Integrand size = 10, antiderivative size = 74 \[ \int (a+b \text {arctanh}(c x))^2 \, dx=\frac {(a+b \text {arctanh}(c x))^2}{c}+x (a+b \text {arctanh}(c x))^2-\frac {2 b (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c} \]
(a+b*arctanh(c*x))^2/c+x*(a+b*arctanh(c*x))^2-2*b*(a+b*arctanh(c*x))*ln(2/ (-c*x+1))/c-b^2*polylog(2,1-2/(-c*x+1))/c
Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int (a+b \text {arctanh}(c x))^2 \, dx=\frac {b^2 (-1+c x) \text {arctanh}(c x)^2+2 b \text {arctanh}(c x) \left (a c x-b \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )+a \left (a c x+b \log \left (1-c^2 x^2\right )\right )+b^2 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )}{c} \]
(b^2*(-1 + c*x)*ArcTanh[c*x]^2 + 2*b*ArcTanh[c*x]*(a*c*x - b*Log[1 + E^(-2 *ArcTanh[c*x])]) + a*(a*c*x + b*Log[1 - c^2*x^2]) + b^2*PolyLog[2, -E^(-2* ArcTanh[c*x])])/c
Time = 0.49 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6436, 6546, 6470, 2849, 2752}
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.
\(\displaystyle \int (a+b \text {arctanh}(c x))^2 \, dx\) |
\(\Big \downarrow \) 6436 |
\(\displaystyle x (a+b \text {arctanh}(c x))^2-2 b c \int \frac {x (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx\) |
\(\Big \downarrow \) 6546 |
\(\displaystyle x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\int \frac {a+b \text {arctanh}(c x)}{1-c x}dx}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )\) |
\(\Big \downarrow \) 6470 |
\(\displaystyle x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}-b \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2}dx}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )\) |
\(\Big \downarrow \) 2849 |
\(\displaystyle x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\frac {b \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-\frac {2}{1-c x}}d\frac {1}{1-c x}}{c}+\frac {\log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c}}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )\) |
x*(a + b*ArcTanh[c*x])^2 - 2*b*c*(-1/2*(a + b*ArcTanh[c*x])^2/(b*c^2) + (( (a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/c + (b*PolyLog[2, 1 - 2/(1 - c*x)]) /(2*c))/c)
3.1.18.3.1 Defintions of rubi rules used
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-e/g Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ {c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Simp[b*c*n*p Int[x^n*((a + b*ArcTanh[c*x^n]) ^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0] && (EqQ[n, 1] || EqQ[p, 1])
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol ] :> Simp[(-(a + b*ArcTanh[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c *(p/e) Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^2*x^ 2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2 , 0]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*e*(p + 1)), x] + Simp[1/ (c*d) Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]
Time = 0.96 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.55
method | result | size |
derivativedivides | \(\frac {c x \,a^{2}+b^{2} \left (\operatorname {arctanh}\left (c x \right )^{2} \left (c x -1\right )+2 \operatorname {arctanh}\left (c x \right )^{2}-2 \,\operatorname {arctanh}\left (c x \right ) \ln \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )\right )+2 c x a b \,\operatorname {arctanh}\left (c x \right )+a b \ln \left (-c^{2} x^{2}+1\right )}{c}\) | \(115\) |
default | \(\frac {c x \,a^{2}+b^{2} \left (\operatorname {arctanh}\left (c x \right )^{2} \left (c x -1\right )+2 \operatorname {arctanh}\left (c x \right )^{2}-2 \,\operatorname {arctanh}\left (c x \right ) \ln \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )\right )+2 c x a b \,\operatorname {arctanh}\left (c x \right )+a b \ln \left (-c^{2} x^{2}+1\right )}{c}\) | \(115\) |
parts | \(a^{2} x +\frac {b^{2} \left (\operatorname {arctanh}\left (c x \right )^{2} \left (c x -1\right )+2 \operatorname {arctanh}\left (c x \right )^{2}-2 \,\operatorname {arctanh}\left (c x \right ) \ln \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )\right )}{c}+2 a b x \,\operatorname {arctanh}\left (c x \right )+\frac {a b \ln \left (-c^{2} x^{2}+1\right )}{c}\) | \(115\) |
risch | \(-\frac {b^{2} \operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{c}-\frac {b^{2} \ln \left (c x -1\right )}{c}+\frac {b^{2} \ln \left (c x +1\right )^{2} x}{4}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{4 c}+\frac {\ln \left (-c x +1\right )^{2} b^{2} x}{4}-\frac {\ln \left (-c x +1\right )^{2} b^{2}}{4 c}+\frac {\ln \left (-c x +1\right ) b^{2}}{c}+b a \ln \left (c x +1\right ) x +\frac {b a \ln \left (c x +1\right )}{c}-\frac {b^{2} \ln \left (-c x +1\right ) \ln \left (c x +1\right ) x}{2}-\frac {b^{2} \ln \left (-c x +1\right ) \ln \left (c x +1\right )}{2 c}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{c}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{c}+a^{2} x -\ln \left (-c x +1\right ) a b x +\frac {\ln \left (-c x +1\right ) a b}{c}-\frac {2 a b}{c}-\frac {b^{2}}{c}-\frac {a^{2}}{c}\) | \(264\) |
1/c*(c*x*a^2+b^2*(arctanh(c*x)^2*(c*x-1)+2*arctanh(c*x)^2-2*arctanh(c*x)*l n(1+(c*x+1)^2/(-c^2*x^2+1))-polylog(2,-(c*x+1)^2/(-c^2*x^2+1)))+2*c*x*a*b* arctanh(c*x)+a*b*ln(-c^2*x^2+1))
\[ \int (a+b \text {arctanh}(c x))^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} \,d x } \]
\[ \int (a+b \text {arctanh}(c x))^2 \, dx=\int \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}\, dx \]
\[ \int (a+b \text {arctanh}(c x))^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} \,d x } \]
-1/4*(c^2*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3) - 6*c*integrate( x*log(c*x + 1)/(c^2*x^2 - 1), x) - (c*x - 1)*(log(-c*x + 1)^2 - 2*log(-c*x + 1) + 2)/c - (c*x*log(c*x + 1)^2 + 2*(c*x - (c*x + 1)*log(c*x + 1))*log( -c*x + 1))/c + log(c^2*x^2 - 1)/c - 2*integrate(log(c*x + 1)/(c^2*x^2 - 1) , x))*b^2 + a^2*x + (2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a*b/c
\[ \int (a+b \text {arctanh}(c x))^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int (a+b \text {arctanh}(c x))^2 \, dx=\int {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2 \,d x \]