Integrand size = 16, antiderivative size = 244 \[ \int (d+e x) (a+b \text {arctanh}(c x))^3 \, dx=\frac {3 b e (a+b \text {arctanh}(c x))^2}{2 c^2}+\frac {3 b e x (a+b \text {arctanh}(c x))^2}{2 c}+\frac {d (a+b \text {arctanh}(c x))^3}{c}-\frac {\left (d^2+\frac {e^2}{c^2}\right ) (a+b \text {arctanh}(c x))^3}{2 e}+\frac {(d+e x)^2 (a+b \text {arctanh}(c x))^3}{2 e}-\frac {3 b^2 e (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1-c x}\right )}{c^2}-\frac {3 b d (a+b \text {arctanh}(c x))^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^3 e \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c^2}-\frac {3 b^2 d (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 d \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 c} \]
3/2*b*e*(a+b*arctanh(c*x))^2/c^2+3/2*b*e*x*(a+b*arctanh(c*x))^2/c+d*(a+b*a rctanh(c*x))^3/c-1/2*(d^2+e^2/c^2)*(a+b*arctanh(c*x))^3/e+1/2*(e*x+d)^2*(a +b*arctanh(c*x))^3/e-3*b^2*e*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^2-3*b*d*( a+b*arctanh(c*x))^2*ln(2/(-c*x+1))/c-3/2*b^3*e*polylog(2,1-2/(-c*x+1))/c^2 -3*b^2*d*(a+b*arctanh(c*x))*polylog(2,1-2/(-c*x+1))/c+3/2*b^3*d*polylog(3, 1-2/(-c*x+1))/c
Time = 0.72 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.36 \[ \int (d+e x) (a+b \text {arctanh}(c x))^3 \, dx=\frac {2 a^2 c (2 a c d+3 b e) x+2 a^3 c^2 e x^2+6 a^2 b c^2 x (2 d+e x) \text {arctanh}(c x)+3 a^2 b (2 c d+e) \log (1-c x)+3 a^2 b (2 c d-e) \log (1+c x)+6 a b^2 e \left (2 c x \text {arctanh}(c x)+\left (-1+c^2 x^2\right ) \text {arctanh}(c x)^2+\log \left (1-c^2 x^2\right )\right )-2 b^3 e \left (\text {arctanh}(c x) \left ((3-3 c x) \text {arctanh}(c x)+\left (1-c^2 x^2\right ) \text {arctanh}(c x)^2+6 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )-3 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )\right )+12 a b^2 c d \left (\text {arctanh}(c x) \left ((-1+c x) \text {arctanh}(c x)-2 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )\right )+4 b^3 c d \left (\text {arctanh}(c x)^2 \left ((-1+c x) \text {arctanh}(c x)-3 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )+3 \text {arctanh}(c x) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+\frac {3}{2} \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c x)}\right )\right )}{4 c^2} \]
(2*a^2*c*(2*a*c*d + 3*b*e)*x + 2*a^3*c^2*e*x^2 + 6*a^2*b*c^2*x*(2*d + e*x) *ArcTanh[c*x] + 3*a^2*b*(2*c*d + e)*Log[1 - c*x] + 3*a^2*b*(2*c*d - e)*Log [1 + c*x] + 6*a*b^2*e*(2*c*x*ArcTanh[c*x] + (-1 + c^2*x^2)*ArcTanh[c*x]^2 + Log[1 - c^2*x^2]) - 2*b^3*e*(ArcTanh[c*x]*((3 - 3*c*x)*ArcTanh[c*x] + (1 - c^2*x^2)*ArcTanh[c*x]^2 + 6*Log[1 + E^(-2*ArcTanh[c*x])]) - 3*PolyLog[2 , -E^(-2*ArcTanh[c*x])]) + 12*a*b^2*c*d*(ArcTanh[c*x]*((-1 + c*x)*ArcTanh[ c*x] - 2*Log[1 + E^(-2*ArcTanh[c*x])]) + PolyLog[2, -E^(-2*ArcTanh[c*x])]) + 4*b^3*c*d*(ArcTanh[c*x]^2*((-1 + c*x)*ArcTanh[c*x] - 3*Log[1 + E^(-2*Ar cTanh[c*x])]) + 3*ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] + (3*PolyL og[3, -E^(-2*ArcTanh[c*x])])/2))/(4*c^2)
Time = 0.80 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int (d+e x) (a+b \text {arctanh}(c x))^3 \, dx\) |
\(\Big \downarrow \) 6480 |
\(\displaystyle \frac {(d+e x)^2 (a+b \text {arctanh}(c x))^3}{2 e}-\frac {3 b c \int \left (\frac {\left (d^2 c^2+2 d e x c^2+e^2\right ) (a+b \text {arctanh}(c x))^2}{c^2 \left (1-c^2 x^2\right )}-\frac {e^2 (a+b \text {arctanh}(c x))^2}{c^2}\right )dx}{2 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(d+e x)^2 (a+b \text {arctanh}(c x))^3}{2 e}-\frac {3 b c \left (-\frac {e^2 (a+b \text {arctanh}(c x))^2}{c^3}+\frac {2 b e^2 \log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c^3}+\frac {2 b d e \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c^2}-\frac {2 d e (a+b \text {arctanh}(c x))^3}{3 b c^2}+\frac {2 d e \log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))^2}{c^2}-\frac {e^2 x (a+b \text {arctanh}(c x))^2}{c^2}+\frac {\left (c^2 d^2+e^2\right ) (a+b \text {arctanh}(c x))^3}{3 b c^3}+\frac {b^2 e^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^3}-\frac {b^2 d e \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{c^2}\right )}{2 e}\) |
((d + e*x)^2*(a + b*ArcTanh[c*x])^3)/(2*e) - (3*b*c*(-((e^2*(a + b*ArcTanh [c*x])^2)/c^3) - (e^2*x*(a + b*ArcTanh[c*x])^2)/c^2 - (2*d*e*(a + b*ArcTan h[c*x])^3)/(3*b*c^2) + ((c^2*d^2 + e^2)*(a + b*ArcTanh[c*x])^3)/(3*b*c^3) + (2*b*e^2*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/c^3 + (2*d*e*(a + b*ArcT anh[c*x])^2*Log[2/(1 - c*x)])/c^2 + (b^2*e^2*PolyLog[2, 1 - 2/(1 - c*x)])/ c^3 + (2*b*d*e*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)])/c^2 - (b^ 2*d*e*PolyLog[3, 1 - 2/(1 - c*x)])/c^2))/(2*e)
3.1.17.3.1 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 = 3.43 (sec) , antiderivative size = 6324, normalized size of antiderivative = 25.92
method | result | size |
parts | \(\text {Expression too large to display}\) | \(6324\) |
derivativedivides | \(\text {Expression too large to display}\) | \(6333\) |
default | \(\text {Expression too large to display}\) | \(6333\) |
\[ \int (d+e x) (a+b \text {arctanh}(c x))^3 \, dx=\int { {\left (e x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3} \,d x } \]
integral(a^3*e*x + a^3*d + (b^3*e*x + b^3*d)*arctanh(c*x)^3 + 3*(a*b^2*e*x + a*b^2*d)*arctanh(c*x)^2 + 3*(a^2*b*e*x + a^2*b*d)*arctanh(c*x), x)
\[ \int (d+e x) (a+b \text {arctanh}(c x))^3 \, dx=\int \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3} \left (d + e x\right )\, dx \]
\[ \int (d+e x) (a+b \text {arctanh}(c x))^3 \, dx=\int { {\left (e x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3} \,d x } \]
1/2*a^3*e*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*e + a^3*d*x + 3/2*(2*c*x*arctanh(c*x) + log(-c^2* x^2 + 1))*a^2*b*d/c - 1/16*((b^3*c^2*e*x^2 + 2*b^3*c^2*d*x - (2*c*d + e)*b ^3)*log(-c*x + 1)^3 - 3*(2*a*b^2*c^2*e*x^2 + 2*(2*a*b^2*c^2*d + b^3*c*e)*x + (b^3*c^2*e*x^2 + 2*b^3*c^2*d*x + (2*c*d - e)*b^3)*log(c*x + 1))*log(-c* x + 1)^2)/c^2 - integrate(-1/8*((b^3*c^2*e*x^2 - b^3*c*d + (c^2*d - c*e)*b ^3*x)*log(c*x + 1)^3 + 6*(a*b^2*c^2*e*x^2 - a*b^2*c*d + (c^2*d - c*e)*a*b^ 2*x)*log(c*x + 1)^2 - 3*(2*a*b^2*c^2*e*x^2 + (b^3*c^2*e*x^2 - b^3*c*d + (c ^2*d - c*e)*b^3*x)*log(c*x + 1)^2 + 2*(2*a*b^2*c^2*d + b^3*c*e)*x - (4*a*b ^2*c*d - (2*c*d - e)*b^3 - (4*a*b^2*c^2*e + b^3*c^2*e)*x^2 - 2*(b^3*c^2*d + 2*(c^2*d - c*e)*a*b^2)*x)*log(c*x + 1))*log(-c*x + 1))/(c^2*x - c), x)
\[ \int (d+e x) (a+b \text {arctanh}(c x))^3 \, dx=\int { {\left (e x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (d+e x) (a+b \text {arctanh}(c x))^3 \, dx=\int {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3\,\left (d+e\,x\right ) \,d x \]