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

Optimal. Leaf size=156 \[ \frac {d \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac {a x}{e}+\frac {b \log \left (1-c^2 x^2\right )}{2 c e}-\frac {b d \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 e^2}+\frac {b d \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 e^2}+\frac {b x \tanh ^{-1}(c x)}{e} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {5940, 5910, 260, 5920, 2402, 2315, 2447} \[ -\frac {b d \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 e^2}+\frac {b d \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^2}+\frac {d \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac {a x}{e}+\frac {b \log \left (1-c^2 x^2\right )}{2 c e}+\frac {b x \tanh ^{-1}(c x)}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 260

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

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 5910

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

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

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Mathematica [C]  time = 2.47, size = 315, normalized size = 2.02 \[ \frac {-2 a d \log (d+e x)+2 a e x+\frac {b \left (e \sqrt {1-\frac {c^2 d^2}{e^2}} \tanh ^{-1}(c x)^2 e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )}+\frac {1}{2} i \pi c d \log \left (1-c^2 x^2\right )+e \log \left (1-c^2 x^2\right )+c d \text {Li}_2\left (e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 c d \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac {c d}{e}\right )-2 c d \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+2 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-c d \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+c d \tanh ^{-1}(c x)^2-i \pi c d \tanh ^{-1}(c x)+2 c d \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+i \pi c d \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )-e \tanh ^{-1}(c x)^2+2 c e x \tanh ^{-1}(c x)\right )}{c}}{2 e^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a*e*x - 2*a*d*Log[d + e*x] + (b*((-I)*c*d*Pi*ArcTanh[c*x] + 2*c*e*x*ArcTanh[c*x] - 2*c*d*ArcTanh[(c*d)/e]*A
rcTanh[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*ArcTanh[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]))] + e*Log[1 - c^2*x^2] + (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*e^2)

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x \operatorname {artanh}\left (c x\right ) + a x}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 217, normalized size = 1.39 \[ \frac {a x}{e}-\frac {a d \ln \left (c x e +c d \right )}{e^{2}}+\frac {b x \arctanh \left (c x \right )}{e}-\frac {b \arctanh \left (c x \right ) d \ln \left (c x e +c d \right )}{e^{2}}+\frac {b d \ln \left (c x e +c d \right ) \ln \left (\frac {c x e +e}{-c d +e}\right )}{2 e^{2}}+\frac {b d \dilog \left (\frac {c x e +e}{-c d +e}\right )}{2 e^{2}}-\frac {b d \ln \left (c x e +c d \right ) \ln \left (\frac {c x e -e}{-c d -e}\right )}{2 e^{2}}-\frac {b d \dilog \left (\frac {c x e -e}{-c d -e}\right )}{2 e^{2}}+\frac {b \ln \left (c^{2} d^{2}-2 c d \left (c x e +c d \right )+\left (c x e +c d \right )^{2}-e^{2}\right )}{2 c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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