3.49 \(\int \frac{a+b \tanh ^{-1}(c x)}{x^3 (d+c d x)} \, dx\)

Optimal. Leaf size=146 \[ -\frac{b c^2 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{2 d}+\frac{c^2 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{a+b \tanh ^{-1}(c x)}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{b c^2 \log \left (1-c^2 x^2\right )}{2 d}-\frac{b c^2 \log (x)}{d}+\frac{b c^2 \tanh ^{-1}(c x)}{2 d}-\frac{b c}{2 d x} \]

[Out]

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

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Rubi [A]  time = 0.234477, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5934, 5916, 325, 206, 266, 36, 29, 31, 5932, 2447} \[ -\frac{b c^2 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{2 d}+\frac{c^2 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{a+b \tanh ^{-1}(c x)}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{b c^2 \log \left (1-c^2 x^2\right )}{2 d}-\frac{b c^2 \log (x)}{d}+\frac{b c^2 \tanh ^{-1}(c x)}{2 d}-\frac{b c}{2 d x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(x^3*(d + c*d*x)),x]

[Out]

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

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]

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 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 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 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 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]]

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}(c x)}{x^3 (d+c d x)} \, dx &=-\left (c \int \frac{a+b \tanh ^{-1}(c x)}{x^2 (d+c d x)} \, dx\right )+\frac{\int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx}{d}\\ &=-\frac{a+b \tanh ^{-1}(c x)}{2 d x^2}+c^2 \int \frac{a+b \tanh ^{-1}(c x)}{x (d+c d x)} \, dx-\frac{c \int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx}{d}+\frac{(b c) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx}{2 d}\\ &=-\frac{b c}{2 d x}-\frac{a+b \tanh ^{-1}(c x)}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{\left (b c^2\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx}{d}+\frac{\left (b c^3\right ) \int \frac{1}{1-c^2 x^2} \, dx}{2 d}-\frac{\left (b c^3\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac{b c}{2 d x}+\frac{b c^2 \tanh ^{-1}(c x)}{2 d}-\frac{a+b \tanh ^{-1}(c x)}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{b c^2 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{2 d}-\frac{\left (b c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d}\\ &=-\frac{b c}{2 d x}+\frac{b c^2 \tanh ^{-1}(c x)}{2 d}-\frac{a+b \tanh ^{-1}(c x)}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{b c^2 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{2 d}-\frac{\left (b c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{2 d}-\frac{\left (b c^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac{b c}{2 d x}+\frac{b c^2 \tanh ^{-1}(c x)}{2 d}-\frac{a+b \tanh ^{-1}(c x)}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{b c^2 \log (x)}{d}+\frac{b c^2 \log \left (1-c^2 x^2\right )}{2 d}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{b c^2 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.344723, size = 133, normalized size = 0.91 \[ -\frac{b c^2 x^2 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-2 a c^2 x^2 \log (x)+2 a c^2 x^2 \log (c x+1)-2 a c x+a+2 b c^2 x^2 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )-b \tanh ^{-1}(c x) \left (c^2 x^2+2 c^2 x^2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+2 c x-1\right )+b c x}{2 d x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])/(x^3*(d + c*d*x)),x]

[Out]

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

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Maple [B]  time = 0.058, size = 286, normalized size = 2. \begin{align*} -{\frac{a}{2\,d{x}^{2}}}+{\frac{a{c}^{2}\ln \left ( cx \right ) }{d}}+{\frac{ac}{dx}}-{\frac{a{c}^{2}\ln \left ( cx+1 \right ) }{d}}-{\frac{b{\it Artanh} \left ( cx \right ) }{2\,d{x}^{2}}}+{\frac{{c}^{2}b{\it Artanh} \left ( cx \right ) \ln \left ( cx \right ) }{d}}+{\frac{bc{\it Artanh} \left ( cx \right ) }{dx}}-{\frac{{c}^{2}b{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{d}}-{\frac{{c}^{2}b{\it dilog} \left ( cx \right ) }{2\,d}}-{\frac{{c}^{2}b{\it dilog} \left ( cx+1 \right ) }{2\,d}}-{\frac{{c}^{2}b\ln \left ( cx \right ) \ln \left ( cx+1 \right ) }{2\,d}}+{\frac{{c}^{2}b}{2\,d}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{c}^{2}b\ln \left ( cx+1 \right ) }{2\,d}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }+{\frac{{c}^{2}b}{2\,d}{\it dilog} \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{c}^{2}b \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{4\,d}}+{\frac{{c}^{2}b\ln \left ( cx-1 \right ) }{4\,d}}-{\frac{bc}{2\,dx}}-{\frac{{c}^{2}b\ln \left ( cx \right ) }{d}}+{\frac{3\,{c}^{2}b\ln \left ( cx+1 \right ) }{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d/x^2+c^2*a/d*ln(c*x)+c*a/d/x-c^2*a/d*ln(c*x+1)-1/2*b/d*arctanh(c*x)/x^2+c^2*b/d*arctanh(c*x)*ln(c*x)+c
*b/d*arctanh(c*x)/x-c^2*b/d*arctanh(c*x)*ln(c*x+1)-1/2*c^2*b/d*dilog(c*x)-1/2*c^2*b/d*dilog(c*x+1)-1/2*c^2*b/d
*ln(c*x)*ln(c*x+1)+1/2*c^2*b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-1/2*c^2*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/2*c^2
*b/d*dilog(1/2+1/2*c*x)+1/4*c^2*b/d*ln(c*x+1)^2+1/4*c^2*b/d*ln(c*x-1)-1/2*b*c/d/x-c^2*b/d*ln(c*x)+3/4*c^2*b/d*
ln(c*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \,{\left (\frac{2 \, c^{2} \log \left (c x + 1\right )}{d} - \frac{2 \, c^{2} \log \left (x\right )}{d} - \frac{2 \, c x - 1}{d x^{2}}\right )} a + \frac{1}{2} \, b \int \frac{\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{c d x^{4} + d x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^3/(c*d*x+d),x, algorithm="maxima")

[Out]

-1/2*(2*c^2*log(c*x + 1)/d - 2*c^2*log(x)/d - (2*c*x - 1)/(d*x^2))*a + 1/2*b*integrate((log(c*x + 1) - log(-c*
x + 1))/(c*d*x^4 + d*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{artanh}\left (c x\right ) + a}{c d x^{4} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^3/(c*d*x+d),x, algorithm="fricas")

[Out]

integral((b*arctanh(c*x) + a)/(c*d*x^4 + d*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c x^{4} + x^{3}}\, dx + \int \frac{b \operatorname{atanh}{\left (c x \right )}}{c x^{4} + x^{3}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c*x**4 + x**3), x) + Integral(b*atanh(c*x)/(c*x**4 + x**3), x))/d

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{artanh}\left (c x\right ) + a}{{\left (c d x + d\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^3/(c*d*x+d),x, algorithm="giac")

[Out]

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