3.1035 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2}{x^3} \, dx\)

Optimal. Leaf size=17 \[ -\frac {c^2 (a x+1)^4}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6150, 74} \[ -\frac {c^2 (a x+1)^4}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2*(1 + a*x)^4)/(2*x^2)

Rule 74

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

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

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \[ -\frac {c^2 (a x+1)^4}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(c^2*(1 + a*x)^4)/x^2

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fricas [B]  time = 0.51, size = 37, normalized size = 2.18 \[ -\frac {a^{4} c^{2} x^{4} + 4 \, a^{3} c^{2} x^{3} + 4 \, a c^{2} x + c^{2}}{2 \, x^{2}} \]

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)^2/x^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.56, size = 37, normalized size = 2.18 \[ -\frac {1}{2} \, a^{4} c^{2} x^{2} - 2 \, a^{3} c^{2} x - \frac {4 \, a c^{2} x + c^{2}}{2 \, x^{2}} \]

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)^2/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 31, normalized size = 1.82 \[ c^{2} \left (-\frac {x^{2} a^{4}}{2}-2 a^{3} x -\frac {2 a}{x}-\frac {1}{2 x^{2}}\right ) \]

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)^2/x^3,x)

[Out]

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

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maxima [B]  time = 0.30, size = 37, normalized size = 2.18 \[ -\frac {1}{2} \, a^{4} c^{2} x^{2} - 2 \, a^{3} c^{2} x - \frac {4 \, a c^{2} x + c^{2}}{2 \, x^{2}} \]

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)^2/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 29, normalized size = 1.71 \[ -\frac {c^2\,\left (a^4\,x^4+4\,a^3\,x^3+4\,a\,x+1\right )}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.14, size = 39, normalized size = 2.29 \[ - \frac {a^{4} c^{2} x^{2}}{2} - 2 a^{3} c^{2} x - \frac {4 a c^{2} x + c^{2}}{2 x^{2}} \]

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)**2/x**3,x)

[Out]

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

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