3.5.23 \(\int \frac {e^{-2 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx\) [423]

Optimal. Leaf size=74 \[ -2 \sqrt {c-a c x}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )+4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6265, 21, 86, 162, 65, 214, 212} \begin {gather*} -2 \sqrt {c-a c x}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )+4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*Sqrt[c - a*c*x] - 2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]] + 4*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(S
qrt[2]*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 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 86

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

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 212

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

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 6265

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\\ &=\frac {\int \frac {(c-a c x)^{3/2}}{x (1+a x)} \, dx}{c}\\ &=-2 \sqrt {c-a c x}+\frac {\int \frac {a c^2-3 a^2 c^2 x}{x (1+a x) \sqrt {c-a c x}} \, dx}{a c}\\ &=-2 \sqrt {c-a c x}+c \int \frac {1}{x \sqrt {c-a c x}} \, dx-(4 a c) \int \frac {1}{(1+a x) \sqrt {c-a c x}} \, dx\\ &=-2 \sqrt {c-a c x}+8 \text {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )-\frac {2 \text {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 )+4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 74, normalized size = 1.00 \begin {gather*} -2 \sqrt {c-a c x}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )+4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*Sqrt[c - a*c*x] - 2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]] + 4*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(S
qrt[2]*Sqrt[c])]

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Maple [A]
time = 0.72, size = 58, normalized size = 0.78

method	result	size
derivativedivides	\(-2 \arctanh \left (\frac {\sqrt {-c x a +c}}{\sqrt {c}}\right ) \sqrt {c}+4 \arctanh \left (\frac {\sqrt {-c x a +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sqrt {c}-2 \sqrt {-c x a +c}\)	\(58\)
default	\(-2 \arctanh \left (\frac {\sqrt {-c x a +c}}{\sqrt {c}}\right ) \sqrt {c}+4 \arctanh \left (\frac {\sqrt {-c x a +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sqrt {c}-2 \sqrt {-c x a +c}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 97, normalized size = 1.31 \begin {gather*} -2 \, \sqrt {2} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.42, size = 157, normalized size = 2.12 \begin {gather*} \left [2 \, \sqrt {2} \sqrt {c} \log \left (\frac {a c x - 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + \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}, -4 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) + 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{c}\right ) - 2 \, \sqrt {-a c x + c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2*sqrt(2)*sqrt(c)*log((a*c*x - 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + sqrt(c)*log((a*c*x + 2*
sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x) - 2*sqrt(-a*c*x + c), -4*sqrt(2)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x +
c)*sqrt(-c)/c) + 2*sqrt(-c)*arctan(sqrt(-a*c*x + c)*sqrt(-c)/c) - 2*sqrt(-a*c*x + c)]

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Sympy [A]
time = 5.33, size = 80, normalized size = 1.08 \begin {gather*} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {- a c x + c}}{\sqrt {- c}} \right )}}{\sqrt {- c}} - \frac {4 \sqrt {2} c \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} - 2 \sqrt {- a c x + c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 67, normalized size = 0.91 \begin {gather*} -\frac {4 \, \sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} + \frac {2 \, c \arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - 2 \, \sqrt {-a c x + c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*sqrt(2)*c*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) + 2*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sq
rt(-c) - 2*sqrt(-a*c*x + c)

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Mupad [B]
time = 0.83, size = 57, normalized size = 0.77 \begin {gather*} 4\,\sqrt {2}\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right )-2\,\sqrt {c-a\,c\,x}-2\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {c-a\,c\,x}}{\sqrt {c}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*2^(1/2)*c^(1/2)*atanh((2^(1/2)*(c - a*c*x)^(1/2))/(2*c^(1/2))) - 2*(c - a*c*x)^(1/2) - 2*c^(1/2)*atanh((c -
a*c*x)^(1/2)/c^(1/2))

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