Integrand size = 18, antiderivative size = 158 \[ \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {2 b e^2 x}{3 c}+\frac {b \left (3 c d^2-e^2\right ) \arctan \left (\sqrt {c} x\right )}{3 c^{3/2}}-\frac {b \left (3 c d^2+e^2\right ) \text {arctanh}\left (\sqrt {c} x\right )}{3 c^{3/2}}+\frac {(d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{3 e}+\frac {b d \left (c d^2+3 e^2\right ) \log \left (1-c x^2\right )}{6 c e}-\frac {b d \left (c d^2-3 e^2\right ) \log \left (1+c x^2\right )}{6 c e} \] Output:
2/3*b*e^2*x/c+1/3*b*(3*c*d^2-e^2)*arctan(c^(1/2)*x)/c^(3/2)-1/3*b*(3*c*d^2 +e^2)*arctanh(c^(1/2)*x)/c^(3/2)+1/3*(e*x+d)^3*(a+b*arctanh(c*x^2))/e+1/6* b*d*(c*d^2+3*e^2)*ln(-c*x^2+1)/c/e-1/6*b*d*(c*d^2-3*e^2)*ln(c*x^2+1)/c/e
Time = 0.10 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.08 \[ \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{6} \left (6 a d^2 x+\frac {4 b e^2 x}{c}+6 a d e x^2+2 a e^2 x^3+\frac {2 b \left (3 c d^2-e^2\right ) \arctan \left (\sqrt {c} x\right )}{c^{3/2}}+2 b x \left (3 d^2+3 d e x+e^2 x^2\right ) \text {arctanh}\left (c x^2\right )+\frac {b \left (3 c d^2+e^2\right ) \log \left (1-\sqrt {c} x\right )}{c^{3/2}}-\frac {b \left (3 c d^2+e^2\right ) \log \left (1+\sqrt {c} x\right )}{c^{3/2}}+\frac {3 b d e \log \left (1-c^2 x^4\right )}{c}\right ) \] Input:
Integrate[(d + e*x)^2*(a + b*ArcTanh[c*x^2]),x]
Output:
(6*a*d^2*x + (4*b*e^2*x)/c + 6*a*d*e*x^2 + 2*a*e^2*x^3 + (2*b*(3*c*d^2 - e ^2)*ArcTan[Sqrt[c]*x])/c^(3/2) + 2*b*x*(3*d^2 + 3*d*e*x + e^2*x^2)*ArcTanh [c*x^2] + (b*(3*c*d^2 + e^2)*Log[1 - Sqrt[c]*x])/c^(3/2) - (b*(3*c*d^2 + e ^2)*Log[1 + Sqrt[c]*x])/c^(3/2) + (3*b*d*e*Log[1 - c^2*x^4])/c)/6
Time = 0.42 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6486, 2370, 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.
| \(\displaystyle \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx\) |
| \(\Big \downarrow \) 6486 |
| \(\displaystyle \frac {(d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{3 e}-\frac {2 b c \int \frac {x (d+e x)^3}{1-c^2 x^4}dx}{3 e}\) |
| \(\Big \downarrow \) 2370 |
| \(\displaystyle \frac {(d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{3 e}-\frac {2 b c \int \left (\frac {\left (x^2 e^3+3 d^2 e\right ) x^2}{1-c^2 x^4}+\frac {\left (d^3+3 e^2 x^2 d\right ) x}{1-c^2 x^4}\right )dx}{3 e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{3 e}-\frac {2 b c \left (-\frac {e \arctan \left (\sqrt {c} x\right ) \left (3 c d^2-e^2\right )}{2 c^{5/2}}+\frac {e \text {arctanh}\left (\sqrt {c} x\right ) \left (3 c d^2+e^2\right )}{2 c^{5/2}}+\frac {d^3 \text {arctanh}\left (c x^2\right )}{2 c}-\frac {3 d e^2 \log \left (1-c^2 x^4\right )}{4 c^2}-\frac {e^3 x}{c^2}\right )}{3 e}\) |
Input:
Int[(d + e*x)^2*(a + b*ArcTanh[c*x^2]),x]
Output:
((d + e*x)^3*(a + b*ArcTanh[c*x^2]))/(3*e) - (2*b*c*(-((e^3*x)/c^2) - (e*( 3*c*d^2 - e^2)*ArcTan[Sqrt[c]*x])/(2*c^(5/2)) + (e*(3*c*d^2 + e^2)*ArcTanh [Sqrt[c]*x])/(2*c^(5/2)) + (d^3*ArcTanh[c*x^2])/(2*c) - (3*d*e^2*Log[1 - c ^2*x^4])/(4*c^2)))/(3*e)
Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[ {v = Sum[(c*x)^(m + ii)*((Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2) )/(c^ii*(a + b*x^n))), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{ a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_ Symbol] :> Simp[(d + e*x)^(m + 1)*((a + b*ArcTanh[c*x^n])/(e*(m + 1))), x] - Simp[b*c*(n/(e*(m + 1))) Int[x^(n - 1)*((d + e*x)^(m + 1)/(1 - c^2*x^(2 *n))), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[m, -1]
Time = 0.37 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.26
| method | result | size |
| default | \(\frac {a \left (e x +d \right )^{3}}{3 e}+b \left (\frac {e^{2} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{3}}{3}+e \,\operatorname {arctanh}\left (c \,x^{2}\right ) x^{2} d +\operatorname {arctanh}\left (c \,x^{2}\right ) x \,d^{2}+\frac {\operatorname {arctanh}\left (c \,x^{2}\right ) d^{3}}{3 e}-\frac {2 c \left (-\frac {e^{3} x}{c^{2}}+\frac {\frac {\left (-c^{2} d^{3}-3 c d \,e^{2}\right ) \ln \left (c \,x^{2}-1\right )}{2 c}-\frac {\left (-3 d^{2} e c -e^{3}\right ) \operatorname {arctanh}\left (\sqrt {c}\, x \right )}{\sqrt {c}}}{2 c^{2}}+\frac {\frac {\left (c^{2} d^{3}-3 c d \,e^{2}\right ) \ln \left (c \,x^{2}+1\right )}{2 c}+\frac {\left (-3 d^{2} e c +e^{3}\right ) \arctan \left (\sqrt {c}\, x \right )}{\sqrt {c}}}{2 c^{2}}\right )}{3 e}\right )\) | \(199\) |
| parts | \(\frac {a \left (e x +d \right )^{3}}{3 e}+b \left (\frac {e^{2} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{3}}{3}+e \,\operatorname {arctanh}\left (c \,x^{2}\right ) x^{2} d +\operatorname {arctanh}\left (c \,x^{2}\right ) x \,d^{2}+\frac {\operatorname {arctanh}\left (c \,x^{2}\right ) d^{3}}{3 e}-\frac {2 c \left (-\frac {e^{3} x}{c^{2}}+\frac {\frac {\left (-c^{2} d^{3}-3 c d \,e^{2}\right ) \ln \left (c \,x^{2}-1\right )}{2 c}-\frac {\left (-3 d^{2} e c -e^{3}\right ) \operatorname {arctanh}\left (\sqrt {c}\, x \right )}{\sqrt {c}}}{2 c^{2}}+\frac {\frac {\left (c^{2} d^{3}-3 c d \,e^{2}\right ) \ln \left (c \,x^{2}+1\right )}{2 c}+\frac {\left (-3 d^{2} e c +e^{3}\right ) \arctan \left (\sqrt {c}\, x \right )}{\sqrt {c}}}{2 c^{2}}\right )}{3 e}\right )\) | \(199\) |
| risch | \(\text {Expression too large to display}\) | \(3662\) |
Input:
int((e*x+d)^2*(a+b*arctanh(c*x^2)),x,method=_RETURNVERBOSE)
Output:
1/3*a*(e*x+d)^3/e+b*(1/3*e^2*arctanh(c*x^2)*x^3+e*arctanh(c*x^2)*x^2*d+arc tanh(c*x^2)*x*d^2+1/3/e*arctanh(c*x^2)*d^3-2/3/e*c*(-e^3/c^2*x+1/2/c^2*(1/ 2*(-c^2*d^3-3*c*d*e^2)/c*ln(c*x^2-1)-(-3*c*d^2*e-e^3)/c^(1/2)*arctanh(c^(1 /2)*x))+1/2/c^2*(1/2*(c^2*d^3-3*c*d*e^2)/c*ln(c*x^2+1)+(-3*c*d^2*e+e^3)/c^ (1/2)*arctan(c^(1/2)*x))))
Time = 0.10 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.54 \[ \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\left [\frac {2 \, a c^{2} e^{2} x^{3} + 6 \, a c^{2} d e x^{2} + 3 \, b c d e \log \left (c x^{2} + 1\right ) + 3 \, b c d e \log \left (c x^{2} - 1\right ) + 2 \, {\left (3 \, b c d^{2} - b e^{2}\right )} \sqrt {c} \arctan \left (\sqrt {c} x\right ) + {\left (3 \, b c d^{2} + b e^{2}\right )} \sqrt {c} \log \left (\frac {c x^{2} - 2 \, \sqrt {c} x + 1}{c x^{2} - 1}\right ) + 2 \, {\left (3 \, a c^{2} d^{2} + 2 \, b c e^{2}\right )} x + {\left (b c^{2} e^{2} x^{3} + 3 \, b c^{2} d e x^{2} + 3 \, b c^{2} d^{2} x\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{6 \, c^{2}}, \frac {2 \, a c^{2} e^{2} x^{3} + 6 \, a c^{2} d e x^{2} + 3 \, b c d e \log \left (c x^{2} + 1\right ) + 3 \, b c d e \log \left (c x^{2} - 1\right ) + 2 \, {\left (3 \, b c d^{2} + b e^{2}\right )} \sqrt {-c} \arctan \left (\sqrt {-c} x\right ) + {\left (3 \, b c d^{2} - b e^{2}\right )} \sqrt {-c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-c} x - 1}{c x^{2} + 1}\right ) + 2 \, {\left (3 \, a c^{2} d^{2} + 2 \, b c e^{2}\right )} x + {\left (b c^{2} e^{2} x^{3} + 3 \, b c^{2} d e x^{2} + 3 \, b c^{2} d^{2} x\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{6 \, c^{2}}\right ] \] Input:
integrate((e*x+d)^2*(a+b*arctanh(c*x^2)),x, algorithm="fricas")
Output:
[1/6*(2*a*c^2*e^2*x^3 + 6*a*c^2*d*e*x^2 + 3*b*c*d*e*log(c*x^2 + 1) + 3*b*c *d*e*log(c*x^2 - 1) + 2*(3*b*c*d^2 - b*e^2)*sqrt(c)*arctan(sqrt(c)*x) + (3 *b*c*d^2 + b*e^2)*sqrt(c)*log((c*x^2 - 2*sqrt(c)*x + 1)/(c*x^2 - 1)) + 2*( 3*a*c^2*d^2 + 2*b*c*e^2)*x + (b*c^2*e^2*x^3 + 3*b*c^2*d*e*x^2 + 3*b*c^2*d^ 2*x)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^2, 1/6*(2*a*c^2*e^2*x^3 + 6*a*c^2*d* e*x^2 + 3*b*c*d*e*log(c*x^2 + 1) + 3*b*c*d*e*log(c*x^2 - 1) + 2*(3*b*c*d^2 + b*e^2)*sqrt(-c)*arctan(sqrt(-c)*x) + (3*b*c*d^2 - b*e^2)*sqrt(-c)*log(( c*x^2 + 2*sqrt(-c)*x - 1)/(c*x^2 + 1)) + 2*(3*a*c^2*d^2 + 2*b*c*e^2)*x + ( b*c^2*e^2*x^3 + 3*b*c^2*d*e*x^2 + 3*b*c^2*d^2*x)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^2]
Leaf count of result is larger than twice the leaf count of optimal. 3465 vs. \(2 (139) = 278\).
Time = 6.06 (sec) , antiderivative size = 3465, normalized size of antiderivative = 21.93 \[ \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\text {Too large to display} \] Input:
integrate((e*x+d)**2*(a+b*atanh(c*x**2)),x)
Output:
Piecewise((12*a*c**2*d**2*x*sqrt(-1/c)/(12*c**3*(-1/c)**(3/2) - 12*c**3*(1 /c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) + 12*a*c**2*d**2*x*sq rt(1/c)/(12*c**3*(-1/c)**(3/2) - 12*c**3*(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) + 12*a*c**2*d*e*x**2*sqrt(-1/c)/(12*c**3*(-1/c)**(3/ 2) - 12*c**3*(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) + 12*a *c**2*d*e*x**2*sqrt(1/c)/(12*c**3*(-1/c)**(3/2) - 12*c**3*(1/c)**(3/2) + 2 4*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) + 4*a*c**2*e**2*x**3*sqrt(-1/c)/(12 *c**3*(-1/c)**(3/2) - 12*c**3*(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2* sqrt(1/c)) + 4*a*c**2*e**2*x**3*sqrt(1/c)/(12*c**3*(-1/c)**(3/2) - 12*c**3 *(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) - 3*b*c**3*d**2*(- 1/c)**(3/2)*sqrt(1/c)*log(x - sqrt(-1/c))/(12*c**3*(-1/c)**(3/2) - 12*c**3 *(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) - 12*b*c**3*d**2*( -1/c)**(3/2)*sqrt(1/c)*log(x + sqrt(-1/c))/(12*c**3*(-1/c)**(3/2) - 12*c** 3*(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) + 3*b*c**3*d**2*( -1/c)**(3/2)*sqrt(1/c)*log(x - sqrt(1/c))/(12*c**3*(-1/c)**(3/2) - 12*c**3 *(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) + 3*b*c**3*d**2*sq rt(-1/c)*(1/c)**(3/2)*log(x - sqrt(-1/c))/(12*c**3*(-1/c)**(3/2) - 12*c**3 *(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) - 12*b*c**3*d**2*s qrt(-1/c)*(1/c)**(3/2)*log(x + sqrt(-1/c))/(12*c**3*(-1/c)**(3/2) - 12*c** 3*(1/c)**(3/2) + 24*c**2*sqrt(-1/c) + 24*c**2*sqrt(1/c)) - 3*b*c**3*d**...
Time = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.08 \[ \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{3} \, a e^{2} x^{3} + a d e x^{2} + \frac {1}{2} \, {\left (c {\left (\frac {2 \, \arctan \left (\sqrt {c} x\right )}{c^{\frac {3}{2}}} + \frac {\log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right )}{c^{\frac {3}{2}}}\right )} + 2 \, x \operatorname {artanh}\left (c x^{2}\right )\right )} b d^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {4 \, x}{c^{2}} - \frac {2 \, \arctan \left (\sqrt {c} x\right )}{c^{\frac {5}{2}}} + \frac {\log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )}\right )} b e^{2} + a d^{2} x + \frac {{\left (2 \, c x^{2} \operatorname {artanh}\left (c x^{2}\right ) + \log \left (-c^{2} x^{4} + 1\right )\right )} b d e}{2 \, c} \] Input:
integrate((e*x+d)^2*(a+b*arctanh(c*x^2)),x, algorithm="maxima")
Output:
1/3*a*e^2*x^3 + a*d*e*x^2 + 1/2*(c*(2*arctan(sqrt(c)*x)/c^(3/2) + log((c*x - sqrt(c))/(c*x + sqrt(c)))/c^(3/2)) + 2*x*arctanh(c*x^2))*b*d^2 + 1/6*(2 *x^3*arctanh(c*x^2) + c*(4*x/c^2 - 2*arctan(sqrt(c)*x)/c^(5/2) + log((c*x - sqrt(c))/(c*x + sqrt(c)))/c^(5/2)))*b*e^2 + a*d^2*x + 1/2*(2*c*x^2*arcta nh(c*x^2) + log(-c^2*x^4 + 1))*b*d*e/c
Time = 1.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.08 \[ \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{3} \, a e^{2} x^{3} + a d e x^{2} + \frac {b d e \log \left (c x^{2} + 1\right )}{2 \, c} + \frac {b d e \log \left (c x^{2} - 1\right )}{2 \, c} + \frac {1}{6} \, {\left (b e^{2} x^{3} + 3 \, b d e x^{2} + 3 \, b d^{2} x\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + \frac {{\left (3 \, a c d^{2} + 2 \, b e^{2}\right )} x}{3 \, c} + \frac {{\left (3 \, b c d^{2} - b e^{2}\right )} \arctan \left (\sqrt {c} x\right )}{3 \, c^{\frac {3}{2}}} + \frac {{\left (3 \, b c d^{2} + b e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {-c}}\right )}{3 \, \sqrt {-c} c} \] Input:
integrate((e*x+d)^2*(a+b*arctanh(c*x^2)),x, algorithm="giac")
Output:
1/3*a*e^2*x^3 + a*d*e*x^2 + 1/2*b*d*e*log(c*x^2 + 1)/c + 1/2*b*d*e*log(c*x ^2 - 1)/c + 1/6*(b*e^2*x^3 + 3*b*d*e*x^2 + 3*b*d^2*x)*log(-(c*x^2 + 1)/(c* x^2 - 1)) + 1/3*(3*a*c*d^2 + 2*b*e^2)*x/c + 1/3*(3*b*c*d^2 - b*e^2)*arctan (sqrt(c)*x)/c^(3/2) + 1/3*(3*b*c*d^2 + b*e^2)*arctan(c*x/sqrt(-c))/(sqrt(- c)*c)
Time = 3.94 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.96 \[ \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\ln \left (c\,x^2+1\right )\,\left (\frac {b\,d^2\,x}{2}+\frac {b\,d\,e\,x^2}{2}+\frac {b\,e^2\,x^3}{6}\right )-\ln \left (1-c\,x^2\right )\,\left (\frac {b\,d^2\,x}{2}+\frac {b\,d\,e\,x^2}{2}+\frac {b\,e^2\,x^3}{6}\right )+\frac {x\,\left (3\,a\,c^2\,d^2+2\,b\,c\,e^2\right )}{3\,c^2}+\frac {a\,e^2\,x^3}{3}-\frac {\ln \left (c+x\,\sqrt {c^3}\right )\,\left (b\,e^2\,\sqrt {c^3}+3\,b\,c\,d^2\,\sqrt {c^3}-3\,b\,c^2\,d\,e\right )}{6\,c^3}+\frac {\ln \left (c-x\,\sqrt {c^3}\right )\,\left (b\,e^2\,\sqrt {c^3}+3\,b\,c\,d^2\,\sqrt {c^3}+3\,b\,c^2\,d\,e\right )}{6\,c^3}+\frac {\ln \left (c+x\,\sqrt {-c^3}\right )\,\left (b\,e^2\,\sqrt {-c^3}+3\,b\,c^2\,d\,e-3\,b\,c\,d^2\,\sqrt {-c^3}\right )}{6\,c^3}+\frac {\ln \left (c-x\,\sqrt {-c^3}\right )\,\left (3\,b\,c^2\,d\,e-b\,e^2\,\sqrt {-c^3}+3\,b\,c\,d^2\,\sqrt {-c^3}\right )}{6\,c^3}+a\,d\,e\,x^2 \] Input:
int((a + b*atanh(c*x^2))*(d + e*x)^2,x)
Output:
log(c*x^2 + 1)*((b*e^2*x^3)/6 + (b*d^2*x)/2 + (b*d*e*x^2)/2) - log(1 - c*x ^2)*((b*e^2*x^3)/6 + (b*d^2*x)/2 + (b*d*e*x^2)/2) + (x*(3*a*c^2*d^2 + 2*b* c*e^2))/(3*c^2) + (a*e^2*x^3)/3 - (log(c + x*(c^3)^(1/2))*(b*e^2*(c^3)^(1/ 2) + 3*b*c*d^2*(c^3)^(1/2) - 3*b*c^2*d*e))/(6*c^3) + (log(c - x*(c^3)^(1/2 ))*(b*e^2*(c^3)^(1/2) + 3*b*c*d^2*(c^3)^(1/2) + 3*b*c^2*d*e))/(6*c^3) + (l og(c + x*(-c^3)^(1/2))*(b*e^2*(-c^3)^(1/2) + 3*b*c^2*d*e - 3*b*c*d^2*(-c^3 )^(1/2)))/(6*c^3) + (log(c - x*(-c^3)^(1/2))*(3*b*c^2*d*e - b*e^2*(-c^3)^( 1/2) + 3*b*c*d^2*(-c^3)^(1/2)))/(6*c^3) + a*d*e*x^2
Time = 0.19 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.58 \[ \int (d+e x)^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {6 \sqrt {c}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}}\right ) b c \,d^{2}-2 \sqrt {c}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}}\right ) b \,e^{2}+6 \sqrt {c}\, \mathit {atanh} \left (c \,x^{2}\right ) b c \,d^{2}+2 \sqrt {c}\, \mathit {atanh} \left (c \,x^{2}\right ) b \,e^{2}+6 \mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} d^{2} x +6 \mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} d e \,x^{2}+2 \mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} e^{2} x^{3}-6 \mathit {atanh} \left (c \,x^{2}\right ) b c d e +6 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}\, x -1\right ) b c \,d^{2}+2 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}\, x -1\right ) b \,e^{2}-3 \sqrt {c}\, \mathrm {log}\left (c \,x^{2}+1\right ) b c \,d^{2}-\sqrt {c}\, \mathrm {log}\left (c \,x^{2}+1\right ) b \,e^{2}+6 \,\mathrm {log}\left (c \,x^{2}+1\right ) b c d e +6 a \,c^{2} d^{2} x +6 a \,c^{2} d e \,x^{2}+2 a \,c^{2} e^{2} x^{3}+4 b c \,e^{2} x}{6 c^{2}} \] Input:
int((e*x+d)^2*(a+b*atanh(c*x^2)),x)
Output:
(6*sqrt(c)*atan((c*x)/sqrt(c))*b*c*d**2 - 2*sqrt(c)*atan((c*x)/sqrt(c))*b* e**2 + 6*sqrt(c)*atanh(c*x**2)*b*c*d**2 + 2*sqrt(c)*atanh(c*x**2)*b*e**2 + 6*atanh(c*x**2)*b*c**2*d**2*x + 6*atanh(c*x**2)*b*c**2*d*e*x**2 + 2*atanh (c*x**2)*b*c**2*e**2*x**3 - 6*atanh(c*x**2)*b*c*d*e + 6*sqrt(c)*log(sqrt(c )*x - 1)*b*c*d**2 + 2*sqrt(c)*log(sqrt(c)*x - 1)*b*e**2 - 3*sqrt(c)*log(c* x**2 + 1)*b*c*d**2 - sqrt(c)*log(c*x**2 + 1)*b*e**2 + 6*log(c*x**2 + 1)*b* c*d*e + 6*a*c**2*d**2*x + 6*a*c**2*d*e*x**2 + 2*a*c**2*e**2*x**3 + 4*b*c*e **2*x)/(6*c**2)