Integrand size = 18, antiderivative size = 154 \[ \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {2 b d e^2 x}{c}+\frac {b e^3 x^2}{4 c}+\frac {b d \left (c d^2-e^2\right ) \arctan \left (\sqrt {c} x\right )}{c^{3/2}}-\frac {b d \left (c d^2+e^2\right ) \text {arctanh}\left (\sqrt {c} x\right )}{c^{3/2}}-\frac {b \left (c^2 d^4+e^4\right ) \text {arctanh}\left (c x^2\right )}{4 c^2 e}+\frac {(d+e x)^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{4 e}+\frac {3 b d^2 e \log \left (1-c^2 x^4\right )}{4 c} \] Output:
2*b*d*e^2*x/c+1/4*b*e^3*x^2/c+b*d*(c*d^2-e^2)*arctan(c^(1/2)*x)/c^(3/2)-b* d*(c*d^2+e^2)*arctanh(c^(1/2)*x)/c^(3/2)-1/4*b*(c^2*d^4+e^4)*arctanh(c*x^2 )/c^2/e+1/4*(e*x+d)^4*(a+b*arctanh(c*x^2))/e+3/4*b*d^2*e*ln(-c^2*x^4+1)/c
Time = 0.14 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.65 \[ \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{8} \left (\frac {8 d \left (a c d^2+2 b e^2\right ) x}{c}+\frac {2 e \left (6 a c d^2+b e^2\right ) x^2}{c}+8 a d e^2 x^3+2 a e^3 x^4+\frac {8 b d \left (c d^2-e^2\right ) \arctan \left (\sqrt {c} x\right )}{c^{3/2}}+2 b x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right ) \text {arctanh}\left (c x^2\right )+\frac {b \left (4 c^{3/2} d^3+4 \sqrt {c} d e^2+e^3\right ) \log \left (1-\sqrt {c} x\right )}{c^2}+\frac {b \left (-4 c^2 d^3-4 c d e^2+\sqrt {c} e^3\right ) \log \left (1+\sqrt {c} x\right )}{c^{5/2}}-\frac {b e^3 \log \left (1+c x^2\right )}{c^2}+\frac {6 b d^2 e \log \left (1-c^2 x^4\right )}{c}\right ) \] Input:
Integrate[(d + e*x)^3*(a + b*ArcTanh[c*x^2]),x]
Output:
((8*d*(a*c*d^2 + 2*b*e^2)*x)/c + (2*e*(6*a*c*d^2 + b*e^2)*x^2)/c + 8*a*d*e ^2*x^3 + 2*a*e^3*x^4 + (8*b*d*(c*d^2 - e^2)*ArcTan[Sqrt[c]*x])/c^(3/2) + 2 *b*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3)*ArcTanh[c*x^2] + (b*(4*c^ (3/2)*d^3 + 4*Sqrt[c]*d*e^2 + e^3)*Log[1 - Sqrt[c]*x])/c^2 + (b*(-4*c^2*d^ 3 - 4*c*d*e^2 + Sqrt[c]*e^3)*Log[1 + Sqrt[c]*x])/c^(5/2) - (b*e^3*Log[1 + c*x^2])/c^2 + (6*b*d^2*e*Log[1 - c^2*x^4])/c)/8
Time = 0.53 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6486, 2372, 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.
| \(\displaystyle \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx\) |
| \(\Big \downarrow \) 6486 |
| \(\displaystyle \frac {(d+e x)^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{4 e}-\frac {b c \int \frac {x (d+e x)^4}{1-c^2 x^4}dx}{2 e}\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \frac {(d+e x)^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{4 e}-\frac {b c \int \left (\frac {\left (4 e d^3+4 e^3 x^2 d\right ) x^2}{1-c^2 x^4}+\frac {\left (d^4+6 e^2 x^2 d^2+e^4 x^4\right ) x}{1-c^2 x^4}\right )dx}{2 e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(d+e x)^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{4 e}-\frac {b c \left (-\frac {2 d e \arctan \left (\sqrt {c} x\right ) \left (c d^2-e^2\right )}{c^{5/2}}+\frac {2 d e \text {arctanh}\left (\sqrt {c} x\right ) \left (c d^2+e^2\right )}{c^{5/2}}+\frac {\text {arctanh}\left (c x^2\right ) \left (c^2 d^4+e^4\right )}{2 c^3}-\frac {3 d^2 e^2 \log \left (1-c^2 x^4\right )}{2 c^2}-\frac {4 d e^3 x}{c^2}-\frac {e^4 x^2}{2 c^2}\right )}{2 e}\) |
Input:
Int[(d + e*x)^3*(a + b*ArcTanh[c*x^2]),x]
Output:
((d + e*x)^4*(a + b*ArcTanh[c*x^2]))/(4*e) - (b*c*((-4*d*e^3*x)/c^2 - (e^4 *x^2)/(2*c^2) - (2*d*e*(c*d^2 - e^2)*ArcTan[Sqrt[c]*x])/c^(5/2) + (2*d*e*( c*d^2 + e^2)*ArcTanh[Sqrt[c]*x])/c^(5/2) + ((c^2*d^4 + e^4)*ArcTanh[c*x^2] )/(2*c^3) - (3*d^2*e^2*Log[1 - c^2*x^4])/(2*c^2)))/(2*e)
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Mo dule[{q = Expon[Pq, x], j, k}, Int[Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0 ] && !PolyQ[Pq, x^(n/2)]
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 = 0.50 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.57
| method | result | size |
| default | \(\frac {a \left (e x +d \right )^{4}}{4 e}+b \left (\frac {e^{3} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{4}}{4}+e^{2} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{3} d +\frac {3 e \,\operatorname {arctanh}\left (c \,x^{2}\right ) x^{2} d^{2}}{2}+\operatorname {arctanh}\left (c \,x^{2}\right ) x \,d^{3}+\frac {\operatorname {arctanh}\left (c \,x^{2}\right ) d^{4}}{4 e}-\frac {c \left (-\frac {e^{3} \left (\frac {1}{2} e \,x^{2}+4 d x \right )}{c^{2}}+\frac {\frac {\left (-c^{2} d^{4}-6 d^{2} e^{2} c -e^{4}\right ) \ln \left (c \,x^{2}-1\right )}{2 c}-\frac {\left (-4 d^{3} e c -4 d \,e^{3}\right ) \operatorname {arctanh}\left (\sqrt {c}\, x \right )}{\sqrt {c}}}{2 c^{2}}+\frac {\frac {\left (c^{2} d^{4}-6 d^{2} e^{2} c +e^{4}\right ) \ln \left (c \,x^{2}+1\right )}{2 c}+\frac {\left (-4 d^{3} e c +4 d \,e^{3}\right ) \arctan \left (\sqrt {c}\, x \right )}{\sqrt {c}}}{2 c^{2}}\right )}{2 e}\right )\) | \(242\) |
| parts | \(\frac {a \left (e x +d \right )^{4}}{4 e}+b \left (\frac {e^{3} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{4}}{4}+e^{2} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{3} d +\frac {3 e \,\operatorname {arctanh}\left (c \,x^{2}\right ) x^{2} d^{2}}{2}+\operatorname {arctanh}\left (c \,x^{2}\right ) x \,d^{3}+\frac {\operatorname {arctanh}\left (c \,x^{2}\right ) d^{4}}{4 e}-\frac {c \left (-\frac {e^{3} \left (\frac {1}{2} e \,x^{2}+4 d x \right )}{c^{2}}+\frac {\frac {\left (-c^{2} d^{4}-6 d^{2} e^{2} c -e^{4}\right ) \ln \left (c \,x^{2}-1\right )}{2 c}-\frac {\left (-4 d^{3} e c -4 d \,e^{3}\right ) \operatorname {arctanh}\left (\sqrt {c}\, x \right )}{\sqrt {c}}}{2 c^{2}}+\frac {\frac {\left (c^{2} d^{4}-6 d^{2} e^{2} c +e^{4}\right ) \ln \left (c \,x^{2}+1\right )}{2 c}+\frac {\left (-4 d^{3} e c +4 d \,e^{3}\right ) \arctan \left (\sqrt {c}\, x \right )}{\sqrt {c}}}{2 c^{2}}\right )}{2 e}\right )\) | \(242\) |
| risch | \(\text {Expression too large to display}\) | \(4746\) |
Input:
int((e*x+d)^3*(a+b*arctanh(c*x^2)),x,method=_RETURNVERBOSE)
Output:
1/4*a*(e*x+d)^4/e+b*(1/4*e^3*arctanh(c*x^2)*x^4+e^2*arctanh(c*x^2)*x^3*d+3 /2*e*arctanh(c*x^2)*x^2*d^2+arctanh(c*x^2)*x*d^3+1/4/e*arctanh(c*x^2)*d^4- 1/2/e*c*(-e^3/c^2*(1/2*e*x^2+4*d*x)+1/2/c^2*(1/2*(-c^2*d^4-6*c*d^2*e^2-e^4 )/c*ln(c*x^2-1)-(-4*c*d^3*e-4*d*e^3)/c^(1/2)*arctanh(c^(1/2)*x))+1/2/c^2*( 1/2*(c^2*d^4-6*c*d^2*e^2+e^4)/c*ln(c*x^2+1)+(-4*c*d^3*e+4*d*e^3)/c^(1/2)*a rctan(c^(1/2)*x))))
Time = 0.15 (sec) , antiderivative size = 519, normalized size of antiderivative = 3.37 \[ \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\left [\frac {2 \, a c^{2} e^{3} x^{4} + 8 \, a c^{2} d e^{2} x^{3} + 2 \, {\left (6 \, a c^{2} d^{2} e + b c e^{3}\right )} x^{2} + 8 \, {\left (b c d^{3} - b d e^{2}\right )} \sqrt {c} \arctan \left (\sqrt {c} x\right ) + 4 \, {\left (b c d^{3} + b d e^{2}\right )} \sqrt {c} \log \left (\frac {c x^{2} - 2 \, \sqrt {c} x + 1}{c x^{2} - 1}\right ) + 8 \, {\left (a c^{2} d^{3} + 2 \, b c d e^{2}\right )} x + {\left (6 \, b c d^{2} e - b e^{3}\right )} \log \left (c x^{2} + 1\right ) + {\left (6 \, b c d^{2} e + b e^{3}\right )} \log \left (c x^{2} - 1\right ) + {\left (b c^{2} e^{3} x^{4} + 4 \, b c^{2} d e^{2} x^{3} + 6 \, b c^{2} d^{2} e x^{2} + 4 \, b c^{2} d^{3} x\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{8 \, c^{2}}, \frac {2 \, a c^{2} e^{3} x^{4} + 8 \, a c^{2} d e^{2} x^{3} + 2 \, {\left (6 \, a c^{2} d^{2} e + b c e^{3}\right )} x^{2} + 8 \, {\left (b c d^{3} + b d e^{2}\right )} \sqrt {-c} \arctan \left (\sqrt {-c} x\right ) + 4 \, {\left (b c d^{3} - b d e^{2}\right )} \sqrt {-c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-c} x - 1}{c x^{2} + 1}\right ) + 8 \, {\left (a c^{2} d^{3} + 2 \, b c d e^{2}\right )} x + {\left (6 \, b c d^{2} e - b e^{3}\right )} \log \left (c x^{2} + 1\right ) + {\left (6 \, b c d^{2} e + b e^{3}\right )} \log \left (c x^{2} - 1\right ) + {\left (b c^{2} e^{3} x^{4} + 4 \, b c^{2} d e^{2} x^{3} + 6 \, b c^{2} d^{2} e x^{2} + 4 \, b c^{2} d^{3} x\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{8 \, c^{2}}\right ] \] Input:
integrate((e*x+d)^3*(a+b*arctanh(c*x^2)),x, algorithm="fricas")
Output:
[1/8*(2*a*c^2*e^3*x^4 + 8*a*c^2*d*e^2*x^3 + 2*(6*a*c^2*d^2*e + b*c*e^3)*x^ 2 + 8*(b*c*d^3 - b*d*e^2)*sqrt(c)*arctan(sqrt(c)*x) + 4*(b*c*d^3 + b*d*e^2 )*sqrt(c)*log((c*x^2 - 2*sqrt(c)*x + 1)/(c*x^2 - 1)) + 8*(a*c^2*d^3 + 2*b* c*d*e^2)*x + (6*b*c*d^2*e - b*e^3)*log(c*x^2 + 1) + (6*b*c*d^2*e + b*e^3)* log(c*x^2 - 1) + (b*c^2*e^3*x^4 + 4*b*c^2*d*e^2*x^3 + 6*b*c^2*d^2*e*x^2 + 4*b*c^2*d^3*x)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^2, 1/8*(2*a*c^2*e^3*x^4 + 8*a*c^2*d*e^2*x^3 + 2*(6*a*c^2*d^2*e + b*c*e^3)*x^2 + 8*(b*c*d^3 + b*d*e^2 )*sqrt(-c)*arctan(sqrt(-c)*x) + 4*(b*c*d^3 - b*d*e^2)*sqrt(-c)*log((c*x^2 + 2*sqrt(-c)*x - 1)/(c*x^2 + 1)) + 8*(a*c^2*d^3 + 2*b*c*d*e^2)*x + (6*b*c* d^2*e - b*e^3)*log(c*x^2 + 1) + (6*b*c*d^2*e + b*e^3)*log(c*x^2 - 1) + (b* c^2*e^3*x^4 + 4*b*c^2*d*e^2*x^3 + 6*b*c^2*d^2*e*x^2 + 4*b*c^2*d^3*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. 2786 vs. \(2 (141) = 282\).
Time = 7.48 (sec) , antiderivative size = 2786, normalized size of antiderivative = 18.09 \[ \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\text {Too large to display} \] Input:
integrate((e*x+d)**3*(a+b*atanh(c*x**2)),x)
Output:
Piecewise((8*a*c**2*d**3*x*sqrt(-1/c)*sqrt(1/c)/(2*c**3*(-1/c)**(3/2)*sqrt (1/c) - 2*c**3*sqrt(-1/c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) + 1 2*a*c**2*d**2*e*x**2*sqrt(-1/c)*sqrt(1/c)/(2*c**3*(-1/c)**(3/2)*sqrt(1/c) - 2*c**3*sqrt(-1/c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) + 8*a*c** 2*d*e**2*x**3*sqrt(-1/c)*sqrt(1/c)/(2*c**3*(-1/c)**(3/2)*sqrt(1/c) - 2*c** 3*sqrt(-1/c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) + 2*a*c**2*e**3* x**4*sqrt(-1/c)*sqrt(1/c)/(2*c**3*(-1/c)**(3/2)*sqrt(1/c) - 2*c**3*sqrt(-1 /c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) + 8*b*c**2*d**3*x*sqrt(-1 /c)*sqrt(1/c)*atanh(c*x**2)/(2*c**3*(-1/c)**(3/2)*sqrt(1/c) - 2*c**3*sqrt( -1/c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) - 2*b*c**2*d**3*(-1/c)* *(3/2)*log(x + sqrt(-1/c))/(2*c**3*(-1/c)**(3/2)*sqrt(1/c) - 2*c**3*sqrt(- 1/c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) + 2*b*c**2*d**3*(1/c)**( 3/2)*log(x + sqrt(-1/c))/(2*c**3*(-1/c)**(3/2)*sqrt(1/c) - 2*c**3*sqrt(-1/ c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) + 12*b*c**2*d**2*e*x**2*sq rt(-1/c)*sqrt(1/c)*atanh(c*x**2)/(2*c**3*(-1/c)**(3/2)*sqrt(1/c) - 2*c**3* sqrt(-1/c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) + 3*b*c**2*d**2*e* (-1/c)**(3/2)*sqrt(1/c)*log(x + sqrt(-1/c))/(2*c**3*(-1/c)**(3/2)*sqrt(1/c ) - 2*c**3*sqrt(-1/c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt(1/c)) - 3*b*c **2*d**2*e*sqrt(-1/c)*(1/c)**(3/2)*log(x + sqrt(-1/c))/(2*c**3*(-1/c)**(3/ 2)*sqrt(1/c) - 2*c**3*sqrt(-1/c)*(1/c)**(3/2) + 12*c**2*sqrt(-1/c)*sqrt...
Time = 0.11 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.56 \[ \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{4} \, a e^{3} x^{4} + a d e^{2} x^{3} + \frac {3}{2} \, a d^{2} e x^{2} + \frac {1}{2} \, {\left (c {\left (\frac {2 \, \arctan \left (\sqrt {c} x\right )}{c^{\frac {3}{2}}} + \frac {\log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right )}{c^{\frac {3}{2}}}\right )} + 2 \, x \operatorname {artanh}\left (c x^{2}\right )\right )} b d^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {4 \, x}{c^{2}} - \frac {2 \, \arctan \left (\sqrt {c} x\right )}{c^{\frac {5}{2}}} + \frac {\log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )}\right )} b d e^{2} + \frac {1}{8} \, {\left (2 \, x^{4} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {\log \left (c x^{2} + 1\right )}{c^{3}} + \frac {\log \left (c x^{2} - 1\right )}{c^{3}}\right )}\right )} b e^{3} + a d^{3} x + \frac {3 \, {\left (2 \, c x^{2} \operatorname {artanh}\left (c x^{2}\right ) + \log \left (-c^{2} x^{4} + 1\right )\right )} b d^{2} e}{4 \, c} \] Input:
integrate((e*x+d)^3*(a+b*arctanh(c*x^2)),x, algorithm="maxima")
Output:
1/4*a*e^3*x^4 + a*d*e^2*x^3 + 3/2*a*d^2*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^3 + 1/2*(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*d*e^2 + 1/8*(2 *x^4*arctanh(c*x^2) + c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c ^3))*b*e^3 + a*d^3*x + 3/4*(2*c*x^2*arctanh(c*x^2) + log(-c^2*x^4 + 1))*b* d^2*e/c
Time = 7.47 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.47 \[ \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{4} \, a e^{3} x^{4} + a d e^{2} x^{3} + \frac {{\left (6 \, a c d^{2} e + b e^{3}\right )} x^{2}}{4 \, c} + \frac {1}{8} \, {\left (b e^{3} x^{4} + 4 \, b d e^{2} x^{3} + 6 \, b d^{2} e x^{2} + 4 \, b d^{3} x\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + \frac {{\left (a c d^{3} + 2 \, b d e^{2}\right )} x}{c} + \frac {{\left (b c d^{3} - b d e^{2}\right )} \arctan \left (\sqrt {c} x\right )}{c^{\frac {3}{2}}} + \frac {{\left (b c d^{3} + b d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {{\left (6 \, b c d^{2} e - b e^{3}\right )} \log \left (c x^{2} + 1\right )}{8 \, c^{2}} + \frac {{\left (6 \, b c d^{2} e + b e^{3}\right )} \log \left (c x^{2} - 1\right )}{8 \, c^{2}} \] Input:
integrate((e*x+d)^3*(a+b*arctanh(c*x^2)),x, algorithm="giac")
Output:
1/4*a*e^3*x^4 + a*d*e^2*x^3 + 1/4*(6*a*c*d^2*e + b*e^3)*x^2/c + 1/8*(b*e^3 *x^4 + 4*b*d*e^2*x^3 + 6*b*d^2*e*x^2 + 4*b*d^3*x)*log(-(c*x^2 + 1)/(c*x^2 - 1)) + (a*c*d^3 + 2*b*d*e^2)*x/c + (b*c*d^3 - b*d*e^2)*arctan(sqrt(c)*x)/ c^(3/2) + (b*c*d^3 + b*d*e^2)*arctan(c*x/sqrt(-c))/(sqrt(-c)*c) + 1/8*(6*b *c*d^2*e - b*e^3)*log(c*x^2 + 1)/c^2 + 1/8*(6*b*c*d^2*e + b*e^3)*log(c*x^2 - 1)/c^2
Time = 4.70 (sec) , antiderivative size = 823, normalized size of antiderivative = 5.34 \[ \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\ln \left (c\,x^2+1\right )\,\left (\frac {b\,d^3\,x}{2}+\frac {3\,b\,d^2\,e\,x^2}{4}+\frac {b\,d\,e^2\,x^3}{2}+\frac {b\,e^3\,x^4}{8}\right )-\ln \left (1-c\,x^2\right )\,\left (\frac {b\,d^3\,x}{2}+\frac {3\,b\,d^2\,e\,x^2}{4}+\frac {b\,d\,e^2\,x^3}{2}+\frac {b\,e^3\,x^4}{8}\right )+\frac {a\,e^3\,x^4}{4}-\frac {\ln \left (8\,c^5\,d^6-c^2\,e^6-4\,d\,e^5\,\sqrt {-c^5}+e^6\,x\,\sqrt {-c^5}+8\,c^3\,d^2\,e^4+4\,c^4\,d^3\,e^3\,x-4\,c^3\,d\,e^5\,x+4\,c\,d^3\,e^3\,\sqrt {-c^5}-8\,c^3\,d^6\,x\,\sqrt {-c^5}-8\,c\,d^2\,e^4\,x\,\sqrt {-c^5}\right )\,\left (b\,c^2\,e^3-4\,b\,c\,d^3\,\sqrt {-c^5}+4\,b\,d\,e^2\,\sqrt {-c^5}-6\,b\,c^3\,d^2\,e\right )}{8\,c^4}-\frac {\ln \left (8\,c^5\,d^6-c^2\,e^6+4\,d\,e^5\,\sqrt {-c^5}-e^6\,x\,\sqrt {-c^5}+8\,c^3\,d^2\,e^4+4\,c^4\,d^3\,e^3\,x-4\,c^3\,d\,e^5\,x-4\,c\,d^3\,e^3\,\sqrt {-c^5}+8\,c^3\,d^6\,x\,\sqrt {-c^5}+8\,c\,d^2\,e^4\,x\,\sqrt {-c^5}\right )\,\left (b\,c^2\,e^3+4\,b\,c\,d^3\,\sqrt {-c^5}-4\,b\,d\,e^2\,\sqrt {-c^5}-6\,b\,c^3\,d^2\,e\right )}{8\,c^4}+\frac {x\,\left (2\,a\,c^2\,d^3+4\,b\,c\,d\,e^2\right )}{2\,c^2}+\frac {\ln \left (64\,c^2\,d^{12}\,{\left (c^5\right )}^{7/2}+128\,d^8\,e^4\,{\left (c^5\right )}^{7/2}-64\,c^{20}\,d^{12}\,x-c^{14}\,e^{12}\,x+c\,e^{12}\,{\left (c^5\right )}^{5/2}-32\,c^{16}\,d^4\,e^8\,x-128\,c^{18}\,d^8\,e^4\,x+32\,c^3\,d^4\,e^8\,{\left (c^5\right )}^{5/2}\right )\,\left (b\,c^2\,e^3+4\,b\,c\,d^3\,\sqrt {c^5}+4\,b\,d\,e^2\,\sqrt {c^5}+6\,b\,c^3\,d^2\,e\right )}{8\,c^4}+\frac {\ln \left (8\,c^{10}\,d^6+c^7\,e^6+8\,c^8\,d^2\,e^4-4\,d\,e^5\,{\left (c^5\right )}^{3/2}+e^6\,x\,{\left (c^5\right )}^{3/2}-4\,c^9\,d^3\,e^3\,x-4\,c\,d^3\,e^3\,{\left (c^5\right )}^{3/2}+8\,c^3\,d^6\,x\,{\left (c^5\right )}^{3/2}-4\,c^8\,d\,e^5\,x+8\,c\,d^2\,e^4\,x\,{\left (c^5\right )}^{3/2}\right )\,\left (b\,c^2\,e^3-4\,b\,c\,d^3\,\sqrt {c^5}-4\,b\,d\,e^2\,\sqrt {c^5}+6\,b\,c^3\,d^2\,e\right )}{8\,c^4}+\frac {x^2\,\left (6\,a\,c^2\,d^2\,e+b\,c\,e^3\right )}{4\,c^2}+a\,d\,e^2\,x^3 \] Input:
int((a + b*atanh(c*x^2))*(d + e*x)^3,x)
Output:
log(c*x^2 + 1)*((b*e^3*x^4)/8 + (b*d^3*x)/2 + (3*b*d^2*e*x^2)/4 + (b*d*e^2 *x^3)/2) - log(1 - c*x^2)*((b*e^3*x^4)/8 + (b*d^3*x)/2 + (3*b*d^2*e*x^2)/4 + (b*d*e^2*x^3)/2) + (a*e^3*x^4)/4 - (log(8*c^5*d^6 - c^2*e^6 - 4*d*e^5*( -c^5)^(1/2) + e^6*x*(-c^5)^(1/2) + 8*c^3*d^2*e^4 + 4*c^4*d^3*e^3*x - 4*c^3 *d*e^5*x + 4*c*d^3*e^3*(-c^5)^(1/2) - 8*c^3*d^6*x*(-c^5)^(1/2) - 8*c*d^2*e ^4*x*(-c^5)^(1/2))*(b*c^2*e^3 - 4*b*c*d^3*(-c^5)^(1/2) + 4*b*d*e^2*(-c^5)^ (1/2) - 6*b*c^3*d^2*e))/(8*c^4) - (log(8*c^5*d^6 - c^2*e^6 + 4*d*e^5*(-c^5 )^(1/2) - e^6*x*(-c^5)^(1/2) + 8*c^3*d^2*e^4 + 4*c^4*d^3*e^3*x - 4*c^3*d*e ^5*x - 4*c*d^3*e^3*(-c^5)^(1/2) + 8*c^3*d^6*x*(-c^5)^(1/2) + 8*c*d^2*e^4*x *(-c^5)^(1/2))*(b*c^2*e^3 + 4*b*c*d^3*(-c^5)^(1/2) - 4*b*d*e^2*(-c^5)^(1/2 ) - 6*b*c^3*d^2*e))/(8*c^4) + (x*(2*a*c^2*d^3 + 4*b*c*d*e^2))/(2*c^2) + (l og(64*c^2*d^12*(c^5)^(7/2) + 128*d^8*e^4*(c^5)^(7/2) - 64*c^20*d^12*x - c^ 14*e^12*x + c*e^12*(c^5)^(5/2) - 32*c^16*d^4*e^8*x - 128*c^18*d^8*e^4*x + 32*c^3*d^4*e^8*(c^5)^(5/2))*(b*c^2*e^3 + 4*b*c*d^3*(c^5)^(1/2) + 4*b*d*e^2 *(c^5)^(1/2) + 6*b*c^3*d^2*e))/(8*c^4) + (log(8*c^10*d^6 + c^7*e^6 + 8*c^8 *d^2*e^4 - 4*d*e^5*(c^5)^(3/2) + e^6*x*(c^5)^(3/2) - 4*c^9*d^3*e^3*x - 4*c *d^3*e^3*(c^5)^(3/2) + 8*c^3*d^6*x*(c^5)^(3/2) - 4*c^8*d*e^5*x + 8*c*d^2*e ^4*x*(c^5)^(3/2))*(b*c^2*e^3 - 4*b*c*d^3*(c^5)^(1/2) - 4*b*d*e^2*(c^5)^(1/ 2) + 6*b*c^3*d^2*e))/(8*c^4) + (x^2*(b*c*e^3 + 6*a*c^2*d^2*e))/(4*c^2) + a *d*e^2*x^3
Time = 0.19 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.04 \[ \int (d+e x)^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {4 \sqrt {c}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}}\right ) b c \,d^{3}-4 \sqrt {c}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}}\right ) b d \,e^{2}+4 \sqrt {c}\, \mathit {atanh} \left (c \,x^{2}\right ) b c \,d^{3}+4 \sqrt {c}\, \mathit {atanh} \left (c \,x^{2}\right ) b d \,e^{2}+4 \mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} d^{3} x +6 \mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} d^{2} e \,x^{2}+4 \mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} d \,e^{2} x^{3}+\mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} e^{3} x^{4}-6 \mathit {atanh} \left (c \,x^{2}\right ) b c \,d^{2} e -\mathit {atanh} \left (c \,x^{2}\right ) b \,e^{3}+4 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}\, x -1\right ) b c \,d^{3}+4 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}\, x -1\right ) b d \,e^{2}-2 \sqrt {c}\, \mathrm {log}\left (c \,x^{2}+1\right ) b c \,d^{3}-2 \sqrt {c}\, \mathrm {log}\left (c \,x^{2}+1\right ) b d \,e^{2}+6 \,\mathrm {log}\left (c \,x^{2}+1\right ) b c \,d^{2} e +4 a \,c^{2} d^{3} x +6 a \,c^{2} d^{2} e \,x^{2}+4 a \,c^{2} d \,e^{2} x^{3}+a \,c^{2} e^{3} x^{4}+8 b c d \,e^{2} x +b c \,e^{3} x^{2}}{4 c^{2}} \] Input:
int((e*x+d)^3*(a+b*atanh(c*x^2)),x)
Output:
(4*sqrt(c)*atan((c*x)/sqrt(c))*b*c*d**3 - 4*sqrt(c)*atan((c*x)/sqrt(c))*b* d*e**2 + 4*sqrt(c)*atanh(c*x**2)*b*c*d**3 + 4*sqrt(c)*atanh(c*x**2)*b*d*e* *2 + 4*atanh(c*x**2)*b*c**2*d**3*x + 6*atanh(c*x**2)*b*c**2*d**2*e*x**2 + 4*atanh(c*x**2)*b*c**2*d*e**2*x**3 + atanh(c*x**2)*b*c**2*e**3*x**4 - 6*at anh(c*x**2)*b*c*d**2*e - atanh(c*x**2)*b*e**3 + 4*sqrt(c)*log(sqrt(c)*x - 1)*b*c*d**3 + 4*sqrt(c)*log(sqrt(c)*x - 1)*b*d*e**2 - 2*sqrt(c)*log(c*x**2 + 1)*b*c*d**3 - 2*sqrt(c)*log(c*x**2 + 1)*b*d*e**2 + 6*log(c*x**2 + 1)*b* c*d**2*e + 4*a*c**2*d**3*x + 6*a*c**2*d**2*e*x**2 + 4*a*c**2*d*e**2*x**3 + a*c**2*e**3*x**4 + 8*b*c*d*e**2*x + b*c*e**3*x**2)/(4*c**2)