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

Optimal. Leaf size=45 \[ \frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )+\frac{1}{12} b c^3 \log \left (c^2-x^4\right )+\frac{1}{12} b c x^4 \]

[Out]

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

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Rubi [A]  time = 0.0329161, antiderivative size = 45, 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, 263, 266, 43} \[ \frac{1}{6} x^6 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )+\frac{1}{12} b c^3 \log \left (c^2-x^4\right )+\frac{1}{12} b c x^4 \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcTanh[c/x^2]),x]

[Out]

(b*c*x^4)/12 + (x^6*(a + b*ArcTanh[c/x^2]))/6 + (b*c^3*Log[c^2 - x^4])/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 263

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

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

Mathematica [A]  time = 0.0119029, size = 50, normalized size = 1.11 \[ \frac{a x^6}{6}+\frac{1}{12} b c^3 \log \left (x^4-c^2\right )+\frac{1}{12} b c x^4+\frac{1}{6} b x^6 \tanh ^{-1}\left (\frac{c}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 65, normalized size = 1.4 \begin{align*}{\frac{{x}^{6}a}{6}}+{\frac{b{x}^{6}}{6}{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) }+{\frac{b{c}^{3}}{12}\ln \left ( 1+{\frac{c}{{x}^{2}}} \right ) }+{\frac{bc{x}^{4}}{12}}-{\frac{b{c}^{3}\ln \left ({x}^{-1} \right ) }{3}}+{\frac{b{c}^{3}}{12}\ln \left ({\frac{c}{{x}^{2}}}-1 \right ) } \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*c^3*ln(1+c/x^2)+1/12*b*c*x^4-1/3*b*c^3*ln(1/x)+1/12*b*c^3*ln(c/x^2-1
)

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Maxima [A]  time = 0.968106, size = 57, normalized size = 1.27 \begin{align*} \frac{1}{6} \, a x^{6} + \frac{1}{12} \,{\left (2 \, x^{6} \operatorname{artanh}\left (\frac{c}{x^{2}}\right ) +{\left (x^{4} + c^{2} \log \left (x^{4} - c^{2}\right )\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(x^4 - c^2))*c)*b

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Fricas [A]  time = 1.71104, size = 124, normalized size = 2.76 \begin{align*} \frac{1}{12} \, b x^{6} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + \frac{1}{6} \, a x^{6} + \frac{1}{12} \, b c x^{4} + \frac{1}{12} \, b c^{3} \log \left (x^{4} - c^{2}\right ) \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*x^6*log((x^2 + c)/(x^2 - c)) + 1/6*a*x^6 + 1/12*b*c*x^4 + 1/12*b*c^3*log(x^4 - c^2)

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Sympy [C]  time = 13.1943, size = 75, normalized size = 1.67 \begin{align*} \frac{a x^{6}}{6} + \frac{b c^{3} \log{\left (- i \sqrt{c} + x \right )}}{6} + \frac{b c^{3} \log{\left (i \sqrt{c} + x \right )}}{6} - \frac{b c^{3} \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}}{6} + \frac{b c x^{4}}{12} + \frac{b x^{6} \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**6/6 + b*c**3*log(-I*sqrt(c) + x)/6 + b*c**3*log(I*sqrt(c) + x)/6 - b*c**3*atanh(c/x**2)/6 + b*c*x**4/12 +
b*x**6*atanh(c/x**2)/6

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Giac [A]  time = 1.24473, size = 70, normalized size = 1.56 \begin{align*} \frac{1}{12} \, b x^{6} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + \frac{1}{6} \, a x^{6} + \frac{1}{12} \, b c x^{4} + \frac{1}{12} \, b c^{3} \log \left (x^{4} - c^{2}\right ) \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((x^2 + c)/(x^2 - c)) + 1/6*a*x^6 + 1/12*b*c*x^4 + 1/12*b*c^3*log(x^4 - c^2)