3.3.10 \(\int e^{-2 \tanh ^{-1}(a x)} (c-a c x) \, dx\) [210]

Optimal. Leaf size=26 \[ -3 c x+\frac {1}{2} a c x^2+\frac {4 c \log (1+a x)}{a} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6264, 45} \begin {gather*} \frac {1}{2} a c x^2+\frac {4 c \log (a x+1)}{a}-3 c x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)/E^(2*ArcTanh[a*x]),x]

[Out]

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

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 6264

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

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Mathematica [A]
time = 0.01, size = 25, normalized size = 0.96 \begin {gather*} c \left (-3 x+\frac {a x^2}{2}+\frac {4 \log (1+a x)}{a}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.38, size = 24, normalized size = 0.92

method	result	size
default	\(c \left (\frac {a \,x^{2}}{2}-3 x +\frac {4 \ln \left (a x +1\right )}{a}\right )\)	\(24\)
risch	\(-3 c x +\frac {a c \,x^{2}}{2}+\frac {4 c \ln \left (a x +1\right )}{a}\)	\(25\)
norman	\(\frac {-3 c x -\frac {5}{2} a c \,x^{2}+\frac {1}{2} c \,x^{3} a^{2}}{a x +1}+\frac {4 c \ln \left (a x +1\right )}{a}\)	\(43\)
meijerg	\(\frac {c \left (-\frac {a x \left (-2 a^{2} x^{2}+6 a x +12\right )}{4 \left (a x +1\right )}+3 \ln \left (a x +1\right )\right )}{a}-\frac {c \left (-\frac {a x}{a x +1}+\ln \left (a x +1\right )\right )}{a}-\frac {c \left (\frac {a x \left (3 a x +6\right )}{3 a x +3}-2 \ln \left (a x +1\right )\right )}{a}+\frac {c x}{a x +1}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 24, normalized size = 0.92 \begin {gather*} \frac {1}{2} \, a c x^{2} - 3 \, c x + \frac {4 \, c \log \left (a x + 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 28, normalized size = 1.08 \begin {gather*} \frac {a^{2} c x^{2} - 6 \, a c x + 8 \, c \log \left (a x + 1\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*c*x^2 - 6*a*c*x + 8*c*log(a*x + 1))/a

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Sympy [A]
time = 0.05, size = 24, normalized size = 0.92 \begin {gather*} \frac {a c x^{2}}{2} - 3 c x + \frac {4 c \log {\left (a x + 1 \right )}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
time = 0.42, size = 50, normalized size = 1.92 \begin {gather*} \frac {{\left (a x + 1\right )}^{2} {\left (c - \frac {8 \, c}{a x + 1}\right )}}{2 \, a} - \frac {4 \, c \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*x + 1)^2*(c - 8*c/(a*x + 1))/a - 4*c*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/a

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Mupad [B]
time = 0.04, size = 26, normalized size = 1.00 \begin {gather*} \frac {c\,\left (8\,\ln \left (a\,x+1\right )-6\,a\,x+a^2\,x^2\right )}{2\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(8*log(a*x + 1) - 6*a*x + a^2*x^2))/(2*a)

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