Integrand size = 20, antiderivative size = 114 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^3} \, dx=-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}-\frac {3}{a^2 \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}+\frac {5}{2 a^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}+\frac {5 \text {Shi}(2 \text {arctanh}(a x))}{16 a^2}+\frac {\text {Shi}(4 \text {arctanh}(a x))}{a^2}+\frac {9 \text {Shi}(6 \text {arctanh}(a x))}{16 a^2} \]
-1/2*x/a/(-a^2*x^2+1)^3/arctanh(a*x)^2-3/a^2/(-a^2*x^2+1)^3/arctanh(a*x)+5 /2/a^2/(-a^2*x^2+1)^2/arctanh(a*x)+5/16*Shi(2*arctanh(a*x))/a^2+Shi(4*arct anh(a*x))/a^2+9/16*Shi(6*arctanh(a*x))/a^2
Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.64 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^3} \, dx=\frac {\frac {8 \left (a x+\left (1+5 a^2 x^2\right ) \text {arctanh}(a x)\right )}{\left (-1+a^2 x^2\right )^3 \text {arctanh}(a x)^2}+5 \text {Shi}(2 \text {arctanh}(a x))+16 \text {Shi}(4 \text {arctanh}(a x))+9 \text {Shi}(6 \text {arctanh}(a x))}{16 a^2} \]
((8*(a*x + (1 + 5*a^2*x^2)*ArcTanh[a*x]))/((-1 + a^2*x^2)^3*ArcTanh[a*x]^2 ) + 5*SinhIntegral[2*ArcTanh[a*x]] + 16*SinhIntegral[4*ArcTanh[a*x]] + 9*S inhIntegral[6*ArcTanh[a*x]])/(16*a^2)
Time = 1.33 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6594, 6528, 6590, 6528, 6596, 5971, 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.
| \(\displaystyle \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^3} \, dx\) |
| \(\Big \downarrow \) 6594 |
| \(\displaystyle \frac {\int \frac {1}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^2}dx}{2 a}+\frac {5}{2} a \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^2}dx-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}\) |
| \(\Big \downarrow \) 6528 |
| \(\displaystyle \frac {5}{2} a \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^2}dx+\frac {6 a \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{2 a}-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}\) |
| \(\Big \downarrow \) 6590 |
| \(\displaystyle \frac {5}{2} a \left (\frac {\int \frac {1}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^2}dx}{a^2}-\frac {\int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}dx}{a^2}\right )+\frac {6 a \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{2 a}-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}\) |
| \(\Big \downarrow \) 6528 |
| \(\displaystyle \frac {6 a \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{2 a}+\frac {5}{2} a \left (\frac {6 a \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{a^2}-\frac {4 a \int \frac {x}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}dx-\frac {1}{a \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}}{a^2}\right )-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}\) |
| \(\Big \downarrow \) 6596 |
| \(\displaystyle \frac {\frac {6 \int \frac {a x}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{2 a}+\frac {5}{2} a \left (\frac {\frac {6 \int \frac {a x}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{a^2}-\frac {\frac {4 \int \frac {a x}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}}{a^2}\right )-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}\) |
| \(\Big \downarrow \) 5971 |
| \(\displaystyle \frac {\frac {6 \int \left (\frac {5 \sinh (2 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}+\frac {\sinh (4 \text {arctanh}(a x))}{8 \text {arctanh}(a x)}+\frac {\sinh (6 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{2 a}+\frac {5}{2} a \left (\frac {\frac {6 \int \left (\frac {5 \sinh (2 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}+\frac {\sinh (4 \text {arctanh}(a x))}{8 \text {arctanh}(a x)}+\frac {\sinh (6 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{a^2}-\frac {\frac {4 \int \left (\frac {\sinh (2 \text {arctanh}(a x))}{4 \text {arctanh}(a x)}+\frac {\sinh (4 \text {arctanh}(a x))}{8 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}}{a^2}\right )-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {5}{2} a \left (\frac {\frac {6 \left (\frac {5}{32} \text {Shi}(2 \text {arctanh}(a x))+\frac {1}{8} \text {Shi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Shi}(6 \text {arctanh}(a x))\right )}{a}-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{a^2}-\frac {\frac {4 \left (\frac {1}{4} \text {Shi}(2 \text {arctanh}(a x))+\frac {1}{8} \text {Shi}(4 \text {arctanh}(a x))\right )}{a}-\frac {1}{a \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}}{a^2}\right )+\frac {\frac {6 \left (\frac {5}{32} \text {Shi}(2 \text {arctanh}(a x))+\frac {1}{8} \text {Shi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Shi}(6 \text {arctanh}(a x))\right )}{a}-\frac {1}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}}{2 a}-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2}\) |
-1/2*x/(a*(1 - a^2*x^2)^3*ArcTanh[a*x]^2) + (5*a*(-((-(1/(a*(1 - a^2*x^2)^ 2*ArcTanh[a*x])) + (4*(SinhIntegral[2*ArcTanh[a*x]]/4 + SinhIntegral[4*Arc Tanh[a*x]]/8))/a)/a^2) + (-(1/(a*(1 - a^2*x^2)^3*ArcTanh[a*x])) + (6*((5*S inhIntegral[2*ArcTanh[a*x]])/32 + SinhIntegral[4*ArcTanh[a*x]]/8 + SinhInt egral[6*ArcTanh[a*x]]/32))/a)/a^2))/2 + (-(1/(a*(1 - a^2*x^2)^3*ArcTanh[a* x])) + (6*((5*SinhIntegral[2*ArcTanh[a*x]])/32 + SinhIntegral[4*ArcTanh[a* x]]/8 + SinhIntegral[6*ArcTanh[a*x]]/32))/a)/(2*a)
3.4.62.3.1 Defintions of rubi rules used
Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] & & IGtQ[p, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1))), x] + Simp[2*c*((q + 1)/(b*(p + 1))) Int[x*(d + e*x^2)^q*(a + b*A rcTanh[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && LtQ[p, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[1/e Int[x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*A rcTanh[c*x])^p, x], x] - Simp[d/e Int[x^(m - 2)*(d + e*x^2)^q*(a + b*ArcT anh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && In tegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m, 1] && NeQ[p, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_) ^2)^(q_), x_Symbol] :> Simp[x^m*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^( p + 1)/(b*c*d*(p + 1))), x] + (Simp[c*((m + 2*q + 2)/(b*(p + 1))) Int[x^( m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x] - Simp[m/(b*c*(p + 1)) Int[x^(m - 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x]) / ; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && LtQ[q, - 1] && LtQ[p, -1] && NeQ[m + 2*q + 2, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_) ^2)^(q_), x_Symbol] :> Simp[d^q/c^(m + 1) Subst[Int[(a + b*x)^p*(Sinh[x]^ m/Cosh[x]^(m + 2*(q + 1))), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (In tegerQ[q] || GtQ[d, 0])
Time = 0.54 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.06
| method | result | size |
| derivativedivides | \(\frac {-\frac {\sinh \left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{16 \operatorname {arctanh}\left (a x \right )^{2}}-\frac {\cosh \left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{4 \,\operatorname {arctanh}\left (a x \right )}+\operatorname {Shi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )-\frac {\sinh \left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{64 \operatorname {arctanh}\left (a x \right )^{2}}-\frac {3 \cosh \left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32 \,\operatorname {arctanh}\left (a x \right )}+\frac {9 \,\operatorname {Shi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{16}-\frac {5 \sinh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{64 \operatorname {arctanh}\left (a x \right )^{2}}-\frac {5 \cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32 \,\operatorname {arctanh}\left (a x \right )}+\frac {5 \,\operatorname {Shi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{16}}{a^{2}}\) | \(121\) |
| default | \(\frac {-\frac {\sinh \left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{16 \operatorname {arctanh}\left (a x \right )^{2}}-\frac {\cosh \left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{4 \,\operatorname {arctanh}\left (a x \right )}+\operatorname {Shi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )-\frac {\sinh \left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{64 \operatorname {arctanh}\left (a x \right )^{2}}-\frac {3 \cosh \left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32 \,\operatorname {arctanh}\left (a x \right )}+\frac {9 \,\operatorname {Shi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{16}-\frac {5 \sinh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{64 \operatorname {arctanh}\left (a x \right )^{2}}-\frac {5 \cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32 \,\operatorname {arctanh}\left (a x \right )}+\frac {5 \,\operatorname {Shi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{16}}{a^{2}}\) | \(121\) |
1/a^2*(-1/16/arctanh(a*x)^2*sinh(4*arctanh(a*x))-1/4/arctanh(a*x)*cosh(4*a rctanh(a*x))+Shi(4*arctanh(a*x))-1/64/arctanh(a*x)^2*sinh(6*arctanh(a*x))- 3/32/arctanh(a*x)*cosh(6*arctanh(a*x))+9/16*Shi(6*arctanh(a*x))-5/64*sinh( 2*arctanh(a*x))/arctanh(a*x)^2-5/32/arctanh(a*x)*cosh(2*arctanh(a*x))+5/16 *Shi(2*arctanh(a*x)))
Leaf count of result is larger than twice the leaf count of optimal. 447 vs. \(2 (103) = 206\).
Time = 0.27 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.92 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^3} \, dx=\frac {{\left (9 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - 9 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 16 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 16 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 64 \, a x + 32 \, {\left (5 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{32 \, {\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \]
1/32*((9*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a^3*x^3 + 3* a^2*x^2 + 3*a*x + 1)/(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)) - 9*(a^6*x^6 - 3*a ^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)/(a ^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)) + 16*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1 )*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) - 16*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) + 5*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a*x + 1)/(a*x - 1)) - 5*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a *x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x - 1))^2 + 64*a*x + 32*(5*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))/((a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)*lo g(-(a*x + 1)/(a*x - 1))^2)
\[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^3} \, dx=\int \frac {x}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname {atanh}^{3}{\left (a x \right )}}\, dx \]
\[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^3} \, dx=\int { \frac {x}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname {artanh}\left (a x\right )^{3}} \,d x } \]
(2*a*x + (5*a^2*x^2 + 1)*log(a*x + 1) - (5*a^2*x^2 + 1)*log(-a*x + 1))/((a ^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)*log(a*x + 1)^2 - 2*(a^8*x^6 - 3*a^6* x^4 + 3*a^4*x^2 - a^2)*log(a*x + 1)*log(-a*x + 1) + (a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)*log(-a*x + 1)^2) - integrate(-4*(5*a^2*x^3 + 4*x)/((a^8* x^8 - 4*a^6*x^6 + 6*a^4*x^4 - 4*a^2*x^2 + 1)*log(a*x + 1) - (a^8*x^8 - 4*a ^6*x^6 + 6*a^4*x^4 - 4*a^2*x^2 + 1)*log(-a*x + 1)), x)
\[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^3} \, dx=\int { \frac {x}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname {artanh}\left (a x\right )^{3}} \,d x } \]
Timed out. \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^3} \, dx=\int \frac {x}{{\mathrm {atanh}\left (a\,x\right )}^3\,{\left (a^2\,x^2-1\right )}^4} \,d x \]