Integrand size = 21, antiderivative size = 61 \[ \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (1+c^2 x^2\right )} \, dx=c^2 \sqrt {1+\frac {1}{c^2 x^2}}-\frac {1}{3} c^2 \left (1+\frac {1}{c^2 x^2}\right )^{3/2}-\frac {1}{3 c x^3}+\frac {c}{x}+c^2 \arctan (c x) \] Output:
c^2*(1+1/c^2/x^2)^(1/2)-1/3*c^2*(1+1/c^2/x^2)^(3/2)-1/3/c/x^3+c/x+c^2*arct an(c*x)
Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (1+c^2 x^2\right )} \, dx=-\frac {1}{3 c x^3}+\frac {c}{x}+\frac {\sqrt {1+\frac {1}{c^2 x^2}} \left (-1+2 c^2 x^2\right )}{3 x^2}+c^2 \arctan (c x) \] Input:
Integrate[E^ArcCsch[c*x]/(x^3*(1 + c^2*x^2)),x]
Output:
-1/3*1/(c*x^3) + c/x + (Sqrt[1 + 1/(c^2*x^2)]*(-1 + 2*c^2*x^2))/(3*x^2) + c^2*ArcTan[c*x]
Time = 0.54 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6896, 264, 264, 216, 798, 53, 2009}
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.
\(\displaystyle \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (c^2 x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 6896 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^5}dx}{c^2}+\frac {\int \frac {1}{x^4 \left (c^2 x^2+1\right )}dx}{c}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^5}dx}{c^2}+\frac {c^2 \left (-\int \frac {1}{x^2 \left (c^2 x^2+1\right )}dx\right )-\frac {1}{3 x^3}}{c}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^5}dx}{c^2}+\frac {-\left (c^2 \left (c^2 \left (-\int \frac {1}{c^2 x^2+1}dx\right )-\frac {1}{x}\right )\right )-\frac {1}{3 x^3}}{c}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^5}dx}{c^2}+\frac {-\left (c^2 \left (-c \arctan (c x)-\frac {1}{x}\right )\right )-\frac {1}{3 x^3}}{c}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {-\left (c^2 \left (-c \arctan (c x)-\frac {1}{x}\right )\right )-\frac {1}{3 x^3}}{c}-\frac {\int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^2}d\frac {1}{x^2}}{2 c^2}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {-\left (c^2 \left (-c \arctan (c x)-\frac {1}{x}\right )\right )-\frac {1}{3 x^3}}{c}-\frac {\int \left (c^2 \sqrt {1+\frac {1}{c^2 x^2}}-\frac {c^2}{\sqrt {1+\frac {1}{c^2 x^2}}}\right )d\frac {1}{x^2}}{2 c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\left (c^2 \left (-c \arctan (c x)-\frac {1}{x}\right )\right )-\frac {1}{3 x^3}}{c}-\frac {\frac {2}{3} c^4 \left (\frac {1}{c^2 x^2}+1\right )^{3/2}-2 c^4 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}\) |
Input:
Int[E^ArcCsch[c*x]/(x^3*(1 + c^2*x^2)),x]
Output:
-1/2*(-2*c^4*Sqrt[1 + 1/(c^2*x^2)] + (2*c^4*(1 + 1/(c^2*x^2))^(3/2))/3)/c^ 2 + (-1/3*1/x^3 - c^2*(-x^(-1) - c*ArcTan[c*x]))/c
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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
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[((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[(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. \(196\) vs. \(2(53)=106\).
Time = 0.39 (sec) , antiderivative size = 197, normalized size of antiderivative = 3.23
method | result | size |
default | \(\frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{2} \left (3 \left (\frac {c^{2} x^{2}+1}{c^{2}}\right )^{\frac {3}{2}} c^{2} x^{2}-3 \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{2} x^{4}-3 \ln \left (x +\sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\right ) x^{3}+3 \ln \left (x +\sqrt {-\frac {\left (-c^{2} x +\sqrt {-c^{2}}\right ) \left (c^{2} x +\sqrt {-c^{2}}\right )}{c^{4}}}\right ) x^{3}-\left (\frac {c^{2} x^{2}+1}{c^{2}}\right )^{\frac {3}{2}}\right )}{3 x^{2} \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}}+\frac {c^{3} \arctan \left (x c \right )-\frac {1}{3 x^{3}}+\frac {c^{2}}{x}}{c}\) | \(197\) |
Input:
int((1/c/x+(1+1/c^2/x^2)^(1/2))/x^3/(c^2*x^2+1),x,method=_RETURNVERBOSE)
Output:
1/3*((c^2*x^2+1)/c^2/x^2)^(1/2)/x^2*c^2*(3*((c^2*x^2+1)/c^2)^(3/2)*c^2*x^2 -3*((c^2*x^2+1)/c^2)^(1/2)*c^2*x^4-3*ln(x+((c^2*x^2+1)/c^2)^(1/2))*x^3+3*l n(x+(-(-c^2*x+(-c^2)^(1/2))*(c^2*x+(-c^2)^(1/2))/c^4)^(1/2))*x^3-((c^2*x^2 +1)/c^2)^(3/2))/((c^2*x^2+1)/c^2)^(1/2)+1/c*(c^3*arctan(x*c)-1/3/x^3+c^2/x )
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (1+c^2 x^2\right )} \, dx=\frac {3 \, c^{3} x^{3} \arctan \left (c x\right ) + 2 \, c^{3} x^{3} + 3 \, c^{2} x^{2} + {\left (2 \, c^{3} x^{3} - c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - 1}{3 \, c x^{3}} \] Input:
integrate((1/c/x+(1+1/c^2/x^2)^(1/2))/x^3/(c^2*x^2+1),x, algorithm="fricas ")
Output:
1/3*(3*c^3*x^3*arctan(c*x) + 2*c^3*x^3 + 3*c^2*x^2 + (2*c^3*x^3 - c*x)*sqr t((c^2*x^2 + 1)/(c^2*x^2)) - 1)/(c*x^3)
Time = 2.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.67 \[ \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (1+c^2 x^2\right )} \, dx=- \frac {c^{3} \operatorname {atan}{\left (\frac {1}{x \sqrt {c^{2}}} \right )}}{\sqrt {c^{2}}} - c \left (\begin {cases} 2 c^{4} \left (\frac {\left (1 + \frac {1}{c^{2} x^{2}}\right )^{\frac {3}{2}}}{6 c^{3}} - \frac {\sqrt {1 + \frac {1}{c^{2} x^{2}}}}{2 c^{3}}\right ) & \text {for}\: \frac {1}{c^{2}} \neq 0 \\- \frac {c \log {\left (c^{2} + \frac {1}{x^{2}} \right )}}{2} + \frac {1}{2 c x^{2}} & \text {otherwise} \end {cases}\right ) + \frac {c}{x} - \frac {1}{3 c x^{3}} \] Input:
integrate((1/c/x+(1+1/c**2/x**2)**(1/2))/x**3/(c**2*x**2+1),x)
Output:
-c**3*atan(1/(x*sqrt(c**2)))/sqrt(c**2) - c*Piecewise((2*c**4*((1 + 1/(c** 2*x**2))**(3/2)/(6*c**3) - sqrt(1 + 1/(c**2*x**2))/(2*c**3)), Ne(c**(-2), 0)), (-c*log(c**2 + x**(-2))/2 + 1/(2*c*x**2), True)) + c/x - 1/(3*c*x**3)
Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (1+c^2 x^2\right )} \, dx=c^{2} \arctan \left (c x\right ) + \frac {{\left (2 \, c^{2} x^{2} - 1\right )} \sqrt {c^{2} x^{2} + 1}}{3 \, c x^{3}} + \frac {3 \, c^{2} x^{2} - 1}{3 \, c x^{3}} \] Input:
integrate((1/c/x+(1+1/c^2/x^2)^(1/2))/x^3/(c^2*x^2+1),x, algorithm="maxima ")
Output:
c^2*arctan(c*x) + 1/3*(2*c^2*x^2 - 1)*sqrt(c^2*x^2 + 1)/(c*x^3) + 1/3*(3*c ^2*x^2 - 1)/(c*x^3)
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.34 \[ \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (1+c^2 x^2\right )} \, dx=c^{2} \arctan \left (c x\right ) + \frac {4 \, {\left (3 \, {\left (x {\left | c \right |} - \sqrt {c^{2} x^{2} + 1}\right )}^{2} - 1\right )} c^{2} \mathrm {sgn}\left (x\right )}{3 \, {\left ({\left (x {\left | c \right |} - \sqrt {c^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{3}} + \frac {3 \, c^{2} x^{2} - 1}{3 \, c x^{3}} \] Input:
integrate((1/c/x+(1+1/c^2/x^2)^(1/2))/x^3/(c^2*x^2+1),x, algorithm="giac")
Output:
c^2*arctan(c*x) + 4/3*(3*(x*abs(c) - sqrt(c^2*x^2 + 1))^2 - 1)*c^2*sgn(x)/ ((x*abs(c) - sqrt(c^2*x^2 + 1))^2 - 1)^3 + 1/3*(3*c^2*x^2 - 1)/(c*x^3)
Time = 24.50 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (1+c^2 x^2\right )} \, dx=\frac {c+\frac {2\,c^2\,x\,\sqrt {\frac {1}{c^2\,x^2}+1}}{3}}{x}-\frac {\frac {x\,\sqrt {\frac {1}{c^2\,x^2}+1}}{3}+\frac {1}{3\,c}}{x^3}+c^2\,\mathrm {atan}\left (c\,x\right ) \] Input:
int(((1/(c^2*x^2) + 1)^(1/2) + 1/(c*x))/(x^3*(c^2*x^2 + 1)),x)
Output:
(c + (2*c^2*x*(1/(c^2*x^2) + 1)^(1/2))/3)/x - ((x*(1/(c^2*x^2) + 1)^(1/2)) /3 + 1/(3*c))/x^3 + c^2*atan(c*x)
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\text {csch}^{-1}(c x)}}{x^3 \left (1+c^2 x^2\right )} \, dx=\frac {3 \mathit {atan} \left (c x \right ) c^{3} x^{3}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}-2 c^{3} x^{3}+3 c^{2} x^{2}-1}{3 c \,x^{3}} \] Input:
int((1/c/x+(1+1/c^2/x^2)^(1/2))/x^3/(c^2*x^2+1),x)
Output:
(3*atan(c*x)*c**3*x**3 + 2*sqrt(c**2*x**2 + 1)*c**2*x**2 - sqrt(c**2*x**2 + 1) - 2*c**3*x**3 + 3*c**2*x**2 - 1)/(3*c*x**3)