Integrand size = 16, antiderivative size = 65 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x) \, dx=2 c \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {1}{2} a c \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {3 c \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \]
-3/2*c*arctanh((1-1/a^2/x^2)^(1/2))/a+2*c*x*(1-1/a^2/x^2)^(1/2)-1/2*a*c*x^ 2*(1-1/a^2/x^2)^(1/2)
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {c \left (a \sqrt {1-\frac {1}{a^2 x^2}} x (-4+a x)+3 \log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )\right )}{2 a} \]
Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6724, 27, 540, 534, 243, 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.
\(\displaystyle \int (c-a c x) e^{-\coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6724 |
\(\displaystyle \frac {\int \frac {c^2 \left (a-\frac {1}{x}\right )^2 x^3}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{a c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c \int \frac {\left (a-\frac {1}{x}\right )^2 x^3}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{a}\) |
\(\Big \downarrow \) 540 |
\(\displaystyle \frac {c \left (-\frac {1}{2} \int \frac {\left (4 a-\frac {3}{x}\right ) x^2}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {1}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {c \left (\frac {1}{2} \left (3 \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+4 a x \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {c \left (\frac {1}{2} \left (\frac {3}{2} \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x^2}+4 a x \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {c \left (\frac {1}{2} \left (4 a x \sqrt {1-\frac {1}{a^2 x^2}}-3 a^2 \int \frac {1}{a^2-a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {c \left (\frac {1}{2} \left (4 a x \sqrt {1-\frac {1}{a^2 x^2}}-3 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-\frac {1}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\) |
(c*(-1/2*(a^2*Sqrt[1 - 1/(a^2*x^2)]*x^2) + (4*a*Sqrt[1 - 1/(a^2*x^2)]*x - 3*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/2))/a
3.2.100.3.1 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_), 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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) , x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[-d^n Subst[Int[(d + c*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(p + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d}, x] && EqQ[a*c + d, 0] && IntegerQ[p] && In tegerQ[n]
Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.52
method | result | size |
risch | \(-\frac {\left (a x -4\right ) \left (a x +1\right ) c \sqrt {\frac {a x -1}{a x +1}}}{2 a}-\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) c \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{2 \sqrt {a^{2}}\, \left (a x -1\right )}\) | \(99\) |
default | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) c \left (\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x -\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -4 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}+4 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )\right )}{2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a \sqrt {a^{2}}}\) | \(153\) |
-1/2*(a*x-4)*(a*x+1)/a*c*((a*x-1)/(a*x+1))^(1/2)-3/2*ln(a^2*x/(a^2)^(1/2)+ (a^2*x^2-1)^(1/2))/(a^2)^(1/2)*c*((a*x-1)/(a*x+1))^(1/2)*((a*x-1)*(a*x+1)) ^(1/2)/(a*x-1)
Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {3 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 3 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{2} c x^{2} - 3 \, a c x - 4 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, a} \]
-1/2*(3*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 3*c*log(sqrt((a*x - 1)/(a*x + 1)) - 1) + (a^2*c*x^2 - 3*a*c*x - 4*c)*sqrt((a*x - 1)/(a*x + 1)))/a
\[ \int e^{-\coth ^{-1}(a x)} (c-a c x) \, dx=- c \left (\int a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}\, dx + \int \left (- \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}\right )\, dx\right ) \]
-c*(Integral(a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)), x) + Integral(-sqrt(a* x/(a*x + 1) - 1/(a*x + 1)), x))
Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (55) = 110\).
Time = 0.21 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.08 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x) \, dx=\frac {1}{2} \, a {\left (\frac {2 \, {\left (5 \, c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 3 \, c \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} - \frac {3 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac {3 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} \]
1/2*a*(2*(5*c*((a*x - 1)/(a*x + 1))^(3/2) - 3*c*sqrt((a*x - 1)/(a*x + 1))) /(2*(a*x - 1)*a^2/(a*x + 1) - (a*x - 1)^2*a^2/(a*x + 1)^2 - a^2) - 3*c*log (sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 + 3*c*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2)
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x) \, dx=\frac {3 \, c \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{2 \, {\left | a \right |}} - \frac {1}{2} \, \sqrt {a^{2} x^{2} - 1} {\left (c x \mathrm {sgn}\left (a x + 1\right ) - \frac {4 \, c \mathrm {sgn}\left (a x + 1\right )}{a}\right )} \]
3/2*c*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sgn(a*x + 1)/abs(a) - 1/2*sq rt(a^2*x^2 - 1)*(c*x*sgn(a*x + 1) - 4*c*sgn(a*x + 1)/a)
Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.48 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x) \, dx=\frac {3\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}-5\,c\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{a-\frac {2\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}}-\frac {3\,c\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \]