3.101 \(\int x^5 (a+b \tanh ^{-1}(c x^3)) \, dx\)

Optimal. Leaf size=43 \[ \frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{b \tanh ^{-1}\left (c x^3\right )}{6 c^2}+\frac{b x^3}{6 c} \]

[Out]

(b*x^3)/(6*c) - (b*ArcTanh[c*x^3])/(6*c^2) + (x^6*(a + b*ArcTanh[c*x^3]))/6

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Rubi [A]  time = 0.030026, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6097, 275, 321, 206} \[ \frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{b \tanh ^{-1}\left (c x^3\right )}{6 c^2}+\frac{b x^3}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcTanh[c*x^3]),x]

[Out]

(b*x^3)/(6*c) - (b*ArcTanh[c*x^3])/(6*c^2) + (x^6*(a + b*ArcTanh[c*x^3]))/6

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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^5 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{1}{2} (b c) \int \frac{x^8}{1-c^2 x^6} \, dx\\ &=\frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{x^2}{1-c^2 x^2} \, dx,x,x^3\right )\\ &=\frac{b x^3}{6 c}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{b \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,x^3\right )}{6 c}\\ &=\frac{b x^3}{6 c}-\frac{b \tanh ^{-1}\left (c x^3\right )}{6 c^2}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0156988, size = 67, normalized size = 1.56 \[ \frac{a x^6}{6}+\frac{b \log \left (1-c x^3\right )}{12 c^2}-\frac{b \log \left (c x^3+1\right )}{12 c^2}+\frac{b x^3}{6 c}+\frac{1}{6} b x^6 \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*ArcTanh[c*x^3]),x]

[Out]

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

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Maple [A]  time = 0.012, size = 57, normalized size = 1.3 \begin{align*}{\frac{{x}^{6}a}{6}}+{\frac{b{x}^{6}{\it Artanh} \left ( c{x}^{3} \right ) }{6}}+{\frac{b{x}^{3}}{6\,c}}+{\frac{b\ln \left ( c{x}^{3}-1 \right ) }{12\,{c}^{2}}}-{\frac{b\ln \left ( c{x}^{3}+1 \right ) }{12\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arctanh(c*x^3)),x)

[Out]

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

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Maxima [A]  time = 0.979988, size = 78, normalized size = 1.81 \begin{align*} \frac{1}{6} \, a x^{6} + \frac{1}{12} \,{\left (2 \, x^{6} \operatorname{artanh}\left (c x^{3}\right ) + c{\left (\frac{2 \, x^{3}}{c^{2}} - \frac{\log \left (c x^{3} + 1\right )}{c^{3}} + \frac{\log \left (c x^{3} - 1\right )}{c^{3}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x^3)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.79863, size = 113, normalized size = 2.63 \begin{align*} \frac{2 \, a c^{2} x^{6} + 2 \, b c x^{3} +{\left (b c^{2} x^{6} - b\right )} \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right )}{12 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x^3)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: KeyError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*atanh(c*x**3)),x)

[Out]

Exception raised: KeyError

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Giac [A]  time = 1.14918, size = 93, normalized size = 2.16 \begin{align*} \frac{1}{12} \, b x^{6} \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + \frac{1}{6} \, a x^{6} + \frac{b x^{3}}{6 \, c} - \frac{b \log \left (c x^{3} + 1\right )}{12 \, c^{2}} + \frac{b \log \left (c x^{3} - 1\right )}{12 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x^3)),x, algorithm="giac")

[Out]

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