Integrand size = 18, antiderivative size = 82 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{a (c-a c x)^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{\sqrt {2} a c^{3/2}} \] Output:
(-a^2*x^2+1)^(1/2)/a/(-a*c*x+c)^(3/2)-1/2*arctanh(1/2*c^(1/2)*(-a^2*x^2+1) ^(1/2)*2^(1/2)/(-a*c*x+c)^(1/2))*2^(1/2)/a/c^(3/2)
Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {\sqrt {c-a c x} \left (2+2 a x+(-1+a x) \sqrt {2+2 a x} \text {arctanh}\left (\frac {\sqrt {1+a x}}{\sqrt {2}}\right )\right )}{2 a c^2 (-1+a x) \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^ArcTanh[a*x]/(c - a*c*x)^(3/2),x]
Output:
-1/2*(Sqrt[c - a*c*x]*(2 + 2*a*x + (-1 + a*x)*Sqrt[2 + 2*a*x]*ArcTanh[Sqrt [1 + a*x]/Sqrt[2]]))/(a*c^2*(-1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.46 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6677, 465, 471, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{(c-a c x)^{5/2}}dx\) |
\(\Big \downarrow \) 465 |
\(\displaystyle c \left (\frac {\sqrt {1-a^2 x^2}}{a c (c-a c x)^{3/2}}-\frac {\int \frac {1}{\sqrt {c-a c x} \sqrt {1-a^2 x^2}}dx}{2 c^2}\right )\) |
\(\Big \downarrow \) 471 |
\(\displaystyle c \left (\frac {a \int \frac {1}{\frac {a^2 c^2 \left (1-a^2 x^2\right )}{c-a c x}-2 a^2 c}d\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}}{c}+\frac {\sqrt {1-a^2 x^2}}{a c (c-a c x)^{3/2}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle c \left (\frac {\sqrt {1-a^2 x^2}}{a c (c-a c x)^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{\sqrt {2} a c^{5/2}}\right )\) |
Input:
Int[E^ArcTanh[a*x]/(c - a*c*x)^(3/2),x]
Output:
c*(Sqrt[1 - a^2*x^2]/(a*c*(c - a*c*x)^(3/2)) - ArcTanh[(Sqrt[c]*Sqrt[1 - a ^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])]/(Sqrt[2]*a*c^(5/2)))
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_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + p + 1))), x] - Simp[b*(p/(d^2*(n + p + 1))) Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n + 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim p[2*d Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] ], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[c^n Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.35
method | result | size |
default | \(\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a c x -\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +2 \sqrt {c \left (a x +1\right )}\, \sqrt {c}\right )}{2 \left (a x -1\right )^{2} \sqrt {c \left (a x +1\right )}\, c^{\frac {5}{2}} a}\) | \(111\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(2^(1/2)*arctanh(1/2*(c*(a*x+1)) ^(1/2)*2^(1/2)/c^(1/2))*a*c*x-2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2 )/c^(1/2))*c+2*(c*(a*x+1))^(1/2)*c^(1/2))/(a*x-1)^2/(c*(a*x+1))^(1/2)/c^(5 /2)/a
Time = 0.08 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.11 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\left [\frac {\sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x + 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{4 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}}, -\frac {\sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{2 \, {\left (a c x - c\right )}}\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{2 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}}\right ] \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(3/2),x, algorithm="fricas ")
Output:
[1/4*(sqrt(2)*(a^2*x^2 - 2*a*x + 1)*sqrt(c)*log(-(a^2*c*x^2 + 2*a*c*x + 2* sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a^2*x^2 - 2*a* x + 1)) + 4*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^3*c^2*x^2 - 2*a^2*c^2* x + a*c^2), -1/2*(sqrt(2)*(a^2*x^2 - 2*a*x + 1)*sqrt(-c)*arctan(1/2*sqrt(2 )*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) - 2*sqrt(-a^2* x^2 + 1)*sqrt(-a*c*x + c))/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2)]
\[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\int \frac {a x + 1}{\left (- c \left (a x - 1\right )\right )^{\frac {3}{2}} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**(3/2),x)
Output:
Integral((a*x + 1)/((-c*(a*x - 1))**(3/2)*sqrt(-(a*x - 1)*(a*x + 1))), x)
\[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\int { \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (-a c x + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(3/2),x, algorithm="maxima ")
Output:
integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(-a*c*x + c)^(3/2)), x)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\frac {\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} - \frac {2 \, \sqrt {a c x + c}}{a c x - c}}{2 \, a {\left | c \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(3/2),x, algorithm="giac")
Output:
1/2*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/sqrt(-c) - 2*sqr t(a*c*x + c)/(a*c*x - c))/(a*abs(c))
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\int \frac {a\,x+1}{\sqrt {1-a^2\,x^2}\,{\left (c-a\,c\,x\right )}^{3/2}} \,d x \] Input:
int((a*x + 1)/((1 - a^2*x^2)^(1/2)*(c - a*c*x)^(3/2)),x)
Output:
int((a*x + 1)/((1 - a^2*x^2)^(1/2)*(c - a*c*x)^(3/2)), x)
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\frac {\sqrt {c}\, \left (-2 \sqrt {a x +1}+\sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a x -\sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )\right )}{2 a \,c^{2} \left (a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(3/2),x)
Output:
(sqrt(c)*( - 2*sqrt(a*x + 1) + sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt( 2))/2))*a*x - sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2))))/(2*a*c* *2*(a*x - 1))