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

Optimal. Leaf size=58 \[ \frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a \sqrt {c}}-\frac {2 \sqrt {c-a c x}}{a c} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6167, 6130, 21, 50, 63, 206} \[ \frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a \sqrt {c}}-\frac {2 \sqrt {c-a c x}}{a c} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcCoth[a*x])*Sqrt[c - a*c*x]),x]

[Out]

(-2*Sqrt[c - a*c*x])/(a*c) + (2*Sqrt[2]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/(a*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 50

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 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 206

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

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]
)

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{-2 \coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx &=-\int \frac {e^{-2 \tanh ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx\\ &=-\int \frac {1-a x}{(1+a x) \sqrt {c-a c x}} \, dx\\ &=-\frac {\int \frac {\sqrt {c-a c x}}{1+a x} \, dx}{c}\\ &=-\frac {2 \sqrt {c-a c x}}{a c}-2 \int \frac {1}{(1+a x) \sqrt {c-a c x}} \, dx\\ &=-\frac {2 \sqrt {c-a c x}}{a c}+\frac {4 \operatorname {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )}{a c}\\ &=-\frac {2 \sqrt {c-a c x}}{a c}+\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 58, normalized size = 1.00 \[ \frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a \sqrt {c}}-\frac {2 \sqrt {c-a c x}}{a c} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*Sqrt[c - a*c*x]),x]

[Out]

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

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fricas [A]  time = 0.56, size = 118, normalized size = 2.03 \[ \left [\frac {\sqrt {2} \sqrt {c} \log \left (\frac {a x - \frac {2 \, \sqrt {2} \sqrt {-a c x + c}}{\sqrt {c}} - 3}{a x + 1}\right ) - 2 \, \sqrt {-a c x + c}}{a c}, \frac {2 \, {\left (\sqrt {2} c \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-\frac {1}{c}}}{a x - 1}\right ) - \sqrt {-a c x + c}\right )}}{a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 51, normalized size = 0.88 \[ -\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c}} - \frac {2 \, \sqrt {-a c x + c}}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 45, normalized size = 0.78 \[ -\frac {2 \left (\sqrt {-a c x +c}-\arctanh \left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sqrt {c}\right )}{c a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 68, normalized size = 1.17 \[ -\frac {\sqrt {2} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 2 \, \sqrt {-a c x + c}}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 47, normalized size = 0.81 \[ \frac {2\,\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right )}{a\,\sqrt {c}}-\frac {2\,\sqrt {c-a\,c\,x}}{a\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 18.12, size = 60, normalized size = 1.03 \[ - \frac {2 \sqrt {- a c x + c}}{a c} - \frac {2 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2}}{\sqrt {- \frac {1}{c}} \sqrt {- a c x + c}} \right )}}{a c \sqrt {- \frac {1}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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