3.266 \(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx\)

Optimal. Leaf size=37 \[ \frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a c^{3/2}} \]

[Out]

arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a/c^(3/2)

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Rubi [A]  time = 0.09, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6167, 6130, 21, 63, 206} \[ \frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^(3/2)),x]

[Out]

(Sqrt[2]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/(a*c^(3/2))

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

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Mathematica [A]  time = 0.02, size = 37, normalized size = 1.00 \[ \frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^(3/2)),x]

[Out]

(Sqrt[2]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/(a*c^(3/2))

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fricas [A]  time = 0.88, size = 88, normalized size = 2.38 \[ \left [\frac {\sqrt {2} \log \left (\frac {a x - \frac {2 \, \sqrt {2} \sqrt {-a c x + c}}{\sqrt {c}} - 3}{a x + 1}\right )}{2 \, a c^{\frac {3}{2}}}, \frac {\sqrt {2} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-\frac {1}{c}}}{a x - 1}\right )}{a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 36, normalized size = 0.97 \[ -\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 29, normalized size = 0.78 \[ \frac {\arctanh \left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}}{a \,c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(a*x-1)/(-a*c*x+c)^(3/2),x)

[Out]

arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a/c^(3/2)

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maxima [A]  time = 0.41, size = 52, normalized size = 1.41 \[ -\frac {\sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right )}{2 \, a c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(3/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c)))/(a*c^(3/2))

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mupad [B]  time = 0.08, size = 28, normalized size = 0.76 \[ \frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right )}{a\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x - 1)/((c - a*c*x)^(3/2)*(a*x + 1)),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*(c - a*c*x)^(1/2))/(2*c^(1/2))))/(a*c^(3/2))

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sympy [A]  time = 15.75, size = 41, normalized size = 1.11 \[ - \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{a c \sqrt {- c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a*c*x+c)**(3/2),x)

[Out]

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

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