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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 45, 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 = {6097, 263, 266, 43} \[ \frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (\frac {c}{x}\right )\right )+\frac {1}{6} b c^3 \log \left (c^2-x^2\right )+\frac {1}{6} b c x^2 \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c/x]),x]

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 50, normalized size = 1.11 \[ \frac {a x^3}{3}+\frac {1}{6} b c^3 \log \left (x^2-c^2\right )+\frac {1}{3} b x^3 \tanh ^{-1}\left (\frac {c}{x}\right )+\frac {1}{6} b c x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTanh[c/x]),x]

[Out]

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

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fricas [A]  time = 0.68, size = 49, normalized size = 1.09 \[ \frac {1}{6} \, b c^{3} \log \left (-c^{2} + x^{2}\right ) + \frac {1}{6} \, b x^{3} \log \left (-\frac {c + x}{c - x}\right ) + \frac {1}{6} \, b c x^{2} + \frac {1}{3} \, a x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.20, size = 227, normalized size = 5.04 \[ -\frac {b c^{4} \log \left (-\frac {c + x}{c - x} - 1\right ) - b c^{4} \log \left (-\frac {c + x}{c - x}\right ) + \frac {{\left (b c^{4} + \frac {3 \, b {\left (c + x\right )}^{2} c^{4}}{{\left (c - x\right )}^{2}}\right )} \log \left (-\frac {c + x}{c - x}\right )}{\frac {{\left (c + x\right )}^{3}}{{\left (c - x\right )}^{3}} + \frac {3 \, {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {3 \, {\left (c + x\right )}}{c - x} + 1} + \frac {2 \, {\left (a c^{4} + \frac {3 \, a {\left (c + x\right )}^{2} c^{4}}{{\left (c - x\right )}^{2}} + \frac {b {\left (c + x\right )}^{2} c^{4}}{{\left (c - x\right )}^{2}} + \frac {b {\left (c + x\right )} c^{4}}{c - x}\right )}}{\frac {{\left (c + x\right )}^{3}}{{\left (c - x\right )}^{3}} + \frac {3 \, {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {3 \, {\left (c + x\right )}}{c - x} + 1}}{3 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 67, normalized size = 1.49 \[ \frac {x^{3} a}{3}+\frac {b \,x^{3} \arctanh \left (\frac {c}{x}\right )}{3}+\frac {b c \,x^{2}}{6}-\frac {c^{3} b \ln \left (\frac {c}{x}\right )}{3}+\frac {c^{3} b \ln \left (\frac {c}{x}-1\right )}{6}+\frac {c^{3} b \ln \left (1+\frac {c}{x}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 42, normalized size = 0.93 \[ \frac {1}{3} \, a x^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (\frac {c}{x}\right ) + {\left (c^{2} \log \left (-c^{2} + x^{2}\right ) + x^{2}\right )} c\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.72, size = 42, normalized size = 0.93 \[ \frac {a\,x^3}{3}+\frac {b\,c^3\,\ln \left (x^2-c^2\right )}{6}+\frac {b\,x^3\,\mathrm {atanh}\left (\frac {c}{x}\right )}{3}+\frac {b\,c\,x^2}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atanh(c/x)),x)

[Out]

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

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sympy [A]  time = 0.45, size = 49, normalized size = 1.09 \[ \frac {a x^{3}}{3} + \frac {b c^{3} \log {\left (- c + x \right )}}{3} + \frac {b c^{3} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{3} + \frac {b c x^{2}}{6} + \frac {b x^{3} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/3 + b*c**3*log(-c + x)/3 + b*c**3*atanh(c/x)/3 + b*c*x**2/6 + b*x**3*atanh(c/x)/3

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