3.1025 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x^2} \, dx\)

Optimal. Leaf size=19 \[ a^2 c x+2 a c \log (x)-\frac {c}{x} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6150, 43} \[ a^2 c x+2 a c \log (x)-\frac {c}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c/x) + a^2*c*x + 2*a*c*Log[x]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 )}{x^2} \, dx &=c \int \frac {(1+a x)^2}{x^2} \, dx\\ &=c \int \left (a^2+\frac {1}{x^2}+\frac {2 a}{x}\right ) \, dx\\ &=-\frac {c}{x}+a^2 c x+2 a c \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.95 \[ c \left (a^2 x+2 a \log (x)-\frac {1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c*(-x^(-1) + a^2*x + 2*a*Log[x])

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fricas [A]  time = 0.61, size = 23, normalized size = 1.21 \[ \frac {a^{2} c x^{2} + 2 \, a c x \log \relax (x) - c}{x} \]

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^2,x, algorithm="fricas")

[Out]

(a^2*c*x^2 + 2*a*c*x*log(x) - c)/x

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giac [A]  time = 0.19, size = 20, normalized size = 1.05 \[ a^{2} c x + 2 \, a c \log \left ({\left | x \right |}\right ) - \frac {c}{x} \]

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^2,x, algorithm="giac")

[Out]

a^2*c*x + 2*a*c*log(abs(x)) - c/x

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maple [A]  time = 0.03, size = 20, normalized size = 1.05 \[ -\frac {c}{x}+a^{2} c x +2 a c \ln \relax (x ) \]

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^2,x)

[Out]

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

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maxima [A]  time = 0.34, size = 19, normalized size = 1.00 \[ a^{2} c x + 2 \, a c \log \relax (x) - \frac {c}{x} \]

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^2,x, algorithm="maxima")

[Out]

a^2*c*x + 2*a*c*log(x) - c/x

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mupad [B]  time = 0.89, size = 20, normalized size = 1.05 \[ \frac {c\,\left (a^2\,x^2+2\,a\,x\,\ln \relax (x)-1\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(a^2*x^2 + 2*a*x*log(x) - 1))/x

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sympy [A]  time = 0.12, size = 17, normalized size = 0.89 \[ a^{2} c x + 2 a c \log {\relax (x )} - \frac {c}{x} \]

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**2,x)

[Out]

a**2*c*x + 2*a*c*log(x) - c/x

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