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

Optimal. Leaf size=45 \[ \frac {2 (a x+1)}{c \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6128, 852, 1805, 12, 266, 63, 208} \[ \frac {2 (a x+1)}{c \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(x*(c - a*c*x)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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

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Mathematica [A]  time = 0.02, size = 55, normalized size = 1.22 \[ \frac {-\sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+2 a x+2}{c \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 56, normalized size = 1.24 \[ \frac {2 \, a x + {\left (a x - 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} - 2}{a c x - c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 80, normalized size = 1.78 \[ -\frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c {\left | a \right |}} + \frac {4 \, a}{c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 61, normalized size = 1.36 \[ -\frac {\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 68, normalized size = 1.51 \[ \frac {2\,a\,\sqrt {1-a^2\,x^2}}{c\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a x}{a x^{2} \sqrt {- a^{2} x^{2} + 1} - x \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a x^{2} \sqrt {- a^{2} x^{2} + 1} - x \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a*x/(a*x**2*sqrt(-a**2*x**2 + 1) - x*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a*x**2*sqrt(-a**2*x**2
+ 1) - x*sqrt(-a**2*x**2 + 1)), x))/c

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