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

Optimal. Leaf size=186 \[ \frac {d \text {Li}_2\left (\frac {c (-a-b x+1)}{-a c+c+b d}\right )}{2 c^2}-\frac {d \text {Li}_2\left (\frac {c (a+b x+1)}{a c+c-b d}\right )}{2 c^2}+\frac {d \log (-a-b x+1) \log \left (\frac {b (c x+d)}{-a c+b d+c}\right )}{2 c^2}-\frac {d \log (a+b x+1) \log \left (-\frac {b (c x+d)}{a c-b d+c}\right )}{2 c^2}+\frac {(-a-b x+1) \log (-a-b x+1)}{2 b c}+\frac {(a+b x+1) \log (a+b x+1)}{2 b c} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6115, 2409, 2389, 2295, 2394, 2393, 2391} \[ \frac {d \text {PolyLog}\left (2,\frac {c (-a-b x+1)}{-a c+b d+c}\right )}{2 c^2}-\frac {d \text {PolyLog}\left (2,\frac {c (a+b x+1)}{a c-b d+c}\right )}{2 c^2}+\frac {d \log (-a-b x+1) \log \left (\frac {b (c x+d)}{-a c+b d+c}\right )}{2 c^2}-\frac {d \log (a+b x+1) \log \left (-\frac {b (c x+d)}{a c-b d+c}\right )}{2 c^2}+\frac {(-a-b x+1) \log (-a-b x+1)}{2 b c}+\frac {(a+b x+1) \log (a+b x+1)}{2 b c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a + b*x]/(c + d/x),x]

[Out]

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

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

Rule 2409

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

Rule 6115

Int[ArcTanh[(c_) + (d_.)*(x_)]/((e_) + (f_.)*(x_)^(n_.)), x_Symbol] :> Dist[1/2, Int[Log[1 + c + d*x]/(e + f*x
^n), x], x] - Dist[1/2, Int[Log[1 - c - d*x]/(e + f*x^n), x], x] /; FreeQ[{c, d, e, f}, x] && RationalQ[n]

Rubi steps

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

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Mathematica [C]  time = 4.15, size = 759, normalized size = 4.08 \[ \frac {b c d \sqrt {-a^2+\frac {2 a b d}{c}-\frac {b^2 d^2}{c^2}+1} \tanh ^{-1}(a+b x)^2 e^{\tanh ^{-1}\left (a-\frac {b d}{c}\right )}-2 a^2 c^2 \tanh ^{-1}(a+b x)+2 b^2 d^2 \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \log \left (1-\exp \left (2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )-2 b^2 d^2 \tanh ^{-1}(a+b x) \log \left (1-\exp \left (2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )+2 b^2 d^2 \tanh ^{-1}(a+b x) \tanh ^{-1}\left (a-\frac {b d}{c}\right )-2 b^2 d^2 \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \log \left (-i \sinh \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )+2 b^2 c d x \tanh ^{-1}(a+b x)-i \pi b^2 d^2 \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+b^2 d^2 \tanh ^{-1}(a+b x)^2-i \pi b^2 d^2 \tanh ^{-1}(a+b x)+2 b^2 d^2 \tanh ^{-1}(a+b x) \log \left (e^{-2 \tanh ^{-1}(a+b x)}+1\right )+i \pi b^2 d^2 \log \left (e^{2 \tanh ^{-1}(a+b x)}+1\right )+2 a c^2 \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )-2 a b c^2 x \tanh ^{-1}(a+b x)+b d (b d-a c) \text {Li}_2\left (\exp \left (2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )-2 a b c d \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \log \left (1-\exp \left (2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )+2 a b c d \tanh ^{-1}(a+b x) \log \left (1-\exp \left (2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )+b d (a c-b d) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a+b x)}\right )-2 b c d \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+i \pi a b c d \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )-a b c d \tanh ^{-1}(a+b x)^2-b c d \tanh ^{-1}(a+b x)^2+2 a b c d \tanh ^{-1}(a+b x)-2 a b c d \tanh ^{-1}(a+b x) \tanh ^{-1}\left (a-\frac {b d}{c}\right )+i \pi a b c d \tanh ^{-1}(a+b x)-2 a b c d \tanh ^{-1}(a+b x) \log \left (e^{-2 \tanh ^{-1}(a+b x)}+1\right )-i \pi a b c d \log \left (e^{2 \tanh ^{-1}(a+b x)}+1\right )+2 a b c d \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \log \left (-i \sinh \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )}{2 b c^2 (b d-a c)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x \operatorname {artanh}\left (b x + a\right )}{c x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 297, normalized size = 1.60 \[ \frac {\arctanh \left (b x +a \right ) x}{c}+\frac {\arctanh \left (b x +a \right ) a}{b c}-\frac {\arctanh \left (b x +a \right ) d \ln \left (c \left (b x +a \right )-a c +b d \right )}{c^{2}}+\frac {d \ln \left (c \left (b x +a \right )-a c +b d \right ) \ln \left (\frac {c \left (b x +a \right )+c}{a c -b d +c}\right )}{2 c^{2}}+\frac {d \dilog \left (\frac {c \left (b x +a \right )+c}{a c -b d +c}\right )}{2 c^{2}}-\frac {d \ln \left (c \left (b x +a \right )-a c +b d \right ) \ln \left (\frac {c \left (b x +a \right )-c}{a c -b d -c}\right )}{2 c^{2}}-\frac {d \dilog \left (\frac {c \left (b x +a \right )-c}{a c -b d -c}\right )}{2 c^{2}}+\frac {\ln \left (a^{2} c^{2}-2 a b c d +b^{2} d^{2}+2 \left (c \left (b x +a \right )-a c +b d \right ) a c -2 \left (c \left (b x +a \right )-a c +b d \right ) b d +\left (c \left (b x +a \right )-a c +b d \right )^{2}-c^{2}\right )}{2 b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 192, normalized size = 1.03 \[ \frac {1}{2} \, b {\left (\frac {{\left (\log \left (c x + d\right ) \log \left (\frac {b c x + b d}{a c - b d + c} + 1\right ) + {\rm Li}_2\left (-\frac {b c x + b d}{a c - b d + c}\right )\right )} d}{b c^{2}} - \frac {{\left (\log \left (c x + d\right ) \log \left (\frac {b c x + b d}{a c - b d - c} + 1\right ) + {\rm Li}_2\left (-\frac {b c x + b d}{a c - b d - c}\right )\right )} d}{b c^{2}} + \frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2} c} - \frac {{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2} c}\right )} + {\left (\frac {x}{c} - \frac {d \log \left (c x + d\right )}{c^{2}}\right )} \operatorname {artanh}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a + b*x)/(c + d/x),x)

[Out]

int(atanh(a + b*x)/(c + d/x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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