3.99 \(\int x^{11} (a+b \tanh ^{-1}(c x^3)) \, dx\)

Optimal. Leaf size=54 \[ \frac{1}{12} x^{12} \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b x^3}{12 c^3}-\frac{b \tanh ^{-1}\left (c x^3\right )}{12 c^4}+\frac{b x^9}{36 c} \]

[Out]

(b*x^3)/(12*c^3) + (b*x^9)/(36*c) - (b*ArcTanh[c*x^3])/(12*c^4) + (x^12*(a + b*ArcTanh[c*x^3]))/12

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Rubi [A]  time = 0.0382175, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6097, 275, 302, 206} \[ \frac{1}{12} x^{12} \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac{b x^3}{12 c^3}-\frac{b \tanh ^{-1}\left (c x^3\right )}{12 c^4}+\frac{b x^9}{36 c} \]

Antiderivative was successfully verified.

[In]

Int[x^11*(a + b*ArcTanh[c*x^3]),x]

[Out]

(b*x^3)/(12*c^3) + (b*x^9)/(36*c) - (b*ArcTanh[c*x^3])/(12*c^4) + (x^12*(a + b*ArcTanh[c*x^3]))/12

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 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^{11} \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\frac{1}{12} x^{12} \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{1}{4} (b c) \int \frac{x^{14}}{1-c^2 x^6} \, dx\\ &=\frac{1}{12} x^{12} \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{1}{12} (b c) \operatorname{Subst}\left (\int \frac{x^4}{1-c^2 x^2} \, dx,x,x^3\right )\\ &=\frac{1}{12} x^{12} \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac{1}{12} (b c) \operatorname{Subst}\left (\int \left (-\frac{1}{c^4}-\frac{x^2}{c^2}+\frac{1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx,x,x^3\right )\\ &=\frac{b x^3}{12 c^3}+\frac{b x^9}{36 c}+\frac{1}{12} x^{12} \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 )}{12 c^3}\\ &=\frac{b x^3}{12 c^3}+\frac{b x^9}{36 c}-\frac{b \tanh ^{-1}\left (c x^3\right )}{12 c^4}+\frac{1}{12} x^{12} \left (a+b \tanh ^{-1}\left (c x^3\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0200856, size = 78, normalized size = 1.44 \[ \frac{a x^{12}}{12}+\frac{b x^3}{12 c^3}+\frac{b \log \left (1-c x^3\right )}{24 c^4}-\frac{b \log \left (c x^3+1\right )}{24 c^4}+\frac{b x^9}{36 c}+\frac{1}{12} b x^{12} \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^11*(a + b*ArcTanh[c*x^3]),x]

[Out]

(b*x^3)/(12*c^3) + (b*x^9)/(36*c) + (a*x^12)/12 + (b*x^12*ArcTanh[c*x^3])/12 + (b*Log[1 - c*x^3])/(24*c^4) - (
b*Log[1 + c*x^3])/(24*c^4)

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Maple [A]  time = 0.009, size = 66, normalized size = 1.2 \begin{align*}{\frac{{x}^{12}a}{12}}+{\frac{b{x}^{12}{\it Artanh} \left ( c{x}^{3} \right ) }{12}}+{\frac{b{x}^{9}}{36\,c}}+{\frac{b{x}^{3}}{12\,{c}^{3}}}+{\frac{b\ln \left ( c{x}^{3}-1 \right ) }{24\,{c}^{4}}}-{\frac{b\ln \left ( c{x}^{3}+1 \right ) }{24\,{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(a+b*arctanh(c*x^3)),x)

[Out]

1/12*x^12*a+1/12*b*x^12*arctanh(c*x^3)+1/36*b*x^9/c+1/12*b*x^3/c^3+1/24*b/c^4*ln(c*x^3-1)-1/24*b/c^4*ln(c*x^3+
1)

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Maxima [A]  time = 0.953359, size = 93, normalized size = 1.72 \begin{align*} \frac{1}{12} \, a x^{12} + \frac{1}{72} \,{\left (6 \, x^{12} \operatorname{artanh}\left (c x^{3}\right ) + c{\left (\frac{2 \,{\left (c^{2} x^{9} + 3 \, x^{3}\right )}}{c^{4}} - \frac{3 \, \log \left (c x^{3} + 1\right )}{c^{5}} + \frac{3 \, \log \left (c x^{3} - 1\right )}{c^{5}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(a+b*arctanh(c*x^3)),x, algorithm="maxima")

[Out]

1/12*a*x^12 + 1/72*(6*x^12*arctanh(c*x^3) + c*(2*(c^2*x^9 + 3*x^3)/c^4 - 3*log(c*x^3 + 1)/c^5 + 3*log(c*x^3 -
1)/c^5))*b

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Fricas [A]  time = 2.04169, size = 138, normalized size = 2.56 \begin{align*} \frac{6 \, a c^{4} x^{12} + 2 \, b c^{3} x^{9} + 6 \, b c x^{3} + 3 \,{\left (b c^{4} x^{12} - b\right )} \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right )}{72 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(a+b*arctanh(c*x^3)),x, algorithm="fricas")

[Out]

1/72*(6*a*c^4*x^12 + 2*b*c^3*x^9 + 6*b*c*x^3 + 3*(b*c^4*x^12 - b)*log(-(c*x^3 + 1)/(c*x^3 - 1)))/c^4

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

[Out]

Exception raised: KeyError

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Giac [A]  time = 1.20153, size = 105, normalized size = 1.94 \begin{align*} \frac{1}{24} \, b x^{12} \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + \frac{1}{12} \, a x^{12} + \frac{b x^{9}}{36 \, c} + \frac{b x^{3}}{12 \, c^{3}} - \frac{b \log \left (c x^{3} + 1\right )}{24 \, c^{4}} + \frac{b \log \left (c x^{3} - 1\right )}{24 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(a+b*arctanh(c*x^3)),x, algorithm="giac")

[Out]

1/24*b*x^12*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 1/12*a*x^12 + 1/36*b*x^9/c + 1/12*b*x^3/c^3 - 1/24*b*log(c*x^3 + 1
)/c^4 + 1/24*b*log(c*x^3 - 1)/c^4