Integrand size = 21, antiderivative size = 369 \[ \int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx=\frac {2 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )}{d}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{d}-\frac {(a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d}-\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{d}-\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 d}+\frac {b^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d} \]
-2*(a+b*arctan(c*x))^2*arctanh(-1+2/(1+I*c*x))/d+(a+b*arctan(c*x))^2*ln(2/ (1-I*c*x))/d-(a+b*arctan(c*x))^2*ln(2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/d-I*b *(a+b*arctan(c*x))*polylog(2,1-2/(1-I*c*x))/d-I*b*(a+b*arctan(c*x))*polylo g(2,1-2/(1+I*c*x))/d+I*b*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))/d+I*b *(a+b*arctan(c*x))*polylog(2,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/d+1/2*b^2* polylog(3,1-2/(1-I*c*x))/d-1/2*b^2*polylog(3,1-2/(1+I*c*x))/d+1/2*b^2*poly log(3,-1+2/(1+I*c*x))/d-1/2*b^2*polylog(3,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x ))/d
\[ \int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx=\int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx \]
Time = 0.65 (sec) , antiderivative size = 369, 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.
\(\displaystyle \int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx\) |
\(\Big \downarrow \) 5411 |
\(\displaystyle \int \left (\frac {(a+b \arctan (c x))^2}{d x}-\frac {e (a+b \arctan (c x))^2}{d (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d}-\frac {(a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{d}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{d}+\frac {i b \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{d}+\frac {\log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2}{d}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{2 d}+\frac {b^2 \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )}{2 d}\) |
(2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)])/d + ((a + b*ArcTan[c* x])^2*Log[2/(1 - I*c*x)])/d - ((a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x))/( (c*d + I*e)*(1 - I*c*x))])/d - (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/( 1 - I*c*x)])/d - (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/d + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/ d + (b^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*d) - (b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*d) + (b^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/(2*d) - (b^2*PolyLo g[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(2*d)
3.2.45.3.1 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 = 2.36 (sec) , antiderivative size = 2337, normalized size of antiderivative = 6.33
method | result | size |
parts | \(\text {Expression too large to display}\) | \(2337\) |
derivativedivides | \(\text {Expression too large to display}\) | \(2345\) |
default | \(\text {Expression too large to display}\) | \(2345\) |
I*b^2*c/(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*b^2*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/(I*c*d+e)-1/2*I*b^2/d*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*Pi*arctan(c*x)^2+1/2*I*b^2/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))*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*Pi*arctan(c*x)^2-1/2*I*b^2/d*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*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*Pi*arctan(c*x) ^2+1/2*I*b^2/d*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*Pi*arctan(c*x)^2+1/2*I*b^2/d*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))*Pi*arctan(c*x)^2-1/2*b^2*c/(c*d-I*e)*polylog(3,(I*e-c*d)/(c *d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))-b^2/d*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x ^2+1)-1)+b^2/d*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^2/d*arcta n(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^2*arctan(c*x)^2/d*ln(c*x)-b^2 *arctan(c*x)^2/d*ln(c*e*x+c*d)+b^2*arctan(c*x)^2/d*ln(-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)+a^2/d*ln(x)-a^2/d*ln(e*x+d)+ 1/2*I*b^2/d*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)...
\[ \int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x} \,d x } \]
\[ \int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{x \left (d + e x\right )}\, dx \]
\[ \int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x} \,d x } \]
-a^2*(log(e*x + d)/d - log(x)/d) + integrate(1/16*(12*b^2*arctan(c*x)^2 + b^2*log(c^2*x^2 + 1)^2 + 32*a*b*arctan(c*x))/(e*x^2 + d*x), x)
\[ \int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x} \,d x } \]
Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x (d+e x)} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x\,\left (d+e\,x\right )} \,d x \]