3.5.88 \(\int e^{-\tanh ^{-1}(a x)} (c-\frac {c}{a x}) \, dx\) [488]

Optimal. Leaf size=50 \[ \frac {c \sqrt {1-a^2 x^2}}{a}+\frac {2 c \text {ArcSin}(a x)}{a}+\frac {c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{a} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6266, 6263, 1823, 858, 222, 272, 65, 214} \begin {gather*} \frac {c \sqrt {1-a^2 x^2}}{a}+\frac {c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{a}+\frac {2 c \text {ArcSin}(a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))/E^ArcTanh[a*x],x]

[Out]

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

Rule 65

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 214

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]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 272

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 858

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

Rule 1823

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 6263

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]

Rule 6266

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*(1 + c*(x/d))^
p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 47, normalized size = 0.94 \begin {gather*} \frac {c \left (\sqrt {1-a^2 x^2}+2 \text {ArcSin}(a x)-\log (x)+\log \left (1+\sqrt {1-a^2 x^2}\right )\right )}{a} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(97\) vs. \(2(46)=92\).
time = 0.75, size = 98, normalized size = 1.96

method	result	size
default	\(\frac {c \left (2 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {2 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}-\sqrt {-a^{2} x^{2}+1}+\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{a}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 70, normalized size = 1.40 \begin {gather*} a c {\left (\frac {\arcsin \left (a x\right )}{a^{2}} + \frac {\log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )}{a^{2}}\right )} + c {\left (\frac {\arcsin \left (a x\right )}{a} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.43, size = 67, normalized size = 1.34 \begin {gather*} -\frac {4 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + c \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} c}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {c \left (\int \left (- \frac {\sqrt {- a^{2} x^{2} + 1}}{a x^{2} + x}\right )\, dx + \int \frac {a x \sqrt {- a^{2} x^{2} + 1}}{a x^{2} + x}\, dx\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 68, normalized size = 1.36 \begin {gather*} \frac {2 \, c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} + \frac {c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 56, normalized size = 1.12 \begin {gather*} \frac {c\,\sqrt {1-a^2\,x^2}}{a}+\frac {c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{a}+\frac {2\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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