Integrand size = 22, antiderivative size = 93 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx=-\frac {x}{c^3}-\frac {\sqrt {1-c x}}{2 c^4 \left (\frac {1}{1+c x}\right )^{3/2}}+\frac {\sqrt {1-c x}}{2 c^4 \sqrt {\frac {1}{1+c x}}}+\frac {\csc ^{-1}\left (\sqrt {2} \sqrt {\frac {1}{1+c x}}\right )}{c^4}+\frac {\text {arctanh}(c x)}{c^4} \] Output:
-x/c^3-1/2*(-c*x+1)^(1/2)/c^4/(1/(c*x+1))^(3/2)+1/2*(-c*x+1)^(1/2)/c^4/(1/ (c*x+1))^(1/2)+arccsc(2^(1/2)*(1/(c*x+1))^(1/2))/c^4+arctanh(c*x)/c^4
Result contains complex when optimal does not.
Time = 0.23 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx=-\frac {2 c x+c x \sqrt {\frac {1-c x}{1+c x}}+c^2 x^2 \sqrt {\frac {1-c x}{1+c x}}+\log (1-c x)-\log (1+c x)-i \log \left (-2 i c x+2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)\right )}{2 c^4} \] Input:
Integrate[(E^ArcSech[c*x]*x^3)/(1 - c^2*x^2),x]
Output:
-1/2*(2*c*x + c*x*Sqrt[(1 - c*x)/(1 + c*x)] + c^2*x^2*Sqrt[(1 - c*x)/(1 + c*x)] + Log[1 - c*x] - Log[1 + c*x] - I*Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/ (1 + c*x)]*(1 + c*x)])/c^4
Time = 0.64 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6895, 262, 219, 2044, 101, 25, 39, 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.
\(\displaystyle \int \frac {x^3 e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx\) |
\(\Big \downarrow \) 6895 |
\(\displaystyle \frac {\int \frac {x^2}{1-c^2 x^2}dx}{c}+\frac {\int \frac {x^2 \sqrt {\frac {1}{c x+1}}}{\sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\int \frac {1}{1-c^2 x^2}dx}{c^2}-\frac {x}{c^2}}{c}+\frac {\int \frac {x^2 \sqrt {\frac {1}{c x+1}}}{\sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\int \frac {x^2 \sqrt {\frac {1}{c x+1}}}{\sqrt {1-c x}}dx}{c}+\frac {\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}}{c}\) |
\(\Big \downarrow \) 2044 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {x^2}{\sqrt {1-c x} \sqrt {c x+1}}dx}{c}+\frac {\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}}{c}\) |
\(\Big \downarrow \) 101 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {\int -\frac {1}{\sqrt {1-c x} \sqrt {c x+1}}dx}{2 c^2}-\frac {x \sqrt {1-c x} \sqrt {c x+1}}{2 c^2}\right )}{c}+\frac {\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\int \frac {1}{\sqrt {1-c x} \sqrt {c x+1}}dx}{2 c^2}-\frac {x \sqrt {1-c x} \sqrt {c x+1}}{2 c^2}\right )}{c}+\frac {\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}}{c}\) |
\(\Big \downarrow \) 39 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\int \frac {1}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c x} \sqrt {c x+1}}{2 c^2}\right )}{c}+\frac {\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}}{c}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\arcsin (c x)}{2 c^3}-\frac {x \sqrt {1-c x} \sqrt {c x+1}}{2 c^2}\right )}{c}+\frac {\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}}{c}\) |
Input:
Int[(E^ArcSech[c*x]*x^3)/(1 - c^2*x^2),x]
Output:
(Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*(-1/2*(x*Sqrt[1 - c*x]*Sqrt[1 + c*x])/ c^2 + ArcSin[c*x]/(2*c^3)))/c + (-(x/c^2) + ArcTanh[c*x]/c^3)/c
Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[( a*c + b*d*x^2)^m, x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 0] && ( IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[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[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_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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 = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.31
method | result | size |
default | \(-\frac {\sqrt {-\frac {x c -1}{x c}}\, x \sqrt {\frac {x c +1}{x c}}\, \left (x \sqrt {-c^{2} x^{2}+1}\, \operatorname {csgn}\left (c \right ) c -\arctan \left (\frac {\operatorname {csgn}\left (c \right ) c x}{\sqrt {-c^{2} x^{2}+1}}\right )\right ) \operatorname {csgn}\left (c \right )}{2 c^{3} \sqrt {-c^{2} x^{2}+1}}+\frac {-\frac {x}{c^{2}}+\frac {\ln \left (x c +1\right )}{2 c^{3}}-\frac {\ln \left (x c -1\right )}{2 c^{3}}}{c}\) | \(122\) |
Input:
int((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^3/(-c^2*x^2+1),x,method=_RET URNVERBOSE)
Output:
-1/2*(-(c*x-1)/x/c)^(1/2)*x*((c*x+1)/x/c)^(1/2)/c^3*(x*(-c^2*x^2+1)^(1/2)* csgn(c)*c-arctan(csgn(c)*c*x/(-c^2*x^2+1)^(1/2)))/(-c^2*x^2+1)^(1/2)*csgn( c)+1/c*(-x/c^2+1/2/c^3*ln(c*x+1)-1/2/c^3*ln(c*x-1))
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx=-\frac {c^{2} x^{2} \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + 2 \, c x + \arctan \left (\sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}}\right ) - \log \left (c x + 1\right ) + \log \left (c x - 1\right )}{2 \, c^{4}} \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^3/(-c^2*x^2+1),x, algo rithm="fricas")
Output:
-1/2*(c^2*x^2*sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c*x)) + 2*c*x + arcta n(sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c*x))) - log(c*x + 1) + log(c*x - 1))/c^4
\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx=- \frac {\int \frac {x^{2}}{c^{2} x^{2} - 1}\, dx + \int \frac {c x^{3} \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{2} - 1}\, dx}{c} \] Input:
integrate((1/c/x+(1/c/x-1)**(1/2)*(1/c/x+1)**(1/2))*x**3/(-c**2*x**2+1),x)
Output:
-(Integral(x**2/(c**2*x**2 - 1), x) + Integral(c*x**3*sqrt(-1 + 1/(c*x))*s qrt(1 + 1/(c*x))/(c**2*x**2 - 1), x))/c
\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx=\int { -\frac {x^{3} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^3/(-c^2*x^2+1),x, algo rithm="maxima")
Output:
-x/c^3 + 1/2*log(c*x + 1)/c^4 - 1/2*log(c*x - 1)/c^4 - integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^2/(c^3*x^2 - c), x)
\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx=\int { -\frac {x^{3} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^3/(-c^2*x^2+1),x, algo rithm="giac")
Output:
integrate(-x^3*(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/(c^2*x^2 - 1), x)
Time = 33.38 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.66 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx=\frac {\mathrm {atanh}\left (c\,x\right )-c\,x}{c^4}-\frac {\ln \left (\frac {\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{c\,x}+1}-1}\right )\,1{}\mathrm {i}}{2\,c^4}-\frac {\frac {1{}\mathrm {i}}{32\,c^4}+\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,c^4\,{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,c^4\,{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^4}}{\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^6}}+\frac {\ln \left (\frac {2\,c\,\sqrt {\frac {c+\frac {1}{x}}{c}}-\frac {2}{x}+c\,\sqrt {-\frac {c-\frac {1}{x}}{c}}\,2{}\mathrm {i}}{2\,c+\frac {1}{x}-2\,c\,\sqrt {\frac {c+\frac {1}{x}}{c}}}\right )\,1{}\mathrm {i}}{2\,c^4}-\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{32\,c^4\,{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^2} \] Input:
int(-(x^3*((1/(c*x) - 1)^(1/2)*(1/(c*x) + 1)^(1/2) + 1/(c*x)))/(c^2*x^2 - 1),x)
Output:
(atanh(c*x) - c*x)/c^4 - (log(((1/(c*x) - 1)^(1/2) - 1i)/((1/(c*x) + 1)^(1 /2) - 1))*1i)/(2*c^4) - (1i/(32*c^4) + (((1/(c*x) - 1)^(1/2) - 1i)^2*1i)/( 16*c^4*((1/(c*x) + 1)^(1/2) - 1)^2) - (((1/(c*x) - 1)^(1/2) - 1i)^4*15i)/( 32*c^4*((1/(c*x) + 1)^(1/2) - 1)^4))/(((1/(c*x) - 1)^(1/2) - 1i)^2/((1/(c* x) + 1)^(1/2) - 1)^2 + (2*((1/(c*x) - 1)^(1/2) - 1i)^4)/((1/(c*x) + 1)^(1/ 2) - 1)^4 + ((1/(c*x) - 1)^(1/2) - 1i)^6/((1/(c*x) + 1)^(1/2) - 1)^6) + (l og((c*(-(c - 1/x)/c)^(1/2)*2i - 2/x + 2*c*((c + 1/x)/c)^(1/2))/(2*c + 1/x - 2*c*((c + 1/x)/c)^(1/2)))*1i)/(2*c^4) - (((1/(c*x) - 1)^(1/2) - 1i)^2*1i )/(32*c^4*((1/(c*x) + 1)^(1/2) - 1)^2)
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx=\frac {-2 \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right )-\sqrt {c x +1}\, \sqrt {-c x +1}\, c x -\mathrm {log}\left (c^{2} x -c \right )+\mathrm {log}\left (c^{2} x +c \right )-2 c x}{2 c^{4}} \] Input:
int((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))*x^3/(-c^2*x^2+1),x)
Output:
( - 2*asin(sqrt( - c*x + 1)/sqrt(2)) - sqrt(c*x + 1)*sqrt( - c*x + 1)*c*x - log(c**2*x - c) + log(c**2*x + c) - 2*c*x)/(2*c**4)