Integrand size = 22, antiderivative size = 162 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 (d+c d x)} \, dx=\frac {c (a+b \text {arctanh}(c x))^2}{d}-\frac {(a+b \text {arctanh}(c x))^2}{d x}+\frac {2 b c (a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {c (a+b \text {arctanh}(c x))^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )}{d}+\frac {b c (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )}{d}+\frac {b^2 c \operatorname {PolyLog}\left (3,-1+\frac {2}{1+c x}\right )}{2 d} \]
c*(a+b*arctanh(c*x))^2/d-(a+b*arctanh(c*x))^2/d/x+2*b*c*(a+b*arctanh(c*x)) *ln(2-2/(c*x+1))/d-c*(a+b*arctanh(c*x))^2*ln(2-2/(c*x+1))/d-b^2*c*polylog( 2,-1+2/(c*x+1))/d+b*c*(a+b*arctanh(c*x))*polylog(2,-1+2/(c*x+1))/d+1/2*b^2 *c*polylog(3,-1+2/(c*x+1))/d
Result contains complex when optimal does not.
Time = 0.64 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 (d+c d x)} \, dx=\frac {-\frac {a^2}{x}-\frac {2 a b \text {arctanh}(c x) \left (1+c x \log \left (1-e^{-2 \text {arctanh}(c x)}\right )\right )}{x}-a^2 c \log (x)+a^2 c \log (1+c x)+a b c \left (2 \log (c x)-\log \left (1-c^2 x^2\right )\right )+a b c \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c x)}\right )+b^2 c \left (-\frac {i \pi ^3}{24}+\text {arctanh}(c x)^2-\frac {\text {arctanh}(c x)^2}{c x}+\frac {2}{3} \text {arctanh}(c x)^3+2 \text {arctanh}(c x) \log \left (1-e^{-2 \text {arctanh}(c x)}\right )-\text {arctanh}(c x)^2 \log \left (1-e^{2 \text {arctanh}(c x)}\right )-\operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c x)}\right )-\text {arctanh}(c x) \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}(c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,e^{2 \text {arctanh}(c x)}\right )\right )}{d} \]
(-(a^2/x) - (2*a*b*ArcTanh[c*x]*(1 + c*x*Log[1 - E^(-2*ArcTanh[c*x])]))/x - a^2*c*Log[x] + a^2*c*Log[1 + c*x] + a*b*c*(2*Log[c*x] - Log[1 - c^2*x^2] ) + a*b*c*PolyLog[2, E^(-2*ArcTanh[c*x])] + b^2*c*((-1/24*I)*Pi^3 + ArcTan h[c*x]^2 - ArcTanh[c*x]^2/(c*x) + (2*ArcTanh[c*x]^3)/3 + 2*ArcTanh[c*x]*Lo g[1 - E^(-2*ArcTanh[c*x])] - ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] - PolyLog[2, E^(-2*ArcTanh[c*x])] - ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x ])] + PolyLog[3, E^(2*ArcTanh[c*x])]/2))/d
Time = 1.44 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {6496, 27, 6452, 6494, 6550, 6494, 2897, 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.
\(\displaystyle \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 (c d x+d)} \, dx\) |
\(\Big \downarrow \) 6496 |
\(\displaystyle \frac {\int \frac {(a+b \text {arctanh}(c x))^2}{x^2}dx}{d}-c \int \frac {(a+b \text {arctanh}(c x))^2}{d x (c x+1)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(a+b \text {arctanh}(c x))^2}{x^2}dx}{d}-\frac {c \int \frac {(a+b \text {arctanh}(c x))^2}{x (c x+1)}dx}{d}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {2 b c \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx-\frac {(a+b \text {arctanh}(c x))^2}{x}}{d}-\frac {c \int \frac {(a+b \text {arctanh}(c x))^2}{x (c x+1)}dx}{d}\) |
\(\Big \downarrow \) 6494 |
\(\displaystyle \frac {2 b c \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx-\frac {(a+b \text {arctanh}(c x))^2}{x}}{d}-\frac {c \left (\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2-2 b c \int \frac {(a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx\right )}{d}\) |
\(\Big \downarrow \) 6550 |
\(\displaystyle \frac {2 b c \left (\int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx+\frac {(a+b \text {arctanh}(c x))^2}{2 b}\right )-\frac {(a+b \text {arctanh}(c x))^2}{x}}{d}-\frac {c \left (\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2-2 b c \int \frac {(a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx\right )}{d}\) |
\(\Big \downarrow \) 6494 |
\(\displaystyle \frac {2 b c \left (-b c \int \frac {\log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx+\frac {(a+b \text {arctanh}(c x))^2}{2 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))\right )-\frac {(a+b \text {arctanh}(c x))^2}{x}}{d}-\frac {c \left (\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2-2 b c \int \frac {(a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx\right )}{d}\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle \frac {2 b c \left (\frac {(a+b \text {arctanh}(c x))^2}{2 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right )\right )-\frac {(a+b \text {arctanh}(c x))^2}{x}}{d}-\frac {c \left (\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2-2 b c \int \frac {(a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx\right )}{d}\) |
\(\Big \downarrow \) 6618 |
\(\displaystyle \frac {2 b c \left (\frac {(a+b \text {arctanh}(c x))^2}{2 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right )\right )-\frac {(a+b \text {arctanh}(c x))^2}{x}}{d}-\frac {c \left (\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right ) (a+b \text {arctanh}(c x))}{2 c}-\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right )}{1-c^2 x^2}dx\right )\right )}{d}\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {2 b c \left (\frac {(a+b \text {arctanh}(c x))^2}{2 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right )\right )-\frac {(a+b \text {arctanh}(c x))^2}{x}}{d}-\frac {c \left (\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2-2 b c \left (\frac {\operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {b \operatorname {PolyLog}\left (3,\frac {2}{c x+1}-1\right )}{4 c}\right )\right )}{d}\) |
(-((a + b*ArcTanh[c*x])^2/x) + 2*b*c*((a + b*ArcTanh[c*x])^2/(2*b) + (a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - (b*PolyLog[2, -1 + 2/(1 + c*x)])/2) )/d - (c*((a + b*ArcTanh[c*x])^2*Log[2 - 2/(1 + c*x)] - 2*b*c*(((a + b*Arc Tanh[c*x])*PolyLog[2, -1 + 2/(1 + c*x)])/(2*c) + (b*PolyLog[3, -1 + 2/(1 + c*x)])/(4*c))))/d
3.1.100.3.1 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[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x _Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Simp[b*c*(p/d) Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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[1/d Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Simp[e/(d*f) 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] && LtQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*d*(p + 1)), x] + Simp[1/ d Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[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 = 2.73 (sec) , antiderivative size = 4168, normalized size of antiderivative = 25.73
method | result | size |
parts | \(\text {Expression too large to display}\) | \(4168\) |
derivativedivides | \(\text {Expression too large to display}\) | \(4170\) |
default | \(\text {Expression too large to display}\) | \(4170\) |
a^2/d*(ln(c*x+1)*c-1/x-c*ln(x))+b^2/d*c*(dilog(1+(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*arctanh(c*x)^2*ln((c*x+ 1)/(-c^2*x^2+1)^(1/2))-arctanh(c*x)^2-1/c/x*arctanh(c*x)^2-dilog((c*x+1)/( -c^2*x^2+1)^(1/2))+2/3*arctanh(c*x)^3+polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2 ))+2*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+polylog(2,(c*x+1)/(-c^2*x^2+1) ^(1/2))+2*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))-ln(c*x)*arctanh(c*x)^2+arc tanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)-arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2 *x^2+1)^(1/2))-2*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))-arcta nh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))-2*arctanh(c*x)*polylog(2,(c*x+1 )/(-c^2*x^2+1)^(1/2))+arctanh(c*x)*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*arct anh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+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)^2-arctanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2 +1)^(1/2))-arctanh(c*x)*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))-polylog(2,-(c*x+1 )/(-c^2*x^2+1)^(1/2))-polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2)))+ln(2)*dilog(( c*x+1)/(-c^2*x^2+1)^(1/2))-ln(2)*dilog(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+ln(2) *polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))+ln(2)*polylog(2,(c*x+1)/(-c^2*x^2+ 1)^(1/2))-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)/(1-(c*x+1)^2/(c^2*x^2-1)))^2*(arctanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+ 1)^(1/2))-dilog((c*x+1)/(-c^2*x^2+1)^(1/2))+dilog(1+(c*x+1)/(-c^2*x^2+1...
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 (d+c d x)} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )} x^{2}} \,d x } \]
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 (d+c d x)} \, dx=\frac {\int \frac {a^{2}}{c x^{3} + x^{2}}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c x^{3} + x^{2}}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{c x^{3} + x^{2}}\, dx}{d} \]
(Integral(a**2/(c*x**3 + x**2), x) + Integral(b**2*atanh(c*x)**2/(c*x**3 + x**2), x) + Integral(2*a*b*atanh(c*x)/(c*x**3 + x**2), x))/d
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 (d+c d x)} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )} x^{2}} \,d x } \]
a^2*(c*log(c*x + 1)/d - c*log(x)/d - 1/(d*x)) + 1/4*(b^2*c*x*log(c*x + 1) - b^2)*log(-c*x + 1)^2/(d*x) - integrate(-1/4*((b^2*c*x - b^2)*log(c*x + 1 )^2 + 4*(a*b*c*x - a*b)*log(c*x + 1) + 2*(b^2*c^2*x^2 + 2*a*b - (2*a*b*c - b^2*c)*x - (b^2*c^3*x^3 + b^2*c^2*x^2 + b^2*c*x - b^2)*log(c*x + 1))*log( -c*x + 1))/(c^2*d*x^4 - d*x^2), x)
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 (d+c d x)} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 (d+c d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{x^2\,\left (d+c\,d\,x\right )} \,d x \]