Integrand size = 21, antiderivative size = 59 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx=\frac {x}{c^3}+\frac {\sqrt {1+\frac {1}{c^2 x^2}} x^2}{2 c^2}-\frac {\arctan (c x)}{c^4}-\frac {\text {arctanh}\left (\sqrt {1+\frac {1}{c^2 x^2}}\right )}{2 c^4} \] Output:
x/c^3+1/2*(1+1/c^2/x^2)^(1/2)*x^2/c^2-arctan(c*x)/c^4-1/2*arctanh((1+1/c^2 /x^2)^(1/2))/c^4
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx=-\frac {-c x \left (2+c \sqrt {1+\frac {1}{c^2 x^2}} x\right )+2 \arctan (c x)+\log \left (\left (1+\sqrt {1+\frac {1}{c^2 x^2}}\right ) x\right )}{2 c^4} \] Input:
Integrate[(E^ArcCsch[c*x]*x^3)/(1 + c^2*x^2),x]
Output:
-1/2*(-(c*x*(2 + c*Sqrt[1 + 1/(c^2*x^2)]*x)) + 2*ArcTan[c*x] + Log[(1 + Sq rt[1 + 1/(c^2*x^2)])*x])/c^4
Time = 0.48 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6896, 262, 216, 798, 52, 73, 221}
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.
\(\displaystyle \int \frac {x^3 e^{\text {csch}^{-1}(c x)}}{c^2 x^2+1} \, dx\) |
\(\Big \downarrow \) 6896 |
\(\displaystyle \frac {\int \frac {x}{\sqrt {1+\frac {1}{c^2 x^2}}}dx}{c^2}+\frac {\int \frac {x^2}{c^2 x^2+1}dx}{c}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\int \frac {x}{\sqrt {1+\frac {1}{c^2 x^2}}}dx}{c^2}+\frac {\frac {x}{c^2}-\frac {\int \frac {1}{c^2 x^2+1}dx}{c^2}}{c}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\int \frac {x}{\sqrt {1+\frac {1}{c^2 x^2}}}dx}{c^2}+\frac {\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}}{c}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}}{c}-\frac {\int \frac {x^4}{\sqrt {1+\frac {1}{c^2 x^2}}}d\frac {1}{x^2}}{2 c^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}}{c}-\frac {x^2 \left (-\sqrt {\frac {1}{c^2 x^2}+1}\right )-\frac {\int \frac {x^2}{\sqrt {1+\frac {1}{c^2 x^2}}}d\frac {1}{x^2}}{2 c^2}}{2 c^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}}{c}-\frac {x^2 \left (-\sqrt {\frac {1}{c^2 x^2}+1}\right )-\int \frac {1}{\frac {c^2}{x^4}-c^2}d\sqrt {1+\frac {1}{c^2 x^2}}}{2 c^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}}{c}-\frac {\frac {\text {arctanh}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{c^2}-x^2 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}\) |
Input:
Int[(E^ArcCsch[c*x]*x^3)/(1 + c^2*x^2),x]
Output:
(x/c^2 - ArcTan[c*x]/c^3)/c - (-(Sqrt[1 + 1/(c^2*x^2)]*x^2) + ArcTanh[Sqrt [1 + 1/(c^2*x^2)]]/c^2)/(2*c^2)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(E^ArcCsch[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Sym bol] :> Simp[d^2/(a*c^2) Int[(d*x)^(m - 2)/Sqrt[1 + 1/(c^2*x^2)], 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]
Leaf count of result is larger than twice the leaf count of optimal. \(137\) vs. \(2(51)=102\).
Time = 0.39 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.34
method | result | size |
default | \(\frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, x \left (x \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{2}+\ln \left (x +\sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\right )-2 \ln \left (x +\sqrt {-\frac {\left (-c^{2} x +\sqrt {-c^{2}}\right ) \left (c^{2} x +\sqrt {-c^{2}}\right )}{c^{4}}}\right )\right )}{2 \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{4}}+\frac {\frac {x}{c^{2}}-\frac {\arctan \left (x c \right )}{c^{3}}}{c}\) | \(138\) |
Input:
int((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x,method=_RETURNVERBOSE)
Output:
1/2*((c^2*x^2+1)/c^2/x^2)^(1/2)*x*(x*((c^2*x^2+1)/c^2)^(1/2)*c^2+ln(x+((c^ 2*x^2+1)/c^2)^(1/2))-2*ln(x+(-(-c^2*x+(-c^2)^(1/2))*(c^2*x+(-c^2)^(1/2))/c ^4)^(1/2)))/((c^2*x^2+1)/c^2)^(1/2)/c^4+1/c*(x/c^2-1/c^3*arctan(x*c))
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx=\frac {c^{2} x^{2} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 2 \, c x - 2 \, \arctan \left (c x\right ) + \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right )}{2 \, c^{4}} \] Input:
integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x, algorithm="fricas ")
Output:
1/2*(c^2*x^2*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 2*c*x - 2*arctan(c*x) + log(c *x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x))/c^4
\[ \int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx=\frac {\int \frac {x^{2}}{c^{2} x^{2} + 1}\, dx + \int \frac {c x^{3} \sqrt {1 + \frac {1}{c^{2} x^{2}}}}{c^{2} x^{2} + 1}\, dx}{c} \] Input:
integrate((1/c/x+(1+1/c**2/x**2)**(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**2*x**2 ))/(c**2*x**2 + 1), x))/c
Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (51) = 102\).
Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx=\frac {x}{c^{3}} + \frac {\frac {2 \, \sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c {\left (\frac {c^{2} x^{2} + 1}{c^{2} x^{2}} - 1\right )}} - \log \left (\frac {\sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} + 1\right ) + \log \left (\frac {\sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} - 1\right )}{4 \, c^{4}} - \frac {\arctan \left (c x\right )}{c^{4}} \] Input:
integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x, algorithm="maxima ")
Output:
x/c^3 + 1/4*(2*sqrt((c^2*x^2 + 1)/x^2)/(c*((c^2*x^2 + 1)/(c^2*x^2) - 1)) - log(sqrt((c^2*x^2 + 1)/x^2)/c + 1) + log(sqrt((c^2*x^2 + 1)/x^2)/c - 1))/ c^4 - arctan(c*x)/c^4
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx=\frac {\sqrt {c^{2} x^{2} + 1} x {\left | c \right |} \mathrm {sgn}\left (x\right )}{2 \, c^{4}} + \frac {x}{c^{3}} + \frac {\log \left (-x {\left | c \right |} + \sqrt {c^{2} x^{2} + 1}\right ) \mathrm {sgn}\left (x\right )}{2 \, c^{4}} - \frac {\arctan \left (c x\right )}{c^{4}} \] Input:
integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x, algorithm="giac")
Output:
1/2*sqrt(c^2*x^2 + 1)*x*abs(c)*sgn(x)/c^4 + x/c^3 + 1/2*log(-x*abs(c) + sq rt(c^2*x^2 + 1))*sgn(x)/c^4 - arctan(c*x)/c^4
Time = 24.74 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx=\frac {x^2\,\sqrt {\frac {1}{c^2\,x^2}+1}}{2\,c^2}-\frac {\mathrm {atan}\left (c\,x\right )-c\,x}{c^4}-\frac {\mathrm {atanh}\left (\sqrt {\frac {1}{c^2\,x^2}+1}\right )}{2\,c^4} \] Input:
int((x^3*((1/(c^2*x^2) + 1)^(1/2) + 1/(c*x)))/(c^2*x^2 + 1),x)
Output:
(x^2*(1/(c^2*x^2) + 1)^(1/2))/(2*c^2) - (atan(c*x) - c*x)/c^4 - atanh((1/( c^2*x^2) + 1)^(1/2))/(2*c^4)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx=\frac {-2 \mathit {atan} \left (c x \right )+\sqrt {c^{2} x^{2}+1}\, c x -\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x \right )+2 c x}{2 c^{4}} \] Input:
int((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x)
Output:
( - 2*atan(c*x) + sqrt(c**2*x**2 + 1)*c*x - log(sqrt(c**2*x**2 + 1) + c*x) + 2*c*x)/(2*c**4)