3.271 \(\int \frac {(a+b \coth ^{-1}(c x)) (d+e \log (1-c^2 x^2))}{x^5} \, dx\)

Optimal. Leaf size=339 \[ \frac {1}{12} c^4 e (3 a+4 b) \log (1-c x)+\frac {1}{12} c^4 e (3 a-4 b) \log (c x+1)-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x^4}-\frac {1}{2} a c^4 e \log (x)+\frac {a c^2 e}{4 x^2}+\frac {1}{4} b c^4 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )+\frac {1}{4} b c^4 e \text {Li}_2\left (\frac {2}{c x+1}-1\right )-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)^2-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)-\frac {1}{4} b c^4 e \coth ^{-1}(c x)^2+\frac {1}{2} b c^4 e \log \left (\frac {2}{1-c x}\right ) \tanh ^{-1}(c x)-\frac {1}{2} b c^4 e \log \left (2-\frac {2}{c x+1}\right ) \coth ^{-1}(c x)+\frac {5 b c^3 e}{12 x}-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{12 x^3}+\frac {b c^2 e \coth ^{-1}(c x)}{4 x^2}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {b c^3 \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x} \]

[Out]

1/4*a*c^2*e/x^2+5/12*b*c^3*e/x+1/4*b*c^2*e*arccoth(c*x)/x^2-1/4*b*c^4*e*arccoth(c*x)^2-1/4*b*c^4*e*arctanh(c*x
)-1/4*b*c^4*e*arctanh(c*x)^2-1/2*a*c^4*e*ln(x)+1/2*b*c^4*e*arctanh(c*x)*ln(2/(-c*x+1))+1/12*(3*a+4*b)*c^4*e*ln
(-c*x+1)+1/12*(3*a-4*b)*c^4*e*ln(c*x+1)-1/12*b*c*(d+e*ln(-c^2*x^2+1))/x^3-1/4*b*c^3*(d+e*ln(-c^2*x^2+1))/x-1/4
*(a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1))/x^4+1/4*b*c^4*arctanh(c*x)*(d+e*ln(-c^2*x^2+1))-1/2*b*c^4*e*arccoth(c
*x)*ln(2-2/(c*x+1))+1/4*b*c^4*e*polylog(2,1-2/(-c*x+1))+1/4*b*c^4*e*polylog(2,-1+2/(c*x+1))

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Rubi [A]  time = 0.72, antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.518, Rules used = {5917, 325, 206, 6086, 6725, 1802, 5983, 5989, 5933, 2447, 5984, 5918, 2402, 2315} \[ \frac {1}{4} b c^4 e \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )+\frac {1}{4} b c^4 e \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x^4}+\frac {1}{12} c^4 e (3 a+4 b) \log (1-c x)+\frac {1}{12} c^4 e (3 a-4 b) \log (c x+1)+\frac {a c^2 e}{4 x^2}-\frac {1}{2} a c^4 e \log (x)-\frac {b c^3 \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x}-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{12 x^3}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )+\frac {b c^2 e \coth ^{-1}(c x)}{4 x^2}+\frac {5 b c^3 e}{12 x}-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)^2-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)-\frac {1}{4} b c^4 e \coth ^{-1}(c x)^2+\frac {1}{2} b c^4 e \log \left (\frac {2}{1-c x}\right ) \tanh ^{-1}(c x)-\frac {1}{2} b c^4 e \log \left (2-\frac {2}{c x+1}\right ) \coth ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]))/x^5,x]

[Out]

(a*c^2*e)/(4*x^2) + (5*b*c^3*e)/(12*x) + (b*c^2*e*ArcCoth[c*x])/(4*x^2) - (b*c^4*e*ArcCoth[c*x]^2)/4 - (b*c^4*
e*ArcTanh[c*x])/4 - (b*c^4*e*ArcTanh[c*x]^2)/4 - (a*c^4*e*Log[x])/2 + (b*c^4*e*ArcTanh[c*x]*Log[2/(1 - c*x)])/
2 + ((3*a + 4*b)*c^4*e*Log[1 - c*x])/12 + ((3*a - 4*b)*c^4*e*Log[1 + c*x])/12 - (b*c*(d + e*Log[1 - c^2*x^2]))
/(12*x^3) - (b*c^3*(d + e*Log[1 - c^2*x^2]))/(4*x) - ((a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]))/(4*x^4) +
(b*c^4*ArcTanh[c*x]*(d + e*Log[1 - c^2*x^2]))/4 - (b*c^4*e*ArcCoth[c*x]*Log[2 - 2/(1 + c*x)])/2 + (b*c^4*e*Po
lyLog[2, 1 - 2/(1 - c*x)])/4 + (b*c^4*e*PolyLog[2, -1 + 2/(1 + c*x)])/4

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

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 5917

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
oth[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCoth[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[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 5933

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcCoth[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcCoth[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 5983

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

Rule 5984

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

Rule 5989

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcCoth[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcCoth[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]

Rule 6086

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Wit
h[{u = IntHide[x^m*(a + b*ArcCoth[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegra
nd[(x*u)/(f + g*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx &=-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\left (2 c^2 e\right ) \int \left (\frac {3 a+b c x+3 b c^3 x^3+3 b \coth ^{-1}(c x)}{12 x^3 \left (-1+c^2 x^2\right )}-\frac {b c^4 x \tanh ^{-1}(c x)}{4 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {1}{6} \left (c^2 e\right ) \int \frac {3 a+b c x+3 b c^3 x^3+3 b \coth ^{-1}(c x)}{x^3 \left (-1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (b c^6 e\right ) \int \frac {x \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx\\ &=-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)^2-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {1}{6} \left (c^2 e\right ) \int \left (\frac {3 a+b c x+3 b c^3 x^3}{x^3 \left (-1+c^2 x^2\right )}+\frac {3 b \coth ^{-1}(c x)}{x^3 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac {1}{2} \left (b c^5 e\right ) \int \frac {\tanh ^{-1}(c x)}{1-c x} \, dx\\ &=-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)^2+\frac {1}{2} b c^4 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {1}{6} \left (c^2 e\right ) \int \frac {3 a+b c x+3 b c^3 x^3}{x^3 \left (-1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (b c^2 e\right ) \int \frac {\coth ^{-1}(c x)}{x^3 \left (-1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (b c^5 e\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)^2+\frac {1}{2} b c^4 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {1}{6} \left (c^2 e\right ) \int \left (-\frac {3 a}{x^3}-\frac {b c}{x^2}-\frac {3 a c^2}{x}+\frac {(3 a+4 b) c^3}{2 (-1+c x)}+\frac {(3 a-4 b) c^3}{2 (1+c x)}\right ) \, dx-\frac {1}{2} \left (b c^2 e\right ) \int \frac {\coth ^{-1}(c x)}{x^3} \, dx+\frac {1}{2} \left (b c^4 e\right ) \int \frac {\coth ^{-1}(c x)}{x \left (-1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (b c^4 e\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )\\ &=\frac {a c^2 e}{4 x^2}+\frac {b c^3 e}{6 x}+\frac {b c^2 e \coth ^{-1}(c x)}{4 x^2}-\frac {1}{4} b c^4 e \coth ^{-1}(c x)^2-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)^2-\frac {1}{2} a c^4 e \log (x)+\frac {1}{2} b c^4 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {1}{4} b c^4 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )-\frac {1}{4} \left (b c^3 e\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx-\frac {1}{2} \left (b c^4 e\right ) \int \frac {\coth ^{-1}(c x)}{x (1+c x)} \, dx\\ &=\frac {a c^2 e}{4 x^2}+\frac {5 b c^3 e}{12 x}+\frac {b c^2 e \coth ^{-1}(c x)}{4 x^2}-\frac {1}{4} b c^4 e \coth ^{-1}(c x)^2-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)^2-\frac {1}{2} a c^4 e \log (x)+\frac {1}{2} b c^4 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {1}{2} b c^4 e \coth ^{-1}(c x) \log \left (2-\frac {2}{1+c x}\right )+\frac {1}{4} b c^4 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )-\frac {1}{4} \left (b c^5 e\right ) \int \frac {1}{1-c^2 x^2} \, dx+\frac {1}{2} \left (b c^5 e\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac {a c^2 e}{4 x^2}+\frac {5 b c^3 e}{12 x}+\frac {b c^2 e \coth ^{-1}(c x)}{4 x^2}-\frac {1}{4} b c^4 e \coth ^{-1}(c x)^2-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)^2-\frac {1}{2} a c^4 e \log (x)+\frac {1}{2} b c^4 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {1}{2} b c^4 e \coth ^{-1}(c x) \log \left (2-\frac {2}{1+c x}\right )+\frac {1}{4} b c^4 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )+\frac {1}{4} b c^4 e \text {Li}_2\left (-1+\frac {2}{1+c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 307, normalized size = 0.91 \[ \frac {1}{12} \log (1-c x) \left (3 a c^4 e+4 b c^4 e\right )+\frac {1}{12} \log (c x+1) \left (3 a c^4 e-4 b c^4 e\right )+\frac {e \log \left (1-c^2 x^2\right ) \left (-3 a+3 b c^4 x^4 \coth ^{-1}(c x)-3 b c^3 x^3-b c x-3 b \coth ^{-1}(c x)\right )}{12 x^4}-\frac {1}{2} a c^4 e \log (x)+\frac {a c^2 e}{4 x^2}-\frac {a d}{4 x^4}-\frac {1}{4} b c^4 e \left (\text {Li}_2\left (-\frac {1}{c x}\right )-\text {Li}_2\left (\frac {1}{c x}\right )\right )+\frac {b c^3 e}{6 x}+b c^4 d \left (\frac {1}{4} \left (-\frac {1}{3 c^3 x^3}-\frac {1}{c x}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (c x+1)\right )-\frac {\coth ^{-1}(c x)}{4 c^4 x^4}\right )-\frac {1}{2} b c^4 e \left (\frac {1}{2} \left (-\frac {1}{c x}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (c x+1)\right )-\frac {\coth ^{-1}(c x)}{2 c^2 x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]))/x^5,x]

[Out]

-1/4*(a*d)/x^4 + (a*c^2*e)/(4*x^2) + (b*c^3*e)/(6*x) - (a*c^4*e*Log[x])/2 + ((3*a*c^4*e + 4*b*c^4*e)*Log[1 - c
*x])/12 - (b*c^4*e*(-1/2*ArcCoth[c*x]/(c^2*x^2) + (-(1/(c*x)) - Log[1 - c*x]/2 + Log[1 + c*x]/2)/2))/2 + b*c^4
*d*(-1/4*ArcCoth[c*x]/(c^4*x^4) + (-1/3*1/(c^3*x^3) - 1/(c*x) - Log[1 - c*x]/2 + Log[1 + c*x]/2)/4) + ((3*a*c^
4*e - 4*b*c^4*e)*Log[1 + c*x])/12 + (e*(-3*a - b*c*x - 3*b*c^3*x^3 - 3*b*ArcCoth[c*x] + 3*b*c^4*x^4*ArcCoth[c*
x])*Log[1 - c^2*x^2])/(12*x^4) - (b*c^4*e*(PolyLog[2, -(1/(c*x))] - PolyLog[2, 1/(c*x)]))/4

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b d \operatorname {arcoth}\left (c x\right ) + a d + {\left (b e \operatorname {arcoth}\left (c x\right ) + a e\right )} \log \left (-c^{2} x^{2} + 1\right )}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1))/x^5,x, algorithm="fricas")

[Out]

integral((b*d*arccoth(c*x) + a*d + (b*e*arccoth(c*x) + a*e)*log(-c^2*x^2 + 1))/x^5, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcoth}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1))/x^5,x, algorithm="giac")

[Out]

integrate((b*arccoth(c*x) + a)*(e*log(-c^2*x^2 + 1) + d)/x^5, x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \,\mathrm {arccoth}\left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1))/x^5,x)

[Out]

int((a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1))/x^5,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{24} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {arcoth}\left (c x\right )}{x^{4}}\right )} b d + \frac {1}{4} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c^{2} - \frac {\log \left (-c^{2} x^{2} + 1\right )}{x^{4}}\right )} a e - \frac {1}{8} \, b e {\left (\frac {\log \left (c x + 1\right )^{2}}{x^{4}} - 4 \, \int -\frac {2 \, {\left (c x + 1\right )} \log \left (c x - 1\right )^{2} - {\left (2 i \, \pi + {\left (2 i \, \pi c + c\right )} x\right )} \log \left (c x + 1\right ) - {\left (-2 i \, \pi - 2 i \, \pi c x\right )} \log \left (c x - 1\right )}{2 \, {\left (c x^{6} + x^{5}\right )}}\,{d x}\right )} - \frac {a d}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1))/x^5,x, algorithm="maxima")

[Out]

1/24*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arccoth(c*x)/x^4)*b*d + 1/4*((c^
2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c^2 - log(-c^2*x^2 + 1)/x^4)*a*e - 1/8*b*e*(log(c*x + 1)^2/x^4 - 4*
integrate(-1/2*(2*(c*x + 1)*log(c*x - 1)^2 - (2*I*pi + (2*I*pi*c + c)*x)*log(c*x + 1) - (-2*I*pi - 2*I*pi*c*x)
*log(c*x - 1))/(c*x^6 + x^5), x)) - 1/4*a*d/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {acoth}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*acoth(c*x))*(d + e*log(1 - c^2*x^2)))/x^5,x)

[Out]

int(((a + b*acoth(c*x))*(d + e*log(1 - c^2*x^2)))/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acoth}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acoth(c*x))*(d+e*ln(-c**2*x**2+1))/x**5,x)

[Out]

Integral((a + b*acoth(c*x))*(d + e*log(-c**2*x**2 + 1))/x**5, x)

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