3.6 \(\int \frac {(d+c d x) (a+b \tanh ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=70 \[ -\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \log \left (1-c^2 x^2\right )-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)+b c d \log (x) \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5940, 5916, 266, 36, 29, 31, 5912} \[ -\frac {1}{2} b c d \text {PolyLog}(2,-c x)+\frac {1}{2} b c d \text {PolyLog}(2,c x)-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \log \left (1-c^2 x^2\right )+b c d \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)*(a + b*ArcTanh[c*x]))/x^2,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 5916

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

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^2} \, dx &=\int \left (\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+\frac {c d \left (a+b \tanh ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+(c d) \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)+(b c d) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)+\frac {1}{2} (b c d) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)+\frac {1}{2} (b c d) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b c^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)+b c d \log (x)-\frac {1}{2} b c d \log \left (1-c^2 x^2\right )-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 71, normalized size = 1.01 \[ a c d \log (x)-\frac {a d}{x}+b c d \left (-\frac {1}{2} \log \left (1-c^2 x^2\right )+\log (c x)-\frac {\tanh ^{-1}(c x)}{c x}\right )+\frac {1}{2} b c d (\text {Li}_2(c x)-\text {Li}_2(-c x)) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + c*d*x)*(a + b*ArcTanh[c*x]))/x^2,x]

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a c d x + a d + {\left (b c d x + b d\right )} \operatorname {artanh}\left (c x\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 105, normalized size = 1.50 \[ c d a \ln \left (c x \right )-\frac {a d}{x}+c d b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d b \arctanh \left (c x \right )}{x}+c d b \ln \left (c x \right )-\frac {c d b \ln \left (c x -1\right )}{2}-\frac {c d b \ln \left (c x +1\right )}{2}-\frac {c d b \dilog \left (c x \right )}{2}-\frac {c d b \dilog \left (c x +1\right )}{2}-\frac {c d b \ln \left (c x \right ) \ln \left (c x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+c\,d\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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