left (-c^2 f-g+2 \sqrt {c^2 f g}\right )}{c^2 f-g}\right )-i b c^2 e f \operatorname {PolyLog}\left (2,-\frac {e^{2 i \arctan (c x)} \left (c^2 f+g+2 \sqrt {c^2 f g}\right )}{c^2 f-g}\right )+i b e g \operatorname {PolyLog}\left (2,-\frac {e^{2 i \arctan (c x)} \left (c^2 f+g+2 \sqrt {c^2 f g}\right )}{c^2 f-g}\right )}{4 c^2 g} \]
(-2*b*c*d*g*x + 6*b*c*e*g*x + 2*a*c^2*d*g*x^2 - 2*a*c^2*e*g*x^2 + 2*b*d*g* ArcTan[c*x] - 2*b*e*g*ArcTan[c*x] + 2*b*c^2*d*g*x^2*ArcTan[c*x] - 2*b*c^2* e*g*x^2*ArcTan[c*x] - 4*b*c*e*Sqrt[f]*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]] + (4*I)*b*c^2*e*f*ArcSin[Sqrt[(c^2*f)/(c^2*f - g)]]*ArcTan[(c*g*x)/Sqrt[c^ 2*f*g]] - (4*I)*b*e*g*ArcSin[Sqrt[(c^2*f)/(c^2*f - g)]]*ArcTan[(c*g*x)/Sqr t[c^2*f*g]] - 4*b*c^2*e*f*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 4*b *e*g*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 2*b*c^2*e*f*ArcSin[Sqrt[ (c^2*f)/(c^2*f - g)]]*Log[(c^2*(1 + E^((2*I)*ArcTan[c*x]))*f + (-1 + E^((2 *I)*ArcTan[c*x]))*g - 2*E^((2*I)*ArcTan[c*x])*Sqrt[c^2*f*g])/(c^2*f - g)] - 2*b*e*g*ArcSin[Sqrt[(c^2*f)/(c^2*f - g)]]*Log[(c^2*(1 + E^((2*I)*ArcTan[ c*x]))*f + (-1 + E^((2*I)*ArcTan[c*x]))*g - 2*E^((2*I)*ArcTan[c*x])*Sqrt[c ^2*f*g])/(c^2*f - g)] + 2*b*c^2*e*f*ArcTan[c*x]*Log[(c^2*(1 + E^((2*I)*Arc Tan[c*x]))*f + (-1 + E^((2*I)*ArcTan[c*x]))*g - 2*E^((2*I)*ArcTan[c*x])*Sq rt[c^2*f*g])/(c^2*f - g)] - 2*b*e*g*ArcTan[c*x]*Log[(c^2*(1 + E^((2*I)*Arc Tan[c*x]))*f + (-1 + E^((2*I)*ArcTan[c*x]))*g - 2*E^((2*I)*ArcTan[c*x])*Sq rt[c^2*f*g])/(c^2*f - g)] - 2*b*c^2*e*f*ArcSin[Sqrt[(c^2*f)/(c^2*f - g)]]* Log[1 + (E^((2*I)*ArcTan[c*x])*(c^2*f + g + 2*Sqrt[c^2*f*g]))/(c^2*f - g)] + 2*b*e*g*ArcSin[Sqrt[(c^2*f)/(c^2*f - g)]]*Log[1 + (E^((2*I)*ArcTan[c*x] )*(c^2*f + g + 2*Sqrt[c^2*f*g]))/(c^2*f - g)] + 2*b*c^2*e*f*ArcTan[c*x]*Lo g[1 + (E^((2*I)*ArcTan[c*x])*(c^2*f + g + 2*Sqrt[c^2*f*g]))/(c^2*f - g)...
Time = 0.89 (sec) , antiderivative size = 554, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5554, 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.
| \(\displaystyle \int x (a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right ) \, dx\) |
| \(\Big \downarrow \) 5554 |
| \(\displaystyle -b c \int \left (\frac {(d-e) x^2}{2 \left (c^2 x^2+1\right )}+\frac {e \left (g x^2+f\right ) \log \left (g x^2+f\right )}{2 g \left (c^2 x^2+1\right )}\right )dx+\frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {e \left (f+g x^2\right ) \log \left (f+g x^2\right ) (a+b \arctan (c x))}{2 g}-\frac {1}{2} e x^2 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {e \left (f+g x^2\right ) \log \left (f+g x^2\right ) (a+b \arctan (c x))}{2 g}-\frac {1}{2} e x^2 (a+b \arctan (c x))-b c \left (-\frac {(d-e) \arctan (c x)}{2 c^3}+\frac {e \sqrt {f} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{c^2 \sqrt {g}}+\frac {e \arctan (c x) \left (c^2 f-g\right ) \log \left (f+g x^2\right )}{2 c^3 g}+\frac {e \arctan (c x) \left (c^2 f-g\right ) \log \left (\frac {2}{1-i c x}\right )}{c^3 g}-\frac {e \arctan (c x) \left (c^2 f-g\right ) \log \left (\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{(1-i c x) \left (c \sqrt {-f}-i \sqrt {g}\right )}\right )}{2 c^3 g}-\frac {e \arctan (c x) \left (c^2 f-g\right ) \log \left (\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{(1-i c x) \left (c \sqrt {-f}+i \sqrt {g}\right )}\right )}{2 c^3 g}+\frac {x (d-e)}{2 c^2}+\frac {e x \log \left (f+g x^2\right )}{2 c^2}-\frac {e x}{c^2}-\frac {i e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 c^3 g}+\frac {i e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-i \sqrt {g}\right ) (1-i c x)}\right )}{4 c^3 g}+\frac {i e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {g} x+\sqrt {-f}\right )}{\left (\sqrt {-f} c+i \sqrt {g}\right ) (1-i c x)}\right )}{4 c^3 g}\right )\) |
(d*x^2*(a + b*ArcTan[c*x]))/2 - (e*x^2*(a + b*ArcTan[c*x]))/2 + (e*(f + g* x^2)*(a + b*ArcTan[c*x])*Log[f + g*x^2])/(2*g) - b*c*(((d - e)*x)/(2*c^2) - (e*x)/c^2 - ((d - e)*ArcTan[c*x])/(2*c^3) + (e*Sqrt[f]*ArcTan[(Sqrt[g]*x )/Sqrt[f]])/(c^2*Sqrt[g]) + (e*(c^2*f - g)*ArcTan[c*x]*Log[2/(1 - I*c*x)]) /(c^3*g) - (e*(c^2*f - g)*ArcTan[c*x]*Log[(2*c*(Sqrt[-f] - Sqrt[g]*x))/((c *Sqrt[-f] - I*Sqrt[g])*(1 - I*c*x))])/(2*c^3*g) - (e*(c^2*f - g)*ArcTan[c* x]*Log[(2*c*(Sqrt[-f] + Sqrt[g]*x))/((c*Sqrt[-f] + I*Sqrt[g])*(1 - I*c*x)) ])/(2*c^3*g) + (e*x*Log[f + g*x^2])/(2*c^2) + (e*(c^2*f - g)*ArcTan[c*x]*L og[f + g*x^2])/(2*c^3*g) - ((I/2)*e*(c^2*f - g)*PolyLog[2, 1 - 2/(1 - I*c* x)])/(c^3*g) + ((I/4)*e*(c^2*f - g)*PolyLog[2, 1 - (2*c*(Sqrt[-f] - Sqrt[g ]*x))/((c*Sqrt[-f] - I*Sqrt[g])*(1 - I*c*x))])/(c^3*g) + ((I/4)*e*(c^2*f - g)*PolyLog[2, 1 - (2*c*(Sqrt[-f] + Sqrt[g]*x))/((c*Sqrt[-f] + I*Sqrt[g])* (1 - I*c*x))])/(c^3*g))
3.13.97.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*( e_.))*(x_)^(m_.), x_Symbol] :> With[{u = IntHide[x^m*(d + e*Log[f + g*x^2]) , x]}, Simp[(a + b*ArcTan[c*x]) u, x] - Simp[b*c Int[ExpandIntegrand[u/ (1 + c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[(m + 1)/2, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 13.54 (sec) , antiderivative size = 10102, normalized size of antiderivative = 17.98
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(10102\) |
| parts | \(\text {Expression too large to display}\) | \(10102\) |
\[ \int x (a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right ) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (g x^{2} + f\right ) + d\right )} x \,d x } \]
Timed out. \[ \int x (a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right ) \, dx=\text {Timed out} \]
\[ \int x (a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right ) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (g x^{2} + f\right ) + d\right )} x \,d x } \]
1/2*a*d*x^2 + 1/2*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b*d - 1/ 2*(g*x^2 - (g*x^2 + f)*log(g*x^2 + f) + f)*a*e/g - 1/2*(2*c*f*arctan(g*x/s qrt(f*g)) + (4*c^4*g*integrate(1/2*x^3*arctan(c*x)/(c^2*g*x^2 + c^2*f), x) + 4*c^2*g*integrate(1/2*x*arctan(c*x)/(c^2*g*x^2 + c^2*f), x) - 2*c*x + ( c*x - (c^2*x^2 + 1)*arctan(c*x))*log(g*x^2 + f))*sqrt(f*g))*b*e/(sqrt(f*g) *c^2)
\[ \int x (a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right ) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (g x^{2} + f\right ) + d\right )} x \,d x } \]
Timed out. \[ \int x (a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right ) \, dx=\int x\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (g\,x^2+f\right )\right ) \,d x \]