3.11.24 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x} \, dx\) [1024]

Optimal. Leaf size=21 \[ 2 a c x+\frac {1}{2} a^2 c x^2+c \log (x) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 21, 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*} \frac {1}{2} a^2 c x^2+2 a c x+c \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 19, normalized size = 0.90 \begin {gather*} c \left (2 a x+\frac {a^2 x^2}{2}+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

c*(2*a*x + (a^2*x^2)/2 + Log[x])

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Maple [A]
time = 0.06, size = 18, normalized size = 0.86

method	result	size
default	\(c \left (\frac {a^{2} x^{2}}{2}+2 a x +\ln \left (x \right )\right )\)	\(18\)
norman	\(2 a c x +\frac {a^{2} c \,x^{2}}{2}+c \ln \left (x \right )\)	\(20\)
risch	\(2 a c x +\frac {a^{2} c \,x^{2}}{2}+c \ln \left (x \right )\)	\(20\)
meijerg	\(-\frac {c \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{2}+\frac {a c \left (-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}+2 c \arctanh \left (a x \right )+\frac {c \left (-\ln \left (-a^{2} x^{2}+1\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right )}{2}\)	\(101\)

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,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 19, normalized size = 0.90 \begin {gather*} \frac {1}{2} \, a^{2} c x^{2} + 2 \, a c x + c \log \left (x\right ) \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,x, algorithm="maxima")

[Out]

1/2*a^2*c*x^2 + 2*a*c*x + c*log(x)

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Fricas [A]
time = 0.33, size = 19, normalized size = 0.90 \begin {gather*} \frac {1}{2} \, a^{2} c x^{2} + 2 \, a c x + c \log \left (x\right ) \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,x, algorithm="fricas")

[Out]

1/2*a^2*c*x^2 + 2*a*c*x + c*log(x)

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Sympy [A]
time = 0.04, size = 20, normalized size = 0.95 \begin {gather*} \frac {a^{2} c x^{2}}{2} + 2 a c x + c \log {\left (x \right )} \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,x)

[Out]

a**2*c*x**2/2 + 2*a*c*x + c*log(x)

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Giac [A]
time = 0.41, size = 20, normalized size = 0.95 \begin {gather*} \frac {1}{2} \, a^{2} c x^{2} + 2 \, a c x + c \log \left ({\left | x \right |}\right ) \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,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.04, size = 19, normalized size = 0.90 \begin {gather*} \frac {c\,\left (2\,\ln \left (x\right )+4\,a\,x+a^2\,x^2\right )}{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*(a^2*x^2 - 1)),x)

[Out]

(c*(2*log(x) + 4*a*x + a^2*x^2))/2

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