Integrand size = 21, antiderivative size = 68 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx=\frac {2 c \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}-2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right ) \] Output:
2*c*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(1/2)-2*c^(1/2)*arctanh(c^(1/2)*(-a^2*x^ 2+1)^(1/2)/(-a*c*x+c)^(1/2))
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx=\frac {\sqrt {c-a c x} \left (2 \sqrt {1+a x}-2 \text {arctanh}\left (\sqrt {1+a x}\right )\right )}{\sqrt {1-a x}} \] Input:
Integrate[(E^ArcTanh[a*x]*Sqrt[c - a*c*x])/x,x]
Output:
(Sqrt[c - a*c*x]*(2*Sqrt[1 + a*x] - 2*ArcTanh[Sqrt[1 + a*x]]))/Sqrt[1 - a* x]
Time = 0.55 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6678, 576, 573, 219}
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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{x \sqrt {c-a c x}}dx\) |
\(\Big \downarrow \) 576 |
\(\displaystyle c \left (\frac {\int \frac {\sqrt {c-a c x}}{x \sqrt {1-a^2 x^2}}dx}{c}+\frac {2 \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right )\) |
\(\Big \downarrow \) 573 |
\(\displaystyle c \left (\frac {2 \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}-2 \int \frac {1}{1-\frac {c \left (1-a^2 x^2\right )}{c-a c x}}d\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle c \left (\frac {2 \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right )}{\sqrt {c}}\right )\) |
Input:
Int[(E^ArcTanh[a*x]*Sqrt[c - a*c*x])/x,x]
Output:
c*((2*Sqrt[1 - a^2*x^2])/Sqrt[c - a*c*x] - (2*ArcTanh[(Sqrt[c]*Sqrt[1 - a^ 2*x^2])/Sqrt[c - a*c*x]])/Sqrt[c])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[(c_) + (d_.)*(x_)]/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp[-2*c Subst[Int[1/(a - c*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_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(e*x)^(m + 1))*(c + d*x)^n*((a + b*x^2)^p/(e*(n - m - 1 ))), x] - Simp[b*c*(n/(d^2*(n - m - 1))) Int[(e*x)^m*(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c^2 + a* d^2, 0] && EqQ[n + p, 0] && GtQ[p, 0] && NeQ[m - n + 1, 0] && !IGtQ[m, 0] && !(IntegerQ[m + p] && LtQ[m + p + 2, 0]) && RationalQ[m]
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 = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {2 \sqrt {-c \left (a x -1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (\sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}}{\sqrt {c}}\right )-\sqrt {c \left (a x +1\right )}\right )}{\left (a x -1\right ) \sqrt {c \left (a x +1\right )}}\) | \(71\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(1/2)/x,x,method=_RETURNVERBOSE)
Output:
2*(-c*(a*x-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(c^(1/2)*arctanh((c*(a*x+1))^(1/2) /c^(1/2))-(c*(a*x+1))^(1/2))/(a*x-1)/(c*(a*x+1))^(1/2)
Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.63 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx=\left [\frac {{\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + a c x + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{a x^{2} - x}\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{a x - 1}, -\frac {2 \, {\left ({\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}\right )}}{a x - 1}\right ] \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(1/2)/x,x, algorithm="fric as")
Output:
[((a*x - 1)*sqrt(c)*log(-(a^2*c*x^2 + a*c*x + 2*sqrt(-a^2*x^2 + 1)*sqrt(-a *c*x + c)*sqrt(c) - 2*c)/(a*x^2 - x)) - 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a*x - 1), -2*((a*x - 1)*sqrt(-c)*arctan(sqrt(-a^2*x^2 + 1)*sqrt(-a*c *x + c)*sqrt(-c)/(a*c*x - c)) + sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a*x - 1)]
\[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx=\int \frac {\sqrt {- c \left (a x - 1\right )} \left (a x + 1\right )}{x \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)**(1/2)/x,x)
Output:
Integral(sqrt(-c*(a*x - 1))*(a*x + 1)/(x*sqrt(-(a*x - 1)*(a*x + 1))), x)
\[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx=\int { \frac {\sqrt {-a c x + c} {\left (a x + 1\right )}}{\sqrt {-a^{2} x^{2} + 1} x} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(1/2)/x,x, algorithm="maxi ma")
Output:
sqrt(c)*integrate(1/(sqrt(a*x + 1)*x), x) + 2*(a*sqrt(c)*x + sqrt(c))/sqrt (a*x + 1)
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.22 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx=\frac {2 \, c^{3} {\left (\frac {\arctan \left (\frac {\sqrt {a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} - \frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c}}{\sqrt {-c}}\right ) + \sqrt {2} \sqrt {-c}}{\sqrt {-c} c^{\frac {3}{2}}} + \frac {\sqrt {a c x + c}}{c^{2}}\right )}}{{\left | c \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(1/2)/x,x, algorithm="giac ")
Output:
2*c^3*(arctan(sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c) - (sqrt(c)*arctan(sqr t(2)*sqrt(c)/sqrt(-c)) + sqrt(2)*sqrt(-c))/(sqrt(-c)*c^(3/2)) + sqrt(a*c*x + c)/c^2)/abs(c)
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx=\int \frac {\sqrt {c-a\,c\,x}\,\left (a\,x+1\right )}{x\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int(((c - a*c*x)^(1/2)*(a*x + 1))/(x*(1 - a^2*x^2)^(1/2)),x)
Output:
int(((c - a*c*x)^(1/2)*(a*x + 1))/(x*(1 - a^2*x^2)^(1/2)), x)
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-a c x}}{x} \, dx=\sqrt {c}\, \left (2 \sqrt {a x +1}+\mathrm {log}\left (\frac {2 \sqrt {a x +1}-2}{\sqrt {2}}\right )-\mathrm {log}\left (\frac {2 \sqrt {a x +1}+2}{\sqrt {2}}\right )\right ) \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(1/2)/x,x)
Output:
sqrt(c)*(2*sqrt(a*x + 1) + log((2*sqrt(a*x + 1) - 2)/sqrt(2)) - log((2*sqr t(a*x + 1) + 2)/sqrt(2)))