Integrand size = 24, antiderivative size = 528 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx=\frac {b c e \sqrt {g} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\sqrt {f}}+\frac {a e g \log (x)}{f}-\frac {b e \left (c^2 f-g\right ) \arctan (c x) \log \left (\frac {2}{1-i c x}\right )}{f}+\frac {b e \left (c^2 f-g\right ) \arctan (c x) \log \left (\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-i \sqrt {g}\right ) (1-i c x)}\right )}{2 f}+\frac {b e \left (c^2 f-g\right ) \arctan (c x) \log \left (\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{\left (c \sqrt {-f}+i \sqrt {g}\right ) (1-i c x)}\right )}{2 f}-\frac {a e g \log \left (f+g x^2\right )}{2 f}-\frac {b c \left (d+e \log \left (f+g x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (f+g x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{2 x^2}+\frac {i b e g \operatorname {PolyLog}(2,-i c x)}{2 f}-\frac {i b e g \operatorname {PolyLog}(2,i c x)}{2 f}+\frac {i b e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 f}-\frac {i b e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-i \sqrt {g}\right ) (1-i c x)}\right )}{4 f}-\frac {i b e \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{\left (c \sqrt {-f}+i \sqrt {g}\right ) (1-i c x)}\right )}{4 f} \]
a*e*g*ln(x)/f-b*e*(c^2*f-g)*arctan(c*x)*ln(2/(1-I*c*x))/f-1/2*a*e*g*ln(g*x ^2+f)/f-1/2*b*c*(d+e*ln(g*x^2+f))/x-1/2*b*c^2*arctan(c*x)*(d+e*ln(g*x^2+f) )-1/2*(a+b*arctan(c*x))*(d+e*ln(g*x^2+f))/x^2+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)))/f+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)))/f+1/2*I*b*e*g*polylog(2,-I*c*x)/f-1/2*I*b*e*g*polylog(2, I*c*x)/f+1/2*I*b*e*(c^2*f-g)*polylog(2,1-2/(1-I*c*x))/f-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) ))/f-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)))/f+b*c*e*arctan(x*g^(1/2)/f^(1/2))*g^(1/2)/f^(1/2 )
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1217\) vs. \(2(528)=1056\).
Time = 5.20 (sec) , antiderivative size = 1217, normalized size of antiderivative = 2.30 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx=-\frac {2 a d f+2 b c d f x+2 b d f \arctan (c x)+2 b c^2 d f x^2 \arctan (c x)-4 b c e \sqrt {f} \sqrt {g} x^2 \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )-4 i b c^2 e f x^2 \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 x^2 \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 e g x^2 \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )+4 b c^2 e f x^2 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-2 b c^2 e f x^2 \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 x^2 \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 x^2 \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 x^2 \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 x^2 \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 x^2 \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 x^2 \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 x^2 \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 )-4 a e g x^2 \log (x)+2 a e f \log \left (f+g x^2\right )+2 b c e f x \log \left (f+g x^2\right )+2 a e g x^2 \log \left (f+g x^2\right )+2 b e f \arctan (c x) \log \left (f+g x^2\right )+2 b c^2 e f x^2 \arctan (c x) \log \left (f+g x^2\right )-2 i b c^2 e f x^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+2 i b e g x^2 \operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )+i b c^2 e f x^2 \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 x^2 \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 x^2 \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 x^2 \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 f x^2} \]
-1/4*(2*a*d*f + 2*b*c*d*f*x + 2*b*d*f*ArcTan[c*x] + 2*b*c^2*d*f*x^2*ArcTan [c*x] - 4*b*c*e*Sqrt[f]*Sqrt[g]*x^2*ArcTan[(Sqrt[g]*x)/Sqrt[f]] - (4*I)*b* c^2*e*f*x^2*ArcSin[Sqrt[(c^2*f)/(c^2*f - g)]]*ArcTan[(c*g*x)/Sqrt[c^2*f*g] ] + (4*I)*b*e*g*x^2*ArcSin[Sqrt[(c^2*f)/(c^2*f - g)]]*ArcTan[(c*g*x)/Sqrt[ c^2*f*g]] - 4*b*e*g*x^2*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + 4*b*c ^2*e*f*x^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - 2*b*c^2*e*f*x^2*Ar cSin[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*x^2*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)*ArcTa n[c*x])*Sqrt[c^2*f*g])/(c^2*f - g)] - 2*b*c^2*e*f*x^2*ArcTan[c*x]*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*x^2*ArcTan[c*x]*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*x^2*ArcSin[S qrt[(c^2*f)/(c^2*f - g)]]*Log[1 + (E^((2*I)*ArcTan[c*x])*(c^2*f + g + 2*Sq rt[c^2*f*g]))/(c^2*f - g)] - 2*b*e*g*x^2*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*x^2*ArcTan[c*x]*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*x^2*ArcTan[c*x]*Log[1 + (E^(...
Time = 0.94 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5556, 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.
| \(\displaystyle \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 5556 |
| \(\displaystyle -2 e g \int \left (-\frac {a+b c x}{2 x \left (g x^2+f\right )}-\frac {b \left (c^2 x^2+1\right ) \arctan (c x)}{2 x \left (g x^2+f\right )}\right )dx-\frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{2 x^2}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (f+g x^2\right )\right )-\frac {b c \left (d+e \log \left (f+g x^2\right )\right )}{2 x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 e g \left (\frac {a \log \left (f+g x^2\right )}{4 f}-\frac {a \log (x)}{2 f}+\frac {b \arctan (c x) \left (c^2 f-g\right ) \log \left (\frac {2}{1-i c x}\right )}{2 f g}-\frac {b \arctan (c x) \left (c^2 f-g\right ) \log \left (\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{(1-i c x) \left (c \sqrt {-f}-i \sqrt {g}\right )}\right )}{4 f g}-\frac {b \arctan (c x) \left (c^2 f-g\right ) \log \left (\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{(1-i c x) \left (c \sqrt {-f}+i \sqrt {g}\right )}\right )}{4 f g}-\frac {b c \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g}}-\frac {i b \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{4 f g}+\frac {i b \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-i \sqrt {g}\right ) (1-i c x)}\right )}{8 f g}+\frac {i b \left (c^2 f-g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {g} x+\sqrt {-f}\right )}{\left (\sqrt {-f} c+i \sqrt {g}\right ) (1-i c x)}\right )}{8 f g}-\frac {i b \operatorname {PolyLog}(2,-i c x)}{4 f}+\frac {i b \operatorname {PolyLog}(2,i c x)}{4 f}\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{2 x^2}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (f+g x^2\right )\right )-\frac {b c \left (d+e \log \left (f+g x^2\right )\right )}{2 x}\) |
-1/2*(b*c*(d + e*Log[f + g*x^2]))/x - (b*c^2*ArcTan[c*x]*(d + e*Log[f + g* x^2]))/2 - ((a + b*ArcTan[c*x])*(d + e*Log[f + g*x^2]))/(2*x^2) - 2*e*g*(- 1/2*(b*c*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(Sqrt[f]*Sqrt[g]) - (a*Log[x])/(2*f) + (b*(c^2*f - g)*ArcTan[c*x]*Log[2/(1 - I*c*x)])/(2*f*g) - (b*(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))])/(4*f*g) - (b*(c^2*f - g)*ArcTan[c*x]*Log[(2*c*(Sqrt[-f] + Sqr t[g]*x))/((c*Sqrt[-f] + I*Sqrt[g])*(1 - I*c*x))])/(4*f*g) + (a*Log[f + g*x ^2])/(4*f) - ((I/4)*b*PolyLog[2, (-I)*c*x])/f + ((I/4)*b*PolyLog[2, I*c*x] )/f - ((I/4)*b*(c^2*f - g)*PolyLog[2, 1 - 2/(1 - I*c*x)])/(f*g) + ((I/8)*b *(c^2*f - g)*PolyLog[2, 1 - (2*c*(Sqrt[-f] - Sqrt[g]*x))/((c*Sqrt[-f] - I* Sqrt[g])*(1 - I*c*x))])/(f*g) + ((I/8)*b*(c^2*f - g)*PolyLog[2, 1 - (2*c*( Sqrt[-f] + Sqrt[g]*x))/((c*Sqrt[-f] + I*Sqrt[g])*(1 - I*c*x))])/(f*g))
3.14.1.3.1 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*(a + b*ArcTan[c*x]), x ]}, Simp[(d + e*Log[f + g*x^2]) u, x] - Simp[2*e*g Int[ExpandIntegrand[ x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Intege rQ[m] && NeQ[m, -1]
\[\int \frac {\left (a +b \arctan \left (c x \right )\right ) \left (d +e \ln \left (g \,x^{2}+f \right )\right )}{x^{3}}d x\]
\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (g x^{2} + f\right ) + d\right )}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (g x^{2} + f\right ) + d\right )}}{x^{3}} \,d x } \]
-1/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*d - 1/2*(g*(log(g*x^2 + f)/f - log(x^2)/f) + log(g*x^2 + f)/x^2)*a*e + 1/2*(2*c*g*x^2*arctan(g*x/ sqrt(f*g)) + (4*c^2*g*x^2*integrate(1/2*x*arctan(c*x)/(g*x^2 + f), x) + 4* g*x^2*integrate(1/2*arctan(c*x)/(g*x^3 + f*x), x) - (c*x + (c^2*x^2 + 1)*a rctan(c*x))*log(g*x^2 + f))*sqrt(f*g))*b*e/(sqrt(f*g)*x^2) - 1/2*a*d/x^2
Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (g\,x^2+f\right )\right )}{x^3} \,d x \]