3.11.34 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2}{x^2} \, dx\) [1034]

Optimal. Leaf size=41 \[ -\frac {c^2}{x}-a^3 c^2 x^2-\frac {1}{3} a^4 c^2 x^3+2 a c^2 \log (x) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 41, 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}{3} a^4 c^2 x^3-a^3 c^2 x^2+2 a c^2 \log (x)-\frac {c^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 41, normalized size = 1.00 \begin {gather*} -\frac {c^2}{x}-a^3 c^2 x^2-\frac {1}{3} a^4 c^2 x^3+2 a c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 32, normalized size = 0.78

method	result	size
default	\(c^{2} \left (-\frac {a^{4} x^{3}}{3}-a^{3} x^{2}-\frac {1}{x}+2 a \ln \left (x \right )\right )\)	\(32\)
risch	\(-\frac {c^{2}}{x}-a^{3} c^{2} x^{2}-\frac {a^{4} c^{2} x^{3}}{3}+2 a \,c^{2} \ln \left (x \right )\)	\(40\)
norman	\(\frac {-c^{2}-a^{3} c^{2} x^{3}-\frac {1}{3} a^{4} c^{2} x^{4}}{x}+2 a \,c^{2} \ln \left (x \right )\)	\(42\)
meijerg	\(\frac {a^{2} c^{2} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}+\frac {a^{2} c^{2} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}-a \,c^{2} \arctanh \left (a x \right )+a \,c^{2} \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )+2 a \,c^{2} \ln \left (-a^{2} x^{2}+1\right )+a \,c^{2} \left (-\ln \left (-a^{2} x^{2}+1\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right )-\frac {a^{2} c^{2} \left (-\frac {2}{x \sqrt {-a^{2}}}+\frac {2 a \arctanh \left (a x \right )}{\sqrt {-a^{2}}}\right )}{2 \sqrt {-a^{2}}}\)	\(227\)

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^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 39, normalized size = 0.95 \begin {gather*} -\frac {1}{3} \, a^{4} c^{2} x^{3} - a^{3} c^{2} x^{2} + 2 \, a c^{2} \log \left (x\right ) - \frac {c^{2}}{x} \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^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 41, normalized size = 1.00 \begin {gather*} -\frac {a^{4} c^{2} x^{4} + 3 \, a^{3} c^{2} x^{3} - 6 \, a c^{2} x \log \left (x\right ) + 3 \, c^{2}}{3 \, x} \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^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.05, size = 36, normalized size = 0.88 \begin {gather*} - \frac {a^{4} c^{2} x^{3}}{3} - a^{3} c^{2} x^{2} + 2 a c^{2} \log {\left (x \right )} - \frac {c^{2}}{x} \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**2,x)

[Out]

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

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Giac [A]
time = 0.41, size = 40, normalized size = 0.98 \begin {gather*} -\frac {1}{3} \, a^{4} c^{2} x^{3} - a^{3} c^{2} x^{2} + 2 \, a c^{2} \log \left ({\left | x \right |}\right ) - \frac {c^{2}}{x} \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^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.04, size = 39, normalized size = 0.95 \begin {gather*} 2\,a\,c^2\,\ln \left (x\right )-\frac {c^2}{x}-a^3\,c^2\,x^2-\frac {a^4\,c^2\,x^3}{3} \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^2*(a^2*x^2 - 1)),x)

[Out]

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

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