3.38 \(\int (c+d x^2) \coth ^{-1}(a x) \, dx\)

Optimal. Leaf size=57 \[ \frac{\left (3 a^2 c+d\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+c x \coth ^{-1}(a x)+\frac{d x^2}{6 a}+\frac{1}{3} d x^3 \coth ^{-1}(a x) \]

[Out]

(d*x^2)/(6*a) + c*x*ArcCoth[a*x] + (d*x^3*ArcCoth[a*x])/3 + ((3*a^2*c + d)*Log[1 - a^2*x^2])/(6*a^3)

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Rubi [A]  time = 0.0659503, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5977, 1593, 444, 43} \[ \frac{\left (3 a^2 c+d\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+c x \coth ^{-1}(a x)+\frac{d x^2}{6 a}+\frac{1}{3} d x^3 \coth ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)*ArcCoth[a*x],x]

[Out]

(d*x^2)/(6*a) + c*x*ArcCoth[a*x] + (d*x^3*ArcCoth[a*x])/3 + ((3*a^2*c + d)*Log[1 - a^2*x^2])/(6*a^3)

Rule 5977

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^q, x]}, Dist[a + b*ArcCoth[c*x], u, x] - Dist[b*c, Int[u/(1 - c^2*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x
] && (IntegerQ[q] || ILtQ[q + 1/2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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])

Rubi steps

\begin{align*} \int \left (c+d x^2\right ) \coth ^{-1}(a x) \, dx &=c x \coth ^{-1}(a x)+\frac{1}{3} d x^3 \coth ^{-1}(a x)-a \int \frac{c x+\frac{d x^3}{3}}{1-a^2 x^2} \, dx\\ &=c x \coth ^{-1}(a x)+\frac{1}{3} d x^3 \coth ^{-1}(a x)-a \int \frac{x \left (c+\frac{d x^2}{3}\right )}{1-a^2 x^2} \, dx\\ &=c x \coth ^{-1}(a x)+\frac{1}{3} d x^3 \coth ^{-1}(a x)-\frac{1}{2} a \operatorname{Subst}\left (\int \frac{c+\frac{d x}{3}}{1-a^2 x} \, dx,x,x^2\right )\\ &=c x \coth ^{-1}(a x)+\frac{1}{3} d x^3 \coth ^{-1}(a x)-\frac{1}{2} a \operatorname{Subst}\left (\int \left (-\frac{d}{3 a^2}+\frac{-3 a^2 c-d}{3 a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac{d x^2}{6 a}+c x \coth ^{-1}(a x)+\frac{1}{3} d x^3 \coth ^{-1}(a x)+\frac{\left (3 a^2 c+d\right ) \log \left (1-a^2 x^2\right )}{6 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0116903, size = 69, normalized size = 1.21 \[ \frac{c \log \left (1-a^2 x^2\right )}{2 a}+\frac{d \log \left (1-a^2 x^2\right )}{6 a^3}+c x \coth ^{-1}(a x)+\frac{d x^2}{6 a}+\frac{1}{3} d x^3 \coth ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)*ArcCoth[a*x],x]

[Out]

(d*x^2)/(6*a) + c*x*ArcCoth[a*x] + (d*x^3*ArcCoth[a*x])/3 + (c*Log[1 - a^2*x^2])/(2*a) + (d*Log[1 - a^2*x^2])/
(6*a^3)

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Maple [A]  time = 0.038, size = 76, normalized size = 1.3 \begin{align*}{\frac{d{x}^{3}{\rm arccoth} \left (ax\right )}{3}}+cx{\rm arccoth} \left (ax\right )+{\frac{d{x}^{2}}{6\,a}}+{\frac{\ln \left ( ax-1 \right ) c}{2\,a}}+{\frac{\ln \left ( ax-1 \right ) d}{6\,{a}^{3}}}+{\frac{\ln \left ( ax+1 \right ) c}{2\,a}}+{\frac{\ln \left ( ax+1 \right ) d}{6\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)*arccoth(a*x),x)

[Out]

1/3*d*x^3*arccoth(a*x)+c*x*arccoth(a*x)+1/6*d*x^2/a+1/2/a*ln(a*x-1)*c+1/6/a^3*ln(a*x-1)*d+1/2/a*ln(a*x+1)*c+1/
6/a^3*ln(a*x+1)*d

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Maxima [A]  time = 0.974406, size = 88, normalized size = 1.54 \begin{align*} \frac{1}{6} \, a{\left (\frac{d x^{2}}{a^{2}} + \frac{{\left (3 \, a^{2} c + d\right )} \log \left (a x + 1\right )}{a^{4}} + \frac{{\left (3 \, a^{2} c + d\right )} \log \left (a x - 1\right )}{a^{4}}\right )} + \frac{1}{3} \,{\left (d x^{3} + 3 \, c x\right )} \operatorname{arcoth}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)*arccoth(a*x),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51526, size = 142, normalized size = 2.49 \begin{align*} \frac{a^{2} d x^{2} +{\left (3 \, a^{2} c + d\right )} \log \left (a^{2} x^{2} - 1\right ) +{\left (a^{3} d x^{3} + 3 \, a^{3} c x\right )} \log \left (\frac{a x + 1}{a x - 1}\right )}{6 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)*arccoth(a*x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.37328, size = 87, normalized size = 1.53 \begin{align*} \begin{cases} c x \operatorname{acoth}{\left (a x \right )} + \frac{d x^{3} \operatorname{acoth}{\left (a x \right )}}{3} + \frac{c \log{\left (x - \frac{1}{a} \right )}}{a} + \frac{c \operatorname{acoth}{\left (a x \right )}}{a} + \frac{d x^{2}}{6 a} + \frac{d \log{\left (x - \frac{1}{a} \right )}}{3 a^{3}} + \frac{d \operatorname{acoth}{\left (a x \right )}}{3 a^{3}} & \text{for}\: a \neq 0 \\\frac{i \pi \left (c x + \frac{d x^{3}}{3}\right )}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)*acoth(a*x),x)

[Out]

Piecewise((c*x*acoth(a*x) + d*x**3*acoth(a*x)/3 + c*log(x - 1/a)/a + c*acoth(a*x)/a + d*x**2/(6*a) + d*log(x -
 1/a)/(3*a**3) + d*acoth(a*x)/(3*a**3), Ne(a, 0)), (I*pi*(c*x + d*x**3/3)/2, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x^{2} + c\right )} \operatorname{arcoth}\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)*arccoth(a*x),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*arccoth(a*x), x)