3.9.17 \(\int e^{-2 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\) [817]

Optimal. Leaf size=21 \[ -\frac {c}{a^2 x}+c x-\frac {2 c \log (x)}{a} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6302, 6292, 6285, 45} \begin {gather*} -\frac {c}{a^2 x}-\frac {2 c \log (x)}{a}+c x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 21, normalized size = 1.00 \begin {gather*} -\frac {c}{a^2 x}+c x-\frac {2 c \log (x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(c/(a^2*x)) + c*x - (2*c*Log[x])/a

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Maple [A]
time = 0.16, size = 22, normalized size = 1.05

method	result	size
default	\(\frac {c \left (a^{2} x -\frac {1}{x}-2 a \ln \left (x \right )\right )}{a^{2}}\)	\(22\)
risch	\(-\frac {c}{a^{2} x}+c x -\frac {2 c \ln \left (x \right )}{a}\)	\(22\)
norman	\(\frac {a c \,x^{2}-\frac {c}{a}}{a x}-\frac {2 c \ln \left (x \right )}{a}\)	\(30\)
meijerg	\(\frac {c \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {c \ln \left (a x +1\right )}{a}-\frac {c \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )\right )}{a}+\frac {c \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{a x}\right )}{a}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 21, normalized size = 1.00 \begin {gather*} c x - \frac {2 \, 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((c-c/a^2/x^2)*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

c*x - 2*c*log(x)/a - c/(a^2*x)

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Fricas [A]
time = 0.49, size = 26, normalized size = 1.24 \begin {gather*} \frac {a^{2} c x^{2} - 2 \, 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((c-c/a^2/x^2)*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.05, size = 20, normalized size = 0.95 \begin {gather*} \frac {a^{2} c x - 2 a c \log {\left (x \right )} - \frac {c}{x}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a**2*c*x - 2*a*c*log(x) - c/x)/a**2

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Giac [A]
time = 0.39, size = 22, normalized size = 1.05 \begin {gather*} c x - \frac {2 \, 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((c-c/a^2/x^2)*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

c*x - 2*c*log(abs(x))/a - c/(a^2*x)

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Mupad [B]
time = 1.23, size = 25, normalized size = 1.19 \begin {gather*} -\frac {c\,\left (2\,a\,x\,\ln \left (x\right )-a^2\,x^2+1\right )}{a^2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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