Integrand size = 14, antiderivative size = 187 \[ \int e^{3 \text {arctanh}(a+b x)} x^3 \, dx=\frac {3 \left (17-44 a+36 a^2-8 a^3\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{8 b^4}+\frac {2 x^3 (1+a+b x)^{3/2}}{b \sqrt {1-a-b x}}+\frac {9 x^2 \sqrt {1-a-b x} (1+a+b x)^{3/2}}{4 b^2}+\frac {\sqrt {1-a-b x} (1+a+b x)^{3/2} \left (29-54 a+22 a^2+2 (11-10 a) b x\right )}{8 b^4}-\frac {3 \left (17-44 a+36 a^2-8 a^3\right ) \arcsin (a+b x)}{8 b^4} \]
-3/8*(-8*a^3+36*a^2-44*a+17)*arcsin(b*x+a)/b^4+2*x^3*(b*x+a+1)^(3/2)/b/(-b *x-a+1)^(1/2)+9/4*x^2*(b*x+a+1)^(3/2)*(-b*x-a+1)^(1/2)/b^2+1/8*(b*x+a+1)^( 3/2)*(29-54*a+22*a^2+2*(11-10*a)*b*x)*(-b*x-a+1)^(1/2)/b^4+3/8*(-8*a^3+36* a^2-44*a+17)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/b^4
Time = 0.21 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.09 \[ \int e^{3 \text {arctanh}(a+b x)} x^3 \, dx=\frac {-\frac {\sqrt {b} \sqrt {1+a+b x} \left (-80+78 a^3-2 a^4+29 b x+11 b^2 x^2+6 b^3 x^3+2 b^4 x^4+a^2 (-233+22 b x)+a \left (237-54 b x-10 b^2 x^2\right )\right )}{\sqrt {1-a-b x}}+24 a \left (11+2 a^2\right ) \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )+6 \left (17+36 a^2\right ) \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {-b}}\right )}{8 b^{9/2}} \]
(-((Sqrt[b]*Sqrt[1 + a + b*x]*(-80 + 78*a^3 - 2*a^4 + 29*b*x + 11*b^2*x^2 + 6*b^3*x^3 + 2*b^4*x^4 + a^2*(-233 + 22*b*x) + a*(237 - 54*b*x - 10*b^2*x ^2)))/Sqrt[1 - a - b*x]) + 24*a*(11 + 2*a^2)*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sq rt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])] + 6*(17 + 36*a^2)*Sqrt[-b]*ArcSinh[(Sq rt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-b])])/(8*b^(9/2))
Time = 0.42 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6713, 108, 27, 170, 25, 27, 164, 60, 62, 1090, 223}
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.
| \(\displaystyle \int x^3 e^{3 \text {arctanh}(a+b x)} \, dx\) |
| \(\Big \downarrow \) 6713 |
| \(\displaystyle \int \frac {x^3 (a+b x+1)^{3/2}}{(-a-b x+1)^{3/2}}dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {2 \int \frac {3 x^2 \sqrt {a+b x+1} (2 (a+1)+3 b x)}{2 \sqrt {-a-b x+1}}dx}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \int \frac {x^2 \sqrt {a+b x+1} (2 (a+1)+3 b x)}{\sqrt {-a-b x+1}}dx}{b}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \left (-\frac {\int -\frac {b x \sqrt {a+b x+1} (6 (1-a) (a+1)+(11-10 a) b x)}{\sqrt {-a-b x+1}}dx}{4 b^2}-\frac {3 x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b}\right )}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \left (\frac {\int \frac {b x \sqrt {a+b x+1} \left (6 \left (1-a^2\right )+(11-10 a) b x\right )}{\sqrt {-a-b x+1}}dx}{4 b^2}-\frac {3 x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b}\right )}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \left (\frac {\int \frac {x \sqrt {a+b x+1} \left (6 \left (1-a^2\right )+(11-10 a) b x\right )}{\sqrt {-a-b x+1}}dx}{4 b}-\frac {3 x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b}\right )}{b}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \left (\frac {\frac {\left (-8 a^3+36 a^2-44 a+17\right ) \int \frac {\sqrt {a+b x+1}}{\sqrt {-a-b x+1}}dx}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2} \left (22 a^2+2 (11-10 a) b x-54 a+29\right )}{6 b^2}}{4 b}-\frac {3 x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b}\right )}{b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \left (\frac {\frac {\left (-8 a^3+36 a^2-44 a+17\right ) \left (\int \frac {1}{\sqrt {-a-b x+1} \sqrt {a+b x+1}}dx-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2} \left (22 a^2+2 (11-10 a) b x-54 a+29\right )}{6 b^2}}{4 b}-\frac {3 x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b}\right )}{b}\) |
| \(\Big \downarrow \) 62 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \left (\frac {\frac {\left (-8 a^3+36 a^2-44 a+17\right ) \left (\int \frac {1}{\sqrt {-b^2 x^2-2 a b x+(1-a) (a+1)}}dx-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2} \left (22 a^2+2 (11-10 a) b x-54 a+29\right )}{6 b^2}}{4 b}-\frac {3 x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b}\right )}{b}\) |
| \(\Big \downarrow \) 1090 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \left (\frac {\frac {\left (-8 a^3+36 a^2-44 a+17\right ) \left (-\frac {\int \frac {1}{\sqrt {1-\frac {\left (-2 x b^2-2 a b\right )^2}{4 b^2}}}d\left (-2 x b^2-2 a b\right )}{2 b^2}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2} \left (22 a^2+2 (11-10 a) b x-54 a+29\right )}{6 b^2}}{4 b}-\frac {3 x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b}\right )}{b}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {2 x^3 (a+b x+1)^{3/2}}{b \sqrt {-a-b x+1}}-\frac {3 \left (\frac {\frac {\left (-8 a^3+36 a^2-44 a+17\right ) \left (-\frac {\arcsin \left (\frac {-2 a b-2 b^2 x}{2 b}\right )}{b}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2} \left (22 a^2+2 (11-10 a) b x-54 a+29\right )}{6 b^2}}{4 b}-\frac {3 x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b}\right )}{b}\) |
(2*x^3*(1 + a + b*x)^(3/2))/(b*Sqrt[1 - a - b*x]) - (3*((-3*x^2*Sqrt[1 - a - b*x]*(1 + a + b*x)^(3/2))/(4*b) + (-1/6*(Sqrt[1 - a - b*x]*(1 + a + b*x )^(3/2)*(29 - 54*a + 22*a^2 + 2*(11 - 10*a)*b*x))/b^2 + ((17 - 44*a + 36*a ^2 - 8*a^3)*(-((Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/b) - ArcSin[(-2*a*b - 2*b^2*x)/(2*b)]/b))/(2*b))/(4*b)))/b
3.9.35.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Int[ 1/Sqrt[a*c - b*(a - c)*x - b^2*x^2], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 *n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.) , x_Symbol] :> Int[(d + e*x)^m*((1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^( n/2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(347\) vs. \(2(163)=326\).
Time = 0.71 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.86
| method | result | size |
| risch | \(\frac {\left (-2 b^{3} x^{3}+2 a \,b^{2} x^{2}-2 a^{2} b x -8 b^{2} x^{2}+2 a^{3}+20 a b x -44 a^{2}-19 b x +93 a -48\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{8 b^{4} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {-\frac {51 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}+\frac {132 a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}+\frac {24 a^{3} \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}-\frac {108 a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}+\frac {\left (32 a^{3}-96 a^{2}+96 a -32\right ) \sqrt {-\left (x +\frac {-1+a}{b}\right )^{2} b^{2}-2 \left (x +\frac {-1+a}{b}\right ) b}}{b^{2} \left (x +\frac {-1+a}{b}\right )}}{8 b^{3}}\) | \(348\) |
| default | \(\text {Expression too large to display}\) | \(3196\) |
1/8*(-2*b^3*x^3+2*a*b^2*x^2-2*a^2*b*x-8*b^2*x^2+2*a^3+20*a*b*x-44*a^2-19*b *x+93*a-48)*(b^2*x^2+2*a*b*x+a^2-1)/b^4/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/8 /b^3*(-51/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^ (1/2))+132*a/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+ 1)^(1/2))+24*a^3/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x- a^2+1)^(1/2))-108*a^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a *b*x-a^2+1)^(1/2))+(32*a^3-96*a^2+96*a-32)/b^2/(x+(-1+a)/b)*(-(x+(-1+a)/b) ^2*b^2-2*(x+(-1+a)/b)*b)^(1/2))
Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.03 \[ \int e^{3 \text {arctanh}(a+b x)} x^3 \, dx=-\frac {3 \, {\left (8 \, a^{4} - 44 \, a^{3} + {\left (8 \, a^{3} - 36 \, a^{2} + 44 \, a - 17\right )} b x + 80 \, a^{2} - 61 \, a + 17\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - {\left (2 \, b^{4} x^{4} + 6 \, b^{3} x^{3} - {\left (10 \, a - 11\right )} b^{2} x^{2} - 2 \, a^{4} + 78 \, a^{3} + {\left (22 \, a^{2} - 54 \, a + 29\right )} b x - 233 \, a^{2} + 237 \, a - 80\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{8 \, {\left (b^{5} x + {\left (a - 1\right )} b^{4}\right )}} \]
-1/8*(3*(8*a^4 - 44*a^3 + (8*a^3 - 36*a^2 + 44*a - 17)*b*x + 80*a^2 - 61*a + 17)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a* b*x + a^2 - 1)) - (2*b^4*x^4 + 6*b^3*x^3 - (10*a - 11)*b^2*x^2 - 2*a^4 + 7 8*a^3 + (22*a^2 - 54*a + 29)*b*x - 233*a^2 + 237*a - 80)*sqrt(-b^2*x^2 - 2 *a*b*x - a^2 + 1))/(b^5*x + (a - 1)*b^4)
\[ \int e^{3 \text {arctanh}(a+b x)} x^3 \, dx=\int \frac {x^{3} \left (a + b x + 1\right )^{3}}{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 2349 vs. \(2 (163) = 326\).
Time = 0.31 (sec) , antiderivative size = 2349, normalized size of antiderivative = 12.56 \[ \int e^{3 \text {arctanh}(a+b x)} x^3 \, dx=\text {Too large to display} \]
-1/4*b*x^5/sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1) + 315/4*a^6*x/((a^2*b^2 - (a ^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) + 3/4*a*x^4/sqrt(-b^2*x ^2 - 2*a*b*x - a^2 + 1) - 945/8*(a^2 - 1)*a^4*x/((a^2*b^2 - (a^2 - 1)*b^2) *sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) - 21/8*a^2*x^3/(sqrt(-b^2*x^2 - 2*a *b*x - a^2 + 1)*b) + 105/8*(a^2 - 1)*a^5/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(- b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2) - 105*(a*b^2 + b^2)*a^5*x/((a^2*b^2 - (a ^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^3) + 45*(a^2*b + 2*a*b + b)*a^4*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^ 2) + 169/4*(a^2 - 1)^2*a^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2* a*b*x - a^2 + 1)*b) - 6*(a^3 + 3*a^2 + 3*a + 1)*a^3*x/((a^2*b^2 - (a^2 - 1 )*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) + 105/8*a^3*x^2/(sqrt(-b^2*x^ 2 - 2*a*b*x - a^2 + 1)*b^2) + 5/8*(a^2 - 1)*x^3/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) - (a*b^2 + b^2)*x^4/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2) - 14*(a^2 - 1)^2*a^3/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2) + 265/2*(a*b^2 + b^2)*(a^2 - 1)*a^3*x/((a^2*b^2 - (a^2 - 1)* b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^3) - 93/2*(a^2*b + 2*a*b + b)*(a ^2 - 1)*a^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1 )*b^2) - 15/8*(a^2 - 1)^3*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a *b*x - a^2 + 1)*b) + 5*(a^3 + 3*a^2 + 3*a + 1)*(a^2 - 1)*a*x/((a^2*b^2 - ( a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) - 49/8*(a^2 - 1)*a*...
Time = 0.29 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13 \[ \int e^{3 \text {arctanh}(a+b x)} x^3 \, dx=\frac {1}{8} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left ({\left (2 \, x {\left (\frac {x}{b} - \frac {a b^{11} - 4 \, b^{11}}{b^{13}}\right )} + \frac {2 \, a^{2} b^{10} - 20 \, a b^{10} + 19 \, b^{10}}{b^{13}}\right )} x - \frac {2 \, a^{3} b^{9} - 44 \, a^{2} b^{9} + 93 \, a b^{9} - 48 \, b^{9}}{b^{13}}\right )} - \frac {3 \, {\left (8 \, a^{3} - 36 \, a^{2} + 44 \, a - 17\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{8 \, b^{3} {\left | b \right |}} - \frac {8 \, {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )}}{b^{3} {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )} {\left | b \right |}} \]
1/8*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*((2*x*(x/b - (a*b^11 - 4*b^11)/b^13 ) + (2*a^2*b^10 - 20*a*b^10 + 19*b^10)/b^13)*x - (2*a^3*b^9 - 44*a^2*b^9 + 93*a*b^9 - 48*b^9)/b^13) - 3/8*(8*a^3 - 36*a^2 + 44*a - 17)*arcsin(-b*x - a)*sgn(b)/(b^3*abs(b)) - 8*(a^3 - 3*a^2 + 3*a - 1)/(b^3*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) - 1)*abs(b))
Timed out. \[ \int e^{3 \text {arctanh}(a+b x)} x^3 \, dx=\int \frac {x^3\,{\left (a+b\,x+1\right )}^3}{{\left (1-{\left (a+b\,x\right )}^2\right )}^{3/2}} \,d x \]