3.585 \(\int e^{4 \coth ^{-1}(a x)} (c-a^2 c x^2) \, dx\)

Optimal. Leaf size=46 \[ -\frac {c (a x+1)^3}{3 a}-\frac {c (a x+1)^2}{a}-\frac {8 c \log (1-a x)}{a}-4 c x \]

[Out]

-4*c*x-c*(a*x+1)^2/a-1/3*c*(a*x+1)^3/a-8*c*ln(-a*x+1)/a

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Rubi [A]  time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6167, 6140, 43} \[ -\frac {c (a x+1)^3}{3 a}-\frac {c (a x+1)^2}{a}-\frac {8 c \log (1-a x)}{a}-4 c x \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[a*x])*(c - a^2*c*x^2),x]

[Out]

-4*c*x - (c*(1 + a*x)^2)/a - (c*(1 + a*x)^3)/(3*a) - (8*c*Log[1 - a*x])/a

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 6140

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx &=\int e^{4 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx\\ &=c \int \frac {(1+a x)^3}{1-a x} \, dx\\ &=c \int \left (-4+\frac {8}{1-a x}-2 (1+a x)-(1+a x)^2\right ) \, dx\\ &=-4 c x-\frac {c (1+a x)^2}{a}-\frac {c (1+a x)^3}{3 a}-\frac {8 c \log (1-a x)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.78 \[ -\frac {1}{3} a^2 c x^3-2 a c x^2-\frac {8 c \log (1-a x)}{a}-7 c x \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - a^2*c*x^2),x]

[Out]

-7*c*x - 2*a*c*x^2 - (a^2*c*x^3)/3 - (8*c*Log[1 - a*x])/a

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fricas [A]  time = 0.64, size = 37, normalized size = 0.80 \[ -\frac {a^{3} c x^{3} + 6 \, a^{2} c x^{2} + 21 \, a c x + 24 \, c \log \left (a x - 1\right )}{3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a^3*c*x^3 + 6*a^2*c*x^2 + 21*a*c*x + 24*c*log(a*x - 1))/a

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giac [A]  time = 0.14, size = 60, normalized size = 1.30 \[ -\frac {{\left (a x - 1\right )}^{3} {\left (c + \frac {9 \, c}{a x - 1} + \frac {36 \, c}{{\left (a x - 1\right )}^{2}}\right )}}{3 \, a} + \frac {8 \, c \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a*x - 1)^3*(c + 9*c/(a*x - 1) + 36*c/(a*x - 1)^2)/a + 8*c*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a

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maple [A]  time = 0.04, size = 34, normalized size = 0.74 \[ -\frac {a^{2} c \,x^{3}}{3}-2 a c \,x^{2}-7 c x -\frac {8 c \ln \left (a x -1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*c*x^3-2*a*c*x^2-7*c*x-8*c/a*ln(a*x-1)

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maxima [A]  time = 0.30, size = 33, normalized size = 0.72 \[ -\frac {1}{3} \, a^{2} c x^{3} - 2 \, a c x^{2} - 7 \, c x - \frac {8 \, c \log \left (a x - 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*c*x^3 - 2*a*c*x^2 - 7*c*x - 8*c*log(a*x - 1)/a

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mupad [B]  time = 0.05, size = 33, normalized size = 0.72 \[ -7\,c\,x-\frac {a^2\,c\,x^3}{3}-\frac {8\,c\,\ln \left (a\,x-1\right )}{a}-2\,a\,c\,x^2 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 7*c*x - (a^2*c*x^3)/3 - (8*c*log(a*x - 1))/a - 2*a*c*x^2

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sympy [A]  time = 0.17, size = 36, normalized size = 0.78 \[ - \frac {a^{2} c x^{3}}{3} - 2 a c x^{2} - 7 c x - \frac {8 c \log {\left (a x - 1 \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*c*x**3/3 - 2*a*c*x**2 - 7*c*x - 8*c*log(a*x - 1)/a

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