3.6.68 \(\int e^{2 \coth ^{-1}(a x)} (c-a^2 c x^2) \, dx\) [568]

Optimal. Leaf size=15 \[ -\frac {c (1+a x)^3}{3 a} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6302, 6275, 32} \begin {gather*} -\frac {c (a x+1)^3}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(c*(1 + a*x)^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 6275

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

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Mathematica [A]
time = 0.01, size = 23, normalized size = 1.53 \begin {gather*} -c x-a c x^2-\frac {1}{3} a^2 c x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*x) - a*c*x^2 - (a^2*c*x^3)/3

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Maple [A]
time = 0.10, size = 14, normalized size = 0.93

method	result	size
default	\(-\frac {c \left (a x +1\right )^{3}}{3 a}\)	\(14\)
gosper	\(-\frac {c x \left (a^{2} x^{2}+3 a x +3\right )}{3}\)	\(18\)
norman	\(-c x -a c \,x^{2}-\frac {1}{3} a^{2} c \,x^{3}\)	\(22\)
risch	\(-\frac {a^{2} c \,x^{3}}{3}-a c \,x^{2}-c x -\frac {c}{3 a}\)	\(28\)
meijerg	\(\frac {c \left (-\frac {a x \left (4 a^{2} x^{2}+6 a x +12\right )}{12}-\ln \left (-a x +1\right )\right )}{a}-\frac {c \left (-a x -\ln \left (-a x +1\right )\right )}{a}-\frac {c \left (\frac {a x \left (3 a x +6\right )}{6}+\ln \left (-a x +1\right )\right )}{a}+\frac {c \ln \left (-a x +1\right )}{a}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 21, normalized size = 1.40 \begin {gather*} -\frac {1}{3} \, a^{2} c x^{3} - a c x^{2} - c x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*c*x^3 - a*c*x^2 - c*x

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Fricas [A]
time = 0.35, size = 21, normalized size = 1.40 \begin {gather*} -\frac {1}{3} \, a^{2} c x^{3} - a c x^{2} - c x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*c*x^3 - a*c*x^2 - c*x

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Sympy [A]
time = 0.01, size = 20, normalized size = 1.33 \begin {gather*} - \frac {a^{2} c x^{3}}{3} - a c x^{2} - c x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*c*x**3/3 - a*c*x**2 - c*x

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Giac [A]
time = 0.41, size = 21, normalized size = 1.40 \begin {gather*} -\frac {1}{3} \, a^{2} c x^{3} - a c x^{2} - c x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*c*x^3 - a*c*x^2 - c*x

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Mupad [B]
time = 0.03, size = 17, normalized size = 1.13 \begin {gather*} -\frac {c\,x\,\left (a^2\,x^2+3\,a\,x+3\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*x*(3*a*x + a^2*x^2 + 3))/3

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