Integrand size = 19, antiderivative size = 77 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x} \, dx=-\frac {b e x}{2 c}+\frac {b e \arctan (c x)}{2 c^2}+\frac {1}{2} e x^2 (a+b \arctan (c x))+a d \log (x)+\frac {1}{2} i b d \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d \operatorname {PolyLog}(2,i c x) \]
-1/2*b*e*x/c+1/2*b*e*arctan(c*x)/c^2+1/2*e*x^2*(a+b*arctan(c*x))+a*d*ln(x) +1/2*I*b*d*polylog(2,-I*c*x)-1/2*I*b*d*polylog(2,I*c*x)
Time = 0.00 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x} \, dx=-\frac {b e x}{2 c}+\frac {1}{2} a e x^2+\frac {b e \arctan (c x)}{2 c^2}+\frac {1}{2} b e x^2 \arctan (c x)+a d \log (x)+\frac {1}{2} i b d \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d \operatorname {PolyLog}(2,i c x) \]
-1/2*(b*e*x)/c + (a*e*x^2)/2 + (b*e*ArcTan[c*x])/(2*c^2) + (b*e*x^2*ArcTan [c*x])/2 + a*d*Log[x] + (I/2)*b*d*PolyLog[2, (-I)*c*x] - (I/2)*b*d*PolyLog [2, I*c*x]
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5515, 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.
\(\displaystyle \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x} \, dx\) |
\(\Big \downarrow \) 5515 |
\(\displaystyle \int \left (\frac {d (a+b \arctan (c x))}{x}+e x (a+b \arctan (c x))\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} e x^2 (a+b \arctan (c x))+a d \log (x)+\frac {b e \arctan (c x)}{2 c^2}+\frac {1}{2} i b d \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d \operatorname {PolyLog}(2,i c x)-\frac {b e x}{2 c}\) |
-1/2*(b*e*x)/c + (b*e*ArcTan[c*x])/(2*c^2) + (e*x^2*(a + b*ArcTan[c*x]))/2 + a*d*Log[x] + (I/2)*b*d*PolyLog[2, (-I)*c*x] - (I/2)*b*d*PolyLog[2, I*c* x]
3.12.18.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*ArcTan[c*x] )^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d , e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])
Time = 0.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.56
method | result | size |
derivativedivides | \(\frac {a e \,x^{2}}{2}+a d \ln \left (c x \right )+\frac {b \left (\frac {\arctan \left (c x \right ) e \,c^{2} x^{2}}{2}+\arctan \left (c x \right ) d \,c^{2} \ln \left (c x \right )-\frac {e \left (c x -\arctan \left (c x \right )\right )}{2}-d \,c^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{2}}\) | \(120\) |
default | \(\frac {a e \,x^{2}}{2}+a d \ln \left (c x \right )+\frac {b \left (\frac {\arctan \left (c x \right ) e \,c^{2} x^{2}}{2}+\arctan \left (c x \right ) d \,c^{2} \ln \left (c x \right )-\frac {e \left (c x -\arctan \left (c x \right )\right )}{2}-d \,c^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{2}}\) | \(120\) |
parts | \(\frac {a e \,x^{2}}{2}+a d \ln \left (x \right )+b \left (\frac {\arctan \left (c x \right ) x^{2} e}{2}+\arctan \left (c x \right ) d \ln \left (c x \right )-\frac {e \left (c x -\arctan \left (c x \right )\right )-i c^{2} d \ln \left (c x \right ) \ln \left (i c x +1\right )+i c^{2} d \ln \left (c x \right ) \ln \left (-i c x +1\right )-i c^{2} d \operatorname {dilog}\left (i c x +1\right )+i c^{2} d \operatorname {dilog}\left (-i c x +1\right )}{2 c^{2}}\right )\) | \(123\) |
risch | \(\frac {i b d \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i b e \ln \left (c^{2} x^{2}+1\right )}{8 c^{2}}+\frac {b e \arctan \left (c x \right )}{4 c^{2}}-\frac {i b e \ln \left (i c x +1\right )}{4 c^{2}}-\frac {b e x}{2 c}-\frac {i b e \ln \left (i c x +1\right ) x^{2}}{4}-\frac {i b d \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {a e \,x^{2}}{2}+\frac {a e}{2 c^{2}}+a d \ln \left (-i c x \right )+\frac {i b e \ln \left (-i c x +1\right ) x^{2}}{4}\) | \(136\) |
1/2*a*e*x^2+a*d*ln(c*x)+b/c^2*(1/2*arctan(c*x)*e*c^2*x^2+arctan(c*x)*d*c^2 *ln(c*x)-1/2*e*(c*x-arctan(c*x))-d*c^2*(-1/2*I*ln(c*x)*ln(1+I*c*x)+1/2*I*l n(c*x)*ln(1-I*c*x)-1/2*I*dilog(1+I*c*x)+1/2*I*dilog(1-I*c*x)))
\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]
\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x}\, dx \]
Time = 0.41 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.35 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x} \, dx=\frac {1}{2} \, a e x^{2} + a d \log \left (x\right ) - \frac {\pi b c^{2} d \log \left (c^{2} x^{2} + 1\right ) - 4 \, b c^{2} d \arctan \left (c x\right ) \log \left (c x\right ) + 2 i \, b c^{2} d {\rm Li}_2\left (i \, c x + 1\right ) - 2 i \, b c^{2} d {\rm Li}_2\left (-i \, c x + 1\right ) + 2 \, b c e x - 2 \, {\left (b c^{2} e x^{2} + b e\right )} \arctan \left (c x\right )}{4 \, c^{2}} \]
1/2*a*e*x^2 + a*d*log(x) - 1/4*(pi*b*c^2*d*log(c^2*x^2 + 1) - 4*b*c^2*d*ar ctan(c*x)*log(c*x) + 2*I*b*c^2*d*dilog(I*c*x + 1) - 2*I*b*c^2*d*dilog(-I*c *x + 1) + 2*b*c*e*x - 2*(b*c^2*e*x^2 + b*e)*arctan(c*x))/c^2
\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]
Time = 0.82 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x} \, dx=\left \{\begin {array}{cl} \frac {a\,\left (e\,x^2+2\,d\,\ln \left (x\right )\right )}{2} & \text {\ if\ \ }c=0\\ \frac {a\,\left (e\,x^2+2\,d\,\ln \left (x\right )\right )}{2}-b\,e\,\left (\frac {x}{2\,c}-\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\right )-\frac {b\,d\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]