3.397 \(\int \frac {e^{2 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx\)

Optimal. Leaf size=39 \[ -2 \sqrt {c-a c x}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6130, 21, 80, 63, 208} \[ -2 \sqrt {c-a c x}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x,x]

[Out]

-2*Sqrt[c - a*c*x] - 2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 6130

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 39, normalized size = 1.00 \[ -2 \sqrt {c-a c x}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x,x]

[Out]

-2*Sqrt[c - a*c*x] - 2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]]

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fricas [A]  time = 0.71, size = 82, normalized size = 2.10 \[ \left [\sqrt {c} \log \left (\frac {a c x + 2 \, \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{x}\right ) - 2 \, \sqrt {-a c x + c}, 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{c}\right ) - 2 \, \sqrt {-a c x + c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[sqrt(c)*log((a*c*x + 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x) - 2*sqrt(-a*c*x + c), 2*sqrt(-c)*arctan(sqrt(-a*c*x
+ c)*sqrt(-c)/c) - 2*sqrt(-a*c*x + c)]

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giac [A]  time = 0.20, size = 40, normalized size = 1.03 \[ 2 \, c {\left (\frac {\arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {\sqrt {-a c x + c}}{c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*(arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - sqrt(-a*c*x + c)/c)

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maple [A]  time = 0.03, size = 32, normalized size = 0.82 \[ -2 \arctanh \left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right ) \sqrt {c}-2 \sqrt {-a c x +c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 48, normalized size = 1.23 \[ \sqrt {c} \log \left (\frac {\sqrt {-a c x + c} - \sqrt {c}}{\sqrt {-a c x + c} + \sqrt {c}}\right ) - 2 \, \sqrt {-a c x + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.81, size = 31, normalized size = 0.79 \[ -2\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {c-a\,c\,x}}{\sqrt {c}}\right )-2\,\sqrt {c-a\,c\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.28, size = 39, normalized size = 1.00 \[ \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {- a c x + c}}{\sqrt {- c}} \right )}}{\sqrt {- c}} - 2 \sqrt {- a c x + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*atan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - 2*sqrt(-a*c*x + c)

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