3.3 \(\int x (d+c d x) (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=84 \[ \frac{1}{3} c d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{5 b d \log (1-c x)}{12 c^2}-\frac{b d \log (c x+1)}{12 c^2}+\frac{b d x}{2 c}+\frac{1}{6} b d x^2 \]

[Out]

(b*d*x)/(2*c) + (b*d*x^2)/6 + (d*x^2*(a + b*ArcTanh[c*x]))/2 + (c*d*x^3*(a + b*ArcTanh[c*x]))/3 + (5*b*d*Log[1
- c*x])/(12*c^2) - (b*d*Log[1 + c*x])/(12*c^2)

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Rubi [A]  time = 0.074793, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {43, 5936, 12, 801, 633, 31} \[ \frac{1}{3} c d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{5 b d \log (1-c x)}{12 c^2}-\frac{b d \log (c x+1)}{12 c^2}+\frac{b d x}{2 c}+\frac{1}{6} b d x^2 \]

Antiderivative was successfully verified.

[In]

Int[x*(d + c*d*x)*(a + b*ArcTanh[c*x]),x]

[Out]

(b*d*x)/(2*c) + (b*d*x^2)/6 + (d*x^2*(a + b*ArcTanh[c*x]))/2 + (c*d*x^3*(a + b*ArcTanh[c*x]))/3 + (5*b*d*Log[1
- c*x])/(12*c^2) - (b*d*Log[1 + c*x])/(12*c^2)

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 5936

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int x (d+c d x) \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{3} c d x^3 \left (a+b \tanh ^{-1}(c x)\right )-(b c) \int \frac{d x^2 (3+2 c x)}{6-6 c^2 x^2} \, dx\\ &=\frac{1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{3} c d x^3 \left (a+b \tanh ^{-1}(c x)\right )-(b c d) \int \frac{x^2 (3+2 c x)}{6-6 c^2 x^2} \, dx\\ &=\frac{1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{3} c d x^3 \left (a+b \tanh ^{-1}(c x)\right )-(b c d) \int \left (-\frac{1}{2 c^2}-\frac{x}{3 c}+\frac{3+2 c x}{c^2 \left (6-6 c^2 x^2\right )}\right ) \, dx\\ &=\frac{b d x}{2 c}+\frac{1}{6} b d x^2+\frac{1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{3} c d x^3 \left (a+b \tanh ^{-1}(c x)\right )-\frac{(b d) \int \frac{3+2 c x}{6-6 c^2 x^2} \, dx}{c}\\ &=\frac{b d x}{2 c}+\frac{1}{6} b d x^2+\frac{1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{3} c d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} (b d) \int \frac{1}{-6 c-6 c^2 x} \, dx-\frac{1}{2} (5 b d) \int \frac{1}{6 c-6 c^2 x} \, dx\\ &=\frac{b d x}{2 c}+\frac{1}{6} b d x^2+\frac{1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{3} c d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{5 b d \log (1-c x)}{12 c^2}-\frac{b d \log (1+c x)}{12 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0554376, size = 79, normalized size = 0.94 \[ \frac{d \left (4 a c^3 x^3+6 a c^2 x^2+2 b c^2 x^2+2 b c^2 x^2 (2 c x+3) \tanh ^{-1}(c x)+6 b c x+5 b \log (1-c x)-b \log (c x+1)\right )}{12 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + c*d*x)*(a + b*ArcTanh[c*x]),x]

[Out]

(d*(6*b*c*x + 6*a*c^2*x^2 + 2*b*c^2*x^2 + 4*a*c^3*x^3 + 2*b*c^2*x^2*(3 + 2*c*x)*ArcTanh[c*x] + 5*b*Log[1 - c*x
] - b*Log[1 + c*x]))/(12*c^2)

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Maple [A]  time = 0.031, size = 81, normalized size = 1. \begin{align*}{\frac{cda{x}^{3}}{3}}+{\frac{da{x}^{2}}{2}}+{\frac{cdb{\it Artanh} \left ( cx \right ){x}^{3}}{3}}+{\frac{db{\it Artanh} \left ( cx \right ){x}^{2}}{2}}+{\frac{bd{x}^{2}}{6}}+{\frac{bdx}{2\,c}}+{\frac{5\,db\ln \left ( cx-1 \right ) }{12\,{c}^{2}}}-{\frac{db\ln \left ( cx+1 \right ) }{12\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*d*x+d)*(a+b*arctanh(c*x)),x)

[Out]

1/3*c*d*a*x^3+1/2*d*a*x^2+1/3*c*d*b*arctanh(c*x)*x^3+1/2*d*b*arctanh(c*x)*x^2+1/6*b*d*x^2+1/2*b*d*x/c+5/12/c^2
*d*b*ln(c*x-1)-1/12*b*d*ln(c*x+1)/c^2

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Maxima [A]  time = 0.97236, size = 134, normalized size = 1.6 \begin{align*} \frac{1}{3} \, a c d x^{3} + \frac{1}{6} \,{\left (2 \, x^{3} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{x^{2}}{c^{2}} + \frac{\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c d + \frac{1}{2} \, a d x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \, x}{c^{2}} - \frac{\log \left (c x + 1\right )}{c^{3}} + \frac{\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*d*x+d)*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.96771, size = 219, normalized size = 2.61 \begin{align*} \frac{4 \, a c^{3} d x^{3} + 2 \,{\left (3 \, a + b\right )} c^{2} d x^{2} + 6 \, b c d x - b d \log \left (c x + 1\right ) + 5 \, b d \log \left (c x - 1\right ) +{\left (2 \, b c^{3} d x^{3} + 3 \, b c^{2} d x^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{12 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*d*x+d)*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/12*(4*a*c^3*d*x^3 + 2*(3*a + b)*c^2*d*x^2 + 6*b*c*d*x - b*d*log(c*x + 1) + 5*b*d*log(c*x - 1) + (2*b*c^3*d*x
^3 + 3*b*c^2*d*x^2)*log(-(c*x + 1)/(c*x - 1)))/c^2

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Sympy [A]  time = 1.61303, size = 100, normalized size = 1.19 \begin{align*} \begin{cases} \frac{a c d x^{3}}{3} + \frac{a d x^{2}}{2} + \frac{b c d x^{3} \operatorname{atanh}{\left (c x \right )}}{3} + \frac{b d x^{2} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{b d x^{2}}{6} + \frac{b d x}{2 c} + \frac{b d \log{\left (x - \frac{1}{c} \right )}}{3 c^{2}} - \frac{b d \operatorname{atanh}{\left (c x \right )}}{6 c^{2}} & \text{for}\: c \neq 0 \\\frac{a d x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*d*x+d)*(a+b*atanh(c*x)),x)

[Out]

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

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Giac [A]  time = 1.22319, size = 120, normalized size = 1.43 \begin{align*} \frac{1}{3} \, a c d x^{3} + \frac{1}{6} \,{\left (3 \, a d + b d\right )} x^{2} + \frac{b d x}{2 \, c} + \frac{1}{12} \,{\left (2 \, b c d x^{3} + 3 \, b d x^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) - \frac{b d \log \left (c x + 1\right )}{12 \, c^{2}} + \frac{5 \, b d \log \left (c x - 1\right )}{12 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*d*x+d)*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

1/3*a*c*d*x^3 + 1/6*(3*a*d + b*d)*x^2 + 1/2*b*d*x/c + 1/12*(2*b*c*d*x^3 + 3*b*d*x^2)*log(-(c*x + 1)/(c*x - 1))
- 1/12*b*d*log(c*x + 1)/c^2 + 5/12*b*d*log(c*x - 1)/c^2