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

Optimal. Leaf size=93 \[ -\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {c \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d}+\frac {b c \text {Li}_2\left (\frac {2}{c x+1}-1\right )}{2 d}+\frac {b c \log (x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5934, 5916, 266, 36, 29, 31, 5932, 2447} \[ \frac {b c \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )}{2 d}-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {c \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d}+\frac {b c \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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]

Rule 5934

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 93, normalized size = 1.00 \[ \frac {b c x \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )-2 \left (a c x \log (x)-a c x \log (c x+1)+a-b c x \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+b \tanh ^{-1}(c x) \left (c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+1\right )\right )}{2 d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{c d x^{3} + d x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctanh(c*x) + a)/(c*d*x^3 + d*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 225, normalized size = 2.42 \[ -\frac {a}{d x}-\frac {c a \ln \left (c x \right )}{d}+\frac {c a \ln \left (c x +1\right )}{d}-\frac {b \arctanh \left (c x \right )}{d x}-\frac {c b \arctanh \left (c x \right ) \ln \left (c x \right )}{d}+\frac {c b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d}+\frac {c b \ln \left (c x \right )}{d}-\frac {c b \ln \left (c x -1\right )}{2 d}-\frac {c b \ln \left (c x +1\right )}{2 d}+\frac {c b \dilog \left (c x \right )}{2 d}+\frac {c b \dilog \left (c x +1\right )}{2 d}+\frac {c b \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d}-\frac {c b \ln \left (c x +1\right )^{2}}{4 d}+\frac {c b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 d}-\frac {c b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 d}-\frac {c b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d/x-c*a/d*ln(c*x)+c*a/d*ln(c*x+1)-b/d*arctanh(c*x)/x-c*b*arctanh(c*x)/d*ln(c*x)+c*b/d*arctanh(c*x)*ln(c*x+1
)+c*b/d*ln(c*x)-1/2*c*b/d*ln(c*x-1)-1/2*c*b/d*ln(c*x+1)+1/2*c*b/d*dilog(c*x)+1/2*c*b/d*dilog(c*x+1)+1/2*c*b/d*
ln(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 \[ a {\left (\frac {c \log \left (c x + 1\right )}{d} - \frac {c \log \relax (x)}{d} - \frac {1}{d x}\right )} + \frac {1}{2} \, b \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{c d x^{3} + d x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(c*log(c*x + 1)/d - c*log(x)/d - 1/(d*x)) + 1/2*b*integrate((log(c*x + 1) - log(-c*x + 1))/(c*d*x^3 + d*x^2)
, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c x^{3} + x^{2}}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{c x^{3} + x^{2}}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c*x**3 + x**2), x) + Integral(b*atanh(c*x)/(c*x**3 + x**2), x))/d

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