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

Optimal. Leaf size=114 \[ \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e}-\frac {b \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 e}+\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5920, 2402, 2315, 2447} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e}+\frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 e}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(d + e*x),x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.24, size = 257, normalized size = 2.25 \[ \frac {a \log (d+e x)-\frac {1}{2} i b \left (-\log \left (\frac {2}{\sqrt {1-c^2 x^2}}\right ) \left (\pi -2 i \tanh ^{-1}(c x)\right )-i \text {Li}_2\left (e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+i \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )^2+2 i \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 i \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right ) \log \left (2 i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-i \text {Li}_2\left (-e^{2 \tanh ^{-1}(c x)}\right )-\frac {1}{4} i \left (\pi -2 i \tanh ^{-1}(c x)\right )^2+\left (\pi -2 i \tanh ^{-1}(c x)\right ) \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )\right )+b \tanh ^{-1}(c x) \left (\frac {1}{2} \log \left (1-c^2 x^2\right )+\log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )\right )}{e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 148, normalized size = 1.30 \[ \frac {a \ln \left (c x e +c d \right )}{e}+\frac {b \ln \left (c x e +c d \right ) \arctanh \left (c x \right )}{e}-\frac {b \ln \left (c x e +c d \right ) \ln \left (\frac {c x e +e}{-c d +e}\right )}{2 e}-\frac {b \dilog \left (\frac {c x e +e}{-c d +e}\right )}{2 e}+\frac {b \ln \left (c x e +c d \right ) \ln \left (\frac {c x e -e}{-c d -e}\right )}{2 e}+\frac {b \dilog \left (\frac {c x e -e}{-c d -e}\right )}{2 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(d + e*x),x)

[Out]

int((a + b*atanh(c*x))/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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