Integrand size = 21, antiderivative size = 228 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x} \, dx=-\frac {3 b d^2 e x}{2 c}+\frac {3 b d e^2 x}{4 c^3}-\frac {b e^3 x}{6 c^5}-\frac {b d e^2 x^3}{4 c}+\frac {b e^3 x^3}{18 c^3}-\frac {b e^3 x^5}{30 c}+\frac {3 b d^2 e \arctan (c x)}{2 c^2}-\frac {3 b d e^2 \arctan (c x)}{4 c^4}+\frac {b e^3 \arctan (c x)}{6 c^6}+\frac {3}{2} d^2 e x^2 (a+b \arctan (c x))+\frac {3}{4} d e^2 x^4 (a+b \arctan (c x))+\frac {1}{6} e^3 x^6 (a+b \arctan (c x))+a d^3 \log (x)+\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,i c x) \]
-3/2*b*d^2*e*x/c+3/4*b*d*e^2*x/c^3-1/6*b*e^3*x/c^5-1/4*b*d*e^2*x^3/c+1/18* b*e^3*x^3/c^3-1/30*b*e^3*x^5/c+3/2*b*d^2*e*arctan(c*x)/c^2-3/4*b*d*e^2*arc tan(c*x)/c^4+1/6*b*e^3*arctan(c*x)/c^6+3/2*d^2*e*x^2*(a+b*arctan(c*x))+3/4 *d*e^2*x^4*(a+b*arctan(c*x))+1/6*e^3*x^6*(a+b*arctan(c*x))+a*d^3*ln(x)+1/2 *I*b*d^3*polylog(2,-I*c*x)-1/2*I*b*d^3*polylog(2,I*c*x)
Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.83 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x} \, dx=-\frac {b e^3 \left (15 c x-5 c^3 x^3+3 c^5 x^5-15 \arctan (c x)\right )}{90 c^6}-\frac {3 b d^2 e (c x-\arctan (c x))}{2 c^2}-\frac {b d e^2 \left (-3 c x+c^3 x^3+3 \arctan (c x)\right )}{4 c^4}+\frac {3}{2} d^2 e x^2 (a+b \arctan (c x))+\frac {3}{4} d e^2 x^4 (a+b \arctan (c x))+\frac {1}{6} e^3 x^6 (a+b \arctan (c x))+a d^3 \log (x)+\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,i c x) \]
-1/90*(b*e^3*(15*c*x - 5*c^3*x^3 + 3*c^5*x^5 - 15*ArcTan[c*x]))/c^6 - (3*b *d^2*e*(c*x - ArcTan[c*x]))/(2*c^2) - (b*d*e^2*(-3*c*x + c^3*x^3 + 3*ArcTa n[c*x]))/(4*c^4) + (3*d^2*e*x^2*(a + b*ArcTan[c*x]))/2 + (3*d*e^2*x^4*(a + b*ArcTan[c*x]))/4 + (e^3*x^6*(a + b*ArcTan[c*x]))/6 + a*d^3*Log[x] + (I/2 )*b*d^3*PolyLog[2, (-I)*c*x] - (I/2)*b*d^3*PolyLog[2, I*c*x]
Time = 0.43 (sec) , antiderivative size = 228, 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {d^3 (a+b \arctan (c x))}{x}+3 d^2 e x (a+b \arctan (c x))+3 d e^2 x^3 (a+b \arctan (c x))+e^3 x^5 (a+b \arctan (c x))\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3}{2} d^2 e x^2 (a+b \arctan (c x))+\frac {3}{4} d e^2 x^4 (a+b \arctan (c x))+\frac {1}{6} e^3 x^6 (a+b \arctan (c x))+a d^3 \log (x)+\frac {b e^3 \arctan (c x)}{6 c^6}-\frac {3 b d e^2 \arctan (c x)}{4 c^4}+\frac {3 b d^2 e \arctan (c x)}{2 c^2}-\frac {b e^3 x}{6 c^5}+\frac {3 b d e^2 x}{4 c^3}+\frac {b e^3 x^3}{18 c^3}+\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,i c x)-\frac {3 b d^2 e x}{2 c}-\frac {b d e^2 x^3}{4 c}-\frac {b e^3 x^5}{30 c}\) |
(-3*b*d^2*e*x)/(2*c) + (3*b*d*e^2*x)/(4*c^3) - (b*e^3*x)/(6*c^5) - (b*d*e^ 2*x^3)/(4*c) + (b*e^3*x^3)/(18*c^3) - (b*e^3*x^5)/(30*c) + (3*b*d^2*e*ArcT an[c*x])/(2*c^2) - (3*b*d*e^2*ArcTan[c*x])/(4*c^4) + (b*e^3*ArcTan[c*x])/( 6*c^6) + (3*d^2*e*x^2*(a + b*ArcTan[c*x]))/2 + (3*d*e^2*x^4*(a + b*ArcTan[ c*x]))/4 + (e^3*x^6*(a + b*ArcTan[c*x]))/6 + a*d^3*Log[x] + (I/2)*b*d^3*Po lyLog[2, (-I)*c*x] - (I/2)*b*d^3*PolyLog[2, I*c*x]
3.12.41.3.1 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 = 0.34 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.13
| method | result | size |
| parts | \(a \left (\frac {e^{3} x^{6}}{6}+\frac {3 x^{4} e^{2} d}{4}+\frac {3 e \,d^{2} x^{2}}{2}+d^{3} \ln \left (x \right )\right )+b \left (\frac {\arctan \left (c x \right ) e^{3} x^{6}}{6}+\frac {3 \arctan \left (c x \right ) e^{2} d \,x^{4}}{4}+\frac {3 \arctan \left (c x \right ) d^{2} e \,x^{2}}{2}+\arctan \left (c x \right ) d^{3} \ln \left (c x \right )-\frac {e \left (18 c^{5} x \,d^{2}+3 d \,c^{5} e \,x^{3}+\frac {2 e^{2} c^{5} x^{5}}{5}-9 c^{3} d e x -\frac {2 e^{2} c^{3} x^{3}}{3}+2 c x \,e^{2}+\left (-18 c^{4} d^{2}+9 c^{2} d e -2 e^{2}\right ) \arctan \left (c x \right )\right )-6 i c^{6} d^{3} \ln \left (c x \right ) \ln \left (i c x +1\right )+6 i c^{6} d^{3} \ln \left (c x \right ) \ln \left (-i c x +1\right )-6 i c^{6} d^{3} \operatorname {dilog}\left (i c x +1\right )+6 i c^{6} d^{3} \operatorname {dilog}\left (-i c x +1\right )}{12 c^{6}}\right )\) | \(257\) |
| derivativedivides | \(\frac {a \left (\frac {3 d^{2} c^{6} e \,x^{2}}{2}+\frac {3 d \,c^{6} e^{2} x^{4}}{4}+\frac {e^{3} c^{6} x^{6}}{6}+c^{6} d^{3} \ln \left (c x \right )\right )}{c^{6}}+\frac {b \left (\frac {3 \arctan \left (c x \right ) d^{2} c^{6} e \,x^{2}}{2}+\frac {3 \arctan \left (c x \right ) d \,c^{6} e^{2} x^{4}}{4}+\frac {\arctan \left (c x \right ) e^{3} c^{6} x^{6}}{6}+\arctan \left (c x \right ) c^{6} d^{3} \ln \left (c x \right )-\frac {e \left (18 c^{5} x \,d^{2}+3 d \,c^{5} e \,x^{3}+\frac {2 e^{2} c^{5} x^{5}}{5}-9 c^{3} d e x -\frac {2 e^{2} c^{3} x^{3}}{3}+2 c x \,e^{2}+\left (-18 c^{4} d^{2}+9 c^{2} d e -2 e^{2}\right ) \arctan \left (c x \right )\right )}{12}-c^{6} d^{3} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{6}}\) | \(269\) |
| default | \(\frac {a \left (\frac {3 d^{2} c^{6} e \,x^{2}}{2}+\frac {3 d \,c^{6} e^{2} x^{4}}{4}+\frac {e^{3} c^{6} x^{6}}{6}+c^{6} d^{3} \ln \left (c x \right )\right )}{c^{6}}+\frac {b \left (\frac {3 \arctan \left (c x \right ) d^{2} c^{6} e \,x^{2}}{2}+\frac {3 \arctan \left (c x \right ) d \,c^{6} e^{2} x^{4}}{4}+\frac {\arctan \left (c x \right ) e^{3} c^{6} x^{6}}{6}+\arctan \left (c x \right ) c^{6} d^{3} \ln \left (c x \right )-\frac {e \left (18 c^{5} x \,d^{2}+3 d \,c^{5} e \,x^{3}+\frac {2 e^{2} c^{5} x^{5}}{5}-9 c^{3} d e x -\frac {2 e^{2} c^{3} x^{3}}{3}+2 c x \,e^{2}+\left (-18 c^{4} d^{2}+9 c^{2} d e -2 e^{2}\right ) \arctan \left (c x \right )\right )}{12}-c^{6} d^{3} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{6}}\) | \(269\) |
| risch | \(\frac {3 a d \,e^{2} x^{4}}{4}+\frac {3 a \,d^{2} e \,x^{2}}{2}-\frac {3 i b e \,d^{2} \ln \left (i c x +1\right ) x^{2}}{4}-\frac {3 i b \,e^{2} d \ln \left (i c x +1\right ) x^{4}}{8}+\frac {3 i b \,d^{2} e \ln \left (-i c x +1\right ) x^{2}}{4}+\frac {3 i b \,e^{2} d \ln \left (-i c x +1\right ) x^{4}}{8}-\frac {b \,e^{3} x}{6 c^{5}}+\frac {b \,e^{3} x^{3}}{18 c^{3}}-\frac {b \,e^{3} x^{5}}{30 c}+\frac {b \,e^{3} \arctan \left (c x \right )}{6 c^{6}}-\frac {3 b \,d^{2} e x}{2 c}+\frac {3 b d \,e^{2} x}{4 c^{3}}-\frac {b d \,e^{2} x^{3}}{4 c}+\frac {3 b \,d^{2} e \arctan \left (c x \right )}{2 c^{2}}-\frac {3 b d \,e^{2} \arctan \left (c x \right )}{4 c^{4}}+\frac {3 a \,d^{2} e}{2 c^{2}}-\frac {3 a \,e^{2} d}{4 c^{4}}+\frac {a \,e^{3} x^{6}}{6}+\frac {a \,e^{3}}{6 c^{6}}+\frac {i b \,d^{3} \operatorname {dilog}\left (i c x +1\right )}{2}+a \,d^{3} \ln \left (-i c x \right )-\frac {i b \,d^{3} \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {i b \,e^{3} \ln \left (i c x +1\right ) x^{6}}{12}+\frac {i b \,e^{3} \ln \left (-i c x +1\right ) x^{6}}{12}\) | \(323\) |
a*(1/6*e^3*x^6+3/4*x^4*e^2*d+3/2*e*d^2*x^2+d^3*ln(x))+b*(1/6*arctan(c*x)*e ^3*x^6+3/4*arctan(c*x)*e^2*d*x^4+3/2*arctan(c*x)*d^2*e*x^2+arctan(c*x)*d^3 *ln(c*x)-1/12/c^6*(e*(18*c^5*x*d^2+3*d*c^5*e*x^3+2/5*e^2*c^5*x^5-9*c^3*d*e *x-2/3*e^2*c^3*x^3+2*c*x*e^2+(-18*c^4*d^2+9*c^2*d*e-2*e^2)*arctan(c*x))-6* I*c^6*d^3*ln(c*x)*ln(1+I*c*x)+6*I*c^6*d^3*ln(c*x)*ln(1-I*c*x)-6*I*c^6*d^3* dilog(1+I*c*x)+6*I*c^6*d^3*dilog(1-I*c*x)))
\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]
integral((a*e^3*x^6 + 3*a*d*e^2*x^4 + 3*a*d^2*e*x^2 + a*d^3 + (b*e^3*x^6 + 3*b*d*e^2*x^4 + 3*b*d^2*e*x^2 + b*d^3)*arctan(c*x))/x, x)
\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x}\, dx \]
Time = 0.46 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.10 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x} \, dx=\frac {1}{6} \, a e^{3} x^{6} + \frac {3}{4} \, a d e^{2} x^{4} + \frac {3}{2} \, a d^{2} e x^{2} + a d^{3} \log \left (x\right ) - \frac {6 \, b c^{5} e^{3} x^{5} + 45 \, \pi b c^{6} d^{3} \log \left (c^{2} x^{2} + 1\right ) - 180 \, b c^{6} d^{3} \arctan \left (c x\right ) \log \left (c x\right ) + 90 i \, b c^{6} d^{3} {\rm Li}_2\left (i \, c x + 1\right ) - 90 i \, b c^{6} d^{3} {\rm Li}_2\left (-i \, c x + 1\right ) + 5 \, {\left (9 \, b c^{5} d e^{2} - 2 \, b c^{3} e^{3}\right )} x^{3} + 15 \, {\left (18 \, b c^{5} d^{2} e - 9 \, b c^{3} d e^{2} + 2 \, b c e^{3}\right )} x - 15 \, {\left (2 \, b c^{6} e^{3} x^{6} + 9 \, b c^{6} d e^{2} x^{4} + 18 \, b c^{6} d^{2} e x^{2} + 18 \, b c^{4} d^{2} e - 9 \, b c^{2} d e^{2} + 2 \, b e^{3}\right )} \arctan \left (c x\right )}{180 \, c^{6}} \]
1/6*a*e^3*x^6 + 3/4*a*d*e^2*x^4 + 3/2*a*d^2*e*x^2 + a*d^3*log(x) - 1/180*( 6*b*c^5*e^3*x^5 + 45*pi*b*c^6*d^3*log(c^2*x^2 + 1) - 180*b*c^6*d^3*arctan( c*x)*log(c*x) + 90*I*b*c^6*d^3*dilog(I*c*x + 1) - 90*I*b*c^6*d^3*dilog(-I* c*x + 1) + 5*(9*b*c^5*d*e^2 - 2*b*c^3*e^3)*x^3 + 15*(18*b*c^5*d^2*e - 9*b* c^3*d*e^2 + 2*b*c*e^3)*x - 15*(2*b*c^6*e^3*x^6 + 9*b*c^6*d*e^2*x^4 + 18*b* c^6*d^2*e*x^2 + 18*b*c^4*d^2*e - 9*b*c^2*d*e^2 + 2*b*e^3)*arctan(c*x))/c^6
\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]
Time = 0.90 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.02 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x} \, dx=\left \{\begin {array}{cl} \frac {a\,e^3\,x^6}{6}+a\,d^3\,\ln \left (x\right )+\frac {3\,a\,d^2\,e\,x^2}{2}+\frac {3\,a\,d\,e^2\,x^4}{4} & \text {\ if\ \ }c=0\\ \frac {a\,e^3\,x^6}{6}+a\,d^3\,\ln \left (x\right )-\frac {b\,e^3\,\left (\frac {x}{c^4}-\frac {\mathrm {atan}\left (c\,x\right )}{c^5}+\frac {x^5}{5}-\frac {x^3}{3\,c^2}\right )}{6\,c}-3\,b\,d^2\,e\,\left (\frac {x}{2\,c}-\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\right )+\frac {3\,a\,d^2\,e\,x^2}{2}+\frac {3\,a\,d\,e^2\,x^4}{4}-3\,b\,d\,e^2\,\left (\frac {3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3}{12\,c^4}-\frac {x^4\,\mathrm {atan}\left (c\,x\right )}{4}\right )+\frac {b\,e^3\,x^6\,\mathrm {atan}\left (c\,x\right )}{6}-\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
piecewise(c == 0, (a*e^3*x^6)/6 + a*d^3*log(x) + (3*a*d^2*e*x^2)/2 + (3*a* d*e^2*x^4)/4, c ~= 0, (a*e^3*x^6)/6 + a*d^3*log(x) - (b*d^3*dilog(- c*x*1i + 1)*1i)/2 + (b*d^3*dilog(c*x*1i + 1)*1i)/2 - (b*e^3*(x/c^4 - atan(c*x)/c ^5 + x^5/5 - x^3/(3*c^2)))/(6*c) - 3*b*d^2*e*(x/(2*c) - atan(c*x)*(1/(2*c^ 2) + x^2/2)) + (3*a*d^2*e*x^2)/2 + (3*a*d*e^2*x^4)/4 - 3*b*d*e^2*((3*atan( c*x) - 3*c*x + c^3*x^3)/(12*c^4) - (x^4*atan(c*x))/4) + (b*e^3*x^6*atan(c* x))/6)