Integrand size = 21, antiderivative size = 319 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x (d+e x)} \, dx=\frac {2 (a+b \text {arctanh}(c x))^2 \text {arctanh}\left (1-\frac {2}{1-c x}\right )}{d}+\frac {(a+b \text {arctanh}(c x))^2 \log \left (\frac {2}{1+c x}\right )}{d}-\frac {(a+b \text {arctanh}(c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{d}+\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-c x}\right )}{d}-\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{d}+\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-c x}\right )}{2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d} \]
-2*(a+b*arctanh(c*x))^2*arctanh(-1+2/(-c*x+1))/d+(a+b*arctanh(c*x))^2*ln(2 /(c*x+1))/d-(a+b*arctanh(c*x))^2*ln(2*c*(e*x+d)/(c*d+e)/(c*x+1))/d-b*(a+b* arctanh(c*x))*polylog(2,1-2/(-c*x+1))/d+b*(a+b*arctanh(c*x))*polylog(2,-1+ 2/(-c*x+1))/d-b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/d+b*(a+b*arctanh (c*x))*polylog(2,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/d+1/2*b^2*polylog(3,1-2/(- c*x+1))/d-1/2*b^2*polylog(3,-1+2/(-c*x+1))/d-1/2*b^2*polylog(3,1-2/(c*x+1) )/d+1/2*b^2*polylog(3,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/d
Result contains complex when optimal does not.
Time = 9.55 (sec) , antiderivative size = 1151, normalized size of antiderivative = 3.61 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x (d+e x)} \, dx =\text {Too large to display} \]
(a^2*Log[x])/d - (a^2*Log[d + e*x])/d + (a*b*((-I)*c*d*Pi*ArcTanh[c*x] - 2 *c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x] + c*d*ArcTanh[c*x]^2 - e*ArcTanh[c*x]^2 + (Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^ArcTanh[(c*d)/e] + 2*c*d*A rcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] + I*c*d*Pi*Log[1 + E^(2*ArcTanh[c *x])] - 2*c*d*ArcTanh[(c*d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c *x]))] - 2*c*d*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x] ))] + (I/2)*c*d*Pi*Log[1 - c^2*x^2] + 2*c*d*ArcTanh[(c*d)/e]*Log[I*Sinh[Ar cTanh[(c*d)/e] + ArcTanh[c*x]]] - c*d*PolyLog[2, E^(-2*ArcTanh[c*x])] + c* d*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/(c*d^2) + (b^2*(I *c*d*Pi^3 - 8*c*d*ArcTanh[c*x]^3 - 8*e*ArcTanh[c*x]^3 + 24*c*d*ArcTanh[c*x ]^2*Log[1 - E^(2*ArcTanh[c*x])] + 24*c*d*ArcTanh[c*x]*PolyLog[2, E^(2*ArcT anh[c*x])] - 12*c*d*PolyLog[3, E^(2*ArcTanh[c*x])] - (24*(c*d - e)*(c*d + e)*(-6*c*d*ArcTanh[c*x]^3 + 2*e*ArcTanh[c*x]^3 - (4*Sqrt[1 - (c^2*d^2)/e^2 ]*e*ArcTanh[c*x]^3)/E^ArcTanh[(c*d)/e] - (6*I)*c*d*Pi*ArcTanh[c*x]*Log[(E^ (-ArcTanh[c*x]) + E^ArcTanh[c*x])/2] - 6*c*d*ArcTanh[c*x]^2*Log[1 - (Sqrt[ c*d + e]*E^ArcTanh[c*x])/Sqrt[-(c*d) + e]] - 6*c*d*ArcTanh[c*x]^2*Log[1 + (Sqrt[c*d + e]*E^ArcTanh[c*x])/Sqrt[-(c*d) + e]] + 6*c*d*ArcTanh[c*x]^2*Lo g[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 + E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(2*( ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*...
Time = 0.74 (sec) , antiderivative size = 319, 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 = {6502, 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.
\(\displaystyle \int \frac {(a+b \text {arctanh}(c x))^2}{x (d+e x)} \, dx\) |
\(\Big \downarrow \) 6502 |
\(\displaystyle \int \left (\frac {(a+b \text {arctanh}(c x))^2}{d x}-\frac {e (a+b \text {arctanh}(c x))^2}{d (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{d}-\frac {(a+b \text {arctanh}(c x))^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{d}+\frac {b \operatorname {PolyLog}\left (2,\frac {2}{1-c x}-1\right ) (a+b \text {arctanh}(c x))}{d}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{d}+\frac {2 \text {arctanh}\left (1-\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))^2}{d}+\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,\frac {2}{1-c x}-1\right )}{2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 d}\) |
(2*(a + b*ArcTanh[c*x])^2*ArcTanh[1 - 2/(1 - c*x)])/d + ((a + b*ArcTanh[c* x])^2*Log[2/(1 + c*x)])/d - ((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/(( c*d + e)*(1 + c*x))])/d - (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c* x)])/d + (b*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)])/d - (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/d + (b*(a + b*ArcTanh[c*x])* PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/d + (b^2*PolyLog[3, 1 - 2/(1 - c*x)])/(2*d) - (b^2*PolyLog[3, -1 + 2/(1 - c*x)])/(2*d) - (b^2 *PolyLog[3, 1 - 2/(1 + c*x)])/(2*d) + (b^2*PolyLog[3, 1 - (2*c*(d + e*x))/ ((c*d + e)*(1 + c*x))])/(2*d)
3.2.57.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e _.)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTanh[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 = 11.26 (sec) , antiderivative size = 1787, normalized size of antiderivative = 5.60
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1787\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1795\) |
default | \(\text {Expression too large to display}\) | \(1795\) |
b^2/d*ln(c*x)*arctanh(c*x)^2+a^2/d*ln(x)-2*b^2/d*polylog(3,-(c*x+1)/(-c^2* x^2+1)^(1/2))-2*b^2/d*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))-b^2/d*arctanh( c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+b^2/d*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^ 2*x^2+1)^(1/2))+2*b^2/d*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2) )+b^2/d*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^2/d*arctanh(c* x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*I*b^2/d*Pi*arctanh(c*x)^2*csg n(I*(-(c*x+1)^2/(c^2*x^2-1)-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^3-1/2*I*b^2/d*Pi *arctanh(c*x)^2*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(-(c*x+1)^2/(c^2* x^2-1)-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^2-1/2*I*b^2/d*Pi*arctanh(c*x)^2*csgn( I*(-(c*x+1)^2/(c^2*x^2-1)-1))*csgn(I*(-(c*x+1)^2/(c^2*x^2-1)-1)/(1-(c*x+1) ^2/(c^2*x^2-1)))^2+1/2*I*b^2/d*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I *(d*c*(1-(c*x+1)^2/(c^2*x^2-1))+e*(-(c*x+1)^2/(c^2*x^2-1)-1))/(1-(c*x+1)^2 /(c^2*x^2-1)))^2*arctanh(c*x)^2+1/2*I*b^2/d*Pi*csgn(I*(d*c*(1-(c*x+1)^2/(c ^2*x^2-1))+e*(-(c*x+1)^2/(c^2*x^2-1)-1)))*csgn(I*(d*c*(1-(c*x+1)^2/(c^2*x^ 2-1))+e*(-(c*x+1)^2/(c^2*x^2-1)-1))/(1-(c*x+1)^2/(c^2*x^2-1)))^2*arctanh(c *x)^2-b^2*arctanh(c*x)^2/d*ln(c*e*x+c*d)+b^2*arctanh(c*x)^2/d*ln(d*c*(1+(c *x+1)^2/(-c^2*x^2+1))+e*((c*x+1)^2/(-c^2*x^2+1)-1))+1/2*b^2*c/(c*d+e)*poly log(3,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-b^2/d*e/(c*d+e)*arctanh(c*x )^2*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-b^2/d*e/(c*d+e)*arctanh( c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-1/2*I*b^2/d*Pi*...
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{x (d+e x)} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x} \,d x } \]
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{x (d+e x)} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}}{x \left (d + e x\right )}\, dx \]
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{x (d+e x)} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x} \,d x } \]
-a^2*(log(e*x + d)/d - log(x)/d) + integrate(1/4*b^2*(log(c*x + 1) - log(- c*x + 1))^2/(e*x^2 + d*x) + a*b*(log(c*x + 1) - log(-c*x + 1))/(e*x^2 + d* x), x)
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{x (d+e x)} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )} x} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x (d+e x)} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{x\,\left (d+e\,x\right )} \,d x \]