3.1.99 \(\int x^{11} (a+b \tanh ^{-1}(c x^3)) \, dx\) [99]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, 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 = {6037, 281, 308, 212} \begin {gather*} \frac {1}{12} x^{12} \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {b \tanh ^{-1}\left (c x^3\right )}{12 c^4}+\frac {b x^3}{12 c^3}+\frac {b x^9}{36 c} \end {gather*}

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 212

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

Rule 281

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 308

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 6037

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

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) \text {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) \text {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 \text {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*}

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Mathematica [A]
time = 0.02, size = 78, normalized size = 1.44 \begin {gather*} \frac {b x^3}{12 c^3}+\frac {b x^9}{36 c}+\frac {a x^{12}}{12}+\frac {1}{12} b x^{12} \tanh ^{-1}\left (c x^3\right )+\frac {b \log \left (1-c x^3\right )}{24 c^4}-\frac {b \log \left (1+c x^3\right )}{24 c^4} \end {gather*}

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 = 0.03, size = 66, normalized size = 1.22

method	result	size
default	\(\frac {x^{12} a}{12}+\frac {b \,x^{12} \arctanh \left (c \,x^{3}\right )}{12}+\frac {b \,x^{9}}{36 c}+\frac {b \,x^{3}}{12 c^{3}}+\frac {b \ln \left (c \,x^{3}-1\right )}{24 c^{4}}-\frac {b \ln \left (c \,x^{3}+1\right )}{24 c^{4}}\)	\(66\)
risch	\(\frac {x^{12} b \ln \left (c \,x^{3}+1\right )}{24}-\frac {x^{12} b \ln \left (-c \,x^{3}+1\right )}{24}+\frac {x^{12} a}{12}+\frac {b \,x^{9}}{36 c}+\frac {b \,x^{3}}{12 c^{3}}-\frac {b \ln \left (c \,x^{3}+1\right )}{24 c^{4}}+\frac {b \ln \left (c \,x^{3}-1\right )}{24 c^{4}}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.25, size = 69, normalized size = 1.28 \begin {gather*} \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 {gather*}

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 = 0.37, size = 64, normalized size = 1.19 \begin {gather*} \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 {gather*}

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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.43, size = 78, normalized size = 1.44 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 1.11, size = 69, normalized size = 1.28 \begin {gather*} \frac {a\,x^{12}}{12}+\frac {b\,x^3}{12\,c^3}+\frac {b\,x^9}{36\,c}+\frac {b\,x^{12}\,\ln \left (c\,x^3+1\right )}{24}-\frac {b\,x^{12}\,\ln \left (1-c\,x^3\right )}{24}+\frac {b\,\mathrm {atan}\left (c\,x^3\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{12\,c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(a + b*atanh(c*x^3)),x)

[Out]

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

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