3.293 \(\int \frac {e^{\tanh ^{-1}(a x)} (c-a c x)}{x^3} \, dx\)

Optimal. Leaf size=46 \[ \frac {1}{2} a^2 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {c \sqrt {1-a^2 x^2}}{2 x^2} \]

[Out]

1/2*a^2*c*arctanh((-a^2*x^2+1)^(1/2))-1/2*c*(-a^2*x^2+1)^(1/2)/x^2

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6128, 266, 47, 63, 208} \[ \frac {1}{2} a^2 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {c \sqrt {1-a^2 x^2}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x))/x^3,x]

[Out]

-(c*Sqrt[1 - a^2*x^2])/(2*x^2) + (a^2*c*ArcTanh[Sqrt[1 - a^2*x^2]])/2

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[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]

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)} (c-a c x)}{x^3} \, dx &=c \int \frac {\sqrt {1-a^2 x^2}}{x^3} \, dx\\ &=\frac {1}{2} c \operatorname {Subst}\left (\int \frac {\sqrt {1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {1}{4} \left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {1-a^2 x^2}}{2 x^2}+\frac {1}{2} c \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )\\ &=-\frac {c \sqrt {1-a^2 x^2}}{2 x^2}+\frac {1}{2} a^2 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 67, normalized size = 1.46 \[ \frac {c \left (a^2 x^2+a^2 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-1\right )}{2 x^2 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*(c - a*c*x))/x^3,x]

[Out]

(c*(-1 + a^2*x^2 + a^2*x^2*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]]))/(2*x^2*Sqrt[1 - a^2*x^2])

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 47, normalized size = 1.02 \[ -\frac {a^{2} c x^{2} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1} c}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="fricas")

[Out]

-1/2*(a^2*c*x^2*log((sqrt(-a^2*x^2 + 1) - 1)/x) + sqrt(-a^2*x^2 + 1)*c)/x^2

________________________________________________________________________________________

giac [A]  time = 0.16, size = 70, normalized size = 1.52 \[ \frac {a^{4} c \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - a^{4} c \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right ) - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{2} c}{x^{2}}}{4 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="giac")

[Out]

1/4*(a^4*c*log(sqrt(-a^2*x^2 + 1) + 1) - a^4*c*log(-sqrt(-a^2*x^2 + 1) + 1) - 2*sqrt(-a^2*x^2 + 1)*a^2*c/x^2)/
a^2

________________________________________________________________________________________

maple [A]  time = 0.04, size = 40, normalized size = 0.87 \[ -c \left (-\frac {a^{2} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x)

[Out]

-c*(-1/2*a^2*arctanh(1/(-a^2*x^2+1)^(1/2))+1/2*(-a^2*x^2+1)^(1/2)/x^2)

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 51, normalized size = 1.11 \[ \frac {1}{2} \, a^{2} c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {\sqrt {-a^{2} x^{2} + 1} c}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="maxima")

[Out]

1/2*a^2*c*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) - 1/2*sqrt(-a^2*x^2 + 1)*c/x^2

________________________________________________________________________________________

mupad [B]  time = 0.78, size = 38, normalized size = 0.83 \[ \frac {a^2\,c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{2}-\frac {c\,\sqrt {1-a^2\,x^2}}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - a*c*x)*(a*x + 1))/(x^3*(1 - a^2*x^2)^(1/2)),x)

[Out]

(a^2*c*atanh((1 - a^2*x^2)^(1/2)))/2 - (c*(1 - a^2*x^2)^(1/2))/(2*x^2)

________________________________________________________________________________________

sympy [A]  time = 11.45, size = 73, normalized size = 1.59 \[ - a^{2} c \left (\frac {\log {\left (\sqrt {- a^{2} x^{2} + 1} - 1 \right )}}{4} - \frac {\log {\left (\sqrt {- a^{2} x^{2} + 1} + 1 \right )}}{4} - \frac {1}{4 \left (\sqrt {- a^{2} x^{2} + 1} + 1\right )} - \frac {1}{4 \left (\sqrt {- a^{2} x^{2} + 1} - 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)/x**3,x)

[Out]

-a**2*c*(log(sqrt(-a**2*x**2 + 1) - 1)/4 - log(sqrt(-a**2*x**2 + 1) + 1)/4 - 1/(4*(sqrt(-a**2*x**2 + 1) + 1))
- 1/(4*(sqrt(-a**2*x**2 + 1) - 1)))

________________________________________________________________________________________