3.3.20 \(\int \frac {e^{-3 \coth ^{-1}(a x)}}{c-a c x} \, dx\) [220]

Optimal. Leaf size=53 \[ \frac {2 \left (a-\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6310, 6313, 1819, 272, 65, 214} \begin {gather*} \frac {2 \left (a-\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 6310

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

Rule 6313

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[-c^n, Subst[Int[(c +
d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 54, normalized size = 1.02 \begin {gather*} \frac {\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{1+a x}-\frac {\log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{a}}{c} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(247\) vs. \(2(49)=98\).
time = 0.12, size = 248, normalized size = 4.68

method	result	size
default	\(-\frac {\left (-\sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a^{2} x^{2}+\ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+\left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-2 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a x +2 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}+a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right )\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a \sqrt {a^{2}}\, c \left (a x -1\right ) \sqrt {\left (a x +1\right ) \left (a x -1\right )}}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 78, normalized size = 1.47 \begin {gather*} -a {\left (\frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2} c} - \frac {2 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.41, size = 63, normalized size = 1.19 \begin {gather*} \frac {2 \, \sqrt {\frac {a x - 1}{a x + 1}} - \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \left (- \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{2} x^{2} - 1}\right )\, dx + \int \frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{2} x^{2} - 1}\, dx}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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Mupad [B]
time = 1.17, size = 48, normalized size = 0.91 \begin {gather*} \frac {2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a\,c}-\frac {2\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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