3.1027 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x^4} \, dx\)

Optimal. Leaf size=15 \[ -\frac {c (a x+1)^3}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6150, 37} \[ -\frac {c (a x+1)^3}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(1 + a*x)^3)/(3*x^3)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 6150

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])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \[ -\frac {c (a x+1)^3}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(c*(1 + a*x)^3)/x^3

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fricas [A]  time = 0.48, size = 21, normalized size = 1.40 \[ -\frac {3 \, a^{2} c x^{2} + 3 \, a c x + c}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*a^2*c*x^2 + 3*a*c*x + c)/x^3

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giac [A]  time = 0.35, size = 21, normalized size = 1.40 \[ -\frac {3 \, a^{2} c x^{2} + 3 \, a c x + c}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*a^2*c*x^2 + 3*a*c*x + c)/x^3

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maple [A]  time = 0.03, size = 23, normalized size = 1.53 \[ c \left (-\frac {a^{2}}{x}-\frac {1}{3 x^{3}}-\frac {a}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 21, normalized size = 1.40 \[ -\frac {3 \, a^{2} c x^{2} + 3 \, a c x + c}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*a^2*c*x^2 + 3*a*c*x + c)/x^3

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mupad [B]  time = 0.04, size = 21, normalized size = 1.40 \[ -\frac {c\,a^2\,x^2+c\,a\,x+\frac {c}{3}}{x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c/3 + a^2*c*x^2 + a*c*x)/x^3

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sympy [A]  time = 0.16, size = 24, normalized size = 1.60 \[ \frac {- 3 a^{2} c x^{2} - 3 a c x - c}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*a**2*c*x**2 - 3*a*c*x - c)/(3*x**3)

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