Integrand size = 10, antiderivative size = 86 \[ \int e^{-3 \text {arctanh}(a x)} x \, dx=\frac {9 \sqrt {1-a^2 x^2}}{2 a^2}+\frac {3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (1+a x)}+\frac {\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac {9 \arcsin (a x)}{2 a^2} \]
3/2*(-a^2*x^2+1)^(3/2)/a^2/(a*x+1)+(-a^2*x^2+1)^(5/2)/a^2/(a*x+1)^3+9/2*ar csin(a*x)/a^2+9/2*(-a^2*x^2+1)^(1/2)/a^2
Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.51 \[ \int e^{-3 \text {arctanh}(a x)} x \, dx=\frac {\sqrt {1-a^2 x^2} \left (6-a x+\frac {8}{1+a x}\right )+9 \arcsin (a x)}{2 a^2} \]
Time = 0.58 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.100, Rules used = {6674, 2164, 2027, 2164, 27, 563, 25, 2346, 27, 455, 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.
\(\displaystyle \int x e^{-3 \text {arctanh}(a x)} \, dx\) |
\(\Big \downarrow \) 6674 |
\(\displaystyle \int \frac {x (1-a x)^2}{(a x+1) \sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 2164 |
\(\displaystyle a \int \frac {\left (\frac {x}{a}-x^2\right ) \sqrt {1-a^2 x^2}}{(a x+1)^2}dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle a \int \frac {\left (\frac {1}{a}-x\right ) x \sqrt {1-a^2 x^2}}{(a x+1)^2}dx\) |
\(\Big \downarrow \) 2164 |
\(\displaystyle a^2 \int \frac {x \left (1-a^2 x^2\right )^{3/2}}{a^2 (a x+1)^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x \left (1-a^2 x^2\right )^{3/2}}{(a x+1)^3}dx\) |
\(\Big \downarrow \) 563 |
\(\displaystyle \frac {4 \sqrt {1-a^2 x^2}}{a^2 (a x+1)}-\frac {\int -\frac {a^2 x^2-3 a x+4}{\sqrt {1-a^2 x^2}}dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a^2 x^2-3 a x+4}{\sqrt {1-a^2 x^2}}dx}{a}+\frac {4 \sqrt {1-a^2 x^2}}{a^2 (a x+1)}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {-\frac {\int -\frac {3 a^2 (3-2 a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a}+\frac {4 \sqrt {1-a^2 x^2}}{a^2 (a x+1)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3}{2} \int \frac {3-2 a x}{\sqrt {1-a^2 x^2}}dx-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a}+\frac {4 \sqrt {1-a^2 x^2}}{a^2 (a x+1)}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {3}{2} \left (3 \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {2 \sqrt {1-a^2 x^2}}{a}\right )-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a}+\frac {4 \sqrt {1-a^2 x^2}}{a^2 (a x+1)}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {3}{2} \left (\frac {2 \sqrt {1-a^2 x^2}}{a}+\frac {3 \arcsin (a x)}{a}\right )-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a}+\frac {4 \sqrt {1-a^2 x^2}}{a^2 (a x+1)}\) |
(4*Sqrt[1 - a^2*x^2])/(a^2*(1 + a*x)) + (-1/2*(x*Sqrt[1 - a^2*x^2]) + (3*( (2*Sqrt[1 - a^2*x^2])/a + (3*ArcSin[a*x])/a))/2)/a
3.1.53.3.1 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[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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)* b^(n + 2)*(c + d*x))), x] - Simp[d^(2*n - m + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(-n - 1)*(-c)^(m - n - 1) - d^m*x^m*(-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2 , 0] && IGtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*e Int[(d + e*x)^(m - 1)*PolynomialQuotient[Pq, a*e + b*d*x, x]* (a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + b*d*x, x], 0 ]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x )^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/2)*Sqrt[1 - a^2*x^2])), x] / ; FreeQ[{a, c, m}, x] && IntegerQ[(n - 1)/2]
Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {\left (a x -6\right ) \left (a^{2} x^{2}-1\right )}{2 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {9 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a \sqrt {a^{2}}}+\frac {4 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a^{3} \left (x +\frac {1}{a}\right )}\) | \(98\) |
default | \(\frac {\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )}{a^{3}}-\frac {-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{3}}-2 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{4}}\) | \(344\) |
1/2*(a*x-6)*(a^2*x^2-1)/a^2/(-a^2*x^2+1)^(1/2)+9/2/a/(a^2)^(1/2)*arctan((a ^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+4/a^3/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a)) ^(1/2)
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int e^{-3 \text {arctanh}(a x)} x \, dx=\frac {14 \, a x - 18 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (a^{2} x^{2} - 5 \, a x - 14\right )} \sqrt {-a^{2} x^{2} + 1} + 14}{2 \, {\left (a^{3} x + a^{2}\right )}} \]
1/2*(14*a*x - 18*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (a^2*x ^2 - 5*a*x - 14)*sqrt(-a^2*x^2 + 1) + 14)/(a^3*x + a^2)
\[ \int e^{-3 \text {arctanh}(a x)} x \, dx=\int \frac {x \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (a x + 1\right )^{3}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.28 \[ \int e^{-3 \text {arctanh}(a x)} x \, dx=-\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4} x^{2} + 2 \, a^{3} x + a^{2}} + \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{2 \, {\left (a^{3} x + a^{2}\right )}} + \frac {6 \, \sqrt {-a^{2} x^{2} + 1}}{a^{3} x + a^{2}} + \frac {9 \, \arcsin \left (a x\right )}{2 \, a^{2}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1}}{2 \, a^{2}} \]
-(-a^2*x^2 + 1)^(3/2)/(a^4*x^2 + 2*a^3*x + a^2) + 1/2*(-a^2*x^2 + 1)^(3/2) /(a^3*x + a^2) + 6*sqrt(-a^2*x^2 + 1)/(a^3*x + a^2) + 9/2*arcsin(a*x)/a^2 + 3/2*sqrt(-a^2*x^2 + 1)/a^2
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91 \[ \int e^{-3 \text {arctanh}(a x)} x \, dx=-\frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {x}{a} - \frac {6}{a^{2}}\right )} + \frac {9 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{2 \, a {\left | a \right |}} - \frac {8}{a {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]
-1/2*sqrt(-a^2*x^2 + 1)*(x/a - 6/a^2) + 9/2*arcsin(a*x)*sgn(a)/(a*abs(a)) - 8/(a*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))
Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int e^{-3 \text {arctanh}(a x)} x \, dx=-\frac {\left (\frac {3}{\sqrt {-a^2}}+\frac {x\,\sqrt {-a^2}}{2\,a}\right )\,\sqrt {1-a^2\,x^2}-\frac {9\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a}+\frac {4\,\sqrt {1-a^2\,x^2}}{a\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )}}{\sqrt {-a^2}} \]