\(\int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx\) [191]

Optimal result

Integrand size = 18, antiderivative size = 17 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx=\frac {c^2 (1+a x)^3}{3 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6302, 6264, 32} \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx=\frac {c^2 (a x+1)^3}{3 a} \]

[In]

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

[Out]

(c^2*(1 + a*x)^3)/(3*a)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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])

Rule 6302

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*} \text {integral}& = \int e^{4 \text {arctanh}(a x)} (c-a c x)^2 \, dx \\ & = c^2 \int (1+a x)^2 \, dx \\ & = \frac {c^2 (1+a x)^3}{3 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx=c^2 \left (x+a x^2+\frac {a^2 x^3}{3}\right ) \]

[In]

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

[Out]

c^2*(x + a*x^2 + (a^2*x^3)/3)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {c^{2} \left (a x +1\right )^{3}}{3 a}\)	\(16\)
gosper	\(\frac {x \left (a^{2} x^{2}+3 a x +3\right ) c^{2}}{3}\)	\(20\)
parallelrisch	\(\frac {1}{3} a^{2} c^{2} x^{3}+a \,c^{2} x^{2}+c^{2} x\)	\(26\)
risch	\(\frac {a^{2} c^{2} x^{3}}{3}+a \,c^{2} x^{2}+c^{2} x +\frac {c^{2}}{3 a}\)	\(34\)
norman	\(\frac {-\frac {c^{2}}{a}+\frac {2 a^{2} c^{2} x^{3}}{3}+\frac {a^{3} c^{2} x^{4}}{3}}{a x -1}\)	\(40\)
meijerg	\(-\frac {c^{2} \left (-\frac {a x \left (-5 a^{3} x^{3}-10 a^{2} x^{2}-30 a x +60\right )}{15 \left (-a x +1\right )}-4 \ln \left (-a x +1\right )\right )}{a}+\frac {2 c^{2} \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}+\frac {c^{2} x}{-a x +1}\)	\(103\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx=\frac {1}{3} \, a^{2} c^{2} x^{3} + a c^{2} x^{2} + c^{2} x \]

[In]

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

[Out]

1/3*a^2*c^2*x^3 + a*c^2*x^2 + c^2*x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx=\frac {a^{2} c^{2} x^{3}}{3} + a c^{2} x^{2} + c^{2} x \]

[In]

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

[Out]

a**2*c**2*x**3/3 + a*c**2*x**2 + c**2*x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx=\frac {1}{3} \, a^{2} c^{2} x^{3} + a c^{2} x^{2} + c^{2} x \]

[In]

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

[Out]

1/3*a^2*c^2*x^3 + a*c^2*x^2 + c^2*x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (15) = 30\).

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.35 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx=\frac {{\left (c^{2} + \frac {6 \, c^{2}}{a x - 1} + \frac {12 \, c^{2}}{{\left (a x - 1\right )}^{2}}\right )} {\left (a x - 1\right )}^{3}}{3 \, a} \]

[In]

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

[Out]

1/3*(c^2 + 6*c^2/(a*x - 1) + 12*c^2/(a*x - 1)^2)*(a*x - 1)^3/a

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^2 \, dx=\frac {c^2\,x\,\left (a^2\,x^2+3\,a\,x+3\right )}{3} \]

[In]

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

[Out]

(c^2*x*(3*a*x + a^2*x^2 + 3))/3