3.3 \(\int x^3 (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=48 \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b x}{4 c^3}-\frac{b \tanh ^{-1}(c x)}{4 c^4}+\frac{b x^3}{12 c} \]

[Out]

(b*x)/(4*c^3) + (b*x^3)/(12*c) - (b*ArcTanh[c*x])/(4*c^4) + (x^4*(a + b*ArcTanh[c*x]))/4

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Rubi [A]  time = 0.0289794, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5916, 302, 206} \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b x}{4 c^3}-\frac{b \tanh ^{-1}(c x)}{4 c^4}+\frac{b x^3}{12 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x)/(4*c^3) + (b*x^3)/(12*c) - (b*ArcTanh[c*x])/(4*c^4) + (x^4*(a + b*ArcTanh[c*x]))/4

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int x^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{4} (b c) \int \frac{x^4}{1-c^2 x^2} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{4} (b c) \int \left (-\frac{1}{c^4}-\frac{x^2}{c^2}+\frac{1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac{b x}{4 c^3}+\frac{b x^3}{12 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{b \int \frac{1}{1-c^2 x^2} \, dx}{4 c^3}\\ &=\frac{b x}{4 c^3}+\frac{b x^3}{12 c}-\frac{b \tanh ^{-1}(c x)}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0086165, size = 70, normalized size = 1.46 \[ \frac{a x^4}{4}+\frac{b x}{4 c^3}+\frac{b \log (1-c x)}{8 c^4}-\frac{b \log (c x+1)}{8 c^4}+\frac{b x^3}{12 c}+\frac{1}{4} b x^4 \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x)/(4*c^3) + (b*x^3)/(12*c) + (a*x^4)/4 + (b*x^4*ArcTanh[c*x])/4 + (b*Log[1 - c*x])/(8*c^4) - (b*Log[1 + c*
x])/(8*c^4)

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Maple [A]  time = 0.007, size = 58, normalized size = 1.2 \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}{\it Artanh} \left ( cx \right ) }{4}}+{\frac{b{x}^{3}}{12\,c}}+{\frac{bx}{4\,{c}^{3}}}+{\frac{b\ln \left ( cx-1 \right ) }{8\,{c}^{4}}}-{\frac{b\ln \left ( cx+1 \right ) }{8\,{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x)),x)

[Out]

1/4*x^4*a+1/4*b*x^4*arctanh(c*x)+1/12*b*x^3/c+1/4*b*x/c^3+1/8/c^4*b*ln(c*x-1)-1/8/c^4*b*ln(c*x+1)

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Maxima [A]  time = 0.959693, size = 82, normalized size = 1.71 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{24} \,{\left (6 \, x^{4} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \,{\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac{3 \, \log \left (c x + 1\right )}{c^{5}} + \frac{3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/4*a*x^4 + 1/24*(6*x^4*arctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*b

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Fricas [A]  time = 2.02546, size = 127, normalized size = 2.65 \begin{align*} \frac{6 \, a c^{4} x^{4} + 2 \, b c^{3} x^{3} + 6 \, b c x + 3 \,{\left (b c^{4} x^{4} - b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{24 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/24*(6*a*c^4*x^4 + 2*b*c^3*x^3 + 6*b*c*x + 3*(b*c^4*x^4 - b)*log(-(c*x + 1)/(c*x - 1)))/c^4

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Sympy [A]  time = 1.16716, size = 53, normalized size = 1.1 \begin{align*} \begin{cases} \frac{a x^{4}}{4} + \frac{b x^{4} \operatorname{atanh}{\left (c x \right )}}{4} + \frac{b x^{3}}{12 c} + \frac{b x}{4 c^{3}} - \frac{b \operatorname{atanh}{\left (c x \right )}}{4 c^{4}} & \text{for}\: c \neq 0 \\\frac{a x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*x**4/4 + b*x**4*atanh(c*x)/4 + b*x**3/(12*c) + b*x/(4*c**3) - b*atanh(c*x)/(4*c**4), Ne(c, 0)), (
a*x**4/4, True))

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Giac [A]  time = 1.2352, size = 92, normalized size = 1.92 \begin{align*} \frac{1}{8} \, b x^{4} \log \left (-\frac{c x + 1}{c x - 1}\right ) + \frac{1}{4} \, a x^{4} + \frac{b x^{3}}{12 \, c} + \frac{b x}{4 \, c^{3}} - \frac{b \log \left (c x + 1\right )}{8 \, c^{4}} + \frac{b \log \left (c x - 1\right )}{8 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

1/8*b*x^4*log(-(c*x + 1)/(c*x - 1)) + 1/4*a*x^4 + 1/12*b*x^3/c + 1/4*b*x/c^3 - 1/8*b*log(c*x + 1)/c^4 + 1/8*b*
log(c*x - 1)/c^4