Integrand size = 16, antiderivative size = 191 \[ \int (1+c x) (a+b \text {arctanh}(c x))^3 \, dx=\frac {3 b (a+b \text {arctanh}(c x))^2}{2 c}+\frac {3}{2} b x (a+b \text {arctanh}(c x))^2+\frac {(1+c x)^2 (a+b \text {arctanh}(c x))^3}{2 c}-\frac {3 b^2 (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b (a+b \text {arctanh}(c x))^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c}-\frac {3 b^2 (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 c} \]
3/2*b*(a+b*arctanh(c*x))^2/c+3/2*b*x*(a+b*arctanh(c*x))^2+1/2*(c*x+1)^2*(a +b*arctanh(c*x))^3/c-3*b^2*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c-3*b*(a+b*ar ctanh(c*x))^2*ln(2/(-c*x+1))/c-3/2*b^3*polylog(2,1-2/(-c*x+1))/c-3*b^2*(a+ b*arctanh(c*x))*polylog(2,1-2/(-c*x+1))/c+3/2*b^3*polylog(3,1-2/(-c*x+1))/ c
Time = 0.48 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.75 \[ \int (1+c x) (a+b \text {arctanh}(c x))^3 \, dx=\frac {4 a^3 c x+6 a^2 b c x+2 a^3 c^2 x^2+12 a^2 b c x \text {arctanh}(c x)+12 a b^2 c x \text {arctanh}(c x)+6 a^2 b c^2 x^2 \text {arctanh}(c x)-18 a b^2 \text {arctanh}(c x)^2-6 b^3 \text {arctanh}(c x)^2+12 a b^2 c x \text {arctanh}(c x)^2+6 b^3 c x \text {arctanh}(c x)^2+6 a b^2 c^2 x^2 \text {arctanh}(c x)^2-6 b^3 \text {arctanh}(c x)^3+4 b^3 c x \text {arctanh}(c x)^3+2 b^3 c^2 x^2 \text {arctanh}(c x)^3-24 a b^2 \text {arctanh}(c x) \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-12 b^3 \text {arctanh}(c x) \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-12 b^3 \text {arctanh}(c x)^2 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )+9 a^2 b \log (1-c x)+3 a^2 b \log (1+c x)+6 a b^2 \log \left (1-c^2 x^2\right )+6 b^2 (2 a+b+2 b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+6 b^3 \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c x)}\right )}{4 c} \]
(4*a^3*c*x + 6*a^2*b*c*x + 2*a^3*c^2*x^2 + 12*a^2*b*c*x*ArcTanh[c*x] + 12* a*b^2*c*x*ArcTanh[c*x] + 6*a^2*b*c^2*x^2*ArcTanh[c*x] - 18*a*b^2*ArcTanh[c *x]^2 - 6*b^3*ArcTanh[c*x]^2 + 12*a*b^2*c*x*ArcTanh[c*x]^2 + 6*b^3*c*x*Arc Tanh[c*x]^2 + 6*a*b^2*c^2*x^2*ArcTanh[c*x]^2 - 6*b^3*ArcTanh[c*x]^3 + 4*b^ 3*c*x*ArcTanh[c*x]^3 + 2*b^3*c^2*x^2*ArcTanh[c*x]^3 - 24*a*b^2*ArcTanh[c*x ]*Log[1 + E^(-2*ArcTanh[c*x])] - 12*b^3*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh [c*x])] - 12*b^3*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 9*a^2*b*Log [1 - c*x] + 3*a^2*b*Log[1 + c*x] + 6*a*b^2*Log[1 - c^2*x^2] + 6*b^2*(2*a + b + 2*b*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 6*b^3*PolyLog[3, -E^(-2*ArcTanh[c*x])])/(4*c)
Time = 0.55 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6480, 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.
\(\displaystyle \int (c x+1) (a+b \text {arctanh}(c x))^3 \, dx\) |
\(\Big \downarrow \) 6480 |
\(\displaystyle \frac {(c x+1)^2 (a+b \text {arctanh}(c x))^3}{2 c}-\frac {3}{2} b \int \left (\frac {2 (c x+1) (a+b \text {arctanh}(c x))^2}{1-c^2 x^2}-(a+b \text {arctanh}(c x))^2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(c x+1)^2 (a+b \text {arctanh}(c x))^3}{2 c}-\frac {3}{2} b \left (\frac {2 b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}-x (a+b \text {arctanh}(c x))^2-\frac {(a+b \text {arctanh}(c x))^2}{c}+\frac {2 b \log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}+\frac {2 \log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))^2}{c}+\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{c}\right )\) |
((1 + c*x)^2*(a + b*ArcTanh[c*x])^3)/(2*c) - (3*b*(-((a + b*ArcTanh[c*x])^ 2/c) - x*(a + b*ArcTanh[c*x])^2 + (2*b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x )])/c + (2*(a + b*ArcTanh[c*x])^2*Log[2/(1 - c*x)])/c + (b^2*PolyLog[2, 1 - 2/(1 - c*x)])/c + (2*b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)]) /c - (b^2*PolyLog[3, 1 - 2/(1 - c*x)])/c))/2
3.2.22.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_S ymbol] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTanh[c*x])^p/(e*(q + 1))), x] - Simp[b*c*(p/(e*(q + 1))) Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^(p - 1 ), (d + e*x)^(q + 1)/(1 - c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && NeQ[q, -1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 4.46 (sec) , antiderivative size = 3401, normalized size of antiderivative = 17.81
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(3401\) |
default | \(\text {Expression too large to display}\) | \(3401\) |
parts | \(\text {Expression too large to display}\) | \(3402\) |
1/c*(a^3*(1/2*c^2*x^2+c*x)+b^3*(3/2*c*x*arctanh(c*x)^2-3/2*arctanh(c*x)*ln (1+(c*x+1)^2/(-c^2*x^2+1))-3/2*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-3/2*d ilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+9/4*arctanh(c*x)^2*ln(c*x-1)+3/4*arct anh(c*x)^2*ln(c*x+1)-3*arctanh(c*x)^2*ln(2)-3/2*arctanh(c*x)^2*ln((c*x+1)/ (-c^2*x^2+1)^(1/2))+3/2*arctanh(c*x)^2-3/4*polylog(2,-(c*x+1)^2/(-c^2*x^2+ 1))+1/2*arctanh(c*x)^3-3/2*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2)) -3/2*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+3/2*polylog(3,-(c*x+1 )^2/(-c^2*x^2+1))-3*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-3*ln(2 )*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-3*ln(2)*arctanh(c*x)*ln( 1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+3*ln(2)*arctanh(c*x)*ln(1+(c*x+1)^2/(-c^2* x^2+1))-3*ln(2)*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-3*ln(2)*dilog(1-I*(c *x+1)/(-c^2*x^2+1)^(1/2))+3/2*ln(2)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+1/2 *arctanh(c*x)^3*c^2*x^2+arctanh(c*x)^3*c*x+3/8*I*Pi*csgn(I/(1-(c*x+1)^2/(c ^2*x^2-1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1- (c*x+1)^2/(c^2*x^2-1)))*(arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+a rctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+dilog(1+I*(c*x+1)/(-c^2*x^2 +1)^(1/2))+dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2)))-3/8*I*Pi*csgn(I*(c*x+1)/ (-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*(arctanh(c*x)*ln(1+I*( c*x+1)/(-c^2*x^2+1)^(1/2))+arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2)) +dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(...
\[ \int (1+c x) (a+b \text {arctanh}(c x))^3 \, dx=\int { {\left (c x + 1\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3} \,d x } \]
integral(a^3*c*x + (b^3*c*x + b^3)*arctanh(c*x)^3 + a^3 + 3*(a*b^2*c*x + a *b^2)*arctanh(c*x)^2 + 3*(a^2*b*c*x + a^2*b)*arctanh(c*x), x)
\[ \int (1+c x) (a+b \text {arctanh}(c x))^3 \, dx=\int \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3} \left (c x + 1\right )\, dx \]
\[ \int (1+c x) (a+b \text {arctanh}(c x))^3 \, dx=\int { {\left (c x + 1\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3} \,d x } \]
1/2*a^3*c*x^2 + 3/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*a^2*b*c + a^3*x + 3/2*(2*c*x*arctanh(c*x) + log(-c^2*x^ 2 + 1))*a^2*b/c - 1/16*((b^3*c^2*x^2 + 2*b^3*c*x - 3*b^3)*log(-c*x + 1)^3 - 3*(2*a*b^2*c^2*x^2 + 2*(2*a*b^2*c + b^3*c)*x + (b^3*c^2*x^2 + 2*b^3*c*x + b^3)*log(c*x + 1))*log(-c*x + 1)^2)/c - integrate(-1/8*((b^3*c^2*x^2 - b ^3)*log(c*x + 1)^3 + 6*(a*b^2*c^2*x^2 - a*b^2)*log(c*x + 1)^2 - 3*(2*a*b^2 *c^2*x^2 + (b^3*c^2*x^2 - b^3)*log(c*x + 1)^2 + 2*(2*a*b^2*c + b^3*c)*x + (2*b^3*c*x - 4*a*b^2 + b^3 + (4*a*b^2*c^2 + b^3*c^2)*x^2)*log(c*x + 1))*lo g(-c*x + 1))/(c*x - 1), x)
\[ \int (1+c x) (a+b \text {arctanh}(c x))^3 \, dx=\int { {\left (c x + 1\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (1+c x) (a+b \text {arctanh}(c x))^3 \, dx=\int {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3\,\left (c\,x+1\right ) \,d x \]