3.5.78 \(\int e^{4 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^2 \, dx\) [478]

Optimal. Leaf size=27 \[ -\frac {c^2}{a^2 x}+c^2 x+\frac {2 c^2 \log (x)}{a} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6266, 6264, 45} \begin {gather*} -\frac {c^2}{a^2 x}+\frac {2 c^2 \log (x)}{a}+c^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 29, normalized size = 1.07 \begin {gather*} -\frac {c^2}{a^2 x}+c^2 x+\frac {2 c^2 \log (a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.84, size = 24, normalized size = 0.89

method	result	size
default	\(\frac {c^{2} \left (a^{2} x -\frac {1}{x}+2 a \ln \left (x \right )\right )}{a^{2}}\)	\(24\)
risch	\(-\frac {c^{2}}{a^{2} x}+x \,c^{2}+\frac {2 c^{2} \ln \left (x \right )}{a}\)	\(28\)
norman	\(\frac {\frac {c^{2}}{a}+a^{3} c^{2} x^{4}-2 a \,c^{2} x^{2}}{x \left (a^{2} x^{2}-1\right ) a}+\frac {2 c^{2} \ln \left (x \right )}{a}\)	\(57\)
meijerg	\(\frac {c^{2} \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}+\frac {c^{2} \left (\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}+\frac {c^{2} \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}-\frac {2 a \,c^{2} x^{2}}{-a^{2} x^{2}+1}-\frac {c^{2} \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{2 \sqrt {-a^{2}}}+\frac {c^{2} \left (\frac {2 a^{2} x^{2}}{-2 a^{2} x^{2}+2}-\ln \left (-a^{2} x^{2}+1\right )+1+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right )}{a}-\frac {c^{2} \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \arctanh \left (a x \right )}{\sqrt {-a^{2}}}\right )}{2 \sqrt {-a^{2}}}\)	\(341\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 27, normalized size = 1.00 \begin {gather*} c^{2} x + \frac {2 \, c^{2} \log \left (x\right )}{a} - \frac {c^{2}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^2,x, algorithm="maxima")

[Out]

c^2*x + 2*c^2*log(x)/a - c^2/(a^2*x)

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Fricas [A]
time = 0.35, size = 32, normalized size = 1.19 \begin {gather*} \frac {a^{2} c^{2} x^{2} + 2 \, a c^{2} x \log \left (x\right ) - c^{2}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.06, size = 26, normalized size = 0.96 \begin {gather*} \frac {a^{2} c^{2} x + 2 a c^{2} \log {\left (x \right )} - \frac {c^{2}}{x}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2*(c-c/a/x)**2,x)

[Out]

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

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Giac [A]
time = 0.41, size = 28, normalized size = 1.04 \begin {gather*} c^{2} x + \frac {2 \, c^{2} \log \left ({\left | x \right |}\right )}{a} - \frac {c^{2}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.81, size = 25, normalized size = 0.93 \begin {gather*} \frac {c^2\,\left (a^2\,x^2+2\,a\,x\,\ln \left (x\right )-1\right )}{a^2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - c/(a*x))^2*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)

[Out]

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

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