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

Optimal. Leaf size=25 \[ \frac {(a x+1)^3}{6 a c^2 (1-a x)^3} \]

[Out]

1/6*(a*x+1)^3/a/c^2/(-a*x+1)^3

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Rubi [A]  time = 0.03, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6129, 37} \[ \frac {(a x+1)^3}{6 a c^2 (1-a x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^3/(6*a*c^2*(1 - a*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 6129

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])

Rubi steps

\begin {align*} \int \frac {e^{4 \tanh ^{-1}(a x)}}{(c-a c x)^2} \, dx &=\frac {\int \frac {(1+a x)^2}{(1-a x)^4} \, dx}{c^2}\\ &=\frac {(1+a x)^3}{6 a c^2 (1-a x)^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ \frac {(a x+1)^3}{6 a c^2 (1-a x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^3/(6*a*c^2*(1 - a*x)^3)

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fricas [B]  time = 0.54, size = 51, normalized size = 2.04 \[ -\frac {3 \, a^{2} x^{2} + 1}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.19, size = 50, normalized size = 2.00 \[ -\frac {2}{{\left (a c x - c\right )}^{2} a} - \frac {1}{{\left (a c x - c\right )} a c} - \frac {4 \, c}{3 \, {\left (a c x - c\right )}^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 42, normalized size = 1.68 \[ \frac {-\frac {2}{a \left (a x -1\right )^{2}}-\frac {4}{3 a \left (a x -1\right )^{3}}-\frac {1}{a \left (a x -1\right )}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.33, size = 51, normalized size = 2.04 \[ -\frac {3 \, a^{2} x^{2} + 1}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 25, normalized size = 1.00 \[ -\frac {3\,a^2\,x^2+1}{3\,a\,c^2\,{\left (a\,x-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.26, size = 51, normalized size = 2.04 \[ \frac {- 3 a^{2} x^{2} - 1}{3 a^{4} c^{2} x^{3} - 9 a^{3} c^{2} x^{2} + 9 a^{2} c^{2} x - 3 a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*a**2*x**2 - 1)/(3*a**4*c**2*x**3 - 9*a**3*c**2*x**2 + 9*a**2*c**2*x - 3*a*c**2)

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