3.2.36 \(\int x^2 (a+b \tanh ^{-1}(\frac {c}{x})) \, dx\) [136]

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

[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 = 0.02, 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 = {6037, 269, 272, 45} \begin {gather*} \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 \end {gather*}

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 45

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 269

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 272

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 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^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) \text {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) \text {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 \begin {gather*} \frac {1}{6} b c x^2+\frac {a x^3}{3}+\frac {1}{3} b x^3 \tanh ^{-1}\left (\frac {c}{x}\right )+\frac {1}{6} b c^3 \log \left (-c^2+x^2\right ) \end {gather*}

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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Maple [A]
time = 0.11, size = 71, normalized size = 1.58

method	result	size
derivativedivides	\(-c^{3} \left (-\frac {a \,x^{3}}{3 c^{3}}-\frac {b \,x^{3} \arctanh \left (\frac {c}{x}\right )}{3 c^{3}}-\frac {b \ln \left (1+\frac {c}{x}\right )}{6}-\frac {b \,x^{2}}{6 c^{2}}+\frac {b \ln \left (\frac {c}{x}\right )}{3}-\frac {b \ln \left (\frac {c}{x}-1\right )}{6}\right )\)	\(71\)
default	\(-c^{3} \left (-\frac {a \,x^{3}}{3 c^{3}}-\frac {b \,x^{3} \arctanh \left (\frac {c}{x}\right )}{3 c^{3}}-\frac {b \ln \left (1+\frac {c}{x}\right )}{6}-\frac {b \,x^{2}}{6 c^{2}}+\frac {b \ln \left (\frac {c}{x}\right )}{3}-\frac {b \ln \left (\frac {c}{x}-1\right )}{6}\right )\)	\(71\)
risch	\(\frac {x^{3} b \ln \left (x +c \right )}{6}-\frac {x^{3} b \ln \left (c -x \right )}{6}+\frac {i \pi b \,x^{3} \mathrm {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{6}-\frac {i \pi b \,x^{3} \mathrm {csgn}\left (i \left (c -x \right )\right ) \mathrm {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{12}-\frac {i \pi b \,x^{3}}{6}+\frac {i \pi b \,x^{3} \mathrm {csgn}\left (i \left (x +c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}}{12}-\frac {i \pi b \,x^{3} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x +c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x +c \right )}{x}\right )}{12}-\frac {i \pi b \,x^{3} \mathrm {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{3}}{12}-\frac {i \pi b \,x^{3} \mathrm {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{3}}{12}+\frac {i \pi b \,x^{3} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (c -x \right )\right ) \mathrm {csgn}\left (\frac {i \left (c -x \right )}{x}\right )}{12}+\frac {i \pi b \,x^{3} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}}{12}-\frac {i \pi b \,x^{3} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{12}+\frac {x^{3} a}{3}+\frac {b c \,x^{2}}{6}+\frac {b \,c^{3} \ln \left (-c^{2}+x^{2}\right )}{6}\)	\(307\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 42, normalized size = 0.93 \begin {gather*} \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 \end {gather*}

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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Fricas [A]
time = 0.37, size = 49, normalized size = 1.09 \begin {gather*} \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} \end {gather*}

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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Sympy [A]
time = 0.15, size = 49, normalized size = 1.09 \begin {gather*} \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} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (39) = 78\).
time = 0.44, size = 227, normalized size = 5.04 \begin {gather*} -\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} \end {gather*}

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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Mupad [B]
time = 0.72, size = 42, normalized size = 0.93 \begin {gather*} \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} \end {gather*}

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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