3.153 \(\int \frac {a+b \tanh ^{-1}(c x)}{x^3 (d+e x)} \, dx\)

Optimal. Leaf size=261 \[ \frac {e^2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac {e^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^3}+\frac {e \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}-\frac {a+b \tanh ^{-1}(c x)}{2 d x^2}+\frac {a e^2 \log (x)}{d^3}+\frac {b c e \log \left (1-c^2 x^2\right )}{2 d^2}+\frac {b c^2 \tanh ^{-1}(c x)}{2 d}-\frac {b e^2 \text {Li}_2(-c x)}{2 d^3}+\frac {b e^2 \text {Li}_2(c x)}{2 d^3}-\frac {b e^2 \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 d^3}+\frac {b e^2 \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 d^3}-\frac {b c e \log (x)}{d^2}-\frac {b c}{2 d x} \]

[Out]

-1/2*b*c/d/x+1/2*b*c^2*arctanh(c*x)/d+1/2*(-a-b*arctanh(c*x))/d/x^2+e*(a+b*arctanh(c*x))/d^2/x-b*c*e*ln(x)/d^2
+a*e^2*ln(x)/d^3+e^2*(a+b*arctanh(c*x))*ln(2/(c*x+1))/d^3-e^2*(a+b*arctanh(c*x))*ln(2*c*(e*x+d)/(c*d+e)/(c*x+1
))/d^3+1/2*b*c*e*ln(-c^2*x^2+1)/d^2-1/2*b*e^2*polylog(2,-c*x)/d^3+1/2*b*e^2*polylog(2,c*x)/d^3-1/2*b*e^2*polyl
og(2,1-2/(c*x+1))/d^3+1/2*b*e^2*polylog(2,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/d^3

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Rubi [A]  time = 0.25, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 13, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {5940, 5916, 325, 206, 266, 36, 29, 31, 5912, 5920, 2402, 2315, 2447} \[ -\frac {b e^2 \text {PolyLog}(2,-c x)}{2 d^3}+\frac {b e^2 \text {PolyLog}(2,c x)}{2 d^3}-\frac {b e^2 \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 d^3}+\frac {b e^2 \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d^3}+\frac {e^2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac {e^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^3}+\frac {e \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}-\frac {a+b \tanh ^{-1}(c x)}{2 d x^2}+\frac {a e^2 \log (x)}{d^3}+\frac {b c e \log \left (1-c^2 x^2\right )}{2 d^2}+\frac {b c^2 \tanh ^{-1}(c x)}{2 d}-\frac {b c e \log (x)}{d^2}-\frac {b c}{2 d x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(x^3*(d + e*x)),x]

[Out]

-(b*c)/(2*d*x) + (b*c^2*ArcTanh[c*x])/(2*d) - (a + b*ArcTanh[c*x])/(2*d*x^2) + (e*(a + b*ArcTanh[c*x]))/(d^2*x
) - (b*c*e*Log[x])/d^2 + (a*e^2*Log[x])/d^3 + (e^2*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d^3 - (e^2*(a + b*Ar
cTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/d^3 + (b*c*e*Log[1 - c^2*x^2])/(2*d^2) - (b*e^2*PolyLo
g[2, -(c*x)])/(2*d^3) + (b*e^2*PolyLog[2, c*x])/(2*d^3) - (b*e^2*PolyLog[2, 1 - 2/(1 + c*x)])/(2*d^3) + (b*e^2
*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*d^3)

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 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 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 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 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 5912

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[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 5920

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

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

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

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Mathematica [C]  time = 6.09, size = 435, normalized size = 1.67 \[ \frac {a e^2 \log (x)}{d^3}-\frac {a e^2 \log (d+e x)}{d^3}+\frac {a e}{d^2 x}-\frac {a}{2 d x^2}-\frac {b \left (\frac {c d^3 \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)}{x^2}+\frac {c^2 d^3}{x}-e^3 \sqrt {1-\frac {c^2 d^2}{e^2}} \tanh ^{-1}(c x)^2 e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )}+2 c^2 d^2 e \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+i \pi c d e^2 \log \left (\frac {1}{\sqrt {1-c^2 x^2}}\right )-\frac {2 c d^2 e \tanh ^{-1}(c x)}{x}+c d e^2 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )-c d e^2 \text {Li}_2\left (e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-c d e^2 \tanh ^{-1}(c x)^2+2 c d e^2 \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac {c d}{e}\right )+i \pi c d e^2 \tanh ^{-1}(c x)-2 c d e^2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-i \pi c d e^2 \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )+2 c d e^2 \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+2 c d e^2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 c d e^2 \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )+e^3 \tanh ^{-1}(c x)^2\right )}{2 c d^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*a/(d*x^2) + (a*e)/(d^2*x) + (a*e^2*Log[x])/d^3 - (a*e^2*Log[d + e*x])/d^3 - (b*((c^2*d^3)/x + I*c*d*e^2*P
i*ArcTanh[c*x] - (2*c*d^2*e*ArcTanh[c*x])/x + (c*d^3*(1 - c^2*x^2)*ArcTanh[c*x])/x^2 + 2*c*d*e^2*ArcTanh[(c*d)
/e]*ArcTanh[c*x] - c*d*e^2*ArcTanh[c*x]^2 + e^3*ArcTanh[c*x]^2 - (Sqrt[1 - (c^2*d^2)/e^2]*e^3*ArcTanh[c*x]^2)/
E^ArcTanh[(c*d)/e] - 2*c*d*e^2*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - I*c*d*e^2*Pi*Log[1 + E^(2*ArcTanh[c
*x])] + 2*c*d*e^2*ArcTanh[(c*d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + 2*c*d*e^2*ArcTanh[c*x]*
Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + I*c*d*e^2*Pi*Log[1/Sqrt[1 - c^2*x^2]] + 2*c^2*d^2*e*Log[(c
*x)/Sqrt[1 - c^2*x^2]] - 2*c*d*e^2*ArcTanh[(c*d)/e]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] + c*d*e^2*Pol
yLog[2, E^(-2*ArcTanh[c*x])] - c*d*e^2*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/(2*c*d^4)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{e x^{4} + d x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^3/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*arctanh(c*x) + a)/(e*x^4 + d*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {artanh}\left (c x\right ) + a}{{\left (e x + d\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^3/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)/((e*x + d)*x^3), x)

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maple [A]  time = 0.07, size = 367, normalized size = 1.41 \[ -\frac {a}{2 d \,x^{2}}+\frac {a \,e^{2} \ln \left (c x \right )}{d^{3}}+\frac {a e}{d^{2} x}-\frac {a \,e^{2} \ln \left (c x e +c d \right )}{d^{3}}-\frac {b \arctanh \left (c x \right )}{2 d \,x^{2}}+\frac {b \arctanh \left (c x \right ) e^{2} \ln \left (c x \right )}{d^{3}}+\frac {b \arctanh \left (c x \right ) e}{d^{2} x}-\frac {b \arctanh \left (c x \right ) e^{2} \ln \left (c x e +c d \right )}{d^{3}}-\frac {b \,e^{2} \dilog \left (c x \right )}{2 d^{3}}-\frac {b \,e^{2} \dilog \left (c x +1\right )}{2 d^{3}}-\frac {b \,e^{2} \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d^{3}}+\frac {b \,e^{2} \ln \left (c x e +c d \right ) \ln \left (\frac {c x e +e}{-c d +e}\right )}{2 d^{3}}+\frac {b \,e^{2} \dilog \left (\frac {c x e +e}{-c d +e}\right )}{2 d^{3}}-\frac {b \,e^{2} \ln \left (c x e +c d \right ) \ln \left (\frac {c x e -e}{-c d -e}\right )}{2 d^{3}}-\frac {b \,e^{2} \dilog \left (\frac {c x e -e}{-c d -e}\right )}{2 d^{3}}-\frac {b c}{2 d x}-\frac {c b e \ln \left (c x \right )}{d^{2}}-\frac {c^{2} b \ln \left (c x -1\right )}{4 d}+\frac {c b \ln \left (c x -1\right ) e}{2 d^{2}}+\frac {c^{2} b \ln \left (c x +1\right )}{4 d}+\frac {c b \ln \left (c x +1\right ) e}{2 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x^3/(e*x+d),x)

[Out]

-1/2*a/d/x^2+a/d^3*e^2*ln(c*x)+a/d^2*e/x-a/d^3*e^2*ln(c*e*x+c*d)-1/2*b*arctanh(c*x)/d/x^2+b*arctanh(c*x)/d^3*e
^2*ln(c*x)+b*arctanh(c*x)/d^2*e/x-b*arctanh(c*x)/d^3*e^2*ln(c*e*x+c*d)-1/2*b/d^3*e^2*dilog(c*x)-1/2*b/d^3*e^2*
dilog(c*x+1)-1/2*b/d^3*e^2*ln(c*x)*ln(c*x+1)+1/2*b/d^3*e^2*ln(c*e*x+c*d)*ln((c*e*x+e)/(-c*d+e))+1/2*b/d^3*e^2*
dilog((c*e*x+e)/(-c*d+e))-1/2*b/d^3*e^2*ln(c*e*x+c*d)*ln((c*e*x-e)/(-c*d-e))-1/2*b/d^3*e^2*dilog((c*e*x-e)/(-c
*d-e))-1/2*b*c/d/x-c*b/d^2*e*ln(c*x)-1/4*c^2*b/d*ln(c*x-1)+1/2*c*b/d^2*ln(c*x-1)*e+1/4*c^2*b/d*ln(c*x+1)+1/2*c
*b/d^2*ln(c*x+1)*e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^3/(e*x+d),x, algorithm="maxima")

[Out]

-1/2*a*(2*e^2*log(e*x + d)/d^3 - 2*e^2*log(x)/d^3 - (2*e*x - d)/(d^2*x^2)) + 1/2*b*integrate((log(c*x + 1) - l
og(-c*x + 1))/(e*x^4 + d*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{x^3\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(x^3*(d + e*x)),x)

[Out]

int((a + b*atanh(c*x))/(x^3*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{x^{3} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x**3/(e*x+d),x)

[Out]

Integral((a + b*atanh(c*x))/(x**3*(d + e*x)), x)

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