Integrand size = 19, antiderivative size = 103 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx=-\frac {\sqrt {1-a^2 x^2}}{2 c x^2}-\frac {2 a \sqrt {1-a^2 x^2}}{c x}+\frac {2 a^2 \sqrt {1-a^2 x^2}}{c (1-a x)}-\frac {5 a^2 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{2 c} \] Output:
-1/2*(-a^2*x^2+1)^(1/2)/c/x^2-2*a*(-a^2*x^2+1)^(1/2)/c/x+2*a^2*(-a^2*x^2+1 )^(1/2)/c/(-a*x+1)-5/2*a^2*arctanh((-a^2*x^2+1)^(1/2))/c
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx=-\frac {1+4 a x-5 a^2 x^2-8 a^3 x^3+5 a^2 x^2 \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{2 c x^2 \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^ArcTanh[a*x]/(x^3*(c - a*c*x)),x]
Output:
-1/2*(1 + 4*a*x - 5*a^2*x^2 - 8*a^3*x^3 + 5*a^2*x^2*Sqrt[1 - a^2*x^2]*ArcT anh[Sqrt[1 - a^2*x^2]])/(c*x^2*Sqrt[1 - a^2*x^2])
Time = 0.43 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {6678, 27, 564, 2338, 25, 27, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{c^2 x^3 (1-a x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {1-a^2 x^2}}{x^3 (1-a x)^2}dx}{c}\) |
\(\Big \downarrow \) 564 |
\(\displaystyle \frac {\int \frac {2 a^2 x^2+2 a x+1}{x^3 \sqrt {1-a^2 x^2}}dx+\frac {2 a^2 \sqrt {1-a^2 x^2}}{1-a x}}{c}\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle \frac {-\frac {1}{2} \int -\frac {a (5 a x+4)}{x^2 \sqrt {1-a^2 x^2}}dx+\frac {2 a^2 \sqrt {1-a^2 x^2}}{1-a x}-\frac {\sqrt {1-a^2 x^2}}{2 x^2}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {a (5 a x+4)}{x^2 \sqrt {1-a^2 x^2}}dx+\frac {2 a^2 \sqrt {1-a^2 x^2}}{1-a x}-\frac {\sqrt {1-a^2 x^2}}{2 x^2}}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{2} a \int \frac {5 a x+4}{x^2 \sqrt {1-a^2 x^2}}dx+\frac {2 a^2 \sqrt {1-a^2 x^2}}{1-a x}-\frac {\sqrt {1-a^2 x^2}}{2 x^2}}{c}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {\frac {1}{2} a \left (5 a \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\frac {4 \sqrt {1-a^2 x^2}}{x}\right )+\frac {2 a^2 \sqrt {1-a^2 x^2}}{1-a x}-\frac {\sqrt {1-a^2 x^2}}{2 x^2}}{c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{2} a \left (\frac {5}{2} a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\frac {4 \sqrt {1-a^2 x^2}}{x}\right )+\frac {2 a^2 \sqrt {1-a^2 x^2}}{1-a x}-\frac {\sqrt {1-a^2 x^2}}{2 x^2}}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {1}{2} a \left (-\frac {5 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a}-\frac {4 \sqrt {1-a^2 x^2}}{x}\right )+\frac {2 a^2 \sqrt {1-a^2 x^2}}{1-a x}-\frac {\sqrt {1-a^2 x^2}}{2 x^2}}{c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {1}{2} a \left (-5 a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {4 \sqrt {1-a^2 x^2}}{x}\right )+\frac {2 a^2 \sqrt {1-a^2 x^2}}{1-a x}-\frac {\sqrt {1-a^2 x^2}}{2 x^2}}{c}\) |
Input:
Int[E^ArcTanh[a*x]/(x^3*(c - a*c*x)),x]
Output:
(-1/2*Sqrt[1 - a^2*x^2]/x^2 + (2*a^2*Sqrt[1 - a^2*x^2])/(1 - a*x) + (a*((- 4*Sqrt[1 - a^2*x^2])/x - 5*a*ArcTanh[Sqrt[1 - a^2*x^2]]))/2)/c
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 ] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b ^(n + 2)*(c + d*x))), x] - Simp[d^(2*n + 2)/b^(n + 1) Int[(x^m/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] && EqQ[b*c^2 + a*d^ 2, 0] && ILtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( m + 1)) Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96
method | result | size |
default | \(-\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}+\frac {5 a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {2 a \sqrt {-a^{2} x^{2}+1}}{x}+\frac {2 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}}{c}\) | \(99\) |
risch | \(\frac {4 a^{3} x^{3}+a^{2} x^{2}-4 a x -1}{2 x^{2} \sqrt {-a^{2} x^{2}+1}\, c}+\frac {a^{2} \left (-5 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {4 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}\right )}{2 c}\) | \(108\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^3/(-a*c*x+c),x,method=_RETURNVERBOSE)
Output:
-1/c*(1/2*(-a^2*x^2+1)^(1/2)/x^2+5/2*a^2*arctanh(1/(-a^2*x^2+1)^(1/2))+2*a *(-a^2*x^2+1)^(1/2)/x+2*a/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx=\frac {4 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + 5 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (8 \, a^{2} x^{2} - 3 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1}}{2 \, {\left (a c x^{3} - c x^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^3/(-a*c*x+c),x, algorithm="fricas")
Output:
1/2*(4*a^3*x^3 - 4*a^2*x^2 + 5*(a^3*x^3 - a^2*x^2)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (8*a^2*x^2 - 3*a*x - 1)*sqrt(-a^2*x^2 + 1))/(a*c*x^3 - c*x^2)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx=- \frac {\int \frac {a x}{a x^{4} \sqrt {- a^{2} x^{2} + 1} - x^{3} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a x^{4} \sqrt {- a^{2} x^{2} + 1} - x^{3} \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x**3/(-a*c*x+c),x)
Output:
-(Integral(a*x/(a*x**4*sqrt(-a**2*x**2 + 1) - x**3*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a*x**4*sqrt(-a**2*x**2 + 1) - x**3*sqrt(-a**2*x**2 + 1)), x))/c
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx=\int { -\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )} x^{3}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^3/(-a*c*x+c),x, algorithm="maxima")
Output:
-integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(a*c*x - c)*x^3), x)
Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (90) = 180\).
Time = 0.14 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.17 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx=-\frac {{\left (a^{3} + \frac {7 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a}{x} - \frac {40 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a x^{2}}\right )} a^{4} x^{2}}{8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} - \frac {5 \, a^{3} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, c {\left | a \right |}} - \frac {\frac {8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a c {\left | a \right |}}{x} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c {\left | a \right |}}{a x^{2}}}{8 \, a^{2} c^{2}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^3/(-a*c*x+c),x, algorithm="giac")
Output:
-1/8*(a^3 + 7*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a/x - 40*(sqrt(-a^2*x^2 + 1) *abs(a) + a)^2/(a*x^2))*a^4*x^2/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c*((sqr t(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a)) - 5/2*a^3*log(1/2*abs(-2* sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs(a)) - 1/8*(8*(sqrt(- a^2*x^2 + 1)*abs(a) + a)*a*c*abs(a)/x + (sqrt(-a^2*x^2 + 1)*abs(a) + a)^2* c*abs(a)/(a*x^2))/(a^2*c^2)
Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx=-\frac {\sqrt {1-a^2\,x^2}}{2\,c\,x^2}-\frac {2\,a\,\sqrt {1-a^2\,x^2}}{c\,x}-\frac {2\,a^3\,\sqrt {1-a^2\,x^2}}{\left (\frac {c\,\sqrt {-a^2}}{a}-c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {a^2\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{2\,c} \] Input:
int((a*x + 1)/(x^3*(1 - a^2*x^2)^(1/2)*(c - a*c*x)),x)
Output:
(a^2*atan((1 - a^2*x^2)^(1/2)*1i)*5i)/(2*c) - (1 - a^2*x^2)^(1/2)/(2*c*x^2 ) - (2*a*(1 - a^2*x^2)^(1/2))/(c*x) - (2*a^3*(1 - a^2*x^2)^(1/2))/(((c*(-a ^2)^(1/2))/a - c*x*(-a^2)^(1/2))*(-a^2)^(1/2))
Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)} \, dx=\frac {a^{2} \left (20 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-20 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{5}+7 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}-48 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}+7 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )+1\right )}{8 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2} c \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^3/(-a*c*x+c),x)
Output:
(a**2*(20*log(tan(asin(a*x)/2))*tan(asin(a*x)/2)**3 - 20*log(tan(asin(a*x) /2))*tan(asin(a*x)/2)**2 + tan(asin(a*x)/2)**5 + 7*tan(asin(a*x)/2)**4 - 4 8*tan(asin(a*x)/2)**3 + 7*tan(asin(a*x)/2) + 1))/(8*tan(asin(a*x)/2)**2*c* (tan(asin(a*x)/2) - 1))