Integrand size = 21, antiderivative size = 591 \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x)} \, dx=-\frac {b c (a+b \arctan (c x))}{d x}-\frac {c^2 (a+b \arctan (c x))^2}{2 d}+\frac {i c e (a+b \arctan (c x))^2}{d^2}-\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {e (a+b \arctan (c x))^2}{d^2 x}+\frac {2 e^2 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d}+\frac {e^2 (a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {e^2 (a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}-\frac {b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d}-\frac {2 b c e (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{d^3}+\frac {i b^2 c e \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {b^2 e^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {b^2 e^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {b^2 e^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3} \]
-b*c*(a+b*arctan(c*x))/d/x-1/2*c^2*(a+b*arctan(c*x))^2/d+I*b^2*c*e*polylog (2,-1+2/(1-I*c*x))/d^2-1/2*(a+b*arctan(c*x))^2/d/x^2+e*(a+b*arctan(c*x))^2 /d^2/x-2*e^2*(a+b*arctan(c*x))^2*arctanh(-1+2/(1+I*c*x))/d^3+b^2*c^2*ln(x) /d+e^2*(a+b*arctan(c*x))^2*ln(2/(1-I*c*x))/d^3-e^2*(a+b*arctan(c*x))^2*ln( 2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/d^3-1/2*b^2*c^2*ln(c^2*x^2+1)/d-2*b*c*e*( a+b*arctan(c*x))*ln(2-2/(1-I*c*x))/d^2+I*c*e*(a+b*arctan(c*x))^2/d^2-I*b*e ^2*(a+b*arctan(c*x))*polylog(2,1-2/(1-I*c*x))/d^3+I*b*e^2*(a+b*arctan(c*x) )*polylog(2,-1+2/(1+I*c*x))/d^3-I*b*e^2*(a+b*arctan(c*x))*polylog(2,1-2/(1 +I*c*x))/d^3+I*b*e^2*(a+b*arctan(c*x))*polylog(2,1-2*c*(e*x+d)/(c*d+I*e)/( 1-I*c*x))/d^3+1/2*b^2*e^2*polylog(3,1-2/(1-I*c*x))/d^3-1/2*b^2*e^2*polylog (3,1-2/(1+I*c*x))/d^3+1/2*b^2*e^2*polylog(3,-1+2/(1+I*c*x))/d^3-1/2*b^2*e^ 2*polylog(3,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/d^3
Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x)} \, dx=\text {\$Aborted} \]
Time = 1.06 (sec) , antiderivative size = 591, 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.
| \(\displaystyle \int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x)} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (-\frac {e^3 (a+b \arctan (c x))^2}{d^3 (d+e x)}+\frac {e^2 (a+b \arctan (c x))^2}{d^3 x}-\frac {e (a+b \arctan (c x))^2}{d^2 x^2}+\frac {(a+b \arctan (c x))^2}{d x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 e^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{d^3}-\frac {c^2 (a+b \arctan (c x))^2}{2 d}-\frac {i b e^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{d^3}-\frac {i b e^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{d^3}+\frac {i b e^2 \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{d^3}+\frac {i b e^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {e^2 \log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2}{d^3}-\frac {e^2 (a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac {i c e (a+b \arctan (c x))^2}{d^2}+\frac {e (a+b \arctan (c x))^2}{d^2 x}-\frac {2 b c e \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{d^2}-\frac {(a+b \arctan (c x))^2}{2 d x^2}-\frac {b c (a+b \arctan (c x))}{d x}-\frac {b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d}+\frac {b^2 c^2 \log (x)}{d}+\frac {b^2 e^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{2 d^3}+\frac {b^2 e^2 \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )}{2 d^3}-\frac {b^2 e^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}+\frac {i b^2 c e \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )}{d^2}\) |
-((b*c*(a + b*ArcTan[c*x]))/(d*x)) - (c^2*(a + b*ArcTan[c*x])^2)/(2*d) + ( I*c*e*(a + b*ArcTan[c*x])^2)/d^2 - (a + b*ArcTan[c*x])^2/(2*d*x^2) + (e*(a + b*ArcTan[c*x])^2)/(d^2*x) + (2*e^2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/ (1 + I*c*x)])/d^3 + (b^2*c^2*Log[x])/d + (e^2*(a + b*ArcTan[c*x])^2*Log[2/ (1 - I*c*x)])/d^3 - (e^2*(a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/d^3 - (b^2*c^2*Log[1 + c^2*x^2])/(2*d) - (2*b*c*e*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/d^2 - (I*b*e^2*(a + b*ArcTan[c*x] )*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^3 + (I*b^2*c*e*PolyLog[2, -1 + 2/(1 - I *c*x)])/d^2 - (I*b*e^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/ d^3 + (I*b*e^2*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d^3 + ( I*b*e^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/d^3 + (b^2*e^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*d^3) - (b^2* e^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*d^3) + (b^2*e^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/(2*d^3) - (b^2*e^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e) *(1 - I*c*x))])/(2*d^3)
3.2.47.3.1 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 = 53.89 (sec) , antiderivative size = 2804, normalized size of antiderivative = 4.74
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(2804\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(2853\) |
| default | \(\text {Expression too large to display}\) | \(2853\) |
-1/2*I*b^2/d^3*e^2*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))*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))*arctan(c*x)^2+1/2*I*b^2/d^3*e^2*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*b^2/d^3*e^2* 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)^2-1/2*I*b^2/d^3*e^2*Pi*csgn (I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c *x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I*b^2/d^3*e^2*Pi*csgn(I*((1+I*c* x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^ 2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-1/2*I*b^2/d^3*e^2*Pi*cs gn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I *c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-I*b^2*c*e^2*arctan(c*x)^2*ln(1-(I* e-c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))/(I*c*d+e)/d^2-1/2*I*b^2/d^3*e^2* Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((( 1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+a^2 *(-1/2/d/x^2+e^2/d^3*ln(x)+e/d^2/x-e^2/d^3*ln(e*x+d))+1/2*I*b^2/d^3*e^2*Pi *csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)...
\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x^{3}} \,d x } \]
\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x)} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{x^{3} \left (d + e x\right )}\, dx \]
\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x^{3}} \,d x } \]
-1/2*a^2*(2*e^2*log(e*x + d)/d^3 - 2*e^2*log(x)/d^3 - (2*e*x - d)/(d^2*x^2 )) + 1/32*(32*d^2*x^2*integrate(1/16*(12*(b^2*c^2*d^2*x^2 + b^2*d^2)*arcta n(c*x)^2 + (b^2*c^2*d^2*x^2 + b^2*d^2)*log(c^2*x^2 + 1)^2 - 4*(2*b^2*c*e^2 *x^3 - b^2*c*d^2*x - 8*a*b*d^2 - (8*a*b*c^2*d^2 - b^2*c*d*e)*x^2)*arctan(c *x) + 2*(2*b^2*c^2*e^2*x^4 + b^2*c^2*d*e*x^3 - b^2*c^2*d^2*x^2)*log(c^2*x^ 2 + 1))/(c^2*d^2*e*x^6 + c^2*d^3*x^5 + d^2*e*x^4 + d^3*x^3), x) + 4*(2*b^2 *e*x - b^2*d)*arctan(c*x)^2 - (2*b^2*e*x - b^2*d)*log(c^2*x^2 + 1)^2)/(d^2 *x^2)
\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+e x)} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x^3\,\left (d+e\,x\right )} \,d x \]