Integrand size = 22, antiderivative size = 85 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=-\frac {1}{3 c x^3}-\frac {c}{x}-\frac {\sqrt {1-c x}}{3 c x^3 \sqrt {\frac {1}{1+c x}}}-\frac {2 c \sqrt {1-c x}}{3 x \sqrt {\frac {1}{1+c x}}}+c^2 \text {arctanh}(c x) \]
-1/3/c/x^3-c/x+c^2*arctanh(c*x)-1/3*(-c*x+1)^(1/2)/c/x^3/(1/(c*x+1))^(1/2) -2/3*c*(-c*x+1)^(1/2)/x/(1/(c*x+1))^(1/2)
Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=-\frac {2+6 c^2 x^2+2 \sqrt {\frac {1-c x}{1+c x}} \left (1+c x+2 c^2 x^2+2 c^3 x^3\right )+3 c^3 x^3 \log (1-c x)-3 c^3 x^3 \log (1+c x)}{6 c x^3} \]
-1/6*(2 + 6*c^2*x^2 + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x + 2*c^2*x^2 + 2 *c^3*x^3) + 3*c^3*x^3*Log[1 - c*x] - 3*c^3*x^3*Log[1 + c*x])/(c*x^3)
Time = 0.41 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.28, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6895, 264, 264, 219, 2044, 114, 27, 106}
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.
\(\displaystyle \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx\) |
\(\Big \downarrow \) 6895 |
\(\displaystyle \frac {\int \frac {1}{x^4 \left (1-c^2 x^2\right )}dx}{c}+\frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x^4 \sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {c^2 \int \frac {1}{x^2 \left (1-c^2 x^2\right )}dx-\frac {1}{3 x^3}}{c}+\frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x^4 \sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {c^2 \left (c^2 \int \frac {1}{1-c^2 x^2}dx-\frac {1}{x}\right )-\frac {1}{3 x^3}}{c}+\frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x^4 \sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x^4 \sqrt {1-c x}}dx}{c}+\frac {c^2 \left (c \text {arctanh}(c x)-\frac {1}{x}\right )-\frac {1}{3 x^3}}{c}\) |
\(\Big \downarrow \) 2044 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {1}{x^4 \sqrt {1-c x} \sqrt {c x+1}}dx}{c}+\frac {c^2 \left (c \text {arctanh}(c x)-\frac {1}{x}\right )-\frac {1}{3 x^3}}{c}\) |
\(\Big \downarrow \) 114 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {1}{3} \int -\frac {2 c^2}{x^2 \sqrt {1-c x} \sqrt {c x+1}}dx-\frac {\sqrt {1-c x} \sqrt {c x+1}}{3 x^3}\right )}{c}+\frac {c^2 \left (c \text {arctanh}(c x)-\frac {1}{x}\right )-\frac {1}{3 x^3}}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {2}{3} c^2 \int \frac {1}{x^2 \sqrt {1-c x} \sqrt {c x+1}}dx-\frac {\sqrt {1-c x} \sqrt {c x+1}}{3 x^3}\right )}{c}+\frac {c^2 \left (c \text {arctanh}(c x)-\frac {1}{x}\right )-\frac {1}{3 x^3}}{c}\) |
\(\Big \downarrow \) 106 |
\(\displaystyle \frac {c^2 \left (c \text {arctanh}(c x)-\frac {1}{x}\right )-\frac {1}{3 x^3}}{c}+\frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {2 c^2 \sqrt {1-c x} \sqrt {c x+1}}{3 x}-\frac {\sqrt {1-c x} \sqrt {c x+1}}{3 x^3}\right )}{c}\) |
(Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*(-1/3*(Sqrt[1 - c*x]*Sqrt[1 + c*x])/x^ 3 - (2*c^2*Sqrt[1 - c*x]*Sqrt[1 + c*x])/(3*x)))/c + (-1/3*1/x^3 + c^2*(-x^ (-1) + c*ArcTanh[c*x]))/c
3.1.96.3.1 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[((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 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0] && NeQ[m, -1]
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 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.73 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06
method | result | size |
default | \(-\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \operatorname {csgn}\left (c \right )^{2} \left (2 c^{2} x^{2}+1\right )}{3 x^{2}}+\frac {\frac {c^{3} \ln \left (c x +1\right )}{2}-\frac {1}{3 x^{3}}-\frac {c^{2}}{x}-\frac {c^{3} \ln \left (c x -1\right )}{2}}{c}\) | \(90\) |
-1/3*(-(c*x-1)/c/x)^(1/2)/x^2*((c*x+1)/c/x)^(1/2)*csgn(c)^2*(2*c^2*x^2+1)+ 1/c*(1/2*c^3*ln(c*x+1)-1/3/x^3-c^2/x-1/2*c^3*ln(c*x-1))
Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=\frac {3 \, c^{3} x^{3} \log \left (c x + 1\right ) - 3 \, c^{3} x^{3} \log \left (c x - 1\right ) - 6 \, c^{2} x^{2} - 2 \, {\left (2 \, c^{3} x^{3} + c x\right )} \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} - 2}{6 \, c x^{3}} \]
1/6*(3*c^3*x^3*log(c*x + 1) - 3*c^3*x^3*log(c*x - 1) - 6*c^2*x^2 - 2*(2*c^ 3*x^3 + c*x)*sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c*x)) - 2)/(c*x^3)
\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=- \frac {\int \frac {c x \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{6} - x^{4}}\, dx + \int \frac {1}{c^{2} x^{6} - x^{4}}\, dx}{c} \]
-(Integral(c*x*sqrt(-1 + 1/(c*x))*sqrt(1 + 1/(c*x))/(c**2*x**6 - x**4), x) + Integral(1/(c**2*x**6 - x**4), x))/c
\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=\int { -\frac {\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}}{{\left (c^{2} x^{2} - 1\right )} x^{3}} \,d x } \]
1/2*c^2*log(c*x + 1) - 1/2*c^2*log(c*x - 1) + c*integrate(x^(-2), x) + int egrate(x^(-4), x)/c - integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^3*x^6 - c* x^4), x)
\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=\int { -\frac {\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}}{{\left (c^{2} x^{2} - 1\right )} x^{3}} \,d x } \]
Time = 5.48 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=c^2\,\mathrm {atanh}\left (c\,x\right )-\frac {\left (\frac {\sqrt {\frac {1}{c\,x}+1}}{3}+\frac {2\,c^2\,x^2\,\sqrt {\frac {1}{c\,x}+1}}{3}\right )\,\sqrt {\frac {1}{c\,x}-1}}{x^2}-\frac {c^2\,x^2+\frac {1}{3}}{c\,x^3} \]