Integrand size = 22, antiderivative size = 111 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {x^2}{2 c^3}-\frac {\sqrt {1-c x}}{3 c^5 \left (\frac {1}{1+c x}\right )^{5/2}}+\frac {2 \sqrt {1-c x}}{3 c^5 \left (\frac {1}{1+c x}\right )^{3/2}}-\frac {\sqrt {1-c x}}{c^5 \sqrt {\frac {1}{1+c x}}}-\frac {\log \left (1-c^2 x^2\right )}{2 c^5} \] Output:
-1/2*x^2/c^3-1/3*(-c*x+1)^(1/2)/c^5/(1/(c*x+1))^(5/2)+2/3*(-c*x+1)^(1/2)/c ^5/(1/(c*x+1))^(3/2)-(-c*x+1)^(1/2)/c^5/(1/(c*x+1))^(1/2)-1/2*ln(-c^2*x^2+ 1)/c^5
Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.62 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {3 c^2 x^2+2 \sqrt {\frac {1-c x}{1+c x}} \left (2+2 c x+c^2 x^2+c^3 x^3\right )+3 \log \left (1-c^2 x^2\right )}{6 c^5} \] Input:
Integrate[(E^ArcSech[c*x]*x^4)/(1 - c^2*x^2),x]
Output:
-1/6*(3*c^2*x^2 + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(2 + 2*c*x + c^2*x^2 + c^3*x ^3) + 3*Log[1 - c^2*x^2])/c^5
Time = 0.76 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6895, 243, 49, 2009, 2044, 111, 27, 83}
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.
\(\displaystyle \int \frac {x^4 e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx\) |
\(\Big \downarrow \) 6895 |
\(\displaystyle \frac {\int \frac {x^3}{1-c^2 x^2}dx}{c}+\frac {\int \frac {x^3 \sqrt {\frac {1}{c x+1}}}{\sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \frac {x^2}{1-c^2 x^2}dx^2}{2 c}+\frac {\int \frac {x^3 \sqrt {\frac {1}{c x+1}}}{\sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x^2-1\right )}\right )dx^2}{2 c}+\frac {\int \frac {x^3 \sqrt {\frac {1}{c x+1}}}{\sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \frac {x^3 \sqrt {\frac {1}{c x+1}}}{\sqrt {1-c x}}dx}{c}+\frac {-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}}{2 c}\) |
\(\Big \downarrow \) 2044 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {x^3}{\sqrt {1-c x} \sqrt {c x+1}}dx}{c}+\frac {-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}}{2 c}\) |
\(\Big \downarrow \) 111 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {\int -\frac {2 x}{\sqrt {1-c x} \sqrt {c x+1}}dx}{3 c^2}-\frac {x^2 \sqrt {1-c x} \sqrt {c x+1}}{3 c^2}\right )}{c}+\frac {-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}}{2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {2 \int \frac {x}{\sqrt {1-c x} \sqrt {c x+1}}dx}{3 c^2}-\frac {x^2 \sqrt {1-c x} \sqrt {c x+1}}{3 c^2}\right )}{c}+\frac {-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}}{2 c}\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {2 \sqrt {1-c x} \sqrt {c x+1}}{3 c^4}-\frac {x^2 \sqrt {1-c x} \sqrt {c x+1}}{3 c^2}\right )}{c}+\frac {-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}}{2 c}\) |
Input:
Int[(E^ArcSech[c*x]*x^4)/(1 - c^2*x^2),x]
Output:
(Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*((-2*Sqrt[1 - c*x]*Sqrt[1 + c*x])/(3*c ^4) - (x^2*Sqrt[1 - c*x]*Sqrt[1 + c*x])/(3*c^2)))/c + (-(x^2/c^2) - Log[1 - c^2*x^2]/c^4)/(2*c)
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] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_.)*((c_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[S imp[(c*(a + b*x^n)^q)^p/(a + b*x^n)^(p*q)] Int[u*(a + b*x^n)^(p*q), x], x ] /; FreeQ[{a, b, c, n, p, q}, x] && GeQ[a, 0]
Int[(E^ArcSech[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Sym bol] :> Simp[d/(a*c) Int[(d*x)^(m - 1)*(Sqrt[1/(1 + c*x)]/Sqrt[1 - c*x]), x], x] + Simp[d/c Int[(d*x)^(m - 1)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b + a*c^2, 0]
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.67
method | result | size |
default | \(-\frac {\sqrt {-\frac {x c -1}{x c}}\, x \sqrt {\frac {x c +1}{x c}}\, \left (c^{2} x^{2}+2\right )}{3 c^{4}}+\frac {-\frac {x^{2}}{2 c^{2}}-\frac {\ln \left (c^{2} x^{2}-1\right )}{2 c^{4}}}{c}\) | \(74\) |
Input:
int((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^4/(-c^2*x^2+1),x,method=_RET URNVERBOSE)
Output:
-1/3*(-(c*x-1)/x/c)^(1/2)*x*((c*x+1)/x/c)^(1/2)*(c^2*x^2+2)/c^4+1/c*(-1/2* x^2/c^2-1/2/c^4*ln(c^2*x^2-1))
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.62 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {3 \, c^{2} x^{2} + 2 \, {\left (c^{3} x^{3} + 2 \, c x\right )} \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + 3 \, \log \left (c^{2} x^{2} - 1\right )}{6 \, c^{5}} \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^4/(-c^2*x^2+1),x, algo rithm="fricas")
Output:
-1/6*(3*c^2*x^2 + 2*(c^3*x^3 + 2*c*x)*sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1 )/(c*x)) + 3*log(c^2*x^2 - 1))/c^5
\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=- \frac {\int \frac {x^{3}}{c^{2} x^{2} - 1}\, dx + \int \frac {c x^{4} \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{2} - 1}\, dx}{c} \] Input:
integrate((1/c/x+(1/c/x-1)**(1/2)*(1/c/x+1)**(1/2))*x**4/(-c**2*x**2+1),x)
Output:
-(Integral(x**3/(c**2*x**2 - 1), x) + Integral(c*x**4*sqrt(-1 + 1/(c*x))*s qrt(1 + 1/(c*x))/(c**2*x**2 - 1), x))/c
\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=\int { -\frac {x^{4} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^4/(-c^2*x^2+1),x, algo rithm="maxima")
Output:
-integrate(x, x)/c^3 - 1/2*log(c*x + 1)/c^5 - 1/2*log(c*x - 1)/c^5 - integ rate(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^3/(c^3*x^2 - c), x)
\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=\int { -\frac {x^{4} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^4/(-c^2*x^2+1),x, algo rithm="giac")
Output:
integrate(-x^4*(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/(c^2*x^2 - 1), x)
Time = 24.84 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {\ln \left (c^2\,x^2-1\right )+c^2\,x^2}{2\,c^5}-x^3\,\sqrt {\frac {1}{c\,x}-1}\,\left (\frac {\sqrt {\frac {1}{c\,x}+1}}{3\,c^2}+\frac {2\,\sqrt {\frac {1}{c\,x}+1}}{3\,c^4\,x^2}\right ) \] Input:
int(-(x^4*((1/(c*x) - 1)^(1/2)*(1/(c*x) + 1)^(1/2) + 1/(c*x)))/(c^2*x^2 - 1),x)
Output:
- (log(c^2*x^2 - 1) + c^2*x^2)/(2*c^5) - x^3*(1/(c*x) - 1)^(1/2)*((1/(c*x) + 1)^(1/2)/(3*c^2) + (2*(1/(c*x) + 1)^(1/2))/(3*c^4*x^2))
Time = 0.17 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=\frac {-2 \sqrt {c x +1}\, \sqrt {-c x +1}\, c^{2} x^{2}-4 \sqrt {c x +1}\, \sqrt {-c x +1}-3 \,\mathrm {log}\left (c^{2} x -c \right )-3 \,\mathrm {log}\left (c^{2} x +c \right )-3 c^{2} x^{2}}{6 c^{5}} \] Input:
int((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^4/(-c^2*x^2+1),x)
Output:
( - 2*sqrt(c*x + 1)*sqrt( - c*x + 1)*c**2*x**2 - 4*sqrt(c*x + 1)*sqrt( - c *x + 1) - 3*log(c**2*x - c) - 3*log(c**2*x + c) - 3*c**2*x**2)/(6*c**5)