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

Optimal. Leaf size=51 \[ \frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 c d}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {5918, 2402, 2315} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c d}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(c*d)) + (b*PolyLog[2, 1 - 2/(1 + c*x)])/(2*c*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 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[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)}{d+c d x} \, dx &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c d}+\frac {b \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c d}+\frac {b \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{c d}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c d}+\frac {b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 c d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 52, normalized size = 1.02 \[ \frac {2 a \log (c x+1)+b \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )-2 b \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )}{2 c d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{c d x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 112, normalized size = 2.20 \[ \frac {a \ln \left (c x +1\right )}{c d}+\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{c d}-\frac {b \ln \left (c x +1\right )^{2}}{4 c d}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 c d}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c d}-\frac {b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*a/d*ln(c*x+1)+1/c*b/d*arctanh(c*x)*ln(c*x+1)-1/4/c*b/d*ln(c*x+1)^2+1/2/c*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/
2/c*b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-1/2/c*b/d*dilog(1/2+1/2*c*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(d + c*d*x),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c x + 1}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c*x + 1), x) + Integral(b*atanh(c*x)/(c*x + 1), x))/d

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