Integrand size = 21, antiderivative size = 598 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx=\frac {a b d x}{c e^2}+\frac {b^2 x}{3 c^2 e}-\frac {b^2 \arctan (c x)}{3 c^3 e}+\frac {b^2 d x \arctan (c x)}{c e^2}-\frac {b x^2 (a+b \arctan (c x))}{3 c e}+\frac {i d^2 (a+b \arctan (c x))^2}{c e^3}-\frac {d (a+b \arctan (c x))^2}{2 c^2 e^2}-\frac {i (a+b \arctan (c x))^2}{3 c^3 e}+\frac {d^2 x (a+b \arctan (c x))^2}{e^3}-\frac {d x^2 (a+b \arctan (c x))^2}{2 e^2}+\frac {x^3 (a+b \arctan (c x))^2}{3 e}+\frac {d^3 (a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{e^4}+\frac {2 b d^2 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c e^3}-\frac {2 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^3 e}-\frac {d^3 (a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^4}-\frac {b^2 d \log \left (1+c^2 x^2\right )}{2 c^2 e^2}-\frac {i b d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{e^4}+\frac {i b^2 d^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c e^3}-\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^3 e}+\frac {i b d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^4}+\frac {b^2 d^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e^4}-\frac {b^2 d^3 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^4} \]
a*b*d*x/c/e^2+1/3*b^2*x/c^2/e-1/3*b^2*arctan(c*x)/c^3/e+b^2*d*x*arctan(c*x )/c/e^2-1/3*b*x^2*(a+b*arctan(c*x))/c/e-1/3*I*(a+b*arctan(c*x))^2/c^3/e-1/ 2*d*(a+b*arctan(c*x))^2/c^2/e^2-I*b*d^3*(a+b*arctan(c*x))*polylog(2,1-2/(1 -I*c*x))/e^4+d^2*x*(a+b*arctan(c*x))^2/e^3-1/2*d*x^2*(a+b*arctan(c*x))^2/e ^2+1/3*x^3*(a+b*arctan(c*x))^2/e+d^3*(a+b*arctan(c*x))^2*ln(2/(1-I*c*x))/e ^4+2*b*d^2*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c/e^3-2/3*b*(a+b*arctan(c*x)) *ln(2/(1+I*c*x))/c^3/e-d^3*(a+b*arctan(c*x))^2*ln(2*c*(e*x+d)/(c*d+I*e)/(1 -I*c*x))/e^4-1/2*b^2*d*ln(c^2*x^2+1)/c^2/e^2+I*d^2*(a+b*arctan(c*x))^2/c/e ^3-1/3*I*b^2*polylog(2,1-2/(1+I*c*x))/c^3/e+I*b*d^3*(a+b*arctan(c*x))*poly log(2,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/e^4+I*b^2*d^2*polylog(2,1-2/(1+I* c*x))/c/e^3+1/2*b^2*d^3*polylog(3,1-2/(1-I*c*x))/e^4-1/2*b^2*d^3*polylog(3 ,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/e^4
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx=\int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx \]
Time = 0.97 (sec) , antiderivative size = 598, 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 = {5411, 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.
| \(\displaystyle \int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (-\frac {d^3 (a+b \arctan (c x))^2}{e^3 (d+e x)}+\frac {d^2 (a+b \arctan (c x))^2}{e^3}-\frac {d x (a+b \arctan (c x))^2}{e^2}+\frac {x^2 (a+b \arctan (c x))^2}{e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {i (a+b \arctan (c x))^2}{3 c^3 e}-\frac {2 b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{3 c^3 e}-\frac {d (a+b \arctan (c x))^2}{2 c^2 e^2}-\frac {i b d^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{e^4}+\frac {i b d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^4}+\frac {d^3 \log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2}{e^4}-\frac {d^3 (a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^4}+\frac {d^2 x (a+b \arctan (c x))^2}{e^3}+\frac {i d^2 (a+b \arctan (c x))^2}{c e^3}+\frac {2 b d^2 \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c e^3}-\frac {d x^2 (a+b \arctan (c x))^2}{2 e^2}+\frac {x^3 (a+b \arctan (c x))^2}{3 e}-\frac {b x^2 (a+b \arctan (c x))}{3 c e}+\frac {a b d x}{c e^2}-\frac {b^2 \arctan (c x)}{3 c^3 e}+\frac {b^2 d x \arctan (c x)}{c e^2}-\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{3 c^3 e}-\frac {b^2 d \log \left (c^2 x^2+1\right )}{2 c^2 e^2}+\frac {b^2 x}{3 c^2 e}+\frac {b^2 d^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e^4}-\frac {b^2 d^3 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^4}+\frac {i b^2 d^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c e^3}\) |
(a*b*d*x)/(c*e^2) + (b^2*x)/(3*c^2*e) - (b^2*ArcTan[c*x])/(3*c^3*e) + (b^2 *d*x*ArcTan[c*x])/(c*e^2) - (b*x^2*(a + b*ArcTan[c*x]))/(3*c*e) + (I*d^2*( a + b*ArcTan[c*x])^2)/(c*e^3) - (d*(a + b*ArcTan[c*x])^2)/(2*c^2*e^2) - (( I/3)*(a + b*ArcTan[c*x])^2)/(c^3*e) + (d^2*x*(a + b*ArcTan[c*x])^2)/e^3 - (d*x^2*(a + b*ArcTan[c*x])^2)/(2*e^2) + (x^3*(a + b*ArcTan[c*x])^2)/(3*e) + (d^3*(a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/e^4 + (2*b*d^2*(a + b*Arc Tan[c*x])*Log[2/(1 + I*c*x)])/(c*e^3) - (2*b*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3*e) - (d^3*(a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x))/((c* d + I*e)*(1 - I*c*x))])/e^4 - (b^2*d*Log[1 + c^2*x^2])/(2*c^2*e^2) - (I*b* d^3*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e^4 + (I*b^2*d^2*Po lyLog[2, 1 - 2/(1 + I*c*x)])/(c*e^3) - ((I/3)*b^2*PolyLog[2, 1 - 2/(1 + I* c*x)])/(c^3*e) + (I*b*d^3*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(d + e*x ))/((c*d + I*e)*(1 - I*c*x))])/e^4 + (b^2*d^3*PolyLog[3, 1 - 2/(1 - I*c*x) ])/(2*e^4) - (b^2*d^3*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c *x))])/(2*e^4)
3.2.41.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f* x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] & & IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 38.26 (sec) , antiderivative size = 2062, normalized size of antiderivative = 3.45
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2062\) |
| default | \(\text {Expression too large to display}\) | \(2062\) |
| parts | \(\text {Expression too large to display}\) | \(2107\) |
1/c^4*(a^2*c^4/e^3*d^2*x-1/2*a^2*c^4/e^2*d*x^2+1/3*a^2*c^4/e*x^3-a^2*c^4*d ^3/e^4*ln(c*e*x+c*d)+b^2*c*(1/3/e*(I+c*x)+1/3*arctan(c*x)^2/e*c^3*x^3-I/e^ 4*c^3*d^3*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+d^3*c^3/e^3/(c*d -I*e)*arctan(c*x)*polylog(2,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))-d ^4*c^4/e^4/(c*d-I*e)*arctan(c*x)^2*ln(1-(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c ^2*x^2+1))+arctan(c*x)^2/e^3*c^3*d^2*x-1/2*arctan(c*x)^2/e^2*c^3*d*x^2+1/2 *I*d^3*c^3/e^3/(c*d-I*e)*polylog(3,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^ 2+1))+1/2*I/e^4*c^3*d^3*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(-I* e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+c*d)/((1+I*c*x)^ 2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I/e^4*c^3*d^3*Pi*csgn(I*(-I*e*(1+I*c *x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+c*d))*csgn(I*(-I*e*(1+I* c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+c*d)/((1+I*c*x)^2/(c^2* x^2+1)+1))^2*arctan(c*x)^2-I/e^3*c^2*d^2*arctan(c*x)^2-2*I/e^3*c^2*d^2*dil og(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+2/3*I/e*dilog(1+I*(1+I*c*x)/(c^2*x^2+1 )^(1/2))-2/3/e*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-2/3/e*arcta n(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/3*arctan(c*x)*(c*x-I)^2/e+2/3 *I/e*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/3*I/e*arctan(c*x)^2+I*d^4*c^ 4/e^4/(c*d-I*e)*arctan(c*x)*polylog(2,(I*e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2 *x^2+1))+I*d^3*c^3/e^3/(c*d-I*e)*arctan(c*x)^2*ln(1-(I*e-c*d)/(c*d+I*e)*(1 +I*c*x)^2/(c^2*x^2+1))-1/2*I/e^4*c^3*d^3*Pi*csgn(I*(-I*e*(1+I*c*x)^2/(c...
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{e x + d} \,d x } \]
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx=\int \frac {x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{d + e x}\, dx \]
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{e x + d} \,d x } \]
-1/6*a^2*(6*d^3*log(e*x + d)/e^4 - (2*e^2*x^3 - 3*d*e*x^2 + 6*d^2*x)/e^3) + 1/96*(96*e^3*integrate(1/48*(36*(b^2*c^2*e^3*x^5 + b^2*e^3*x^3)*arctan(c *x)^2 + 3*(b^2*c^2*e^3*x^5 + b^2*e^3*x^3)*log(c^2*x^2 + 1)^2 + 4*(24*a*b*c ^2*e^3*x^5 - 2*b^2*c*e^3*x^4 - 3*b^2*c*d^2*e*x^2 - 6*b^2*c*d^3*x + (b^2*c* d*e^2 + 24*a*b*e^3)*x^3)*arctan(c*x) + 2*(2*b^2*c^2*e^3*x^5 - b^2*c^2*d*e^ 2*x^4 + 3*b^2*c^2*d^2*e*x^3 + 6*b^2*c^2*d^3*x^2)*log(c^2*x^2 + 1))/(c^2*e^ 4*x^3 + c^2*d*e^3*x^2 + e^4*x + d*e^3), x) + 4*(2*b^2*e^2*x^3 - 3*b^2*d*e* x^2 + 6*b^2*d^2*x)*arctan(c*x)^2 - (2*b^2*e^2*x^3 - 3*b^2*d*e*x^2 + 6*b^2* d^2*x)*log(c^2*x^2 + 1)^2)/e^3
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+e x} \, dx=\int \frac {x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{d+e\,x} \,d x \]