Integrand size = 20, antiderivative size = 172 \[ \int \frac {x (a+b \text {arctanh}(c x))^2}{d+c d x} \, dx=\frac {(a+b \text {arctanh}(c x))^2}{c^2 d}+\frac {x (a+b \text {arctanh}(c x))^2}{c d}-\frac {2 b (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1-c x}\right )}{c^2 d}+\frac {(a+b \text {arctanh}(c x))^2 \log \left (\frac {2}{1+c x}\right )}{c^2 d}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^2 d}-\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c^2 d} \]
(a+b*arctanh(c*x))^2/c^2/d+x*(a+b*arctanh(c*x))^2/c/d-2*b*(a+b*arctanh(c*x ))*ln(2/(-c*x+1))/c^2/d+(a+b*arctanh(c*x))^2*ln(2/(c*x+1))/c^2/d-b^2*polyl og(2,1-2/(-c*x+1))/c^2/d-b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/c^2/d -1/2*b^2*polylog(3,1-2/(c*x+1))/c^2/d
Time = 0.58 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.81 \[ \int \frac {x (a+b \text {arctanh}(c x))^2}{d+c d x} \, dx=\frac {2 b^2 \text {arctanh}(c x)^2 \left (-1+c x+\log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )+4 b \text {arctanh}(c x) \left (a c x+(a-b) \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )+2 a \left (a c x-a \log (1+c x)+b \log \left (1-c^2 x^2\right )\right )-2 b (a-b+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )-b^2 \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c x)}\right )}{2 c^2 d} \]
(2*b^2*ArcTanh[c*x]^2*(-1 + c*x + Log[1 + E^(-2*ArcTanh[c*x])]) + 4*b*ArcT anh[c*x]*(a*c*x + (a - b)*Log[1 + E^(-2*ArcTanh[c*x])]) + 2*a*(a*c*x - a*L og[1 + c*x] + b*Log[1 - c^2*x^2]) - 2*b*(a - b + b*ArcTanh[c*x])*PolyLog[2 , -E^(-2*ArcTanh[c*x])] - b^2*PolyLog[3, -E^(-2*ArcTanh[c*x])])/(2*c^2*d)
Time = 1.44 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6492, 27, 6436, 6470, 6546, 6470, 2849, 2752, 6618, 7164}
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.
\(\displaystyle \int \frac {x (a+b \text {arctanh}(c x))^2}{c d x+d} \, dx\) |
\(\Big \downarrow \) 6492 |
\(\displaystyle \frac {\int (a+b \text {arctanh}(c x))^2dx}{c d}-\frac {\int \frac {(a+b \text {arctanh}(c x))^2}{d (c x+1)}dx}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a+b \text {arctanh}(c x))^2dx}{c d}-\frac {\int \frac {(a+b \text {arctanh}(c x))^2}{c x+1}dx}{c d}\) |
\(\Big \downarrow \) 6436 |
\(\displaystyle \frac {x (a+b \text {arctanh}(c x))^2-2 b c \int \frac {x (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx}{c d}-\frac {\int \frac {(a+b \text {arctanh}(c x))^2}{c x+1}dx}{c d}\) |
\(\Big \downarrow \) 6470 |
\(\displaystyle \frac {x (a+b \text {arctanh}(c x))^2-2 b c \int \frac {x (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx}{c d}-\frac {2 b \int \frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{c x+1}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c}}{c d}\) |
\(\Big \downarrow \) 6546 |
\(\displaystyle \frac {x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\int \frac {a+b \text {arctanh}(c x)}{1-c x}dx}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )}{c d}-\frac {2 b \int \frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{c x+1}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c}}{c d}\) |
\(\Big \downarrow \) 6470 |
\(\displaystyle \frac {x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}-b \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2}dx}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )}{c d}-\frac {2 b \int \frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{c x+1}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c}}{c d}\) |
\(\Big \downarrow \) 2849 |
\(\displaystyle \frac {x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\frac {b \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-\frac {2}{1-c x}}d\frac {1}{1-c x}}{c}+\frac {\log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )}{c d}-\frac {2 b \int \frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{c x+1}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c}}{c d}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c}}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )}{c d}-\frac {2 b \int \frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{c x+1}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c}}{c d}\) |
\(\Big \downarrow \) 6618 |
\(\displaystyle \frac {x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c}}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )}{c d}-\frac {2 b \left (\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{2 c}-\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx\right )-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c}}{c d}\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {x (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c}}{c}-\frac {(a+b \text {arctanh}(c x))^2}{2 b c^2}\right )}{c d}-\frac {2 b \left (\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {b \operatorname {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{4 c}\right )-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c}}{c d}\) |
(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))/(c*d) - (-(((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/c) + 2*b *(((a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(2*c) + (b*PolyLog[3, 1 - 2/(1 + c*x)])/(4*c)))/(c*d)
3.1.97.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[f/e Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x]) ^p, x], x] - Simp[d*(f/e) Int[(f*x)^(m - 1)*((a + b*ArcTanh[c*x])^p/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && GtQ[m, 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]
Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^ 2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x ] - Simp[b*(p/2) Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 - u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2/(1 + c*x))^2, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 3.01 (sec) , antiderivative size = 2602, normalized size of antiderivative = 15.13
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2602\) |
default | \(\text {Expression too large to display}\) | \(2602\) |
parts | \(\text {Expression too large to display}\) | \(2609\) |
1/c^2*(a^2/d*(c*x-ln(c*x+1))+b^2/d*(c*x*arctanh(c*x)^2-arctanh(c*x)*ln(1+( c*x+1)^2/(-c^2*x^2+1))-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))-arctanh(c*x)^2*ln(c*x+1)+arctanh(c*x)^2*ln(2)+2*a rctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+arctanh(c*x)^2-1/2*polylog(2, -(c*x+1)^2/(-c^2*x^2+1))-2/3*arctanh(c*x)^3-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))-1/2*poly log(3,-(c*x+1)^2/(-c^2*x^2+1))+arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2 +1))-1/4*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)))*(2*arctanh(c*x )^2-2*arctanh(c*x)*ln(1+(c*x+1)^2/(-c^2*x^2+1))-polylog(2,-(c*x+1)^2/(-c^2 *x^2+1)))+ln(2)*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+ln(2)*arct anh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-ln(2)*arctanh(c*x)*ln(1+(c*x+1 )^2/(-c^2*x^2+1))+ln(2)*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+ln(2)*dilog( 1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-1/2*ln(2)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1 ))-1/2*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))+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 )^(1/2)))+1/4*I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*(2*arctanh(c*x)^2-2*arc tanh(c*x)*ln(1+(c*x+1)^2/(-c^2*x^2+1))-polylog(2,-(c*x+1)^2/(-c^2*x^2+1...
\[ \int \frac {x (a+b \text {arctanh}(c x))^2}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x}{c d x + d} \,d x } \]
\[ \int \frac {x (a+b \text {arctanh}(c x))^2}{d+c d x} \, dx=\frac {\int \frac {a^{2} x}{c x + 1}\, dx + \int \frac {b^{2} x \operatorname {atanh}^{2}{\left (c x \right )}}{c x + 1}\, dx + \int \frac {2 a b x \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \]
(Integral(a**2*x/(c*x + 1), x) + Integral(b**2*x*atanh(c*x)**2/(c*x + 1), x) + Integral(2*a*b*x*atanh(c*x)/(c*x + 1), x))/d
\[ \int \frac {x (a+b \text {arctanh}(c x))^2}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x}{c d x + d} \,d x } \]
a^2*(x/(c*d) - log(c*x + 1)/(c^2*d)) + 1/4*(b^2*c*x - b^2*log(c*x + 1))*lo g(-c*x + 1)^2/(c^2*d) - integrate(-1/4*((b^2*c^2*x^2 - b^2*c*x)*log(c*x + 1)^2 + 4*(a*b*c^2*x^2 - a*b*c*x)*log(c*x + 1) - 2*((2*a*b*c^2 + b^2*c^2)*x ^2 - (2*a*b*c - b^2*c)*x + (b^2*c^2*x^2 - 2*b^2*c*x - b^2)*log(c*x + 1))*l og(-c*x + 1))/(c^3*d*x^2 - c*d), x)
\[ \int \frac {x (a+b \text {arctanh}(c x))^2}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x}{c d x + d} \,d x } \]
Timed out. \[ \int \frac {x (a+b \text {arctanh}(c x))^2}{d+c d x} \, dx=\int \frac {x\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{d+c\,d\,x} \,d x \]