3.11.26 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x^3} \, dx\) [1026]

Optimal. Leaf size=23 \[ -\frac {c}{2 x^2}-\frac {2 a c}{x}+a^2 c \log (x) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 23, 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 = {6285, 45} \begin {gather*} a^2 c \log (x)-\frac {2 a c}{x}-\frac {c}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*c/x^2 - (2*a*c)/x + a^2*c*Log[x]

Rule 45

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 6285

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

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Mathematica [A]
time = 0.01, size = 22, normalized size = 0.96 \begin {gather*} c \left (-\frac {1}{2 x^2}-\frac {2 a}{x}+a^2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

c*(-1/2*1/x^2 - (2*a)/x + a^2*Log[x])

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Maple [A]
time = 0.06, size = 21, normalized size = 0.91

method	result	size
default	\(c \left (-\frac {2 a}{x}+a^{2} \ln \left (x \right )-\frac {1}{2 x^{2}}\right )\)	\(21\)
norman	\(\frac {-\frac {1}{2} c -2 a c x}{x^{2}}+a^{2} c \ln \left (x \right )\)	\(22\)
risch	\(\frac {-\frac {1}{2} c -2 a c x}{x^{2}}+a^{2} c \ln \left (x \right )\)	\(22\)
meijerg	\(\frac {a^{2} c \ln \left (-a^{2} x^{2}+1\right )}{2}-2 a^{2} c \arctanh \left (a x \right )-\frac {a^{3} c \left (-\frac {2}{x \sqrt {-a^{2}}}+\frac {2 a \arctanh \left (a x \right )}{\sqrt {-a^{2}}}\right )}{\sqrt {-a^{2}}}-\frac {a^{2} c \left (\ln \left (-a^{2} x^{2}+1\right )-2 \ln \left (x \right )-\ln \left (-a^{2}\right )+\frac {1}{a^{2} x^{2}}\right )}{2}\)	\(106\)

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^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 20, normalized size = 0.87 \begin {gather*} a^{2} c \log \left (x\right ) - \frac {4 \, a c x + c}{2 \, x^{2}} \end {gather*}

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^3,x, algorithm="maxima")

[Out]

a^2*c*log(x) - 1/2*(4*a*c*x + c)/x^2

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Fricas [A]
time = 0.34, size = 25, normalized size = 1.09 \begin {gather*} \frac {2 \, a^{2} c x^{2} \log \left (x\right ) - 4 \, a c x - c}{2 \, x^{2}} \end {gather*}

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^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.06, size = 22, normalized size = 0.96 \begin {gather*} a^{2} c \log {\left (x \right )} + \frac {- 4 a c x - c}{2 x^{2}} \end {gather*}

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**3,x)

[Out]

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

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Giac [A]
time = 0.41, size = 21, normalized size = 0.91 \begin {gather*} a^{2} c \log \left ({\left | x \right |}\right ) - \frac {4 \, a c x + c}{2 \, x^{2}} \end {gather*}

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^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.05, size = 22, normalized size = 0.96 \begin {gather*} a^2\,c\,\ln \left (x\right )-\frac {\frac {c}{2}+2\,a\,c\,x}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*c*log(x) - (c/2 + 2*a*c*x)/x^2

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