3.2.72 \(\int x (a+b \tanh ^{-1}(\frac {c}{x^2}))^2 \, dx\) [172]

Optimal. Leaf size=94 \[ -\frac {1}{2} c \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2+\frac {1}{2} x^2 \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2-b c \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \log \left (2-\frac {2}{1+\frac {c}{x^2}}\right )+\frac {1}{2} b^2 c \text {PolyLog}\left (2,-1+\frac {2}{1+\frac {c}{x^2}}\right ) \]

[Out]

-1/2*c*(a+b*arccoth(x^2/c))^2+1/2*x^2*(a+b*arccoth(x^2/c))^2-b*c*(a+b*arccoth(x^2/c))*ln(2-2/(1+c/x^2))+1/2*b^
2*c*polylog(2,-1+2/(1+c/x^2))

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Rubi [A]
time = 0.12, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6039, 6037, 6135, 6079, 2497} \begin {gather*} \frac {1}{2} x^2 \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2-\frac {1}{2} c \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2-b c \log \left (2-\frac {2}{\frac {c}{x^2}+1}\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )+\frac {1}{2} b^2 c \text {Li}_2\left (\frac {2}{\frac {c}{x^2}+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c/x^2])^2,x]

[Out]

-1/2*(c*(a + b*ArcCoth[x^2/c])^2) + (x^2*(a + b*ArcCoth[x^2/c])^2)/2 - b*c*(a + b*ArcCoth[x^2/c])*Log[2 - 2/(1
+ c/x^2)] + (b^2*c*PolyLog[2, -1 + 2/(1 + c/x^2)])/2

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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
]

Rule 6039

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[S
implify[(m + 1)/n]]

Rule 6079

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTanh[c*x
])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/
d))]/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 6135

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]

Rubi steps

\begin {align*} \int x \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2-\frac {1}{2} b x \left (-2 a+b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{4} b^2 x \log ^2\left (1+\frac {c}{x^2}\right )\right ) \, dx\\ &=\frac {1}{4} \int x \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2 \, dx-\frac {1}{2} b \int x \left (-2 a+b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right ) \, dx+\frac {1}{4} b^2 \int x \log ^2\left (1+\frac {c}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{8} \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x^2} \, dx,x,\frac {1}{x^2}\right )\right )-\frac {1}{4} b \text {Subst}\left (\int \left (-2 a+b \log \left (1-\frac {c}{x}\right )\right ) \log \left (1+\frac {c}{x}\right ) \, dx,x,x^2\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {\log ^2(1+c x)}{x^2} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )-\frac {1}{4} b \text {Subst}\left (\int \left (-2 a \log \left (1+\frac {c}{x}\right )+b \log \left (1-\frac {c}{x}\right ) \log \left (1+\frac {c}{x}\right )\right ) \, dx,x,x^2\right )-\frac {1}{4} (b c) \text {Subst}\left (\int \frac {2 a-b \log (1-c x)}{x} \, dx,x,\frac {1}{x^2}\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )+\frac {1}{2} (a b) \text {Subst}\left (\int \log \left (1+\frac {c}{x}\right ) \, dx,x,x^2\right )-\frac {1}{4} b^2 \text {Subst}\left (\int \log \left (1-\frac {c}{x}\right ) \log \left (1+\frac {c}{x}\right ) \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{2} a b x^2 \log \left (1+\frac {c}{x^2}\right )-\frac {1}{4} b^2 x^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c}{x^2}\right )+\frac {1}{4} b^2 \text {Subst}\left (\int \frac {c \log \left (1-\frac {c}{x}\right )}{-c-x} \, dx,x,x^2\right )+\frac {1}{4} b^2 \text {Subst}\left (\int \frac {c \log \left (1+\frac {c}{x}\right )}{-c+x} \, dx,x,x^2\right )+\frac {1}{2} (a b c) \text {Subst}\left (\int \frac {1}{\left (1+\frac {c}{x}\right ) x} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{2} a b x^2 \log \left (1+\frac {c}{x^2}\right )-\frac {1}{4} b^2 x^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c}{x^2}\right )+\frac {1}{2} (a b c) \text {Subst}\left (\int \frac {1}{c+x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {c}{x}\right )}{-c-x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {c}{x}\right )}{-c+x} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{2} a b x^2 \log \left (1+\frac {c}{x^2}\right )-\frac {1}{4} b^2 x^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)-\frac {1}{4} b^2 c \log \left (1-\frac {c}{x^2}\right ) \log \left (-c-x^2\right )+\frac {1}{4} b^2 c \log \left (1+\frac {c}{x^2}\right ) \log \left (-c+x^2\right )+\frac {1}{2} a b c \log \left (c+x^2\right )+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c}{x^2}\right )+\frac {1}{4} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (-c-x)}{\left (1-\frac {c}{x}\right ) x^2} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (-c+x)}{\left (1+\frac {c}{x}\right ) x^2} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{2} a b x^2 \log \left (1+\frac {c}{x^2}\right )-\frac {1}{4} b^2 x^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)-\frac {1}{4} b^2 c \log \left (1-\frac {c}{x^2}\right ) \log \left (-c-x^2\right )+\frac {1}{4} b^2 c \log \left (1+\frac {c}{x^2}\right ) \log \left (-c+x^2\right )+\frac {1}{2} a b c \log \left (c+x^2\right )+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c}{x^2}\right )+\frac {1}{4} \left (b^2 c^2\right ) \text {Subst}\left (\int \left (-\frac {\log (-c-x)}{c (c-x)}-\frac {\log (-c-x)}{c x}\right ) \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c^2\right ) \text {Subst}\left (\int \left (\frac {\log (-c+x)}{c x}-\frac {\log (-c+x)}{c (c+x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{2} a b x^2 \log \left (1+\frac {c}{x^2}\right )-\frac {1}{4} b^2 x^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)-\frac {1}{4} b^2 c \log \left (1-\frac {c}{x^2}\right ) \log \left (-c-x^2\right )+\frac {1}{4} b^2 c \log \left (1+\frac {c}{x^2}\right ) \log \left (-c+x^2\right )+\frac {1}{2} a b c \log \left (c+x^2\right )+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c}{x^2}\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (-c-x)}{c-x} \, dx,x,x^2\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (-c-x)}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (-c+x)}{x} \, dx,x,x^2\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (-c+x)}{c+x} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{2} a b x^2 \log \left (1+\frac {c}{x^2}\right )-\frac {1}{4} b^2 x^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)-\frac {1}{4} b^2 c \log \left (1-\frac {c}{x^2}\right ) \log \left (-c-x^2\right )-\frac {1}{4} b^2 c \log \left (-\frac {x^2}{c}\right ) \log \left (-c-x^2\right )+\frac {1}{4} b^2 c \log \left (-c-x^2\right ) \log \left (\frac {c-x^2}{2 c}\right )+\frac {1}{4} b^2 c \log \left (1+\frac {c}{x^2}\right ) \log \left (-c+x^2\right )+\frac {1}{4} b^2 c \log \left (\frac {x^2}{c}\right ) \log \left (-c+x^2\right )+\frac {1}{2} a b c \log \left (c+x^2\right )-\frac {1}{4} b^2 c \log \left (-c+x^2\right ) \log \left (\frac {c+x^2}{2 c}\right )+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c}{x^2}\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {x}{c}\right )}{-c-x} \, dx,x,x^2\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (\frac {x}{c}\right )}{-c+x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {-c+x}{2 c}\right )}{-c-x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (\frac {c+x}{2 c}\right )}{-c+x} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{2} a b x^2 \log \left (1+\frac {c}{x^2}\right )-\frac {1}{4} b^2 x^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)-\frac {1}{4} b^2 c \log \left (1-\frac {c}{x^2}\right ) \log \left (-c-x^2\right )-\frac {1}{4} b^2 c \log \left (-\frac {x^2}{c}\right ) \log \left (-c-x^2\right )+\frac {1}{4} b^2 c \log \left (-c-x^2\right ) \log \left (\frac {c-x^2}{2 c}\right )+\frac {1}{4} b^2 c \log \left (1+\frac {c}{x^2}\right ) \log \left (-c+x^2\right )+\frac {1}{4} b^2 c \log \left (\frac {x^2}{c}\right ) \log \left (-c+x^2\right )+\frac {1}{2} a b c \log \left (c+x^2\right )-\frac {1}{4} b^2 c \log \left (-c+x^2\right ) \log \left (\frac {c+x^2}{2 c}\right )+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c+x^2}{c}\right )+\frac {1}{4} b^2 c \text {Li}_2\left (1-\frac {x^2}{c}\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{2 c}\right )}{x} \, dx,x,-c-x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{2 c}\right )}{x} \, dx,x,-c+x^2\right )\\ &=\frac {1}{8} \left (1-\frac {c}{x^2}\right ) x^2 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2+\frac {1}{2} a b x^2 \log \left (1+\frac {c}{x^2}\right )-\frac {1}{4} b^2 x^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )+\frac {1}{8} b^2 \left (1+\frac {c}{x^2}\right ) x^2 \log ^2\left (1+\frac {c}{x^2}\right )+a b c \log (x)-\frac {1}{4} b^2 c \log \left (1-\frac {c}{x^2}\right ) \log \left (-c-x^2\right )-\frac {1}{4} b^2 c \log \left (-\frac {x^2}{c}\right ) \log \left (-c-x^2\right )+\frac {1}{4} b^2 c \log \left (-c-x^2\right ) \log \left (\frac {c-x^2}{2 c}\right )+\frac {1}{4} b^2 c \log \left (1+\frac {c}{x^2}\right ) \log \left (-c+x^2\right )+\frac {1}{4} b^2 c \log \left (\frac {x^2}{c}\right ) \log \left (-c+x^2\right )+\frac {1}{2} a b c \log \left (c+x^2\right )-\frac {1}{4} b^2 c \log \left (-c+x^2\right ) \log \left (\frac {c+x^2}{2 c}\right )+\frac {1}{4} b^2 c \text {Li}_2\left (-\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c}{x^2}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c-x^2}{2 c}\right )+\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c+x^2}{2 c}\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {c+x^2}{c}\right )+\frac {1}{4} b^2 c \text {Li}_2\left (1-\frac {x^2}{c}\right )\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 107, normalized size = 1.14 \begin {gather*} \frac {1}{2} \left (b^2 \left (-c+x^2\right ) \tanh ^{-1}\left (\frac {c}{x^2}\right )^2+2 b \tanh ^{-1}\left (\frac {c}{x^2}\right ) \left (a x^2-b c \log \left (1-e^{-2 \tanh ^{-1}\left (\frac {c}{x^2}\right )}\right )\right )+a \left (a x^2+b c \log \left (1-\frac {c^2}{x^4}\right )-2 b c \log \left (\frac {c}{x^2}\right )\right )+b^2 c \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (\frac {c}{x^2}\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(-c + x^2)*ArcTanh[c/x^2]^2 + 2*b*ArcTanh[c/x^2]*(a*x^2 - b*c*Log[1 - E^(-2*ArcTanh[c/x^2])]) + a*(a*x^2
+ b*c*Log[1 - c^2/x^4] - 2*b*c*Log[c/x^2]) + b^2*c*PolyLog[2, E^(-2*ArcTanh[c/x^2])])/2

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.53, size = 6869, normalized size = 73.07

method	result	size
risch	\(\text {Expression too large to display}\)	\(6869\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*x^2 + 1/2*(2*x^2*arctanh(c/x^2) + c*log(x^4 - c^2))*a*b + 1/8*(x^2*log(x^2 + c)^2 - 2*(x^2 + c)*log(x^
2 + c)*log(x^2 - c) + (x^2 - c)*log(x^2 - c)^2 + 2*integrate(2*(3*c*x^3 + c^2*x)*log(x^2 + c)/(x^4 - c^2), x))
*b^2

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a + b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (a+b\,\mathrm {atanh}\left (\frac {c}{x^2}\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c/x^2))^2,x)

[Out]

int(x*(a + b*atanh(c/x^2))^2, x)

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