3.8.82 \(\int e^{2 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^3 \, dx\) [782]

Optimal. Leaf size=76 \[ -\frac {c^3}{5 a^6 x^5}-\frac {c^3}{2 a^5 x^4}+\frac {c^3}{3 a^4 x^3}+\frac {2 c^3}{a^3 x^2}+\frac {c^3}{a^2 x}+c^3 x+\frac {2 c^3 \log (x)}{a} \]

[Out]

-1/5*c^3/a^6/x^5-1/2*c^3/a^5/x^4+1/3*c^3/a^4/x^3+2*c^3/a^3/x^2+c^3/a^2/x+c^3*x+2*c^3*ln(x)/a

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Rubi [A]
time = 0.11, antiderivative size = 76, 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, 90} \begin {gather*} -\frac {c^3}{5 a^6 x^5}-\frac {c^3}{2 a^5 x^4}+\frac {c^3}{3 a^4 x^3}+\frac {2 c^3}{a^3 x^2}+\frac {c^3}{a^2 x}+\frac {2 c^3 \log (x)}{a}+c^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])*(c - c/(a^2*x^2))^3,x]

[Out]

-1/5*c^3/(a^6*x^5) - c^3/(2*a^5*x^4) + c^3/(3*a^4*x^3) + (2*c^3)/(a^3*x^2) + c^3/(a^2*x) + c^3*x + (2*c^3*Log[
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 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 )^3 \, dx &=-\int e^{2 \tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx\\ &=\frac {c^3 \int \frac {e^{2 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^3}{x^6} \, dx}{a^6}\\ &=\frac {c^3 \int \frac {(1-a x)^2 (1+a x)^4}{x^6} \, dx}{a^6}\\ &=\frac {c^3 \int \left (a^6+\frac {1}{x^6}+\frac {2 a}{x^5}-\frac {a^2}{x^4}-\frac {4 a^3}{x^3}-\frac {a^4}{x^2}+\frac {2 a^5}{x}\right ) \, dx}{a^6}\\ &=-\frac {c^3}{5 a^6 x^5}-\frac {c^3}{2 a^5 x^4}+\frac {c^3}{3 a^4 x^3}+\frac {2 c^3}{a^3 x^2}+\frac {c^3}{a^2 x}+c^3 x+\frac {2 c^3 \log (x)}{a}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 76, normalized size = 1.00 \begin {gather*} -\frac {c^3}{5 a^6 x^5}-\frac {c^3}{2 a^5 x^4}+\frac {c^3}{3 a^4 x^3}+\frac {2 c^3}{a^3 x^2}+\frac {c^3}{a^2 x}+c^3 x+\frac {2 c^3 \log (x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCoth[a*x])*(c - c/(a^2*x^2))^3,x]

[Out]

-1/5*c^3/(a^6*x^5) - c^3/(2*a^5*x^4) + c^3/(3*a^4*x^3) + (2*c^3)/(a^3*x^2) + c^3/(a^2*x) + c^3*x + (2*c^3*Log[
x])/a

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Maple [A]
time = 0.23, size = 55, normalized size = 0.72

method	result	size
default	\(\frac {c^{3} \left (a^{6} x -\frac {a}{2 x^{4}}+\frac {a^{4}}{x}+\frac {2 a^{3}}{x^{2}}+2 a^{5} \ln \left (x \right )+\frac {a^{2}}{3 x^{3}}-\frac {1}{5 x^{5}}\right )}{a^{6}}\)	\(55\)
risch	\(c^{3} x +\frac {a^{4} c^{3} x^{4}+2 a^{3} c^{3} x^{3}+\frac {1}{3} a^{2} c^{3} x^{2}-\frac {1}{2} a \,c^{3} x -\frac {1}{5} c^{3}}{a^{6} x^{5}}+\frac {2 c^{3} \ln \left (x \right )}{a}\)	\(69\)
norman	\(\frac {a^{3} c^{3} x^{4}+a^{5} c^{3} x^{6}-\frac {c^{3}}{5 a}-\frac {c^{3} x}{2}+2 a^{2} c^{3} x^{3}+\frac {c^{3} a \,x^{2}}{3}}{a^{5} x^{5}}+\frac {2 c^{3} \ln \left (x \right )}{a}\)	\(74\)
meijerg	\(-\frac {c^{3} \left (-a x -\ln \left (-a x +1\right )\right )}{a}+\frac {3 c^{3} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}-\frac {3 c^{3} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}+\frac {c^{3} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )-\frac {1}{4 x^{4} a^{4}}-\frac {1}{3 x^{3} a^{3}}-\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}+\frac {c^{3} \ln \left (-a x +1\right )}{a}-\frac {3 c^{3} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{a x}\right )}{a}+\frac {3 c^{3} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}-\frac {c^{3} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{5 x^{5} a^{5}}+\frac {1}{4 x^{4} a^{4}}+\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}\)	\(304\)

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)^3,x,method=_RETURNVERBOSE)

[Out]

c^3/a^6*(a^6*x-1/2*a/x^4+a^4/x+2*a^3/x^2+2*a^5*ln(x)+1/3*a^2/x^3-1/5/x^5)

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Maxima [A]
time = 0.26, size = 70, normalized size = 0.92 \begin {gather*} c^{3} x + \frac {2 \, c^{3} \log \left (x\right )}{a} + \frac {30 \, a^{4} c^{3} x^{4} + 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \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)^3,x, algorithm="maxima")

[Out]

c^3*x + 2*c^3*log(x)/a + 1/30*(30*a^4*c^3*x^4 + 60*a^3*c^3*x^3 + 10*a^2*c^3*x^2 - 15*a*c^3*x - 6*c^3)/(a^6*x^5
)

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Fricas [A]
time = 0.34, size = 78, normalized size = 1.03 \begin {gather*} \frac {30 \, a^{6} c^{3} x^{6} + 60 \, a^{5} c^{3} x^{5} \log \left (x\right ) + 30 \, a^{4} c^{3} x^{4} + 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \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)^3,x, algorithm="fricas")

[Out]

1/30*(30*a^6*c^3*x^6 + 60*a^5*c^3*x^5*log(x) + 30*a^4*c^3*x^4 + 60*a^3*c^3*x^3 + 10*a^2*c^3*x^2 - 15*a*c^3*x -
6*c^3)/(a^6*x^5)

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Sympy [A]
time = 0.19, size = 76, normalized size = 1.00 \begin {gather*} \frac {a^{6} c^{3} x + 2 a^{5} c^{3} \log {\left (x \right )} + \frac {30 a^{4} c^{3} x^{4} + 60 a^{3} c^{3} x^{3} + 10 a^{2} c^{3} x^{2} - 15 a c^{3} x - 6 c^{3}}{30 x^{5}}}{a^{6}} \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)**3,x)

[Out]

(a**6*c**3*x + 2*a**5*c**3*log(x) + (30*a**4*c**3*x**4 + 60*a**3*c**3*x**3 + 10*a**2*c**3*x**2 - 15*a*c**3*x -
6*c**3)/(30*x**5))/a**6

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Giac [A]
time = 0.38, size = 71, normalized size = 0.93 \begin {gather*} c^{3} x + \frac {2 \, c^{3} \log \left ({\left | x \right |}\right )}{a} + \frac {30 \, a^{4} c^{3} x^{4} + 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \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)^3,x, algorithm="giac")

[Out]

c^3*x + 2*c^3*log(abs(x))/a + 1/30*(30*a^4*c^3*x^4 + 60*a^3*c^3*x^3 + 10*a^2*c^3*x^2 - 15*a*c^3*x - 6*c^3)/(a^
6*x^5)

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Mupad [B]
time = 1.22, size = 56, normalized size = 0.74 \begin {gather*} \frac {c^3\,\left (\frac {a^2\,x^2}{3}-\frac {a\,x}{2}+2\,a^3\,x^3+a^4\,x^4+a^6\,x^6+2\,a^5\,x^5\,\ln \left (x\right )-\frac {1}{5}\right )}{a^6\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^3*((a^2*x^2)/3 - (a*x)/2 + 2*a^3*x^3 + a^4*x^4 + a^6*x^6 + 2*a^5*x^5*log(x) - 1/5))/(a^6*x^5)

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