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

Optimal. Leaf size=148 \[ -\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac {a \log (x)}{d}+\frac {b \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 d}-\frac {b \text {Li}_2(-c x)}{2 d}+\frac {b \text {Li}_2(c x)}{2 d}-\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 d} \]

[Out]

a*ln(x)/d+(a+b*arctanh(c*x))*ln(2/(c*x+1))/d-(a+b*arctanh(c*x))*ln(2*c*(e*x+d)/(c*d+e)/(c*x+1))/d-1/2*b*polylo
g(2,-c*x)/d+1/2*b*polylog(2,c*x)/d-1/2*b*polylog(2,1-2/(c*x+1))/d+1/2*b*polylog(2,1-2*c*(e*x+d)/(c*d+e)/(c*x+1
))/d

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Rubi [A]  time = 0.16, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5940, 5912, 5920, 2402, 2315, 2447} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d}-\frac {b \text {PolyLog}(2,-c x)}{2 d}+\frac {b \text {PolyLog}(2,c x)}{2 d}-\frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 d}-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac {a \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[x])/d + ((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d - ((a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e
)*(1 + c*x))])/d - (b*PolyLog[2, -(c*x)])/(2*d) + (b*PolyLog[2, c*x])/(2*d) - (b*PolyLog[2, 1 - 2/(1 + c*x)])/
(2*d) + (b*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*d)

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

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Mathematica [C]  time = 1.63, size = 294, normalized size = 1.99 \[ \frac {-2 a d \log (d+e x)+2 a d \log (x)+\frac {b \left (e \sqrt {1-\frac {c^2 d^2}{e^2}} \tanh ^{-1}(c x)^2 e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )}+\frac {1}{2} i \pi c d \log \left (1-c^2 x^2\right )+c d \text {Li}_2\left (e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 c d \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac {c d}{e}\right )-2 c d \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+2 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-c d \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+c d \tanh ^{-1}(c x)^2-i \pi c d \tanh ^{-1}(c x)+2 c d \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+i \pi c d \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )-e \tanh ^{-1}(c x)^2\right )}{c}}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a*d*Log[x] - 2*a*d*Log[d + e*x] + (b*((-I)*c*d*Pi*ArcTanh[c*x] - 2*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x] + c*d*
ArcTanh[c*x]^2 - e*ArcTanh[c*x]^2 + (Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^ArcTanh[(c*d)/e] + 2*c*d*ArcT
anh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] + I*c*d*Pi*Log[1 + E^(2*ArcTanh[c*x])] - 2*c*d*ArcTanh[(c*d)/e]*Log[1 -
E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - 2*c*d*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x])
)] + (I/2)*c*d*Pi*Log[1 - c^2*x^2] + 2*c*d*ArcTanh[(c*d)/e]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] - c*d
*PolyLog[2, E^(-2*ArcTanh[c*x])] + c*d*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/c)/(2*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 210, normalized size = 1.42 \[ \frac {a \ln \left (c x \right )}{d}-\frac {a \ln \left (c x e +c d \right )}{d}+\frac {b \arctanh \left (c x \right ) \ln \left (c x \right )}{d}-\frac {b \arctanh \left (c x \right ) \ln \left (c x e +c d \right )}{d}+\frac {b \ln \left (c x e +c d \right ) \ln \left (\frac {c x e +e}{-c d +e}\right )}{2 d}+\frac {b \dilog \left (\frac {c x e +e}{-c d +e}\right )}{2 d}-\frac {b \ln \left (c x e +c d \right ) \ln \left (\frac {c x e -e}{-c d -e}\right )}{2 d}-\frac {b \dilog \left (\frac {c x e -e}{-c d -e}\right )}{2 d}-\frac {b \dilog \left (c x \right )}{2 d}-\frac {b \dilog \left (c x +1\right )}{2 d}-\frac {b \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/d*ln(c*x)-a/d*ln(c*e*x+c*d)+b*arctanh(c*x)/d*ln(c*x)-b*arctanh(c*x)/d*ln(c*e*x+c*d)+1/2*b/d*ln(c*e*x+c*d)*ln
((c*e*x+e)/(-c*d+e))+1/2*b/d*dilog((c*e*x+e)/(-c*d+e))-1/2*b/d*ln(c*e*x+c*d)*ln((c*e*x-e)/(-c*d-e))-1/2*b/d*di
log((c*e*x-e)/(-c*d-e))-1/2*b/d*dilog(c*x)-1/2*b/d*dilog(c*x+1)-1/2*b/d*ln(c*x)*ln(c*x+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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