Integrand size = 23, antiderivative size = 99 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {x^3 (1+a x)}{3 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {x (3+4 a x)}{3 a^4 c^2 \sqrt {1-a^2 x^2}}-\frac {8 \sqrt {1-a^2 x^2}}{3 a^5 c^2}+\frac {\arcsin (a x)}{a^5 c^2} \]
1/3*x^3*(a*x+1)/a^2/c^2/(-a^2*x^2+1)^(3/2)+arcsin(a*x)/a^5/c^2-1/3*x*(4*a* x+3)/a^4/c^2/(-a^2*x^2+1)^(1/2)-8/3*(-a^2*x^2+1)^(1/2)/a^5/c^2
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {8-5 a x-7 a^2 x^2+3 a^3 x^3+3 (-1+a x) \sqrt {1-a^2 x^2} \arcsin (a x)}{3 a^5 c^2 (-1+a x) \sqrt {1-a^2 x^2}} \]
(8 - 5*a*x - 7*a^2*x^2 + 3*a^3*x^3 + 3*(-1 + a*x)*Sqrt[1 - a^2*x^2]*ArcSin [a*x])/(3*a^5*c^2*(-1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.39 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6698, 529, 2345, 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.
\(\displaystyle \int \frac {x^4 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle \frac {\int \frac {x^4 (a x+1)}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2}\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {\frac {a x+1}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{3} \int \frac {\frac {3 x^3}{a}+\frac {3 x^2}{a^2}+\frac {3 x}{a^3}+\frac {1}{a^4}}{\left (1-a^2 x^2\right )^{3/2}}dx}{c^2}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {\frac {1}{3} \left (\int \frac {3 (a x+1)}{a^4 \sqrt {1-a^2 x^2}}dx-\frac {2 (2 a x+3)}{a^5 \sqrt {1-a^2 x^2}}\right )+\frac {a x+1}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 \int \frac {a x+1}{\sqrt {1-a^2 x^2}}dx}{a^4}-\frac {2 (2 a x+3)}{a^5 \sqrt {1-a^2 x^2}}\right )+\frac {a x+1}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 \left (\int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^4}-\frac {2 (2 a x+3)}{a^5 \sqrt {1-a^2 x^2}}\right )+\frac {a x+1}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {a x+1}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}+\frac {1}{3} \left (\frac {3 \left (\frac {\arcsin (a x)}{a}-\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^4}-\frac {2 (2 a x+3)}{a^5 \sqrt {1-a^2 x^2}}\right )}{c^2}\) |
((1 + a*x)/(3*a^5*(1 - a^2*x^2)^(3/2)) + ((-2*(3 + 2*a*x))/(a^5*Sqrt[1 - a ^2*x^2]) + (3*(-(Sqrt[1 - a^2*x^2]/a) + ArcSin[a*x]/a))/a^4)/3)/c^2
3.9.98.3.1 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] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ (2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b* x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* c^2 + a*d^2, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0]) && IGtQ[(n + 1)/2, 0] && !IntegerQ[p - n/2]
Leaf count of result is larger than twice the leaf count of optimal. \(180\) vs. \(2(87)=174\).
Time = 0.19 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.83
method | result | size |
risch | \(\frac {a^{2} x^{2}-1}{a^{5} \sqrt {-a^{2} x^{2}+1}\, c^{2}}+\frac {\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{4} \sqrt {a^{2}}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{6} \left (x +\frac {1}{a}\right )}+\frac {19 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{12 a^{6} \left (x -\frac {1}{a}\right )}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{6 a^{7} \left (x -\frac {1}{a}\right )^{2}}}{c^{2}}\) | \(181\) |
default | \(\frac {\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{4} \sqrt {a^{2}}}-\frac {\sqrt {-a^{2} x^{2}+1}}{a^{5}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{6} \left (x +\frac {1}{a}\right )}+\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{3 \left (x -\frac {1}{a}\right )}}{2 a^{6}}+\frac {7 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{4 a^{6} \left (x -\frac {1}{a}\right )}}{c^{2}}\) | \(213\) |
1/a^5*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)/c^2+(1/a^4/(a^2)^(1/2)*arctan((a^2)^( 1/2)*x/(-a^2*x^2+1)^(1/2))-1/4/a^6/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1 /2)+19/12/a^6/(x-1/a)*(-(x-1/a)^2*a^2-2*(x-1/a)*a)^(1/2)+1/6/a^7/(x-1/a)^2 *(-(x-1/a)^2*a^2-2*(x-1/a)*a)^(1/2))/c^2
Time = 0.26 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.45 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {8 \, a^{3} x^{3} - 8 \, a^{2} x^{2} - 8 \, a x + 6 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{3} x^{3} - 7 \, a^{2} x^{2} - 5 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} + 8}{3 \, {\left (a^{8} c^{2} x^{3} - a^{7} c^{2} x^{2} - a^{6} c^{2} x + a^{5} c^{2}\right )}} \]
-1/3*(8*a^3*x^3 - 8*a^2*x^2 - 8*a*x + 6*(a^3*x^3 - a^2*x^2 - a*x + 1)*arct an((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^3*x^3 - 7*a^2*x^2 - 5*a*x + 8)*s qrt(-a^2*x^2 + 1) + 8)/(a^8*c^2*x^3 - a^7*c^2*x^2 - a^6*c^2*x + a^5*c^2)
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\int \frac {x^{4}}{a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{5}}{a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \]
(Integral(x**4/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x* *2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**5/(a**4*x**4*sqrt(-a** 2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) )/c**2
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\int { \frac {{\left (a x + 1\right )} x^{4}}{{\left (a^{2} c x^{2} - c\right )}^{2} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\int { \frac {{\left (a x + 1\right )} x^{4}}{{\left (a^{2} c x^{2} - c\right )}^{2} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
Time = 0.06 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.98 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\sqrt {1-a^2\,x^2}}{6\,\left (a^7\,c^2\,x^2-2\,a^6\,c^2\,x+a^5\,c^2\right )}+\frac {\sqrt {1-a^2\,x^2}}{4\,\left (a^3\,c^2\,\sqrt {-a^2}+a^4\,c^2\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {19\,\sqrt {1-a^2\,x^2}}{12\,\left (a^3\,c^2\,\sqrt {-a^2}-a^4\,c^2\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a^5\,c^2}+\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^4\,c^2\,\sqrt {-a^2}} \]
(1 - a^2*x^2)^(1/2)/(6*(a^5*c^2 - 2*a^6*c^2*x + a^7*c^2*x^2)) + (1 - a^2*x ^2)^(1/2)/(4*(a^3*c^2*(-a^2)^(1/2) + a^4*c^2*x*(-a^2)^(1/2))*(-a^2)^(1/2)) + (19*(1 - a^2*x^2)^(1/2))/(12*(a^3*c^2*(-a^2)^(1/2) - a^4*c^2*x*(-a^2)^( 1/2))*(-a^2)^(1/2)) - (1 - a^2*x^2)^(1/2)/(a^5*c^2) + asinh(x*(-a^2)^(1/2) )/(a^4*c^2*(-a^2)^(1/2))