3.3.91 \(\int \frac {e^{\tanh ^{-1}(a x)} (c-a c x)}{x} \, dx\) [291]

Optimal. Leaf size=35 \[ c \sqrt {1-a^2 x^2}-c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 35, 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 = {6263, 272, 52, 65, 214} \begin {gather*} c \sqrt {1-a^2 x^2}-c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x))/x,x]

[Out]

c*Sqrt[1 - a^2*x^2] - c*ArcTanh[Sqrt[1 - a^2*x^2]]

Rule 52

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

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 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 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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(35)=70\).
time = 0.05, size = 79, normalized size = 2.26 \begin {gather*} c \left (\frac {1}{\sqrt {1-a^2 x^2}}-\frac {a^2 x^2}{\sqrt {1-a^2 x^2}}+\text {ArcSin}(a x)+2 \text {ArcSin}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )-\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.48, size = 32, normalized size = 0.91

method	result	size
default	\(-c \left (-\sqrt {-a^{2} x^{2}+1}+\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )\)	\(32\)
meijerg	\(\frac {c \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{2 \sqrt {\pi }}+\frac {c \left (-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )+\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }\right )}{2 \sqrt {\pi }}\)	\(79\)

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,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.46, size = 44, normalized size = 1.26 \begin {gather*} -c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \sqrt {-a^{2} x^{2} + 1} c \end {gather*}

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,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 36, normalized size = 1.03 \begin {gather*} c \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1} c \end {gather*}

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,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 9.78, size = 66, normalized size = 1.89 \begin {gather*} \frac {a^{2} c \left (\begin {cases} - x^{2} & \text {for}\: a^{2} = 0 \\\frac {2 \sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right )}{2} - \frac {c \left (- \log {\left (-1 + \frac {1}{\sqrt {- a^{2} x^{2} + 1}} \right )} + \log {\left (1 + \frac {1}{\sqrt {- a^{2} x^{2} + 1}} \right )}\right )}{2} \end {gather*}

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,x)

[Out]

a**2*c*Piecewise((-x**2, Eq(a**2, 0)), (2*sqrt(-a**2*x**2 + 1)/a**2, True))/2 - c*(-log(-1 + 1/sqrt(-a**2*x**2
+ 1)) + log(1 + 1/sqrt(-a**2*x**2 + 1)))/2

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Giac [A]
time = 0.39, size = 53, normalized size = 1.51 \begin {gather*} -\frac {1}{2} \, c \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) + \frac {1}{2} \, c \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right ) + \sqrt {-a^{2} x^{2} + 1} c \end {gather*}

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,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.77, size = 31, normalized size = 0.89 \begin {gather*} -c\,\left (\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )-\sqrt {1-a^2\,x^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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