3.2 \(\int x^4 (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=57 \[ \frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b x^2}{10 c^3}+\frac{b \log \left (1-c^2 x^2\right )}{10 c^5}+\frac{b x^4}{20 c} \]

[Out]

(b*x^2)/(10*c^3) + (b*x^4)/(20*c) + (x^5*(a + b*ArcTanh[c*x]))/5 + (b*Log[1 - c^2*x^2])/(10*c^5)

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Rubi [A]  time = 0.0426034, antiderivative size = 57, 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, 266, 43} \[ \frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b x^2}{10 c^3}+\frac{b \log \left (1-c^2 x^2\right )}{10 c^5}+\frac{b x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x^2)/(10*c^3) + (b*x^4)/(20*c) + (x^5*(a + b*ArcTanh[c*x]))/5 + (b*Log[1 - c^2*x^2])/(10*c^5)

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 x^4 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{5} (b c) \int \frac{x^5}{1-c^2 x^2} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{x^2}{1-c^2 x} \, dx,x,x^2\right )\\ &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \left (-\frac{1}{c^4}-\frac{x}{c^2}-\frac{1}{c^4 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac{b x^2}{10 c^3}+\frac{b x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b \log \left (1-c^2 x^2\right )}{10 c^5}\\ \end{align*}

Mathematica [A]  time = 0.0085123, size = 62, normalized size = 1.09 \[ \frac{a x^5}{5}+\frac{b x^2}{10 c^3}+\frac{b \log \left (1-c^2 x^2\right )}{10 c^5}+\frac{b x^4}{20 c}+\frac{1}{5} b x^5 \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x^2)/(10*c^3) + (b*x^4)/(20*c) + (a*x^5)/5 + (b*x^5*ArcTanh[c*x])/5 + (b*Log[1 - c^2*x^2])/(10*c^5)

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Maple [A]  time = 0.007, size = 60, normalized size = 1.1 \begin{align*}{\frac{a{x}^{5}}{5}}+{\frac{b{x}^{5}{\it Artanh} \left ( cx \right ) }{5}}+{\frac{b{x}^{4}}{20\,c}}+{\frac{b{x}^{2}}{10\,{c}^{3}}}+{\frac{b\ln \left ( cx-1 \right ) }{10\,{c}^{5}}}+{\frac{b\ln \left ( cx+1 \right ) }{10\,{c}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c*x)),x)

[Out]

1/5*a*x^5+1/5*b*x^5*arctanh(c*x)+1/20*b*x^4/c+1/10*b*x^2/c^3+1/10/c^5*b*ln(c*x-1)+1/10/c^5*b*ln(c*x+1)

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Maxima [A]  time = 0.966233, size = 74, normalized size = 1.3 \begin{align*} \frac{1}{5} \, a x^{5} + \frac{1}{20} \,{\left (4 \, x^{5} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/5*a*x^5 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*b

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Fricas [A]  time = 2.05552, size = 153, normalized size = 2.68 \begin{align*} \frac{2 \, b c^{5} x^{5} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 4 \, a c^{5} x^{5} + b c^{4} x^{4} + 2 \, b c^{2} x^{2} + 2 \, b \log \left (c^{2} x^{2} - 1\right )}{20 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/20*(2*b*c^5*x^5*log(-(c*x + 1)/(c*x - 1)) + 4*a*c^5*x^5 + b*c^4*x^4 + 2*b*c^2*x^2 + 2*b*log(c^2*x^2 - 1))/c^
5

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Sympy [A]  time = 1.61565, size = 68, normalized size = 1.19 \begin{align*} \begin{cases} \frac{a x^{5}}{5} + \frac{b x^{5} \operatorname{atanh}{\left (c x \right )}}{5} + \frac{b x^{4}}{20 c} + \frac{b x^{2}}{10 c^{3}} + \frac{b \log{\left (x - \frac{1}{c} \right )}}{5 c^{5}} + \frac{b \operatorname{atanh}{\left (c x \right )}}{5 c^{5}} & \text{for}\: c \neq 0 \\\frac{a x^{5}}{5} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*x**5/5 + b*x**5*atanh(c*x)/5 + b*x**4/(20*c) + b*x**2/(10*c**3) + b*log(x - 1/c)/(5*c**5) + b*ata
nh(c*x)/(5*c**5), Ne(c, 0)), (a*x**5/5, True))

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Giac [A]  time = 1.15635, size = 84, normalized size = 1.47 \begin{align*} \frac{1}{10} \, b x^{5} \log \left (-\frac{c x + 1}{c x - 1}\right ) + \frac{1}{5} \, a x^{5} + \frac{b x^{4}}{20 \, c} + \frac{b x^{2}}{10 \, c^{3}} + \frac{b \log \left (c^{2} x^{2} - 1\right )}{10 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

1/10*b*x^5*log(-(c*x + 1)/(c*x - 1)) + 1/5*a*x^5 + 1/20*b*x^4/c + 1/10*b*x^2/c^3 + 1/10*b*log(c^2*x^2 - 1)/c^5