Integrand size = 23, antiderivative size = 143 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\frac {(a+b \text {arctanh}(c+d x))^3}{d e^2}-\frac {(a+b \text {arctanh}(c+d x))^3}{d e^2 (c+d x)}+\frac {3 b (a+b \text {arctanh}(c+d x))^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^2 (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+c+d x}\right )}{2 d e^2} \]
(a+b*arctanh(d*x+c))^3/d/e^2-(a+b*arctanh(d*x+c))^3/d/e^2/(d*x+c)+3*b*(a+b *arctanh(d*x+c))^2*ln(2-2/(d*x+c+1))/d/e^2-3*b^2*(a+b*arctanh(d*x+c))*poly log(2,-1+2/(d*x+c+1))/d/e^2-3/2*b^3*polylog(3,-1+2/(d*x+c+1))/d/e^2
Time = 0.50 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\frac {-\frac {2 a^3}{c+d x}-\frac {6 a^2 b \text {arctanh}(c+d x)}{c+d x}+6 a^2 b \log (c+d x)-3 a^2 b \log \left (1-c^2-2 c d x-d^2 x^2\right )+6 a b^2 \left (\text {arctanh}(c+d x) \left (\left (1-\frac {1}{c+d x}\right ) \text {arctanh}(c+d x)+2 \log \left (1-e^{-2 \text {arctanh}(c+d x)}\right )\right )-\operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c+d x)}\right )\right )+2 b^3 \left (\text {arctanh}(c+d x)^2 \left (\left (1-\frac {1}{c+d x}\right ) \text {arctanh}(c+d x)+3 \log \left (1-e^{-2 \text {arctanh}(c+d x)}\right )\right )-3 \text {arctanh}(c+d x) \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c+d x)}\right )-\frac {3}{2} \operatorname {PolyLog}\left (3,e^{-2 \text {arctanh}(c+d x)}\right )\right )}{2 d e^2} \]
((-2*a^3)/(c + d*x) - (6*a^2*b*ArcTanh[c + d*x])/(c + d*x) + 6*a^2*b*Log[c + d*x] - 3*a^2*b*Log[1 - c^2 - 2*c*d*x - d^2*x^2] + 6*a*b^2*(ArcTanh[c + d*x]*((1 - (c + d*x)^(-1))*ArcTanh[c + d*x] + 2*Log[1 - E^(-2*ArcTanh[c + d*x])]) - PolyLog[2, E^(-2*ArcTanh[c + d*x])]) + 2*b^3*(ArcTanh[c + d*x]^2 *((1 - (c + d*x)^(-1))*ArcTanh[c + d*x] + 3*Log[1 - E^(-2*ArcTanh[c + d*x] )]) - 3*ArcTanh[c + d*x]*PolyLog[2, E^(-2*ArcTanh[c + d*x])] - (3*PolyLog[ 3, E^(-2*ArcTanh[c + d*x])])/2))/(2*d*e^2)
Time = 0.86 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6657, 27, 6452, 6550, 6494, 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.
| \(\displaystyle \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx\) |
| \(\Big \downarrow \) 6657 |
| \(\displaystyle \frac {\int \frac {(a+b \text {arctanh}(c+d x))^3}{e^2 (c+d x)^2}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(a+b \text {arctanh}(c+d x))^3}{(c+d x)^2}d(c+d x)}{d e^2}\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {3 b \int \frac {(a+b \text {arctanh}(c+d x))^2}{(c+d x) \left (1-(c+d x)^2\right )}d(c+d x)-\frac {(a+b \text {arctanh}(c+d x))^3}{c+d x}}{d e^2}\) |
| \(\Big \downarrow \) 6550 |
| \(\displaystyle \frac {3 b \left (\int \frac {(a+b \text {arctanh}(c+d x))^2}{(c+d x) (c+d x+1)}d(c+d x)+\frac {(a+b \text {arctanh}(c+d x))^3}{3 b}\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{c+d x}}{d e^2}\) |
| \(\Big \downarrow \) 6494 |
| \(\displaystyle \frac {3 b \left (-2 b \int \frac {(a+b \text {arctanh}(c+d x)) \log \left (2-\frac {2}{c+d x+1}\right )}{1-(c+d x)^2}d(c+d x)+\frac {(a+b \text {arctanh}(c+d x))^3}{3 b}+\log \left (2-\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))^2\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{c+d x}}{d e^2}\) |
| \(\Big \downarrow \) 6618 |
| \(\displaystyle \frac {3 b \left (-2 b \left (\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {2}{c+d x+1}-1\right ) (a+b \text {arctanh}(c+d x))-\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{c+d x+1}-1\right )}{1-(c+d x)^2}d(c+d x)\right )+\frac {(a+b \text {arctanh}(c+d x))^3}{3 b}+\log \left (2-\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))^2\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{c+d x}}{d e^2}\) |
| \(\Big \downarrow \) 7164 |
| \(\displaystyle \frac {3 b \left (-2 b \left (\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {2}{c+d x+1}-1\right ) (a+b \text {arctanh}(c+d x))+\frac {1}{4} b \operatorname {PolyLog}\left (3,\frac {2}{c+d x+1}-1\right )\right )+\frac {(a+b \text {arctanh}(c+d x))^3}{3 b}+\log \left (2-\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))^2\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{c+d x}}{d e^2}\) |
(-((a + b*ArcTanh[c + d*x])^3/(c + d*x)) + 3*b*((a + b*ArcTanh[c + d*x])^3 /(3*b) + (a + b*ArcTanh[c + d*x])^2*Log[2 - 2/(1 + c + d*x)] - 2*b*(((a + b*ArcTanh[c + d*x])*PolyLog[2, -1 + 2/(1 + c + d*x)])/2 + (b*PolyLog[3, -1 + 2/(1 + c + d*x)])/4)))/(d*e^2)
3.1.26.3.1 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[((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_.)/((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[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcTanh[x])^p, x ], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 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 = 0.46 (sec) , antiderivative size = 1466, normalized size of antiderivative = 10.25
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1466\) |
| default | \(\text {Expression too large to display}\) | \(1466\) |
| parts | \(\text {Expression too large to display}\) | \(1474\) |
1/d*(-a^3/e^2/(d*x+c)+b^3/e^2*(-1/(d*x+c)*arctanh(d*x+c)^3-3/2*arctanh(d*x +c)^2*ln(d*x+c-1)+3*ln(d*x+c)*arctanh(d*x+c)^2-3/2*arctanh(d*x+c)^2*ln(d*x +c+1)+3*arctanh(d*x+c)^2*ln((d*x+c+1)/(1-(d*x+c)^2)^(1/2))-arctanh(d*x+c)^ 3+3/4*(-2*I*Pi*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1)))*csgn(I*(-(d*x+c+1)^2/ ((d*x+c)^2-1)-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^2-I*Pi*csgn(I/(1-(d*x+c+1) ^2/((d*x+c)^2-1)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/(( d*x+c)^2-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))+I*Pi*csgn(I*(d*x+c+1)^2/((d*x+c )^2-1))^3+2*I*Pi*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^3-I*Pi*csgn(I*(d*x+ c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1-(d*x+c+1)^2/((d* x+c)^2-1)))^2+I*Pi*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1)))*csgn(I*(d*x+c+1)^ 2/((d*x+c)^2-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^2+2*I*Pi*csgn(I*(-(d*x+c+1) ^2/((d*x+c)^2-1)-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^3-2*I*Pi*csgn(I/(1-(d*x +c+1)^2/((d*x+c)^2-1)))^2+I*Pi*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1-(d*x+c+ 1)^2/((d*x+c)^2-1)))^3+I*Pi*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))^2*csgn(I *(d*x+c+1)^2/((d*x+c)^2-1))+2*I*Pi*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))*c sgn(I*(d*x+c+1)^2/((d*x+c)^2-1))^2+2*I*Pi-2*I*Pi*csgn(I*(-(d*x+c+1)^2/((d* x+c)^2-1)-1))*csgn(I*(-(d*x+c+1)^2/((d*x+c)^2-1)-1)/(1-(d*x+c+1)^2/((d*x+c )^2-1)))^2+2*I*Pi*csgn(I*(-(d*x+c+1)^2/((d*x+c)^2-1)-1))*csgn(I/(1-(d*x+c+ 1)^2/((d*x+c)^2-1)))*csgn(I*(-(d*x+c+1)^2/((d*x+c)^2-1)-1)/(1-(d*x+c+1)^2/ ((d*x+c)^2-1)))+4*ln(2))*arctanh(d*x+c)^2-3*arctanh(d*x+c)^2*ln((d*x+c+...
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}} \,d x } \]
integral((b^3*arctanh(d*x + c)^3 + 3*a*b^2*arctanh(d*x + c)^2 + 3*a^2*b*ar ctanh(d*x + c) + a^3)/(d^2*e^2*x^2 + 2*c*d*e^2*x + c^2*e^2), x)
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\frac {\int \frac {a^{3}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{3} \operatorname {atanh}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {atanh}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]
(Integral(a**3/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(b**3*atanh(c + d*x)**3/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(3*a*b**2*atanh(c + d*x )**2/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(3*a**2*b*atanh(c + d*x)/( c**2 + 2*c*d*x + d**2*x**2), x))/e**2
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}} \,d x } \]
-3/2*(d*(log(d*x + c + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2*e^2) + log(d*x + c - 1)/(d^2*e^2)) + 2*arctanh(d*x + c)/(d^2*e^2*x + c*d*e^2))*a^2*b - a^3 /(d^2*e^2*x + c*d*e^2) - 1/8*((b^3*d*x + b^3*(c - 1))*log(-d*x - c + 1)^3 + 3*(2*a*b^2 + (b^3*d*x + b^3*(c + 1))*log(d*x + c + 1))*log(-d*x - c + 1) ^2)/(d^2*e^2*x + c*d*e^2) - integrate(-1/8*((b^3*d*x + b^3*(c - 1))*log(d* x + c + 1)^3 + 6*(a*b^2*d*x + a*b^2*(c - 1))*log(d*x + c + 1)^2 + 3*(4*a*b ^2*d*x + 4*a*b^2*c - (b^3*d*x + b^3*(c - 1))*log(d*x + c + 1)^2 + 2*(b^3*d ^2*x^2 + (c^2 + c)*b^3 - 2*a*b^2*(c - 1) + ((2*c*d + d)*b^3 - 2*a*b^2*d)*x )*log(d*x + c + 1))*log(-d*x - c + 1))/(d^3*e^2*x^3 + c^3*e^2 - c^2*e^2 + (3*c*d^2*e^2 - d^2*e^2)*x^2 + (3*c^2*d*e^2 - 2*c*d*e^2)*x), x)
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]