3.11.33 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2}{x} \, dx\) [1033]

Optimal. Leaf size=40 \[ 2 a c^2 x-\frac {2}{3} a^3 c^2 x^3-\frac {1}{4} a^4 c^2 x^4+c^2 \log (x) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6285, 76} \begin {gather*} -\frac {1}{4} a^4 c^2 x^4-\frac {2}{3} a^3 c^2 x^3+2 a c^2 x+c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 33, normalized size = 0.82 \begin {gather*} -\frac {1}{12} c^2 \left (3-24 a x+8 a^3 x^3+3 a^4 x^4-12 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(c^2*(3 - 24*a*x + 8*a^3*x^3 + 3*a^4*x^4 - 12*Log[x]))

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Maple [A]
time = 0.07, size = 28, normalized size = 0.70

method	result	size
default	\(c^{2} \left (-\frac {a^{4} x^{4}}{4}-\frac {2 a^{3} x^{3}}{3}+2 a x +\ln \left (x \right )\right )\)	\(28\)
norman	\(2 a \,c^{2} x -\frac {2 a^{3} c^{2} x^{3}}{3}-\frac {a^{4} c^{2} x^{4}}{4}+c^{2} \ln \left (x \right )\)	\(37\)
risch	\(2 a \,c^{2} x -\frac {2 a^{3} c^{2} x^{3}}{3}-\frac {a^{4} c^{2} x^{4}}{4}+c^{2} \ln \left (x \right )\)	\(37\)
meijerg	\(-\frac {c^{2} \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{6}+\ln \left (-a^{2} x^{2}+1\right )\right )}{2}-\frac {c^{2} \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{2}+\frac {c^{2} \ln \left (-a^{2} x^{2}+1\right )}{2}+\frac {a \,c^{2} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{\sqrt {-a^{2}}}+\frac {2 a \,c^{2} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}+2 c^{2} \arctanh \left (a x \right )+\frac {c^{2} \left (-\ln \left (-a^{2} x^{2}+1\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right )}{2}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 36, normalized size = 0.90 \begin {gather*} -\frac {1}{4} \, a^{4} c^{2} x^{4} - \frac {2}{3} \, a^{3} c^{2} x^{3} + 2 \, a c^{2} x + c^{2} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 36, normalized size = 0.90 \begin {gather*} -\frac {1}{4} \, a^{4} c^{2} x^{4} - \frac {2}{3} \, a^{3} c^{2} x^{3} + 2 \, a c^{2} x + c^{2} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.04, size = 39, normalized size = 0.98 \begin {gather*} - \frac {a^{4} c^{2} x^{4}}{4} - \frac {2 a^{3} c^{2} x^{3}}{3} + 2 a c^{2} x + c^{2} \log {\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 37, normalized size = 0.92 \begin {gather*} -\frac {1}{4} \, a^{4} c^{2} x^{4} - \frac {2}{3} \, a^{3} c^{2} x^{3} + 2 \, a c^{2} x + c^{2} \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 36, normalized size = 0.90 \begin {gather*} c^2\,\ln \left (x\right )-\frac {2\,a^3\,c^2\,x^3}{3}-\frac {a^4\,c^2\,x^4}{4}+2\,a\,c^2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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