3.3.14 \(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) [214]

Optimal. Leaf size=51 \[ -\frac {1}{4 a c^4 (1-a x)^2}-\frac {1}{4 a c^4 (1-a x)}-\frac {\tanh ^{-1}(a x)}{4 a c^4} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6302, 6264, 46, 213} \begin {gather*} -\frac {1}{4 a c^4 (1-a x)}-\frac {1}{4 a c^4 (1-a x)^2}-\frac {\tanh ^{-1}(a x)}{4 a c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^4),x]

[Out]

-1/4*1/(a*c^4*(1 - a*x)^2) - 1/(4*a*c^4*(1 - a*x)) - ArcTanh[a*x]/(4*a*c^4)

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 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 \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx &=-\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-a c x)^4} \, dx\\ &=-\frac {\int \frac {1}{(1-a x)^3 (1+a x)} \, dx}{c^4}\\ &=-\frac {\int \left (-\frac {1}{2 (-1+a x)^3}+\frac {1}{4 (-1+a x)^2}-\frac {1}{4 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^4}\\ &=-\frac {1}{4 a c^4 (1-a x)^2}-\frac {1}{4 a c^4 (1-a x)}+\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{4 c^4}\\ &=-\frac {1}{4 a c^4 (1-a x)^2}-\frac {1}{4 a c^4 (1-a x)}-\frac {\tanh ^{-1}(a x)}{4 a c^4}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 35, normalized size = 0.69 \begin {gather*} \frac {-2+a x-(-1+a x)^2 \tanh ^{-1}(a x)}{4 a c^4 (-1+a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^4),x]

[Out]

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

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Maple [A]
time = 0.13, size = 52, normalized size = 1.02

method	result	size
risch	\(\frac {\frac {x}{4}-\frac {1}{2 a}}{\left (a x -1\right )^{2} c^{4}}+\frac {\ln \left (-a x +1\right )}{8 a \,c^{4}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{4}}\)	\(51\)
default	\(\frac {-\frac {\ln \left (a x +1\right )}{8 a}-\frac {1}{4 a \left (a x -1\right )^{2}}+\frac {1}{4 a \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{8 a}}{c^{4}}\)	\(52\)
norman	\(\frac {\frac {3 x}{4 c}-\frac {5 a \,x^{2}}{4 c}+\frac {a^{2} x^{3}}{2 c}}{c^{3} \left (a x -1\right )^{3}}+\frac {\ln \left (a x -1\right )}{8 c^{4} a}-\frac {\ln \left (a x +1\right )}{8 a \,c^{4}}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 63, normalized size = 1.24 \begin {gather*} \frac {a x - 2}{4 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} - \frac {\log \left (a x + 1\right )}{8 \, a c^{4}} + \frac {\log \left (a x - 1\right )}{8 \, a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*x - 2)/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4) - 1/8*log(a*x + 1)/(a*c^4) + 1/8*log(a*x - 1)/(a*c^4)

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Fricas [A]
time = 0.33, size = 76, normalized size = 1.49 \begin {gather*} \frac {2 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 4}{8 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*a*x - (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + (a^2*x^2 - 2*a*x + 1)*log(a*x - 1) - 4)/(a^3*c^4*x^2 - 2*a^2
*c^4*x + a*c^4)

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Sympy [A]
time = 0.15, size = 54, normalized size = 1.06 \begin {gather*} \frac {a x - 2}{4 a^{3} c^{4} x^{2} - 8 a^{2} c^{4} x + 4 a c^{4}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x - 2)/(4*a**3*c**4*x**2 - 8*a**2*c**4*x + 4*a*c**4) + (log(x - 1/a)/8 - log(x + 1/a)/8)/(a*c**4)

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Giac [A]
time = 0.41, size = 51, normalized size = 1.00 \begin {gather*} -\frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{4}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{4}} + \frac {a x - 2}{4 \, {\left (a x - 1\right )}^{2} a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.07, size = 46, normalized size = 0.90 \begin {gather*} \frac {\frac {x}{4}-\frac {1}{2\,a}}{a^2\,c^4\,x^2-2\,a\,c^4\,x+c^4}-\frac {\mathrm {atanh}\left (a\,x\right )}{4\,a\,c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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