3.13.34 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\) [1234]

Optimal. Leaf size=84 \[ \frac {1}{16 a c^3 (1-a x)}-\frac {1}{12 a c^3 (1+a x)^3}-\frac {1}{8 a c^3 (1+a x)^2}-\frac {3}{16 a c^3 (1+a x)}+\frac {\tanh ^{-1}(a x)}{4 a c^3} \]

[Out]

1/16/a/c^3/(-a*x+1)-1/12/a/c^3/(a*x+1)^3-1/8/a/c^3/(a*x+1)^2-3/16/a/c^3/(a*x+1)+1/4*arctanh(a*x)/a/c^3

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Rubi [A]
time = 0.05, antiderivative size = 84, 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 = {6275, 46, 213} \begin {gather*} \frac {1}{16 a c^3 (1-a x)}-\frac {3}{16 a c^3 (a x+1)}-\frac {1}{8 a c^3 (a x+1)^2}-\frac {1}{12 a c^3 (a x+1)^3}+\frac {\tanh ^{-1}(a x)}{4 a c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^3),x]

[Out]

1/(16*a*c^3*(1 - a*x)) - 1/(12*a*c^3*(1 + a*x)^3) - 1/(8*a*c^3*(1 + a*x)^2) - 3/(16*a*c^3*(1 + a*x)) + ArcTanh
[a*x]/(4*a*c^3)

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 6275

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, 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 )^3} \, dx &=\frac {\int \frac {1}{(1-a x)^2 (1+a x)^4} \, dx}{c^3}\\ &=\frac {\int \left (\frac {1}{16 (-1+a x)^2}+\frac {1}{4 (1+a x)^4}+\frac {1}{4 (1+a x)^3}+\frac {3}{16 (1+a x)^2}-\frac {1}{4 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^3}\\ &=\frac {1}{16 a c^3 (1-a x)}-\frac {1}{12 a c^3 (1+a x)^3}-\frac {1}{8 a c^3 (1+a x)^2}-\frac {3}{16 a c^3 (1+a x)}-\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{4 c^3}\\ &=\frac {1}{16 a c^3 (1-a x)}-\frac {1}{12 a c^3 (1+a x)^3}-\frac {1}{8 a c^3 (1+a x)^2}-\frac {3}{16 a c^3 (1+a x)}+\frac {\tanh ^{-1}(a x)}{4 a c^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 61, normalized size = 0.73 \begin {gather*} -\frac {-4+a x+6 a^2 x^2+3 a^3 x^3-3 (-1+a x) (1+a x)^3 \tanh ^{-1}(a x)}{12 a (-1+a x) (c+a c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^3),x]

[Out]

-1/12*(-4 + a*x + 6*a^2*x^2 + 3*a^3*x^3 - 3*(-1 + a*x)*(1 + a*x)^3*ArcTanh[a*x])/(a*(-1 + a*x)*(c + a*c*x)^3)

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Maple [A]
time = 0.06, size = 76, normalized size = 0.90

method	result	size
default	\(\frac {-\frac {1}{12 a \left (a x +1\right )^{3}}-\frac {1}{8 a \left (a x +1\right )^{2}}-\frac {3}{16 a \left (a x +1\right )}+\frac {\ln \left (a x +1\right )}{8 a}-\frac {1}{16 a \left (a x -1\right )}-\frac {\ln \left (a x -1\right )}{8 a}}{c^{3}}\)	\(76\)
risch	\(\frac {-\frac {a^{2} x^{3}}{4}-\frac {x^{2} a}{2}-\frac {x}{12}+\frac {1}{3 a}}{\left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{3}}+\frac {\ln \left (-a x -1\right )}{8 a \,c^{3}}-\frac {\ln \left (a x -1\right )}{8 a \,c^{3}}\)	\(76\)
norman	\(\frac {\frac {3 x}{4 c}+\frac {a \,x^{2}}{2 c}-\frac {7 a^{2} x^{3}}{6 c}-\frac {5 a^{3} x^{4}}{6 c}+\frac {5 a^{4} x^{5}}{12 c}+\frac {a^{5} x^{6}}{3 c}}{\left (a x -1\right )^{2} c^{2} \left (a x +1\right )^{4}}-\frac {\ln \left (a x -1\right )}{8 a \,c^{3}}+\frac {\ln \left (a x +1\right )}{8 a \,c^{3}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

1/c^3*(-1/12/a/(a*x+1)^3-1/8/a/(a*x+1)^2-3/16/a/(a*x+1)+1/8*ln(a*x+1)/a-1/16/a/(a*x-1)-1/8/a*ln(a*x-1))

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Maxima [A]
time = 0.27, size = 91, normalized size = 1.08 \begin {gather*} -\frac {3 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + a x - 4}{12 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} + \frac {\log \left (a x + 1\right )}{8 \, a c^{3}} - \frac {\log \left (a x - 1\right )}{8 \, a c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

-1/12*(3*a^3*x^3 + 6*a^2*x^2 + a*x - 4)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3) + 1/8*log(a*x + 1)
/(a*c^3) - 1/8*log(a*x - 1)/(a*c^3)

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Fricas [A]
time = 0.36, size = 121, normalized size = 1.44 \begin {gather*} -\frac {6 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 2 \, a x - 3 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x - 1\right ) - 8}{24 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

-1/24*(6*a^3*x^3 + 12*a^2*x^2 + 2*a*x - 3*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*log(a*x + 1) + 3*(a^4*x^4 + 2*a^3*
x^3 - 2*a*x - 1)*log(a*x - 1) - 8)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3)

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Sympy [A]
time = 0.23, size = 83, normalized size = 0.99 \begin {gather*} \frac {- 3 a^{3} x^{3} - 6 a^{2} x^{2} - a x + 4}{12 a^{5} c^{3} x^{4} + 24 a^{4} c^{3} x^{3} - 24 a^{2} c^{3} x - 12 a c^{3}} + \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{8} + \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**2*(-a**2*x**2+1)/(-a**2*c*x**2+c)**3,x)

[Out]

(-3*a**3*x**3 - 6*a**2*x**2 - a*x + 4)/(12*a**5*c**3*x**4 + 24*a**4*c**3*x**3 - 24*a**2*c**3*x - 12*a*c**3) +
(-log(x - 1/a)/8 + log(x + 1/a)/8)/(a*c**3)

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Giac [A]
time = 0.41, size = 97, normalized size = 1.15 \begin {gather*} -\frac {\log \left ({\left | -\frac {2}{a x + 1} + 1 \right |}\right )}{8 \, a c^{3}} + \frac {1}{32 \, a c^{3} {\left (\frac {2}{a x + 1} - 1\right )}} - \frac {\frac {9 \, a^{5} c^{6}}{a x + 1} + \frac {6 \, a^{5} c^{6}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, a^{5} c^{6}}{{\left (a x + 1\right )}^{3}}}{48 \, a^{6} c^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

-1/8*log(abs(-2/(a*x + 1) + 1))/(a*c^3) + 1/32/(a*c^3*(2/(a*x + 1) - 1)) - 1/48*(9*a^5*c^6/(a*x + 1) + 6*a^5*c
^6/(a*x + 1)^2 + 4*a^5*c^6/(a*x + 1)^3)/(a^6*c^9)

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Mupad [B]
time = 0.09, size = 72, normalized size = 0.86 \begin {gather*} \frac {\frac {x}{12}+\frac {a\,x^2}{2}-\frac {1}{3\,a}+\frac {a^2\,x^3}{4}}{-a^4\,c^3\,x^4-2\,a^3\,c^3\,x^3+2\,a\,c^3\,x+c^3}+\frac {\mathrm {atanh}\left (a\,x\right )}{4\,a\,c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^2*x^2 - 1)/((c - a^2*c*x^2)^3*(a*x + 1)^2),x)

[Out]

(x/12 + (a*x^2)/2 - 1/(3*a) + (a^2*x^3)/4)/(c^3 - 2*a^3*c^3*x^3 - a^4*c^3*x^4 + 2*a*c^3*x) + atanh(a*x)/(4*a*c
^3)

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