3.4.88 \(\int e^{2 \coth ^{-1}(a x)} (c-\frac {c}{a x})^4 \, dx\) [388]

Optimal. Leaf size=40 \[ \frac {c^4}{3 a^4 x^3}-\frac {c^4}{a^3 x^2}+c^4 x-\frac {2 c^4 \log (x)}{a} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 40, 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, 6266, 6264, 76} \begin {gather*} \frac {c^4}{3 a^4 x^3}-\frac {c^4}{a^3 x^2}-\frac {2 c^4 \log (x)}{a}+c^4 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])*(c - c/(a*x))^4,x]

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 42, normalized size = 1.05 \begin {gather*} \frac {c^4}{3 a^4 x^3}-\frac {c^4}{a^3 x^2}+c^4 x-\frac {2 c^4 \log (a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCoth[a*x])*(c - c/(a*x))^4,x]

[Out]

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

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Maple [A]
time = 0.17, size = 32, normalized size = 0.80

method	result	size
default	\(\frac {c^{4} \left (a^{4} x -\frac {a}{x^{2}}-2 a^{3} \ln \left (x \right )+\frac {1}{3 x^{3}}\right )}{a^{4}}\)	\(32\)
risch	\(c^{4} x +\frac {-a \,c^{4} x +\frac {1}{3} c^{4}}{a^{4} x^{3}}-\frac {2 c^{4} \ln \left (x \right )}{a}\)	\(37\)
norman	\(\frac {a^{3} c^{4} x^{4}+\frac {c^{4}}{3 a}-c^{4} x}{a^{3} x^{3}}-\frac {2 c^{4} \ln \left (x \right )}{a}\)	\(44\)
meijerg	\(-\frac {c^{4} \left (-a x -\ln \left (-a x +1\right )\right )}{a}-\frac {3 c^{4} \ln \left (-a x +1\right )}{a}-\frac {2 c^{4} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}+\frac {2 c^{4} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{a x}\right )}{a}+\frac {3 c^{4} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}+\frac {c^{4} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}\)	\(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x-1)*(a*x+1)*(c-c/a/x)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 37, normalized size = 0.92 \begin {gather*} c^{4} x - \frac {2 \, c^{4} \log \left (x\right )}{a} - \frac {3 \, a c^{4} x - c^{4}}{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/x)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 43, normalized size = 1.08 \begin {gather*} \frac {3 \, a^{4} c^{4} x^{4} - 6 \, a^{3} c^{4} x^{3} \log \left (x\right ) - 3 \, a c^{4} x + c^{4}}{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/x)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.09, size = 39, normalized size = 0.98 \begin {gather*} \frac {a^{4} c^{4} x - 2 a^{3} c^{4} \log {\left (x \right )} + \frac {- 3 a c^{4} x + c^{4}}{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/x)**4,x)

[Out]

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

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Giac [A]
time = 0.43, size = 38, normalized size = 0.95 \begin {gather*} c^{4} x - \frac {2 \, c^{4} \log \left ({\left | x \right |}\right )}{a} - \frac {3 \, a c^{4} x - c^{4}}{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/x)^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.05, size = 35, normalized size = 0.88 \begin {gather*} -\frac {c^4\,\left (3\,a\,x-3\,a^4\,x^4+6\,a^3\,x^3\,\ln \left (x\right )-1\right )}{3\,a^4\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - c/(a*x))^4*(a*x + 1))/(a*x - 1),x)

[Out]

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

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