Integrand size = 22, antiderivative size = 61 \[ \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{a c}+\frac {\sqrt {1-a^2 x^2}}{a c (1+a x)}+\frac {\arcsin (a x)}{a c} \] Output:
(-a^2*x^2+1)^(1/2)/a/c+(-a^2*x^2+1)^(1/2)/a/c/(a*x+1)+arcsin(a*x)/a/c
Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {2-a x-a^2 x^2+\sqrt {1-a^2 x^2} \arcsin (a x)}{a c \sqrt {1-a^2 x^2}} \] Input:
Integrate[1/(E^ArcTanh[a*x]*(c - c/(a^2*x^2))),x]
Output:
(2 - a*x - a^2*x^2 + Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(a*c*Sqrt[1 - a^2*x^2] )
Time = 0.61 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6707, 6699, 527, 455, 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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle -\frac {a^2 \int \frac {e^{-\text {arctanh}(a x)} x^2}{1-a^2 x^2}dx}{c}\) |
\(\Big \downarrow \) 6699 |
\(\displaystyle -\frac {a^2 \int \frac {x^2 (1-a x)}{\left (1-a^2 x^2\right )^{3/2}}dx}{c}\) |
\(\Big \downarrow \) 527 |
\(\displaystyle -\frac {a^2 \left (-\frac {\int \frac {1-a x}{\sqrt {1-a^2 x^2}}dx}{a^2}-\frac {1-a x}{a^3 \sqrt {1-a^2 x^2}}\right )}{c}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle -\frac {a^2 \left (-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a}}{a^2}-\frac {1-a x}{a^3 \sqrt {1-a^2 x^2}}\right )}{c}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {a^2 \left (-\frac {\frac {\sqrt {1-a^2 x^2}}{a}+\frac {\arcsin (a x)}{a}}{a^2}-\frac {1-a x}{a^3 \sqrt {1-a^2 x^2}}\right )}{c}\) |
Input:
Int[1/(E^ArcTanh[a*x]*(c - c/(a^2*x^2))),x]
Output:
-((a^2*(-((1 - a*x)/(a^3*Sqrt[1 - a^2*x^2])) - (Sqrt[1 - a^2*x^2]/a + ArcS in[a*x]/a)/a^2))/c)
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_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[((x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.))/((a_) + (b_.)*(x_)^2)^(3/2), x_S ymbol] :> Simp[(-2^(n - 1))*c^(m + n - 2)*((c + d*x)/(b*d^(m - 1)*Sqrt[a + b*x^2])), x] + Simp[1/(b*d^(m - 2)) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[( 2^(n - 1)*c^(m + n - 1) - d^m*x^m*(c + d*x)^(n - 1))/(c - d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && EqQ[b*c^2 + a*d^2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^p Int[x^m*((1 - a^2*x^2)^(p + n/2)/(1 - a*x)^n), x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c , 0]) && ILtQ[(n - 1)/2, 0] && !IntegerQ[p - n/2]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.72
method | result | size |
risch | \(-\frac {a^{2} x^{2}-1}{a c \sqrt {-a^{2} x^{2}+1}}-\frac {\left (-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a^{4} \left (x +\frac {1}{a}\right )}\right ) a^{2}}{c}\) | \(105\) |
default | \(\frac {a^{2} \left (-\frac {-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}-a \left (\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{2 a^{4}}+\frac {\frac {3 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4}+\frac {3 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{4 \sqrt {a^{2}}}}{a^{3}}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{4 a^{3}}\right )}{c}\) | \(258\) |
Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2),x,method=_RETURNVERBOSE)
Output:
-1/a/c*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)-(-1/a^2/(a^2)^(1/2)*arctan((a^2)^(1/ 2)*x/(-a^2*x^2+1)^(1/2))-1/a^4/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)) /c*a^2
Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {2 \, a x - 2 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x + 2\right )} + 2}{a^{2} c x + a c} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2),x, algorithm="fricas" )
Output:
(2*a*x - 2*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^ 2 + 1)*(a*x + 2) + 2)/(a^2*c*x + a*c)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {a^{2} \int \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + a^{2} x^{2} - a x - 1}\, dx}{c} \] Input:
integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(c-c/a**2/x**2),x)
Output:
a**2*Integral(x**2*sqrt(-a**2*x**2 + 1)/(a**3*x**3 + a**2*x**2 - a*x - 1), x)/c
\[ \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}} \,d x } \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2),x, algorithm="maxima" )
Output:
integrate(sqrt(-a^2*x^2 + 1)/((a*x + 1)*(c - c/(a^2*x^2))), x)
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {\arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{c {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a c} - \frac {2}{c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2),x, algorithm="giac")
Output:
arcsin(a*x)*sgn(a)/(c*abs(a)) + sqrt(-a^2*x^2 + 1)/(a*c) - 2/(c*((sqrt(-a^ 2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))
Time = 22.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c\,\sqrt {-a^2}}+\frac {\sqrt {1-a^2\,x^2}}{a\,c}-\frac {\sqrt {1-a^2\,x^2}}{c\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \] Input:
int((1 - a^2*x^2)^(1/2)/((c - c/(a^2*x^2))*(a*x + 1)),x)
Output:
asinh(x*(-a^2)^(1/2))/(c*(-a^2)^(1/2)) + (1 - a^2*x^2)^(1/2)/(a*c) - (1 - a^2*x^2)^(1/2)/(c*(x*(-a^2)^(1/2) + (-a^2)^(1/2)/a)*(-a^2)^(1/2))
Time = 0.15 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-\text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {\sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-\mathit {asin} \left (a x \right ) a x -\mathit {asin} \left (a x \right )-\sqrt {-a^{2} x^{2}+1}\, a x -2 \sqrt {-a^{2} x^{2}+1}-a^{2} x^{2}-a x +2}{a c \left (\sqrt {-a^{2} x^{2}+1}-a x -1\right )} \] Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2),x)
Output:
(sqrt( - a**2*x**2 + 1)*asin(a*x) - asin(a*x)*a*x - asin(a*x) - sqrt( - a* *2*x**2 + 1)*a*x - 2*sqrt( - a**2*x**2 + 1) - a**2*x**2 - a*x + 2)/(a*c*(s qrt( - a**2*x**2 + 1) - a*x - 1))