3.1.81 \(\int \frac {(a+b \tanh ^{-1}(c x^2))^3}{x^5} \, dx\) [81]

Optimal. Leaf size=139 \[ \frac {3}{4} b c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {3 b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 x^2}+\frac {1}{4} c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{4 x^4}+\frac {3}{2} b^2 c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \log \left (2-\frac {2}{1+c x^2}\right )-\frac {3}{4} b^3 c^2 \text {PolyLog}\left (2,-1+\frac {2}{1+c x^2}\right ) \]

[Out]

3/4*b*c^2*(a+b*arctanh(c*x^2))^2-3/4*b*c*(a+b*arctanh(c*x^2))^2/x^2+1/4*c^2*(a+b*arctanh(c*x^2))^3-1/4*(a+b*ar
ctanh(c*x^2))^3/x^4+3/2*b^2*c^2*(a+b*arctanh(c*x^2))*ln(2-2/(c*x^2+1))-3/4*b^3*c^2*polylog(2,-1+2/(c*x^2+1))

________________________________________________________________________________________

Rubi [A]
time = 0.25, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6039, 6037, 6129, 6135, 6079, 2497, 6095} \begin {gather*} \frac {3}{2} b^2 c^2 \log \left (2-\frac {2}{c x^2+1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {3}{4} b c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {1}{4} c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {3 b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 x^2}-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{4 x^4}-\frac {3}{4} b^3 c^2 \text {Li}_2\left (\frac {2}{c x^2+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])^3/x^5,x]

[Out]

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

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 6095

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

Rule 6129

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

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 \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{x^5} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{8 x^5}+\frac {3 b \left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{8 x^5}-\frac {3 b^2 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{8 x^5}+\frac {b^3 \log ^3\left (1+c x^2\right )}{8 x^5}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{x^5} \, dx+\frac {1}{8} (3 b) \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{x^5} \, dx-\frac {1}{8} \left (3 b^2\right ) \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{x^5} \, dx+\frac {1}{8} b^3 \int \frac {\log ^3\left (1+c x^2\right )}{x^5} \, dx\\ &=\frac {1}{16} \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^3}{x^3} \, dx,x,x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^2\right )+\frac {1}{16} b^3 \text {Subst}\left (\int \frac {\log ^3(1+c x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 x^4}-\frac {b^3 \log ^3\left (1+c x^2\right )}{32 x^4}+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^2\right )+\frac {1}{32} (3 b c) \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x^2 (1-c x)} \, dx,x,x^2\right )+\frac {1}{32} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(1+c x)}{x^2 (1+c x)} \, dx,x,x^2\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 x^4}-\frac {b^3 \log ^3\left (1+c x^2\right )}{32 x^4}-\frac {1}{32} (3 b) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{x \left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^2\right )+\frac {1}{32} \left (3 b^3\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{x \left (-\frac {1}{c}+\frac {x}{c}\right )^2} \, dx,x,1+c x^2\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 x^4}-\frac {b^3 \log ^3\left (1+c x^2\right )}{32 x^4}-\frac {1}{32} (3 b) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{\left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^2\right )+\frac {1}{32} \left (3 b^3\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{\left (-\frac {1}{c}+\frac {x}{c}\right )^2} \, dx,x,1+c x^2\right )-\frac {1}{32} (3 b c) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{x \left (\frac {1}{c}-\frac {x}{c}\right )} \, dx,x,1-c x^2\right )-\frac {1}{32} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{x \left (-\frac {1}{c}+\frac {x}{c}\right )} \, dx,x,1+c x^2\right )\\ &=-\frac {3 b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{32 x^2}-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 x^4}-\frac {3 b^3 c \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{32 x^2}-\frac {b^3 \log ^3\left (1+c x^2\right )}{32 x^4}+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{32} (3 b c) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^2\right )-\frac {1}{16} \left (3 b^2 c\right ) \text {Subst}\left (\int \frac {2 a-b \log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^2\right )-\frac {1}{32} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{-\frac {1}{c}+\frac {x}{c}} \, dx,x,1+c x^2\right )+\frac {1}{16} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log (x)}{-\frac {1}{c}+\frac {x}{c}} \, dx,x,1+c x^2\right )-\frac {1}{32} \left (3 b c^2\right ) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{x} \, dx,x,1-c x^2\right )+\frac {1}{32} \left (3 b^3 c^2\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {3 b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{32 x^2}+\frac {3}{32} b c^2 \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 x^4}-\frac {3 b^3 c \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{32 x^2}-\frac {3}{32} b^3 c^2 \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \log ^3\left (1+c x^2\right )}{32 x^4}-\frac {3}{16} b^3 c^2 \text {Li}_2\left (-c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^2\right )+\frac {1}{32} \left (3 c^2\right ) \text {Subst}\left (\int x^2 \, dx,x,2 a-b \log \left (1-c x^2\right )\right )+\frac {1}{16} \left (3 b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (1-x) (2 a-b \log (x))}{x} \, dx,x,1-c x^2\right )+\frac {1}{32} \left (3 b^3 c^2\right ) \text {Subst}\left (\int x^2 \, dx,x,\log \left (1+c x^2\right )\right )+\frac {1}{16} \left (3 b^3 c^2\right ) \text {Subst}\left (\int \frac {\log (1-x) \log (x)}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {3 b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{32 x^2}+\frac {3}{32} b c^2 \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2+\frac {1}{32} c^2 \left (2 a-b \log \left (1-c x^2\right )\right )^3-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 x^4}-\frac {3 b^3 c \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{32 x^2}-\frac {3}{32} b^3 c^2 \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )+\frac {1}{32} b^3 c^2 \log ^3\left (1+c x^2\right )-\frac {b^3 \log ^3\left (1+c x^2\right )}{32 x^4}-\frac {3}{16} b^3 c^2 \text {Li}_2\left (-c x^2\right )+\frac {3}{16} b^3 c^2 \text {Li}_2\left (c x^2\right )-\frac {3}{16} b^2 c^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (1-c x^2\right )-\frac {3}{16} b^3 c^2 \log \left (1+c x^2\right ) \text {Li}_2\left (1+c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^3 c^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x^2\right )+\frac {1}{16} \left (3 b^3 c^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {3 b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{32 x^2}+\frac {3}{32} b c^2 \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2+\frac {1}{32} c^2 \left (2 a-b \log \left (1-c x^2\right )\right )^3-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 x^4}-\frac {3 b^3 c \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{32 x^2}-\frac {3}{32} b^3 c^2 \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )+\frac {1}{32} b^3 c^2 \log ^3\left (1+c x^2\right )-\frac {b^3 \log ^3\left (1+c x^2\right )}{32 x^4}-\frac {3}{16} b^3 c^2 \text {Li}_2\left (-c x^2\right )+\frac {3}{16} b^3 c^2 \text {Li}_2\left (c x^2\right )-\frac {3}{16} b^2 c^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (1-c x^2\right )-\frac {3}{16} b^3 c^2 \log \left (1+c x^2\right ) \text {Li}_2\left (1+c x^2\right )-\frac {3}{16} b^3 c^2 \text {Li}_3\left (1-c x^2\right )+\frac {3}{16} b^3 c^2 \text {Li}_3\left (1+c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.20, size = 218, normalized size = 1.57 \begin {gather*} \frac {6 b^2 \left (-1+c x^2\right ) \left (a+a c x^2+b c x^2\right ) \tanh ^{-1}\left (c x^2\right )^2+2 b^3 \left (-1+c^2 x^4\right ) \tanh ^{-1}\left (c x^2\right )^3-6 b \tanh ^{-1}\left (c x^2\right ) \left (a^2+2 a b c x^2-2 b^2 c^2 x^4 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )+a \left (-2 a^2-6 a b c x^2-3 a b c^2 x^4 \log \left (1-c x^2\right )+3 a b c^2 x^4 \log \left (1+c x^2\right )+12 b^2 c^2 x^4 \log \left (\frac {c x^2}{\sqrt {1-c^2 x^4}}\right )\right )-6 b^3 c^2 x^4 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )}{8 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])^3/x^5,x]

[Out]

(6*b^2*(-1 + c*x^2)*(a + a*c*x^2 + b*c*x^2)*ArcTanh[c*x^2]^2 + 2*b^3*(-1 + c^2*x^4)*ArcTanh[c*x^2]^3 - 6*b*Arc
Tanh[c*x^2]*(a^2 + 2*a*b*c*x^2 - 2*b^2*c^2*x^4*Log[1 - E^(-2*ArcTanh[c*x^2])]) + a*(-2*a^2 - 6*a*b*c*x^2 - 3*a
*b*c^2*x^4*Log[1 - c*x^2] + 3*a*b*c^2*x^4*Log[1 + c*x^2] + 12*b^2*c^2*x^4*Log[(c*x^2)/Sqrt[1 - c^2*x^4]]) - 6*
b^3*c^2*x^4*PolyLog[2, E^(-2*ArcTanh[c*x^2])])/(8*x^4)

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctanh \left (c \,x^{2}\right )\right )^{3}}{x^{5}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))^3/x^5,x)

[Out]

int((a+b*arctanh(c*x^2))^3/x^5,x)

________________________________________________________________________________________

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((a+b*arctanh(c*x^2))^3/x^5,x, algorithm="maxima")

[Out]

3/8*((c*log(c*x^2 + 1) - c*log(c*x^2 - 1) - 2/x^2)*c - 2*arctanh(c*x^2)/x^4)*a^2*b + 3/16*((2*(log(c*x^2 - 1)
- 2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 - log(c*x^2 - 1)^2 - 4*log(c*x^2 - 1) + 16*log(x))*c^2 + 4*(c*log(c*x^2
+ 1) - c*log(c*x^2 - 1) - 2/x^2)*c*arctanh(c*x^2))*a*b^2 - 1/32*b^3*(((c^2*x^4 - 1)*log(-c*x^2 + 1)^3 + 3*(2*
c*x^2 - (c^2*x^4 - 1)*log(c*x^2 + 1))*log(-c*x^2 + 1)^2)/x^4 + 4*integrate(-((c*x^2 - 1)*log(c*x^2 + 1)^3 + 3*
(2*c^2*x^4 - (c*x^2 - 1)*log(c*x^2 + 1)^2 - (c^3*x^6 - c*x^2)*log(c*x^2 + 1))*log(-c*x^2 + 1))/(c*x^7 - x^5),
x)) - 3/4*a*b^2*arctanh(c*x^2)^2/x^4 - 1/4*a^3/x^4

________________________________________________________________________________________

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((a+b*arctanh(c*x^2))^3/x^5,x, algorithm="fricas")

[Out]

integral((b^3*arctanh(c*x^2)^3 + 3*a*b^2*arctanh(c*x^2)^2 + 3*a^2*b*arctanh(c*x^2) + a^3)/x^5, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**2))**3/x**5,x)

[Out]

Integral((a + b*atanh(c*x**2))**3/x**5, x)

________________________________________________________________________________________

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((a+b*arctanh(c*x^2))^3/x^5,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^2) + a)^3/x^5, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))^3/x^5,x)

[Out]

int((a + b*atanh(c*x^2))^3/x^5, x)

________________________________________________________________________________________