3.2.82 \(\int \frac {e^{3 \coth ^{-1}(a x)}}{c-a c x} \, dx\) [182]

Optimal. Leaf size=80 \[ \frac {8 \left (a+\frac {1}{x}\right )}{3 a^2 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {4}{3 a^2 c \sqrt {1-\frac {1}{a^2 x^2}} x}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c} \]

[Out]

8/3*(a+1/x)/a^2/c/(1-1/a^2/x^2)^(3/2)-arctanh((1-1/a^2/x^2)^(1/2))/a/c+4/3/a^2/c/x/(1-1/a^2/x^2)^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6310, 6313, 866, 1819, 12, 272, 65, 214} \begin {gather*} \frac {8 \left (a+\frac {1}{x}\right )}{3 a^2 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {4}{3 a^2 c x \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[E^(3*ArcCoth[a*x])/(c - a*c*x),x]

[Out]

(8*(a + x^(-1)))/(3*a^2*c*(1 - 1/(a^2*x^2))^(3/2)) + 4/(3*a^2*c*Sqrt[1 - 1/(a^2*x^2)]*x) - ArcTanh[Sqrt[1 - 1/
(a^2*x^2)]]/(a*c)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 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 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[(f + g*x)^n*((a + c*x^2)^(m + p)/(d - e*x)^m), x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

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^2}{a^2}\right )^{3/2}}{x \left (1-\frac {x}{a}\right )^4} \, dx,x,\frac {1}{x}\right )}{a c}\\ &=\frac {\text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^4}{x \left (1-\frac {x^2}{a^2}\right )^{5/2}} \, dx,x,\frac {1}{x}\right )}{a c}\\ &=\frac {8 \left (a+\frac {1}{x}\right )}{3 a^2 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {-3-\frac {4 x}{a}+\frac {3 x^2}{a^2}}{x \left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{3 a c}\\ &=\frac {8 \left (a+\frac {1}{x}\right )}{3 a^2 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {4}{3 a^2 c \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {\text {Subst}\left (\int \frac {3}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{3 a c}\\ &=\frac {8 \left (a+\frac {1}{x}\right )}{3 a^2 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {4}{3 a^2 c \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a c}\\ &=\frac {8 \left (a+\frac {1}{x}\right )}{3 a^2 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {4}{3 a^2 c \sqrt {1-\frac {1}{a^2 x^2}} x}+\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 {8 \left (a+\frac {1}{x}\right )}{3 a^2 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {4}{3 a^2 c \sqrt {1-\frac {1}{a^2 x^2}} x}-\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 {8 \left (a+\frac {1}{x}\right )}{3 a^2 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {4}{3 a^2 c \sqrt {1-\frac {1}{a^2 x^2}} x}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 63, normalized size = 0.79 \begin {gather*} \frac {\frac {4 \sqrt {1-\frac {1}{a^2 x^2}} x (-1+2 a x)}{(-1+a x)^2}-\frac {3 \log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{a}}{3 c} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((4*Sqrt[1 - 1/(a^2*x^2)]*x*(-1 + 2*a*x))/(-1 + a*x)^2 - (3*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/a)/(3*c)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(344\) vs. \(2(70)=140\).
time = 0.16, size = 345, normalized size = 4.31

method	result	size
default	\(-\frac {3 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{4} x^{3}+3 \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, a^{3} x^{3}-9 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}-3 \sqrt {a^{2}}\, \left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} a x -9 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a^{2} x^{2}+9 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x +\left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+9 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a x -3 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right )-3 \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{3 a \sqrt {a^{2}}\, \left (a x -1\right ) c \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\)	\(345\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/a*(3*ln((a^2*x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))*a^4*x^3+3*(a^2)^(1/2)*((a*x+1)*(a*x-1))^
(1/2)*a^3*x^3-9*ln((a^2*x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))*a^3*x^2-3*(a^2)^(1/2)*((a*x+1)*(a*
x-1))^(3/2)*a*x-9*((a*x+1)*(a*x-1))^(1/2)*(a^2)^(1/2)*a^2*x^2+9*ln((a^2*x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))
/(a^2)^(1/2))*a^2*x+((a*x+1)*(a*x-1))^(3/2)*(a^2)^(1/2)+9*((a*x+1)*(a*x-1))^(1/2)*(a^2)^(1/2)*a*x-3*a*ln((a^2*
x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))-3*(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2)/(a*x-1)
/c/((a*x+1)*(a*x-1))^(1/2)/(a*x+1)/((a*x-1)/(a*x+1))^(3/2)

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Maxima [A]
time = 0.26, size = 95, normalized size = 1.19 \begin {gather*} -\frac {1}{3} \, a {\left (\frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2} c} - \frac {2 \, {\left (\frac {3 \, {\left (a x - 1\right )}}{a x + 1} + 1\right )}}{a^{2} c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 120, normalized size = 1.50 \begin {gather*} -\frac {3 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 3 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - 4 \, {\left (2 \, a^{2} x^{2} + a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{3 \, {\left (a^{3} c x^{2} - 2 \, a^{2} c x + a c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(a^2*x^2 - 2*a*x + 1)*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 3*(a^2*x^2 - 2*a*x + 1)*log(sqrt((a*x - 1)/
(a*x + 1)) - 1) - 4*(2*a^2*x^2 + a*x - 1)*sqrt((a*x - 1)/(a*x + 1)))/(a^3*c*x^2 - 2*a^2*c*x + a*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 35, normalized size = 0.44 \begin {gather*} \frac {\log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right )}{c {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.07, size = 63, normalized size = 0.79 \begin {gather*} \frac {\frac {2\,\left (a\,x-1\right )}{a\,x+1}+\frac {2}{3}}{a\,c\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}-\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(1/((c - a*c*x)*((a*x - 1)/(a*x + 1))^(3/2)),x)

[Out]

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

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