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

Optimal. Leaf size=60 \[ a c d x+a d \log (x)+\frac {1}{2} b d \log \left (1-c^2 x^2\right )-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)+b c d x \tanh ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5940, 5910, 260, 5912} \[ -\frac {1}{2} b d \text {PolyLog}(2,-c x)+\frac {1}{2} b d \text {PolyLog}(2,c x)+a c d x+a d \log (x)+\frac {1}{2} b d \log \left (1-c^2 x^2\right )+b c d x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*c*d*x + b*c*d*x*ArcTanh[c*x] + a*d*Log[x] + (b*d*Log[1 - c^2*x^2])/2 - (b*d*PolyLog[2, -(c*x)])/2 + (b*d*Pol
yLog[2, c*x])/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 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 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, x]

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

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Mathematica [A]  time = 0.07, size = 54, normalized size = 0.90 \[ \frac {1}{2} d \left (2 a c x+2 a \log (x)+b \log \left (1-c^2 x^2\right )-b \text {Li}_2(-c x)+b \text {Li}_2(c x)+2 b c x \tanh ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 86, normalized size = 1.43 \[ d a \ln \left (c x \right )+a c d x +d b \arctanh \left (c x \right ) \ln \left (c x \right )+b c d x \arctanh \left (c x \right )+\frac {d b \ln \left (c x -1\right )}{2}+\frac {d b \ln \left (c x +1\right )}{2}-\frac {d b \dilog \left (c x \right )}{2}-\frac {d b \dilog \left (c x +1\right )}{2}-\frac {d b \ln \left (c x \right ) \ln \left (c x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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