Integrand size = 14, antiderivative size = 110 \[ \int \left (c+d x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {d \left (10 a^2 c+3 d\right ) x^2}{30 a^3}+\frac {d^2 x^4}{20 a}+c^2 x \text {arctanh}(a x)+\frac {2}{3} c d x^3 \text {arctanh}(a x)+\frac {1}{5} d^2 x^5 \text {arctanh}(a x)+\frac {\left (15 a^4 c^2+10 a^2 c d+3 d^2\right ) \log \left (1-a^2 x^2\right )}{30 a^5} \] Output:
1/30*d*(10*a^2*c+3*d)*x^2/a^3+1/20*d^2*x^4/a+c^2*x*arctanh(a*x)+2/3*c*d*x^ 3*arctanh(a*x)+1/5*d^2*x^5*arctanh(a*x)+1/30*(15*a^4*c^2+10*a^2*c*d+3*d^2) *ln(-a^2*x^2+1)/a^5
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.89 \[ \int \left (c+d x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {a^2 d x^2 \left (6 d+a^2 \left (20 c+3 d x^2\right )\right )+4 a^5 x \left (15 c^2+10 c d x^2+3 d^2 x^4\right ) \text {arctanh}(a x)+\left (30 a^4 c^2+20 a^2 c d+6 d^2\right ) \log \left (1-a^2 x^2\right )}{60 a^5} \] Input:
Integrate[(c + d*x^2)^2*ArcTanh[a*x],x]
Output:
(a^2*d*x^2*(6*d + a^2*(20*c + 3*d*x^2)) + 4*a^5*x*(15*c^2 + 10*c*d*x^2 + 3 *d^2*x^4)*ArcTanh[a*x] + (30*a^4*c^2 + 20*a^2*c*d + 6*d^2)*Log[1 - a^2*x^2 ])/(60*a^5)
Time = 0.40 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6538, 27, 1576, 1140, 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.
| \(\displaystyle \int \text {arctanh}(a x) \left (c+d x^2\right )^2 \, dx\) |
| \(\Big \downarrow \) 6538 |
| \(\displaystyle -a \int \frac {x \left (3 d^2 x^4+10 c d x^2+15 c^2\right )}{15 \left (1-a^2 x^2\right )}dx+c^2 x \text {arctanh}(a x)+\frac {2}{3} c d x^3 \text {arctanh}(a x)+\frac {1}{5} d^2 x^5 \text {arctanh}(a x)\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{15} a \int \frac {x \left (3 d^2 x^4+10 c d x^2+15 c^2\right )}{1-a^2 x^2}dx+c^2 x \text {arctanh}(a x)+\frac {2}{3} c d x^3 \text {arctanh}(a x)+\frac {1}{5} d^2 x^5 \text {arctanh}(a x)\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle -\frac {1}{30} a \int \frac {3 d^2 x^4+10 c d x^2+15 c^2}{1-a^2 x^2}dx^2+c^2 x \text {arctanh}(a x)+\frac {2}{3} c d x^3 \text {arctanh}(a x)+\frac {1}{5} d^2 x^5 \text {arctanh}(a x)\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle -\frac {1}{30} a \int \left (-\frac {3 d^2 x^2}{a^2}-\frac {d \left (10 c a^2+3 d\right )}{a^4}+\frac {-15 c^2 a^4-10 c d a^2-3 d^2}{a^4 \left (a^2 x^2-1\right )}\right )dx^2+c^2 x \text {arctanh}(a x)+\frac {2}{3} c d x^3 \text {arctanh}(a x)+\frac {1}{5} d^2 x^5 \text {arctanh}(a x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{30} a \left (-\frac {3 d^2 x^4}{2 a^2}-\frac {d x^2 \left (10 a^2 c+3 d\right )}{a^4}-\frac {\left (15 a^4 c^2+10 a^2 c d+3 d^2\right ) \log \left (1-a^2 x^2\right )}{a^6}\right )+c^2 x \text {arctanh}(a x)+\frac {2}{3} c d x^3 \text {arctanh}(a x)+\frac {1}{5} d^2 x^5 \text {arctanh}(a x)\) |
Input:
Int[(c + d*x^2)^2*ArcTanh[a*x],x]
Output:
c^2*x*ArcTanh[a*x] + (2*c*d*x^3*ArcTanh[a*x])/3 + (d^2*x^5*ArcTanh[a*x])/5 - (a*(-((d*(10*a^2*c + 3*d)*x^2)/a^4) - (3*d^2*x^4)/(2*a^2) - ((15*a^4*c^ 2 + 10*a^2*c*d + 3*d^2)*Log[1 - a^2*x^2])/a^6))/30
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[1/2 Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] , x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Sym bol] :> With[{u = IntHide[(d + e*x^2)^q, x]}, Simp[(a + b*ArcTanh[c*x]) u , x] - Simp[b*c Int[SimplifyIntegrand[u/(1 - c^2*x^2), x], x], x]] /; Fre eQ[{a, b, c, d, e}, x] && (IntegerQ[q] || ILtQ[q + 1/2, 0])
Time = 0.53 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95
| method | result | size |
| parts | \(\frac {d^{2} x^{5} \operatorname {arctanh}\left (a x \right )}{5}+\frac {2 c d \,x^{3} \operatorname {arctanh}\left (a x \right )}{3}+c^{2} x \,\operatorname {arctanh}\left (a x \right )-\frac {a \left (-\frac {d \left (\frac {3}{2} d \,x^{4} a^{2}+10 a^{2} c \,x^{2}+3 d \,x^{2}\right )}{2 a^{4}}+\frac {\left (-15 a^{4} c^{2}-10 a^{2} c d -3 d^{2}\right ) \ln \left (a^{2} x^{2}-1\right )}{2 a^{6}}\right )}{15}\) | \(105\) |
| derivativedivides | \(\frac {\operatorname {arctanh}\left (a x \right ) c^{2} a x +\frac {2 a \,\operatorname {arctanh}\left (a x \right ) c d \,x^{3}}{3}+\frac {a \,\operatorname {arctanh}\left (a x \right ) d^{2} x^{5}}{5}-\frac {-5 a^{4} c d \,x^{2}-\frac {3 a^{4} d^{2} x^{4}}{4}-\frac {3 d^{2} x^{2} a^{2}}{2}-\frac {\left (15 a^{4} c^{2}+10 a^{2} c d +3 d^{2}\right ) \ln \left (a x -1\right )}{2}+\frac {\left (-15 a^{4} c^{2}-10 a^{2} c d -3 d^{2}\right ) \ln \left (a x +1\right )}{2}}{15 a^{4}}}{a}\) | \(137\) |
| default | \(\frac {\operatorname {arctanh}\left (a x \right ) c^{2} a x +\frac {2 a \,\operatorname {arctanh}\left (a x \right ) c d \,x^{3}}{3}+\frac {a \,\operatorname {arctanh}\left (a x \right ) d^{2} x^{5}}{5}-\frac {-5 a^{4} c d \,x^{2}-\frac {3 a^{4} d^{2} x^{4}}{4}-\frac {3 d^{2} x^{2} a^{2}}{2}-\frac {\left (15 a^{4} c^{2}+10 a^{2} c d +3 d^{2}\right ) \ln \left (a x -1\right )}{2}+\frac {\left (-15 a^{4} c^{2}-10 a^{2} c d -3 d^{2}\right ) \ln \left (a x +1\right )}{2}}{15 a^{4}}}{a}\) | \(137\) |
| parallelrisch | \(-\frac {-12 x^{5} \operatorname {arctanh}\left (a x \right ) a^{5} d^{2}-40 x^{3} \operatorname {arctanh}\left (a x \right ) a^{5} c d -3 a^{4} d^{2} x^{4}-60 c^{2} \operatorname {arctanh}\left (a x \right ) x \,a^{5}-20 a^{4} c d \,x^{2}-60 \ln \left (a x -1\right ) a^{4} c^{2}-60 \,\operatorname {arctanh}\left (a x \right ) a^{4} c^{2}-6 d^{2} x^{2} a^{2}-40 \ln \left (a x -1\right ) a^{2} c d -40 \,\operatorname {arctanh}\left (a x \right ) a^{2} c d -20 a^{2} c d -12 \ln \left (a x -1\right ) d^{2}-12 \,\operatorname {arctanh}\left (a x \right ) d^{2}-6 d^{2}}{60 a^{5}}\) | \(163\) |
| risch | \(\left (\frac {1}{10} d^{2} x^{5}+\frac {1}{3} c d \,x^{3}+\frac {1}{2} c^{2} x \right ) \ln \left (a x +1\right )-\frac {d^{2} x^{5} \ln \left (-a x +1\right )}{10}-\frac {c d \,x^{3} \ln \left (-a x +1\right )}{3}+\frac {d^{2} x^{4}}{20 a}-\frac {c^{2} x \ln \left (-a x +1\right )}{2}+\frac {c d \,x^{2}}{3 a}+\frac {\ln \left (a^{2} x^{2}-1\right ) c^{2}}{2 a}+\frac {5 c^{2}}{9 a}+\frac {d^{2} x^{2}}{10 a^{3}}+\frac {\ln \left (a^{2} x^{2}-1\right ) c d}{3 a^{3}}+\frac {c d}{3 a^{3}}+\frac {\ln \left (a^{2} x^{2}-1\right ) d^{2}}{10 a^{5}}+\frac {d^{2}}{20 a^{5}}\) | \(181\) |
| meijerg | \(-\frac {d^{2} \left (-\frac {a^{2} x^{2} \left (3 a^{2} x^{2}+6\right )}{15}+\frac {2 a^{6} x^{6} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{5}\right )}{4 a^{5}}+\frac {d c \left (\frac {2 a^{2} x^{2}}{3}-\frac {2 a^{4} x^{4} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}\right )}{2 a^{3}}-\frac {c^{2} \left (\frac {2 a^{2} x^{2} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{4 a}\) | \(231\) |
Input:
int((d*x^2+c)^2*arctanh(a*x),x,method=_RETURNVERBOSE)
Output:
1/5*d^2*x^5*arctanh(a*x)+2/3*c*d*x^3*arctanh(a*x)+c^2*x*arctanh(a*x)-1/15* a*(-1/2*d/a^4*(3/2*d*x^4*a^2+10*a^2*c*x^2+3*d*x^2)+1/2*(-15*a^4*c^2-10*a^2 *c*d-3*d^2)/a^6*ln(a^2*x^2-1))
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08 \[ \int \left (c+d x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {3 \, a^{4} d^{2} x^{4} + 2 \, {\left (10 \, a^{4} c d + 3 \, a^{2} d^{2}\right )} x^{2} + 2 \, {\left (15 \, a^{4} c^{2} + 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (a^{2} x^{2} - 1\right ) + 2 \, {\left (3 \, a^{5} d^{2} x^{5} + 10 \, a^{5} c d x^{3} + 15 \, a^{5} c^{2} x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{60 \, a^{5}} \] Input:
integrate((d*x^2+c)^2*arctanh(a*x),x, algorithm="fricas")
Output:
1/60*(3*a^4*d^2*x^4 + 2*(10*a^4*c*d + 3*a^2*d^2)*x^2 + 2*(15*a^4*c^2 + 10* a^2*c*d + 3*d^2)*log(a^2*x^2 - 1) + 2*(3*a^5*d^2*x^5 + 10*a^5*c*d*x^3 + 15 *a^5*c^2*x)*log(-(a*x + 1)/(a*x - 1)))/a^5
Time = 0.39 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.41 \[ \int \left (c+d x^2\right )^2 \text {arctanh}(a x) \, dx=\begin {cases} c^{2} x \operatorname {atanh}{\left (a x \right )} + \frac {2 c d x^{3} \operatorname {atanh}{\left (a x \right )}}{3} + \frac {d^{2} x^{5} \operatorname {atanh}{\left (a x \right )}}{5} + \frac {c^{2} \log {\left (x - \frac {1}{a} \right )}}{a} + \frac {c^{2} \operatorname {atanh}{\left (a x \right )}}{a} + \frac {c d x^{2}}{3 a} + \frac {d^{2} x^{4}}{20 a} + \frac {2 c d \log {\left (x - \frac {1}{a} \right )}}{3 a^{3}} + \frac {2 c d \operatorname {atanh}{\left (a x \right )}}{3 a^{3}} + \frac {d^{2} x^{2}}{10 a^{3}} + \frac {d^{2} \log {\left (x - \frac {1}{a} \right )}}{5 a^{5}} + \frac {d^{2} \operatorname {atanh}{\left (a x \right )}}{5 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate((d*x**2+c)**2*atanh(a*x),x)
Output:
Piecewise((c**2*x*atanh(a*x) + 2*c*d*x**3*atanh(a*x)/3 + d**2*x**5*atanh(a *x)/5 + c**2*log(x - 1/a)/a + c**2*atanh(a*x)/a + c*d*x**2/(3*a) + d**2*x* *4/(20*a) + 2*c*d*log(x - 1/a)/(3*a**3) + 2*c*d*atanh(a*x)/(3*a**3) + d**2 *x**2/(10*a**3) + d**2*log(x - 1/a)/(5*a**5) + d**2*atanh(a*x)/(5*a**5), N e(a, 0)), (0, True))
Time = 0.03 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.19 \[ \int \left (c+d x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {1}{60} \, a {\left (\frac {3 \, a^{2} d^{2} x^{4} + 2 \, {\left (10 \, a^{2} c d + 3 \, d^{2}\right )} x^{2}}{a^{4}} + \frac {2 \, {\left (15 \, a^{4} c^{2} + 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (a x + 1\right )}{a^{6}} + \frac {2 \, {\left (15 \, a^{4} c^{2} + 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (a x - 1\right )}{a^{6}}\right )} + \frac {1}{15} \, {\left (3 \, d^{2} x^{5} + 10 \, c d x^{3} + 15 \, c^{2} x\right )} \operatorname {artanh}\left (a x\right ) \] Input:
integrate((d*x^2+c)^2*arctanh(a*x),x, algorithm="maxima")
Output:
1/60*a*((3*a^2*d^2*x^4 + 2*(10*a^2*c*d + 3*d^2)*x^2)/a^4 + 2*(15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*log(a*x + 1)/a^6 + 2*(15*a^4*c^2 + 10*a^2*c*d + 3*d^2 )*log(a*x - 1)/a^6) + 1/15*(3*d^2*x^5 + 10*c*d*x^3 + 15*c^2*x)*arctanh(a*x )
Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (100) = 200\).
Time = 0.14 (sec) , antiderivative size = 527, normalized size of antiderivative = 4.79 \[ \int \left (c+d x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {1}{15} \, a {\left (\frac {{\left (15 \, a^{4} c^{2} + 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{6}} - \frac {{\left (15 \, a^{4} c^{2} + 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{6}} + \frac {4 \, {\left (\frac {{\left (5 \, a^{2} c d + 3 \, d^{2}\right )} {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {{\left (10 \, a^{2} c d + 3 \, d^{2}\right )} {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {{\left (5 \, a^{2} c d + 3 \, d^{2}\right )} {\left (a x + 1\right )}}{a x - 1}\right )}}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{4}} + \frac {{\left (\frac {15 \, {\left (a x + 1\right )}^{4} a^{4} c^{2}}{{\left (a x - 1\right )}^{4}} - \frac {60 \, {\left (a x + 1\right )}^{3} a^{4} c^{2}}{{\left (a x - 1\right )}^{3}} + \frac {90 \, {\left (a x + 1\right )}^{2} a^{4} c^{2}}{{\left (a x - 1\right )}^{2}} - \frac {60 \, {\left (a x + 1\right )} a^{4} c^{2}}{a x - 1} + 15 \, a^{4} c^{2} + \frac {30 \, {\left (a x + 1\right )}^{4} a^{2} c d}{{\left (a x - 1\right )}^{4}} - \frac {60 \, {\left (a x + 1\right )}^{3} a^{2} c d}{{\left (a x - 1\right )}^{3}} + \frac {40 \, {\left (a x + 1\right )}^{2} a^{2} c d}{{\left (a x - 1\right )}^{2}} - \frac {20 \, {\left (a x + 1\right )} a^{2} c d}{a x - 1} + 10 \, a^{2} c d + \frac {15 \, {\left (a x + 1\right )}^{4} d^{2}}{{\left (a x - 1\right )}^{4}} + \frac {30 \, {\left (a x + 1\right )}^{2} d^{2}}{{\left (a x - 1\right )}^{2}} + 3 \, d^{2}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{5}}\right )} \] Input:
integrate((d*x^2+c)^2*arctanh(a*x),x, algorithm="giac")
Output:
1/15*a*((15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*log(abs(-a*x - 1)/abs(a*x - 1))/ a^6 - (15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*log(abs(-(a*x + 1)/(a*x - 1) + 1)) /a^6 + 4*((5*a^2*c*d + 3*d^2)*(a*x + 1)^3/(a*x - 1)^3 - (10*a^2*c*d + 3*d^ 2)*(a*x + 1)^2/(a*x - 1)^2 + (5*a^2*c*d + 3*d^2)*(a*x + 1)/(a*x - 1))/(a^6 *((a*x + 1)/(a*x - 1) - 1)^4) + (15*(a*x + 1)^4*a^4*c^2/(a*x - 1)^4 - 60*( a*x + 1)^3*a^4*c^2/(a*x - 1)^3 + 90*(a*x + 1)^2*a^4*c^2/(a*x - 1)^2 - 60*( a*x + 1)*a^4*c^2/(a*x - 1) + 15*a^4*c^2 + 30*(a*x + 1)^4*a^2*c*d/(a*x - 1) ^4 - 60*(a*x + 1)^3*a^2*c*d/(a*x - 1)^3 + 40*(a*x + 1)^2*a^2*c*d/(a*x - 1) ^2 - 20*(a*x + 1)*a^2*c*d/(a*x - 1) + 10*a^2*c*d + 15*(a*x + 1)^4*d^2/(a*x - 1)^4 + 30*(a*x + 1)^2*d^2/(a*x - 1)^2 + 3*d^2)*log(-(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x - 1) + 1)/(( a*x + 1)*a/(a*x - 1) - a) - 1))/(a^6*((a*x + 1)/(a*x - 1) - 1)^5))
Time = 3.75 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.07 \[ \int \left (c+d x^2\right )^2 \text {arctanh}(a x) \, dx=c^2\,x\,\mathrm {atanh}\left (a\,x\right )+\frac {d^2\,x^5\,\mathrm {atanh}\left (a\,x\right )}{5}+\frac {c^2\,\ln \left (a^2\,x^2-1\right )}{2\,a}+\frac {d^2\,\ln \left (a^2\,x^2-1\right )}{10\,a^5}+\frac {d^2\,x^4}{20\,a}+\frac {d^2\,x^2}{10\,a^3}+\frac {2\,c\,d\,x^3\,\mathrm {atanh}\left (a\,x\right )}{3}+\frac {c\,d\,\ln \left (a^2\,x^2-1\right )}{3\,a^3}+\frac {c\,d\,x^2}{3\,a} \] Input:
int(atanh(a*x)*(c + d*x^2)^2,x)
Output:
c^2*x*atanh(a*x) + (d^2*x^5*atanh(a*x))/5 + (c^2*log(a^2*x^2 - 1))/(2*a) + (d^2*log(a^2*x^2 - 1))/(10*a^5) + (d^2*x^4)/(20*a) + (d^2*x^2)/(10*a^3) + (2*c*d*x^3*atanh(a*x))/3 + (c*d*log(a^2*x^2 - 1))/(3*a^3) + (c*d*x^2)/(3* a)
Time = 0.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.47 \[ \int \left (c+d x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {60 \mathit {atanh} \left (a x \right ) a^{5} c^{2} x +40 \mathit {atanh} \left (a x \right ) a^{5} c d \,x^{3}+12 \mathit {atanh} \left (a x \right ) a^{5} d^{2} x^{5}+60 \mathit {atanh} \left (a x \right ) a^{4} c^{2}+40 \mathit {atanh} \left (a x \right ) a^{2} c d +12 \mathit {atanh} \left (a x \right ) d^{2}+60 \,\mathrm {log}\left (a^{2} x -a \right ) a^{4} c^{2}+40 \,\mathrm {log}\left (a^{2} x -a \right ) a^{2} c d +12 \,\mathrm {log}\left (a^{2} x -a \right ) d^{2}+20 a^{4} c d \,x^{2}+3 a^{4} d^{2} x^{4}+6 a^{2} d^{2} x^{2}}{60 a^{5}} \] Input:
int((d*x^2+c)^2*atanh(a*x),x)
Output:
(60*atanh(a*x)*a**5*c**2*x + 40*atanh(a*x)*a**5*c*d*x**3 + 12*atanh(a*x)*a **5*d**2*x**5 + 60*atanh(a*x)*a**4*c**2 + 40*atanh(a*x)*a**2*c*d + 12*atan h(a*x)*d**2 + 60*log(a**2*x - a)*a**4*c**2 + 40*log(a**2*x - a)*a**2*c*d + 12*log(a**2*x - a)*d**2 + 20*a**4*c*d*x**2 + 3*a**4*d**2*x**4 + 6*a**2*d* *2*x**2)/(60*a**5)