Integrand size = 10, antiderivative size = 103 \[ \int e^{\text {arctanh}(a x)} x^3 \, dx=-\frac {2 \sqrt {1-a^2 x^2}}{3 a^4}-\frac {3 x \sqrt {1-a^2 x^2}}{8 a^3}-\frac {x^2 \sqrt {1-a^2 x^2}}{3 a^2}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {3 \arcsin (a x)}{8 a^4} \] Output:
-2/3*(-a^2*x^2+1)^(1/2)/a^4-3/8*x*(-a^2*x^2+1)^(1/2)/a^3-1/3*x^2*(-a^2*x^2 +1)^(1/2)/a^2-1/4*x^3*(-a^2*x^2+1)^(1/2)/a+3/8*arcsin(a*x)/a^4
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.50 \[ \int e^{\text {arctanh}(a x)} x^3 \, dx=\frac {-\sqrt {1-a^2 x^2} \left (16+9 a x+8 a^2 x^2+6 a^3 x^3\right )+9 \arcsin (a x)}{24 a^4} \] Input:
Integrate[E^ArcTanh[a*x]*x^3,x]
Output:
(-(Sqrt[1 - a^2*x^2]*(16 + 9*a*x + 8*a^2*x^2 + 6*a^3*x^3)) + 9*ArcSin[a*x] )/(24*a^4)
Time = 0.49 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {6674, 533, 27, 533, 27, 533, 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.
\(\displaystyle \int x^3 e^{\text {arctanh}(a x)} \, dx\) |
\(\Big \downarrow \) 6674 |
\(\displaystyle \int \frac {x^3 (a x+1)}{\sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\int \frac {a x^2 (4 a x+3)}{\sqrt {1-a^2 x^2}}dx}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x^2 (4 a x+3)}{\sqrt {1-a^2 x^2}}dx}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\frac {\int \frac {a x (9 a x+8)}{\sqrt {1-a^2 x^2}}dx}{3 a^2}-\frac {4 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {x (9 a x+8)}{\sqrt {1-a^2 x^2}}dx}{3 a}-\frac {4 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\frac {\frac {\int \frac {a (16 a x+9)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {9 x \sqrt {1-a^2 x^2}}{2 a}}{3 a}-\frac {4 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\int \frac {16 a x+9}{\sqrt {1-a^2 x^2}}dx}{2 a}-\frac {9 x \sqrt {1-a^2 x^2}}{2 a}}{3 a}-\frac {4 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {\frac {9 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {16 \sqrt {1-a^2 x^2}}{a}}{2 a}-\frac {9 x \sqrt {1-a^2 x^2}}{2 a}}{3 a}-\frac {4 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {\frac {\frac {9 \arcsin (a x)}{a}-\frac {16 \sqrt {1-a^2 x^2}}{a}}{2 a}-\frac {9 x \sqrt {1-a^2 x^2}}{2 a}}{3 a}-\frac {4 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a}\) |
Input:
Int[E^ArcTanh[a*x]*x^3,x]
Output:
-1/4*(x^3*Sqrt[1 - a^2*x^2])/a + ((-4*x^2*Sqrt[1 - a^2*x^2])/(3*a) + ((-9* x*Sqrt[1 - a^2*x^2])/(2*a) + ((-16*Sqrt[1 - a^2*x^2])/a + (9*ArcSin[a*x])/ a)/(2*a))/(3*a))/(4*a)
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_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
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 = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {\left (6 a^{3} x^{3}+8 a^{2} x^{2}+9 a x +16\right ) \left (a^{2} x^{2}-1\right )}{24 a^{4} \sqrt {-a^{2} x^{2}+1}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3} \sqrt {a^{2}}}\) | \(80\) |
default | \(-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}+a \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )\) | \(116\) |
meijerg | \(\frac {-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}}{2 a^{3} \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}}{2 a^{4} \sqrt {\pi }}\) | \(116\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3,x,method=_RETURNVERBOSE)
Output:
1/24*(6*a^3*x^3+8*a^2*x^2+9*a*x+16)*(a^2*x^2-1)/a^4/(-a^2*x^2+1)^(1/2)+3/8 /a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))
Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.63 \[ \int e^{\text {arctanh}(a x)} x^3 \, dx=-\frac {{\left (6 \, a^{3} x^{3} + 8 \, a^{2} x^{2} + 9 \, a x + 16\right )} \sqrt {-a^{2} x^{2} + 1} + 18 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{24 \, a^{4}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3,x, algorithm="fricas")
Output:
-1/24*((6*a^3*x^3 + 8*a^2*x^2 + 9*a*x + 16)*sqrt(-a^2*x^2 + 1) + 18*arctan ((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^4
Time = 0.45 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99 \[ \int e^{\text {arctanh}(a x)} x^3 \, dx=\begin {cases} \sqrt {- a^{2} x^{2} + 1} \left (- \frac {x^{3}}{4 a} - \frac {x^{2}}{3 a^{2}} - \frac {3 x}{8 a^{3}} - \frac {2}{3 a^{4}}\right ) + \frac {3 \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{8 a^{3} \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {a x^{5}}{5} + \frac {x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**3,x)
Output:
Piecewise((sqrt(-a**2*x**2 + 1)*(-x**3/(4*a) - x**2/(3*a**2) - 3*x/(8*a**3 ) - 2/(3*a**4)) + 3*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(8 *a**3*sqrt(-a**2)), Ne(a**2, 0)), (a*x**5/5 + x**4/4, True))
Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int e^{\text {arctanh}(a x)} x^3 \, dx=-\frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{4 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{3 \, a^{2}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{8 \, a^{3}} + \frac {3 \, \arcsin \left (a x\right )}{8 \, a^{4}} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, a^{4}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3,x, algorithm="maxima")
Output:
-1/4*sqrt(-a^2*x^2 + 1)*x^3/a - 1/3*sqrt(-a^2*x^2 + 1)*x^2/a^2 - 3/8*sqrt( -a^2*x^2 + 1)*x/a^3 + 3/8*arcsin(a*x)/a^4 - 2/3*sqrt(-a^2*x^2 + 1)/a^4
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.57 \[ \int e^{\text {arctanh}(a x)} x^3 \, dx=-\frac {1}{24} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, x {\left (\frac {3 \, x}{a} + \frac {4}{a^{2}}\right )} + \frac {9}{a^{3}}\right )} x + \frac {16}{a^{4}}\right )} + \frac {3 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{8 \, a^{3} {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3,x, algorithm="giac")
Output:
-1/24*sqrt(-a^2*x^2 + 1)*((2*x*(3*x/a + 4/a^2) + 9/a^3)*x + 16/a^4) + 3/8* arcsin(a*x)*sgn(a)/(a^3*abs(a))
Time = 24.40 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.94 \[ \int e^{\text {arctanh}(a x)} x^3 \, dx=\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^3\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2}{3\,{\left (-a^2\right )}^{3/2}}+\frac {3\,x\,\sqrt {-a^2}}{8\,a^3}+\frac {a^2\,x^2}{3\,{\left (-a^2\right )}^{3/2}}-\frac {x^3\,{\left (-a^2\right )}^{3/2}}{4\,a^3}\right )}{\sqrt {-a^2}} \] Input:
int((x^3*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
(3*asinh(x*(-a^2)^(1/2)))/(8*a^3*(-a^2)^(1/2)) - ((1 - a^2*x^2)^(1/2)*(2/( 3*(-a^2)^(3/2)) + (3*x*(-a^2)^(1/2))/(8*a^3) + (a^2*x^2)/(3*(-a^2)^(3/2)) - (x^3*(-a^2)^(3/2))/(4*a^3)))/(-a^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int e^{\text {arctanh}(a x)} x^3 \, dx=\frac {9 \mathit {asin} \left (a x \right )-6 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-8 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-9 \sqrt {-a^{2} x^{2}+1}\, a x -16 \sqrt {-a^{2} x^{2}+1}+16}{24 a^{4}} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3,x)
Output:
(9*asin(a*x) - 6*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 8*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 9*sqrt( - a**2*x**2 + 1)*a*x - 16*sqrt( - a**2*x**2 + 1) + 16)/(24*a**4)