Integrand size = 23, antiderivative size = 81 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {x^4 (1+a x)}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {4}{15 a^5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {4}{5 a^5 c^3 \sqrt {1-a^2 x^2}} \] Output:
1/5*x^4*(a*x+1)/a/c^3/(-a^2*x^2+1)^(5/2)-4/15/a^5/c^3/(-a^2*x^2+1)^(3/2)+4 /5/a^5/c^3/(-a^2*x^2+1)^(1/2)
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {8-8 a x-12 a^2 x^2+12 a^3 x^3+3 a^4 x^4}{15 a^5 c^3 (-1+a x)^2 (1+a x) \sqrt {1-a^2 x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^3,x]
Output:
(8 - 8*a*x - 12*a^2*x^2 + 12*a^3*x^3 + 3*a^4*x^4)/(15*a^5*c^3*(-1 + a*x)^2 *(1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6698, 529, 2345, 27, 453}
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.
\(\displaystyle \int \frac {x^4 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle \frac {\int \frac {x^4 (a x+1)}{\left (1-a^2 x^2\right )^{7/2}}dx}{c^3}\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {\frac {a x+1}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{5} \int \frac {\frac {5 x^3}{a}+\frac {5 x^2}{a^2}+\frac {5 x}{a^3}+\frac {1}{a^4}}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^3}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {3 (5 a x+1)}{a^4 \left (1-a^2 x^2\right )^{3/2}}dx-\frac {2 (3 a x+5)}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {a x+1}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {\int \frac {5 a x+1}{\left (1-a^2 x^2\right )^{3/2}}dx}{a^4}-\frac {2 (3 a x+5)}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {a x+1}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\) |
\(\Big \downarrow \) 453 |
\(\displaystyle \frac {\frac {a x+1}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}+\frac {1}{5} \left (\frac {a x+5}{a^5 \sqrt {1-a^2 x^2}}-\frac {2 (3 a x+5)}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}\right )}{c^3}\) |
Input:
Int[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^3,x]
Output:
((1 + a*x)/(5*a^5*(1 - a^2*x^2)^(5/2)) + ((-2*(5 + 3*a*x))/(3*a^5*(1 - a^2 *x^2)^(3/2)) + (5 + a*x)/(a^5*Sqrt[1 - a^2*x^2]))/5)/c^3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-(a* d - b*c*x)/(a*b*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b, c, d}, x]
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]
Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72
method | result | size |
gosper | \(-\frac {3 a^{4} x^{4}+12 a^{3} x^{3}-12 a^{2} x^{2}-8 a x +8}{15 a^{5} c^{3} \left (a x -1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\) | \(58\) |
trager | \(-\frac {\left (3 a^{4} x^{4}+12 a^{3} x^{3}-12 a^{2} x^{2}-8 a x +8\right ) \sqrt {-a^{2} x^{2}+1}}{15 c^{3} a^{5} \left (a x -1\right )^{3} \left (a x +1\right )^{2}}\) | \(65\) |
orering | \(-\frac {\left (3 a^{4} x^{4}+12 a^{3} x^{3}-12 a^{2} x^{2}-8 a x +8\right ) \left (a x -1\right ) \left (a x +1\right )^{2}}{15 a^{5} \sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} c \,x^{2}+c \right )^{3}}\) | \(73\) |
default | \(-\frac {\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{4 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{4 \left (x -\frac {1}{a}\right )}}{a^{6}}+\frac {11 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{16 a^{6} \left (x -\frac {1}{a}\right )}-\frac {-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 a \left (x +\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 \left (x +\frac {1}{a}\right )}}{8 a^{6}}-\frac {5 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{16 a^{6} \left (x +\frac {1}{a}\right )}+\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {2 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{4 a^{7}}}{c^{3}}\) | \(371\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOS E)
Output:
-1/15/a^5/c^3/(a*x-1)/(-a^2*x^2+1)^(3/2)*(3*a^4*x^4+12*a^3*x^3-12*a^2*x^2- 8*a*x+8)
Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (69) = 138\).
Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.80 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {8 \, a^{5} x^{5} - 8 \, a^{4} x^{4} - 16 \, a^{3} x^{3} + 16 \, a^{2} x^{2} + 8 \, a x - {\left (3 \, a^{4} x^{4} + 12 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 8 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} - 8}{15 \, {\left (a^{10} c^{3} x^{5} - a^{9} c^{3} x^{4} - 2 \, a^{8} c^{3} x^{3} + 2 \, a^{7} c^{3} x^{2} + a^{6} c^{3} x - a^{5} c^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^3,x, algorithm="fr icas")
Output:
1/15*(8*a^5*x^5 - 8*a^4*x^4 - 16*a^3*x^3 + 16*a^2*x^2 + 8*a*x - (3*a^4*x^4 + 12*a^3*x^3 - 12*a^2*x^2 - 8*a*x + 8)*sqrt(-a^2*x^2 + 1) - 8)/(a^10*c^3* x^5 - a^9*c^3*x^4 - 2*a^8*c^3*x^3 + 2*a^7*c^3*x^2 + a^6*c^3*x - a^5*c^3)
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\int \frac {x^{4}}{- a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{5}}{- a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**4/(-a**2*c*x**2+c)**3,x)
Output:
(Integral(x**4/(-a**6*x**6*sqrt(-a**2*x**2 + 1) + 3*a**4*x**4*sqrt(-a**2*x **2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**5/(-a**6*x**6*sqrt(-a**2*x**2 + 1) + 3*a**4*x**4*sqrt(-a**2* x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x))/ c**3
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {{\left (a x + 1\right )} x^{4}}{{\left (a^{2} c x^{2} - c\right )}^{3} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^3,x, algorithm="ma xima")
Output:
-integrate((a*x + 1)*x^4/((a^2*c*x^2 - c)^3*sqrt(-a^2*x^2 + 1)), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {{\left (a x + 1\right )} x^{4}}{{\left (a^{2} c x^{2} - c\right )}^{3} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^3,x, algorithm="gi ac")
Output:
integrate(-(a*x + 1)*x^4/((a^2*c*x^2 - c)^3*sqrt(-a^2*x^2 + 1)), x)
Time = 14.39 (sec) , antiderivative size = 335, normalized size of antiderivative = 4.14 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {a^4\,\sqrt {1-a^2\,x^2}}{30\,\left (a^{11}\,c^3\,x^2-2\,a^{10}\,c^3\,x+a^9\,c^3\right )}-\frac {\sqrt {1-a^2\,x^2}}{24\,\left (a^7\,c^3\,x^2+2\,a^6\,c^3\,x+a^5\,c^3\right )}-\frac {\sqrt {1-a^2\,x^2}}{20\,\sqrt {-a^2}\,\left (a^3\,c^3\,\sqrt {-a^2}+3\,a^5\,c^3\,x^2\,\sqrt {-a^2}-a^6\,c^3\,x^3\,\sqrt {-a^2}-3\,a^4\,c^3\,x\,\sqrt {-a^2}\right )}-\frac {\sqrt {1-a^2\,x^2}}{4\,\left (a^7\,c^3\,x^2-2\,a^6\,c^3\,x+a^5\,c^3\right )}-\frac {13\,\sqrt {1-a^2\,x^2}}{48\,\left (a^3\,c^3\,\sqrt {-a^2}+a^4\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {113\,\sqrt {1-a^2\,x^2}}{240\,\left (a^3\,c^3\,\sqrt {-a^2}-a^4\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}} \] Input:
int((x^4*(a*x + 1))/((c - a^2*c*x^2)^3*(1 - a^2*x^2)^(1/2)),x)
Output:
(a^4*(1 - a^2*x^2)^(1/2))/(30*(a^9*c^3 - 2*a^10*c^3*x + a^11*c^3*x^2)) - ( 1 - a^2*x^2)^(1/2)/(24*(a^5*c^3 + 2*a^6*c^3*x + a^7*c^3*x^2)) - (1 - a^2*x ^2)^(1/2)/(20*(-a^2)^(1/2)*(a^3*c^3*(-a^2)^(1/2) + 3*a^5*c^3*x^2*(-a^2)^(1 /2) - a^6*c^3*x^3*(-a^2)^(1/2) - 3*a^4*c^3*x*(-a^2)^(1/2))) - (1 - a^2*x^2 )^(1/2)/(4*(a^5*c^3 - 2*a^6*c^3*x + a^7*c^3*x^2)) - (13*(1 - a^2*x^2)^(1/2 ))/(48*(a^3*c^3*(-a^2)^(1/2) + a^4*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (11 3*(1 - a^2*x^2)^(1/2))/(240*(a^3*c^3*(-a^2)^(1/2) - a^4*c^3*x*(-a^2)^(1/2) )*(-a^2)^(1/2))
Time = 0.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.73 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {8 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-8 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-8 \sqrt {-a^{2} x^{2}+1}\, a x +8 \sqrt {-a^{2} x^{2}+1}+3 a^{4} x^{4}+12 a^{3} x^{3}-12 a^{2} x^{2}-8 a x +8}{15 \sqrt {-a^{2} x^{2}+1}\, a^{5} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^3,x)
Output:
(8*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 8*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 8*sqrt( - a**2*x**2 + 1)*a*x + 8*sqrt( - a**2*x**2 + 1) + 3*a**4*x**4 + 1 2*a**3*x**3 - 12*a**2*x**2 - 8*a*x + 8)/(15*sqrt( - a**2*x**2 + 1)*a**5*c* *3*(a**3*x**3 - a**2*x**2 - a*x + 1))