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

Optimal. Leaf size=46 \[ \frac {\log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {b \text {Li}_2\left (\frac {2}{c x+1}-1\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5932, 2447} \[ \frac {\log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {b \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5932

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

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 55, normalized size = 1.20 \[ \frac {-2 a \log (c x+1)+2 a \log (x)-b \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+2 b \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{c d x^{2} + d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 156, normalized size = 3.39 \[ \frac {a \ln \left (c x \right )}{d}-\frac {a \ln \left (c x +1\right )}{d}-\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d}+\frac {b \arctanh \left (c x \right ) \ln \left (c x \right )}{d}+\frac {b \ln \left (c x +1\right )^{2}}{4 d}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 d}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 d}+\frac {b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 d}-\frac {b \dilog \left (c x \right )}{2 d}-\frac {b \dilog \left (c x +1\right )}{2 d}-\frac {b \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/d*ln(c*x)-a/d*ln(c*x+1)-b/d*arctanh(c*x)*ln(c*x+1)+b*arctanh(c*x)/d*ln(c*x)+1/4*b/d*ln(c*x+1)^2-1/2*b/d*ln(-
1/2*c*x+1/2)*ln(c*x+1)+1/2*b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/2*b/d*dilog(1/2+1/2*c*x)-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 \[ -a {\left (\frac {\log \left (c x + 1\right )}{d} - \frac {\log \relax (x)}{d}\right )} + \frac {1}{2} \, b \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{c d x^{2} + d x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{x\,\left (d+c\,d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c x^{2} + x}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{c x^{2} + x}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c*x**2 + x), x) + Integral(b*atanh(c*x)/(c*x**2 + x), x))/d

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