3.7.57 \(\int e^{4 \tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\) [657]

Optimal. Leaf size=33 \[ \frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a} \]

[Out]

c/a^2/x+c*x-4*c*ln(x)/a+8*c*ln(-a*x+1)/a

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Rubi [A]
time = 0.05, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6292, 6285, 90} \begin {gather*} \frac {c}{a^2 x}-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a}+c x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int e^{4 \tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx &=-\frac {c \int \frac {e^{4 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )}{x^2} \, dx}{a^2}\\ &=-\frac {c \int \frac {(1+a x)^3}{x^2 (1-a x)} \, dx}{a^2}\\ &=-\frac {c \int \left (-a^2+\frac {1}{x^2}+\frac {4 a}{x}-\frac {8 a^2}{-1+a x}\right ) \, dx}{a^2}\\ &=\frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 33, normalized size = 1.00 \begin {gather*} \frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.87, size = 29, normalized size = 0.88

method	result	size
default	\(\frac {c \left (a^{2} x +\frac {1}{x}-4 a \ln \left (x \right )+8 \ln \left (a x -1\right ) a \right )}{a^{2}}\)	\(29\)
risch	\(\frac {c}{a^{2} x}+c x -\frac {4 c \ln \left (x \right )}{a}+\frac {8 c \ln \left (-a x +1\right )}{a}\)	\(34\)
norman	\(\frac {a^{3} c \,x^{4}-\frac {c}{a}}{x \left (a^{2} x^{2}-1\right ) a}-\frac {4 c \ln \left (x \right )}{a}+\frac {8 c \ln \left (a x -1\right )}{a}\)	\(55\)
meijerg	\(\frac {c \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}-\frac {5 c \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}+\frac {2 c \left (\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {5 c \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{2 \sqrt {-a^{2}}}-\frac {2 c \left (\frac {2 a^{2} x^{2}}{-2 a^{2} x^{2}+2}-\ln \left (-a^{2} x^{2}+1\right )+1+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right )}{a}+\frac {c \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \arctanh \left (a x \right )}{\sqrt {-a^{2}}}\right )}{2 \sqrt {-a^{2}}}\)	\(310\)

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^2/x^2),x,method=_RETURNVERBOSE)

[Out]

c/a^2*(a^2*x+1/x-4*a*ln(x)+8*ln(a*x-1)*a)

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Maxima [A]
time = 0.25, size = 32, normalized size = 0.97 \begin {gather*} c x + \frac {8 \, c \log \left (a x - 1\right )}{a} - \frac {4 \, c \log \left (x\right )}{a} + \frac {c}{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^2/x^2),x, algorithm="maxima")

[Out]

c*x + 8*c*log(a*x - 1)/a - 4*c*log(x)/a + c/(a^2*x)

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Fricas [A]
time = 0.36, size = 35, normalized size = 1.06 \begin {gather*} \frac {a^{2} c x^{2} + 8 \, a c x \log \left (a x - 1\right ) - 4 \, a c x \log \left (x\right ) + c}{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^2/x^2),x, algorithm="fricas")

[Out]

(a^2*c*x^2 + 8*a*c*x*log(a*x - 1) - 4*a*c*x*log(x) + c)/(a^2*x)

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Sympy [A]
time = 0.15, size = 26, normalized size = 0.79 \begin {gather*} c x + \frac {4 c \left (- \log {\left (x \right )} + 2 \log {\left (x - \frac {1}{a} \right )}\right )}{a} + \frac {c}{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**2/x**2),x)

[Out]

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

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Giac [A]
time = 0.41, size = 34, normalized size = 1.03 \begin {gather*} c x + \frac {8 \, c \log \left ({\left | a x - 1 \right |}\right )}{a} - \frac {4 \, c \log \left ({\left | x \right |}\right )}{a} + \frac {c}{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^2/x^2),x, algorithm="giac")

[Out]

c*x + 8*c*log(abs(a*x - 1))/a - 4*c*log(abs(x))/a + c/(a^2*x)

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Mupad [B]
time = 0.07, size = 32, normalized size = 0.97 \begin {gather*} c\,x+\frac {c}{a^2\,x}-\frac {4\,c\,\ln \left (x\right )}{a}+\frac {8\,c\,\ln \left (a\,x-1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x + c/(a^2*x) - (4*c*log(x))/a + (8*c*log(a*x - 1))/a

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