3.8.83 \(\int e^{2 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^2 \, dx\) [783]

Optimal. Leaf size=39 \[ \frac {c^2}{3 a^4 x^3}+\frac {c^2}{a^3 x^2}+c^2 x+\frac {2 c^2 \log (x)}{a} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6302, 6292, 6285, 76} \begin {gather*} \frac {c^2}{3 a^4 x^3}+\frac {c^2}{a^3 x^2}+\frac {2 c^2 \log (x)}{a}+c^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 6285

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

Rule 6292

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

Rule 6302

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

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Mathematica [A]
time = 0.02, size = 39, normalized size = 1.00 \begin {gather*} \frac {c^2}{3 a^4 x^3}+\frac {c^2}{a^3 x^2}+c^2 x+\frac {2 c^2 \log (x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.18, size = 31, normalized size = 0.79

method	result	size
default	\(\frac {c^{2} \left (a^{4} x +\frac {a}{x^{2}}+2 a^{3} \ln \left (x \right )+\frac {1}{3 x^{3}}\right )}{a^{4}}\)	\(31\)
risch	\(c^{2} x +\frac {a \,c^{2} x +\frac {1}{3} c^{2}}{a^{4} x^{3}}+\frac {2 c^{2} \ln \left (x \right )}{a}\)	\(36\)
norman	\(\frac {c^{2} x +a^{3} c^{2} x^{4}+\frac {c^{2}}{3 a}}{a^{3} x^{3}}+\frac {2 c^{2} \ln \left (x \right )}{a}\)	\(43\)
meijerg	\(-\frac {c^{2} \left (-a x -\ln \left (-a x +1\right )\right )}{a}+\frac {2 c^{2} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}-\frac {c^{2} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}+\frac {c^{2} \ln \left (-a x +1\right )}{a}-\frac {2 c^{2} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{a x}\right )}{a}+\frac {c^{2} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}\)	\(183\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 35, normalized size = 0.90 \begin {gather*} c^{2} x + \frac {2 \, c^{2} \log \left (x\right )}{a} + \frac {3 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 43, normalized size = 1.10 \begin {gather*} \frac {3 \, a^{4} c^{2} x^{4} + 6 \, a^{3} c^{2} x^{3} \log \left (x\right ) + 3 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^4*c^2*x^4 + 6*a^3*c^2*x^3*log(x) + 3*a*c^2*x + c^2)/(a^4*x^3)

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Sympy [A]
time = 0.11, size = 39, normalized size = 1.00 \begin {gather*} \frac {a^{4} c^{2} x + 2 a^{3} c^{2} \log {\left (x \right )} + \frac {3 a c^{2} x + c^{2}}{3 x^{3}}}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a**4*c**2*x + 2*a**3*c**2*log(x) + (3*a*c**2*x + c**2)/(3*x**3))/a**4

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Giac [A]
time = 0.41, size = 36, normalized size = 0.92 \begin {gather*} c^{2} x + \frac {2 \, c^{2} \log \left ({\left | x \right |}\right )}{a} + \frac {3 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 32, normalized size = 0.82 \begin {gather*} \frac {c^2\,\left (a\,x+a^4\,x^4+2\,a^3\,x^3\,\ln \left (x\right )+\frac {1}{3}\right )}{a^4\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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