3.2.92 \(\int e^{4 \tanh ^{-1}(a x)} (c-a c x) \, dx\) [192]

Optimal. Leaf size=27 \[ -3 c x-\frac {1}{2} a c x^2-\frac {4 c \log (1-a x)}{a} \]

[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 = 27, 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 (1-a x)}{a}-3 c x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])*(c - a*c*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^{4 \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 = 26, 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[E^(4*ArcTanh[a*x])*(c - a*c*x),x]

[Out]

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

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Maple [A]
time = 0.75, size = 24, normalized size = 0.89

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	\(-\frac {a c \,x^{2}}{2}-3 c x -\frac {4 c \ln \left (a x -1\right )}{a}\)	\(25\)
norman	\(\frac {\frac {1}{2} a c \,x^{2}+3 c x -\frac {1}{2} a^{3} c \,x^{4}-3 c \,x^{3} a^{2}}{a^{2} x^{2}-1}-\frac {4 c \ln \left (a x -1\right )}{a}\)	\(56\)
meijerg	\(\frac {c \left (-\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+6\right )}{3 \left (-a^{2} x^{2}+1\right )}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{2 a}-\frac {3 c \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}-\frac {c \left (\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {c \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}+\frac {3 a c \,x^{2}}{2 \left (-a^{2} x^{2}+1\right )}+\frac {c \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{2 \sqrt {-a^{2}}}\)	\(269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c),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 = 0.89 \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*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c),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 = 0.36, size = 28, normalized size = 1.04 \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*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c),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 = 0.06, size = 26, normalized size = 0.96 \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*x+1)**4/(-a**2*x**2+1)**2*(-a*c*x+c),x)

[Out]

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

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Giac [A]
time = 0.41, size = 35, normalized size = 1.30 \begin {gather*} -\frac {4 \, c \log \left ({\left | a x - 1 \right |}\right )}{a} - \frac {a^{3} c x^{2} + 6 \, a^{2} c x}{2 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c),x, algorithm="giac")

[Out]

-4*c*log(abs(a*x - 1))/a - 1/2*(a^3*c*x^2 + 6*a^2*c*x)/a^2

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Mupad [B]
time = 0.04, size = 26, normalized size = 0.96 \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(((c - a*c*x)*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)

[Out]

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

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