3.5.96 \(\int e^{-2 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^2 \, dx\) [496]

Optimal. Leaf size=42 \[ -\frac {c^2}{a^2 x}-c^2 x-\frac {4 c^2 \log (x)}{a}+\frac {8 c^2 \log (1+a x)}{a} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6266, 6264, 90} \begin {gather*} -\frac {c^2}{a^2 x}-\frac {4 c^2 \log (x)}{a}+\frac {8 c^2 \log (a x+1)}{a}+c^2 (-x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6264

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

Rule 6266

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 44, normalized size = 1.05 \begin {gather*} -\frac {c^2}{a^2 x}-c^2 x-\frac {4 c^2 \log (a x)}{a}+\frac {8 c^2 \log (1+a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.68, size = 34, normalized size = 0.81

method	result	size
default	\(\frac {c^{2} \left (-a^{2} x +8 a \ln \left (a x +1\right )-\frac {1}{x}-4 a \ln \left (x \right )\right )}{a^{2}}\)	\(34\)
risch	\(-\frac {c^{2}}{a^{2} x}-x \,c^{2}+\frac {8 c^{2} \ln \left (-a x -1\right )}{a}-\frac {4 c^{2} \ln \left (x \right )}{a}\)	\(44\)
norman	\(\frac {-\frac {c^{2}}{a}-a^{2} c^{2} x^{3}}{a x \left (a x +1\right )}-\frac {4 c^{2} \ln \left (x \right )}{a}+\frac {8 c^{2} \ln \left (a x +1\right )}{a}\)	\(60\)
meijerg	\(-\frac {c^{2} \left (\frac {a x \left (3 a x +6\right )}{3 a x +3}-2 \ln \left (a x +1\right )\right )}{a}+\frac {2 c^{2} \left (-\frac {a x}{a x +1}+\ln \left (a x +1\right )\right )}{a}-\frac {2 c^{2} \left (-\frac {2 a x}{2 a x +2}-\ln \left (a x +1\right )+1+\ln \left (x \right )+\ln \left (a \right )\right )}{a}+\frac {c^{2} \left (\frac {3 a x}{3 a x +3}+2 \ln \left (a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (a \right )-\frac {1}{a x}\right )}{a}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 42, normalized size = 1.00 \begin {gather*} -c^{2} x + \frac {8 \, c^{2} \log \left (a x + 1\right )}{a} - \frac {4 \, c^{2} \log \left (x\right )}{a} - \frac {c^{2}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.41, size = 44, normalized size = 1.05 \begin {gather*} -\frac {a^{2} c^{2} x^{2} - 8 \, a c^{2} x \log \left (a x + 1\right ) + 4 \, a c^{2} x \log \left (x\right ) + c^{2}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.16, size = 32, normalized size = 0.76 \begin {gather*} - c^{2} x - \frac {4 c^{2} \left (\log {\left (x \right )} - 2 \log {\left (x + \frac {1}{a} \right )}\right )}{a} - \frac {c^{2}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.44, size = 72, normalized size = 1.71 \begin {gather*} -\frac {4 \, c^{2} \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {4 \, c^{2} \log \left ({\left | -\frac {1}{a x + 1} + 1 \right |}\right )}{a} + \frac {{\left (a x + 1\right )} c^{2}}{a {\left (\frac {1}{a x + 1} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.87, size = 42, normalized size = 1.00 \begin {gather*} \frac {8\,c^2\,\ln \left (a\,x+1\right )}{a}-\frac {c^2}{a^2\,x}-\frac {4\,c^2\,\ln \left (x\right )}{a}-c^2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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