3.276 \(\int \frac {(a+b \coth ^{-1}(c x)) (d+e \log (1-c^2 x^2))}{x^4} \, dx\)

Optimal. Leaf size=197 \[ -\frac {c^3 e \left (a+b \coth ^{-1}(c x)\right )^2}{3 b}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{3 x^3}+\frac {2 c^2 e \left (a+b \coth ^{-1}(c x)\right )}{3 x}-b c^3 e \log (x)-\frac {b c \left (1-c^2 x^2\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 x^2}+\frac {1}{6} b c^3 \log \left (1-\frac {1}{1-c^2 x^2}\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {1}{6} b c^3 e \text {Li}_2\left (\frac {1}{1-c^2 x^2}\right )+\frac {1}{3} b c^3 e \log \left (1-c^2 x^2\right ) \]

[Out]

2/3*c^2*e*(a+b*arccoth(c*x))/x-1/3*c^3*e*(a+b*arccoth(c*x))^2/b-b*c^3*e*ln(x)+1/3*b*c^3*e*ln(-c^2*x^2+1)-1/6*b
*c*(-c^2*x^2+1)*(d+e*ln(-c^2*x^2+1))/x^2-1/3*(a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1))/x^3+1/6*b*c^3*(d+e*ln(-c^
2*x^2+1))*ln(1-1/(-c^2*x^2+1))-1/6*b*c^3*e*polylog(2,1/(-c^2*x^2+1))

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Rubi [A]  time = 0.46, antiderivative size = 191, normalized size of antiderivative = 0.97, number of steps used = 17, number of rules used = 16, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {6082, 2475, 2411, 2347, 2344, 2301, 2316, 2315, 2314, 31, 5983, 5917, 266, 36, 29, 5949} \[ -\frac {1}{6} b c^3 e \text {PolyLog}\left (2,c^2 x^2\right )-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{3 x^3}-\frac {c^3 e \left (a+b \coth ^{-1}(c x)\right )^2}{3 b}+\frac {2 c^2 e \left (a+b \coth ^{-1}(c x)\right )}{3 x}-\frac {b c^3 \left (e \log \left (1-c^2 x^2\right )+d\right )^2}{12 e}-\frac {b c \left (1-c^2 x^2\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 x^2}+\frac {1}{3} b c^3 d \log (x)+\frac {1}{3} b c^3 e \log \left (1-c^2 x^2\right )-b c^3 e \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

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 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2344

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

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

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 5949

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[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 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 6082

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Sim
p[(x^(m + 1)*(d + e*Log[f + g*x^2])*(a + b*ArcCoth[c*x]))/(m + 1), x] + (-Dist[(b*c)/(m + 1), Int[(x^(m + 1)*(
d + e*Log[f + g*x^2]))/(1 - c^2*x^2), x], x] - Dist[(2*e*g)/(m + 1), Int[(x^(m + 2)*(a + b*ArcCoth[c*x]))/(f +
g*x^2), x], x]) /; FreeQ[{a, b, c, d, e, f, g}, x] && ILtQ[m/2, 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^4} \, dx &=-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{3 x^3}+\frac {1}{3} (b c) \int \frac {d+e \log \left (1-c^2 x^2\right )}{x^3 \left (1-c^2 x^2\right )} \, dx-\frac {1}{3} \left (2 c^2 e\right ) \int \frac {a+b \coth ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{3 x^3}+\frac {1}{6} (b c) \operatorname {Subst}\left (\int \frac {d+e \log \left (1-c^2 x\right )}{x^2 \left (1-c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{3} \left (2 c^2 e\right ) \int \frac {a+b \coth ^{-1}(c x)}{x^2} \, dx-\frac {1}{3} \left (2 c^4 e\right ) \int \frac {a+b \coth ^{-1}(c x)}{1-c^2 x^2} \, dx\\ &=\frac {2 c^2 e \left (a+b \coth ^{-1}(c x)\right )}{3 x}-\frac {c^3 e \left (a+b \coth ^{-1}(c x)\right )^2}{3 b}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{3 x^3}-\frac {b \operatorname {Subst}\left (\int \frac {d+e \log (x)}{x \left (\frac {1}{c^2}-\frac {x}{c^2}\right )^2} \, dx,x,1-c^2 x^2\right )}{6 c}-\frac {1}{3} \left (2 b c^3 e\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx\\ &=\frac {2 c^2 e \left (a+b \coth ^{-1}(c x)\right )}{3 x}-\frac {c^3 e \left (a+b \coth ^{-1}(c x)\right )^2}{3 b}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{3 x^3}-\frac {b \operatorname {Subst}\left (\int \frac {d+e \log (x)}{\left (\frac {1}{c^2}-\frac {x}{c^2}\right )^2} \, dx,x,1-c^2 x^2\right )}{6 c}-\frac {1}{6} (b c) \operatorname {Subst}\left (\int \frac {d+e \log (x)}{x \left (\frac {1}{c^2}-\frac {x}{c^2}\right )} \, dx,x,1-c^2 x^2\right )-\frac {1}{3} \left (b c^3 e\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=\frac {2 c^2 e \left (a+b \coth ^{-1}(c x)\right )}{3 x}-\frac {c^3 e \left (a+b \coth ^{-1}(c x)\right )^2}{3 b}-\frac {b c \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 x^2}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{3 x^3}-\frac {1}{6} (b c) \operatorname {Subst}\left (\int \frac {d+e \log (x)}{\frac {1}{c^2}-\frac {x}{c^2}} \, dx,x,1-c^2 x^2\right )-\frac {1}{6} \left (b c^3\right ) \operatorname {Subst}\left (\int \frac {d+e \log (x)}{x} \, dx,x,1-c^2 x^2\right )+\frac {1}{6} (b c e) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x}{c^2}} \, dx,x,1-c^2 x^2\right )-\frac {1}{3} \left (b c^3 e\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{3} \left (b c^5 e\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=\frac {2 c^2 e \left (a+b \coth ^{-1}(c x)\right )}{3 x}-\frac {c^3 e \left (a+b \coth ^{-1}(c x)\right )^2}{3 b}+\frac {1}{3} b c^3 d \log (x)-b c^3 e \log (x)+\frac {1}{3} b c^3 e \log \left (1-c^2 x^2\right )-\frac {b c \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 x^2}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{3 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{12 e}-\frac {1}{6} (b c e) \operatorname {Subst}\left (\int \frac {\log (x)}{\frac {1}{c^2}-\frac {x}{c^2}} \, dx,x,1-c^2 x^2\right )\\ &=\frac {2 c^2 e \left (a+b \coth ^{-1}(c x)\right )}{3 x}-\frac {c^3 e \left (a+b \coth ^{-1}(c x)\right )^2}{3 b}+\frac {1}{3} b c^3 d \log (x)-b c^3 e \log (x)+\frac {1}{3} b c^3 e \log \left (1-c^2 x^2\right )-\frac {b c \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 x^2}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{3 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{12 e}-\frac {1}{6} b c^3 e \text {Li}_2\left (c^2 x^2\right )\\ \end {align*}

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Mathematica [B]  time = 0.37, size = 457, normalized size = 2.32 \[ \frac {1}{6} \left (-4 a c^3 e \tanh ^{-1}(c x)-\frac {2 a e \log \left (1-c^2 x^2\right )}{x^3}+\frac {4 a c^2 e}{x}-\frac {2 a d}{x^3}+2 b c^3 d \log (x)-2 b c^3 e \text {Li}_2(-c x)-2 b c^3 e \text {Li}_2(c x)+b c^3 e \text {Li}_2\left (\frac {1}{2}-\frac {c x}{2}\right )+b c^3 e \text {Li}_2\left (\frac {1}{2} (c x+1)\right )+\frac {1}{2} b c^3 e \log ^2\left (x-\frac {1}{c}\right )+\frac {1}{2} b c^3 e \log ^2\left (\frac {1}{c}+x\right )-2 b c^3 e \log (x)+b c^3 e \log \left (\frac {1}{c}+x\right ) \log \left (\frac {1}{2} (1-c x)\right )-2 b c^3 e \log (x) \log (1-c x)+b c^3 e \log \left (x-\frac {1}{c}\right ) \log \left (\frac {1}{2} (c x+1)\right )-2 b c^3 e \log (x) \log (c x+1)-2 b c^3 e \coth ^{-1}(c x)^2-\frac {b c e \log \left (1-c^2 x^2\right )}{x^2}-\frac {2 b e \log \left (1-c^2 x^2\right ) \coth ^{-1}(c x)}{x^3}+\frac {4 b c^2 e \coth ^{-1}(c x)}{x}-b c^3 d \log \left (1-c^2 x^2\right )-4 b c^3 e \log \left (\frac {1}{\sqrt {1-\frac {1}{c^2 x^2}}}\right )+b c^3 e \log \left (1-c^2 x^2\right )+2 b c^3 e \log (x) \log \left (1-c^2 x^2\right )-b c^3 e \log \left (x-\frac {1}{c}\right ) \log \left (1-c^2 x^2\right )-b c^3 e \log \left (\frac {1}{c}+x\right ) \log \left (1-c^2 x^2\right )-\frac {2 b d \coth ^{-1}(c x)}{x^3}-\frac {b c d}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b d \operatorname {arcoth}\left (c x\right ) + a d + {\left (b e \operatorname {arcoth}\left (c x\right ) + a e\right )} \log \left (-c^{2} x^{2} + 1\right )}{x^{4}}, x\right ) \]

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^4,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^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcoth}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{4}}\,{d x} \]

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^4,x, algorithm="giac")

[Out]

integrate((b*arccoth(c*x) + a)*(e*log(-c^2*x^2 + 1) + d)/x^4, x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \,\mathrm {arccoth}\left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{4}}\, dx \]

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^4,x)

[Out]

int((a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1))/x^4,x)

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

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^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {acoth}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^4} \,d x \]

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^4,x)

[Out]

int(((a + b*acoth(c*x))*(d + e*log(1 - c^2*x^2)))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acoth}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{4}}\, dx \]

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**4,x)

[Out]

Integral((a + b*acoth(c*x))*(d + e*log(-c**2*x**2 + 1))/x**4, x)

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