3.51 \(\int x^5 (a+b \tanh ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=48 \[ \frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac{b \log \left (1-c^2 x^4\right )}{12 c^3}+\frac{b x^4}{12 c} \]

[Out]

(b*x^4)/(12*c) + (x^6*(a + b*ArcTanh[c*x^2]))/6 + (b*Log[1 - c^2*x^4])/(12*c^3)

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Rubi [A]  time = 0.0347672, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6097, 266, 43} \[ \frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac{b \log \left (1-c^2 x^4\right )}{12 c^3}+\frac{b x^4}{12 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcTanh[c*x^2]),x]

[Out]

(b*x^4)/(12*c) + (x^6*(a + b*ArcTanh[c*x^2]))/6 + (b*Log[1 - c^2*x^4])/(12*c^3)

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 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^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{1}{3} (b c) \int \frac{x^7}{1-c^2 x^4} \, dx\\ &=\frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{1}{12} (b c) \operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^4\right )\\ &=\frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{1}{12} (b c) \operatorname{Subst}\left (\int \left (-\frac{1}{c^2}-\frac{1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^4\right )\\ &=\frac{b x^4}{12 c}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac{b \log \left (1-c^2 x^4\right )}{12 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0139929, size = 53, normalized size = 1.1 \[ \frac{a x^6}{6}+\frac{b \log \left (1-c^2 x^4\right )}{12 c^3}+\frac{b x^4}{12 c}+\frac{1}{6} b x^6 \tanh ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*ArcTanh[c*x^2]),x]

[Out]

(b*x^4)/(12*c) + (a*x^6)/6 + (b*x^6*ArcTanh[c*x^2])/6 + (b*Log[1 - c^2*x^4])/(12*c^3)

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Maple [A]  time = 0.009, size = 45, normalized size = 0.9 \begin{align*}{\frac{{x}^{6}a}{6}}+{\frac{b{x}^{6}{\it Artanh} \left ( c{x}^{2} \right ) }{6}}+{\frac{b{x}^{4}}{12\,c}}+{\frac{b\ln \left ({c}^{2}{x}^{4}-1 \right ) }{12\,{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arctanh(c*x^2)),x)

[Out]

1/6*x^6*a+1/6*b*x^6*arctanh(c*x^2)+1/12*b*x^4/c+1/12*b/c^3*ln(c^2*x^4-1)

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Maxima [A]  time = 0.98235, size = 62, normalized size = 1.29 \begin{align*} \frac{1}{6} \, a x^{6} + \frac{1}{12} \,{\left (2 \, x^{6} \operatorname{artanh}\left (c x^{2}\right ) +{\left (\frac{x^{4}}{c^{2}} + \frac{\log \left (c^{2} x^{4} - 1\right )}{c^{4}}\right )} c\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x^2)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.05896, size = 134, normalized size = 2.79 \begin{align*} \frac{b c^{3} x^{6} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a c^{3} x^{6} + b c^{2} x^{4} + b \log \left (c^{2} x^{4} - 1\right )}{12 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x^2)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 21.0352, size = 85, normalized size = 1.77 \begin{align*} \begin{cases} \frac{a x^{6}}{6} + \frac{b x^{6} \operatorname{atanh}{\left (c x^{2} \right )}}{6} + \frac{b x^{4}}{12 c} + \frac{b \log{\left (x - i \sqrt{\frac{1}{c}} \right )}}{6 c^{3}} + \frac{b \log{\left (x + i \sqrt{\frac{1}{c}} \right )}}{6 c^{3}} - \frac{b \operatorname{atanh}{\left (c x^{2} \right )}}{6 c^{3}} & \text{for}\: c \neq 0 \\\frac{a x^{6}}{6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*atanh(c*x**2)),x)

[Out]

Piecewise((a*x**6/6 + b*x**6*atanh(c*x**2)/6 + b*x**4/(12*c) + b*log(x - I*sqrt(1/c))/(6*c**3) + b*log(x + I*s
qrt(1/c))/(6*c**3) - b*atanh(c*x**2)/(6*c**3), Ne(c, 0)), (a*x**6/6, True))

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Giac [A]  time = 1.17006, size = 77, normalized size = 1.6 \begin{align*} \frac{1}{12} \, b x^{6} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + \frac{1}{6} \, a x^{6} + \frac{b x^{4}}{12 \, c} + \frac{b \log \left (c^{2} x^{4} - 1\right )}{12 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x^2)),x, algorithm="giac")

[Out]

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