Integrand size = 21, antiderivative size = 137 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=-\frac {b d e x}{c}+\frac {b e^2 x}{4 c^3}-\frac {b e^2 x^3}{12 c}+\frac {b d e \arctan (c x)}{c^2}-\frac {b e^2 \arctan (c x)}{4 c^4}+d e x^2 (a+b \arctan (c x))+\frac {1}{4} e^2 x^4 (a+b \arctan (c x))+a d^2 \log (x)+\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,i c x) \]
-b*d*e*x/c+1/4*b*e^2*x/c^3-1/12*b*e^2*x^3/c+b*d*e*arctan(c*x)/c^2-1/4*b*e^ 2*arctan(c*x)/c^4+d*e*x^2*(a+b*arctan(c*x))+1/4*e^2*x^4*(a+b*arctan(c*x))+ a*d^2*ln(x)+1/2*I*b*d^2*polylog(2,-I*c*x)-1/2*I*b*d^2*polylog(2,I*c*x)
Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.90 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=-\frac {b d e (c x-\arctan (c x))}{c^2}-\frac {b e^2 \left (-3 c x+c^3 x^3+3 \arctan (c x)\right )}{12 c^4}+d e x^2 (a+b \arctan (c x))+\frac {1}{4} e^2 x^4 (a+b \arctan (c x))+a d^2 \log (x)+\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,i c x) \]
-((b*d*e*(c*x - ArcTan[c*x]))/c^2) - (b*e^2*(-3*c*x + c^3*x^3 + 3*ArcTan[c *x]))/(12*c^4) + d*e*x^2*(a + b*ArcTan[c*x]) + (e^2*x^4*(a + b*ArcTan[c*x] ))/4 + a*d^2*Log[x] + (I/2)*b*d^2*PolyLog[2, (-I)*c*x] - (I/2)*b*d^2*PolyL og[2, I*c*x]
Time = 0.34 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {d^2 (a+b \arctan (c x))}{x}+2 d e x (a+b \arctan (c x))+e^2 x^3 (a+b \arctan (c x))\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle d e x^2 (a+b \arctan (c x))+\frac {1}{4} e^2 x^4 (a+b \arctan (c x))+a d^2 \log (x)-\frac {b e^2 \arctan (c x)}{4 c^4}+\frac {b d e \arctan (c x)}{c^2}+\frac {b e^2 x}{4 c^3}+\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^2 \operatorname {PolyLog}(2,i c x)-\frac {b d e x}{c}-\frac {b e^2 x^3}{12 c}\) |
-((b*d*e*x)/c) + (b*e^2*x)/(4*c^3) - (b*e^2*x^3)/(12*c) + (b*d*e*ArcTan[c* x])/c^2 - (b*e^2*ArcTan[c*x])/(4*c^4) + d*e*x^2*(a + b*ArcTan[c*x]) + (e^2 *x^4*(a + b*ArcTan[c*x]))/4 + a*d^2*Log[x] + (I/2)*b*d^2*PolyLog[2, (-I)*c *x] - (I/2)*b*d^2*PolyLog[2, I*c*x]
3.12.29.3.1 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 = 0.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.28
| method | result | size |
| derivativedivides | \(a d e \,x^{2}+\frac {a \,e^{2} x^{4}}{4}+a \,d^{2} \ln \left (c x \right )+\frac {b \left (\arctan \left (c x \right ) d \,c^{4} e \,x^{2}+\frac {\arctan \left (c x \right ) e^{2} c^{4} x^{4}}{4}+\arctan \left (c x \right ) c^{4} d^{2} \ln \left (c x \right )-\frac {e \left (4 c^{3} x d +\frac {e \,c^{3} x^{3}}{3}-e c x +\left (-4 c^{2} d +e \right ) \arctan \left (c x \right )\right )}{4}-c^{4} d^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{4}}\) | \(175\) |
| default | \(a d e \,x^{2}+\frac {a \,e^{2} x^{4}}{4}+a \,d^{2} \ln \left (c x \right )+\frac {b \left (\arctan \left (c x \right ) d \,c^{4} e \,x^{2}+\frac {\arctan \left (c x \right ) e^{2} c^{4} x^{4}}{4}+\arctan \left (c x \right ) c^{4} d^{2} \ln \left (c x \right )-\frac {e \left (4 c^{3} x d +\frac {e \,c^{3} x^{3}}{3}-e c x +\left (-4 c^{2} d +e \right ) \arctan \left (c x \right )\right )}{4}-c^{4} d^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{4}}\) | \(175\) |
| parts | \(a \left (\frac {x^{4} e^{2}}{4}+x^{2} e d +d^{2} \ln \left (x \right )\right )+b \left (\frac {\arctan \left (c x \right ) e^{2} x^{4}}{4}+\arctan \left (c x \right ) d e \,x^{2}+\arctan \left (c x \right ) d^{2} \ln \left (c x \right )-\frac {e \left (4 c^{3} x d +\frac {e \,c^{3} x^{3}}{3}-e c x +\left (-4 c^{2} d +e \right ) \arctan \left (c x \right )\right )-2 i c^{4} d^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )+2 i c^{4} d^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )+2 i c^{4} d^{2} \operatorname {dilog}\left (-i c x +1\right )-2 i c^{4} d^{2} \operatorname {dilog}\left (i c x +1\right )}{4 c^{4}}\right )\) | \(181\) |
| risch | \(\frac {b \,e^{2} x}{4 c^{3}}-\frac {b \,e^{2} x^{3}}{12 c}-\frac {b d e x}{c}+a \,d^{2} \ln \left (-i c x \right )+\frac {a \,e^{2} x^{4}}{4}-\frac {i b \,d^{2} \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i b \,d^{2} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {a d e}{c^{2}}-\frac {a \,e^{2}}{4 c^{4}}+a d e \,x^{2}-\frac {i b e d \ln \left (i c x +1\right ) x^{2}}{2}-\frac {b \,e^{2} \arctan \left (c x \right )}{4 c^{4}}+\frac {i b \,e^{2} \ln \left (-i c x +1\right ) x^{4}}{8}-\frac {i b \,e^{2} \ln \left (i c x +1\right ) x^{4}}{8}+\frac {i b d e \ln \left (-i c x +1\right ) x^{2}}{2}+\frac {b d e \arctan \left (c x \right )}{c^{2}}\) | \(200\) |
a*d*e*x^2+1/4*a*e^2*x^4+a*d^2*ln(c*x)+b/c^4*(arctan(c*x)*d*c^4*e*x^2+1/4*a rctan(c*x)*e^2*c^4*x^4+arctan(c*x)*c^4*d^2*ln(c*x)-1/4*e*(4*c^3*x*d+1/3*e* c^3*x^3-e*c*x+(-4*c^2*d+e)*arctan(c*x))-c^4*d^2*(-1/2*I*ln(c*x)*ln(1+I*c*x )+1/2*I*ln(c*x)*ln(1-I*c*x)-1/2*I*dilog(1+I*c*x)+1/2*I*dilog(1-I*c*x)))
\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]
integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d ^2)*arctan(c*x))/x, x)
\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x}\, dx \]
Time = 0.44 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.26 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=\frac {1}{4} \, a e^{2} x^{4} + a d e x^{2} + a d^{2} \log \left (x\right ) - \frac {b c^{3} e^{2} x^{3} + 3 \, \pi b c^{4} d^{2} \log \left (c^{2} x^{2} + 1\right ) - 12 \, b c^{4} d^{2} \arctan \left (c x\right ) \log \left (c x\right ) + 6 i \, b c^{4} d^{2} {\rm Li}_2\left (i \, c x + 1\right ) - 6 i \, b c^{4} d^{2} {\rm Li}_2\left (-i \, c x + 1\right ) + 3 \, {\left (4 \, b c^{3} d e - b c e^{2}\right )} x - 3 \, {\left (b c^{4} e^{2} x^{4} + 4 \, b c^{4} d e x^{2} + 4 \, b c^{2} d e - b e^{2}\right )} \arctan \left (c x\right )}{12 \, c^{4}} \]
1/4*a*e^2*x^4 + a*d*e*x^2 + a*d^2*log(x) - 1/12*(b*c^3*e^2*x^3 + 3*pi*b*c^ 4*d^2*log(c^2*x^2 + 1) - 12*b*c^4*d^2*arctan(c*x)*log(c*x) + 6*I*b*c^4*d^2 *dilog(I*c*x + 1) - 6*I*b*c^4*d^2*dilog(-I*c*x + 1) + 3*(4*b*c^3*d*e - b*c *e^2)*x - 3*(b*c^4*e^2*x^4 + 4*b*c^4*d*e*x^2 + 4*b*c^2*d*e - b*e^2)*arctan (c*x))/c^4
\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]
Time = 0.82 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x} \, dx=\left \{\begin {array}{cl} \frac {a\,\left (4\,d^2\,\ln \left (x\right )+e^2\,x^4+4\,d\,e\,x^2\right )}{4} & \text {\ if\ \ }c=0\\ \frac {a\,\left (4\,d^2\,\ln \left (x\right )+e^2\,x^4+4\,d\,e\,x^2\right )}{4}-2\,b\,d\,e\,\left (\frac {x}{2\,c}-\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\right )-\frac {b\,e^2\,\left (3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3\right )}{12\,c^4}+\frac {b\,e^2\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}-\frac {b\,d^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
piecewise(c == 0, (a*(4*d^2*log(x) + e^2*x^4 + 4*d*e*x^2))/4, c ~= 0, (a*( 4*d^2*log(x) + e^2*x^4 + 4*d*e*x^2))/4 - (b*d^2*dilog(- c*x*1i + 1)*1i)/2 + (b*d^2*dilog(c*x*1i + 1)*1i)/2 - 2*b*d*e*(x/(2*c) - atan(c*x)*(1/(2*c^2) + x^2/2)) - (b*e^2*(3*atan(c*x) - 3*c*x + c^3*x^3))/(12*c^4) + (b*e^2*x^4 *atan(c*x))/4)