Integrand size = 20, antiderivative size = 43 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\frac {5 \text {Shi}(2 \text {arctanh}(a x))}{32 a^2}+\frac {\text {Shi}(4 \text {arctanh}(a x))}{8 a^2}+\frac {\text {Shi}(6 \text {arctanh}(a x))}{32 a^2} \] Output:
5/32*Shi(2*arctanh(a*x))/a^2+1/8*Shi(4*arctanh(a*x))/a^2+1/32*Shi(6*arctan h(a*x))/a^2
Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\frac {5 \text {Shi}(2 \text {arctanh}(a x))}{32 a^2}+\frac {\text {Shi}(4 \text {arctanh}(a x))}{8 a^2}+\frac {\text {Shi}(6 \text {arctanh}(a x))}{32 a^2} \] Input:
Integrate[x/((1 - a^2*x^2)^4*ArcTanh[a*x]),x]
Output:
(5*SinhIntegral[2*ArcTanh[a*x]])/(32*a^2) + SinhIntegral[4*ArcTanh[a*x]]/( 8*a^2) + SinhIntegral[6*ArcTanh[a*x]]/(32*a^2)
Time = 0.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {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.
\(\displaystyle \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx\) |
\(\Big \downarrow \) 6596 |
\(\displaystyle \frac {\int \frac {a x}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \frac {\int \left (\frac {5 \sinh (2 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}+\frac {\sinh (4 \text {arctanh}(a x))}{8 \text {arctanh}(a x)}+\frac {\sinh (6 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {5}{32} \text {Shi}(2 \text {arctanh}(a x))+\frac {1}{8} \text {Shi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Shi}(6 \text {arctanh}(a x))}{a^2}\) |
Input:
Int[x/((1 - a^2*x^2)^4*ArcTanh[a*x]),x]
Output:
((5*SinhIntegral[2*ArcTanh[a*x]])/32 + SinhIntegral[4*ArcTanh[a*x]]/8 + Si nhIntegral[6*ArcTanh[a*x]]/32)/a^2
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_.)*(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 = 0.72 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {Shi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{8}+\frac {\operatorname {Shi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32}+\frac {5 \,\operatorname {Shi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32}}{a^{2}}\) | \(33\) |
default | \(\frac {\frac {\operatorname {Shi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{8}+\frac {\operatorname {Shi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32}+\frac {5 \,\operatorname {Shi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32}}{a^{2}}\) | \(33\) |
Input:
int(x/(-a^2*x^2+1)^4/arctanh(a*x),x,method=_RETURNVERBOSE)
Output:
1/a^2*(1/8*Shi(4*arctanh(a*x))+1/32*Shi(6*arctanh(a*x))+5/32*Shi(2*arctanh (a*x)))
Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (37) = 74\).
Time = 0.08 (sec) , antiderivative size = 200, normalized size of antiderivative = 4.65 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\frac {\operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 4 \, \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 4 \, \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \, \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - 5 \, \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )}{64 \, a^{2}} \] Input:
integrate(x/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="fricas")
Output:
1/64*(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)) - 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)) + 4*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) - 4*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1) ) + 5*log_integral(-(a*x + 1)/(a*x - 1)) - 5*log_integral(-(a*x - 1)/(a*x + 1)))/a^2
\[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\int \frac {x}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname {atanh}{\left (a x \right )}}\, dx \] Input:
integrate(x/(-a**2*x**2+1)**4/atanh(a*x),x)
Output:
Integral(x/((a*x - 1)**4*(a*x + 1)**4*atanh(a*x)), x)
\[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\int { \frac {x}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname {artanh}\left (a x\right )} \,d x } \] Input:
integrate(x/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="maxima")
Output:
integrate(x/((a^2*x^2 - 1)^4*arctanh(a*x)), x)
\[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\int { \frac {x}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname {artanh}\left (a x\right )} \,d x } \] Input:
integrate(x/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="giac")
Output:
integrate(x/((a^2*x^2 - 1)^4*arctanh(a*x)), x)
Timed out. \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\int \frac {x}{\mathrm {atanh}\left (a\,x\right )\,{\left (a^2\,x^2-1\right )}^4} \,d x \] Input:
int(x/(atanh(a*x)*(a^2*x^2 - 1)^4),x)
Output:
int(x/(atanh(a*x)*(a^2*x^2 - 1)^4), x)
\[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\int \frac {x}{\mathit {atanh} \left (a x \right ) a^{8} x^{8}-4 \mathit {atanh} \left (a x \right ) a^{6} x^{6}+6 \mathit {atanh} \left (a x \right ) a^{4} x^{4}-4 \mathit {atanh} \left (a x \right ) a^{2} x^{2}+\mathit {atanh} \left (a x \right )}d x \] Input:
int(x/(-a^2*x^2+1)^4/atanh(a*x),x)
Output:
int(x/(atanh(a*x)*a**8*x**8 - 4*atanh(a*x)*a**6*x**6 + 6*atanh(a*x)*a**4*x **4 - 4*atanh(a*x)*a**2*x**2 + atanh(a*x)),x)