3.16 \(\int \frac{(d+c d x)^2 (a+b \tanh ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=137 \[ -\frac{1}{2} b c^2 d^2 \text{PolyLog}(2,-c x)+\frac{1}{2} b c^2 d^2 \text{PolyLog}(2,c x)-\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^2 d^2 \log (x)-b c^2 d^2 \log \left (1-c^2 x^2\right )+2 b c^2 d^2 \log (x)+\frac{1}{2} b c^2 d^2 \tanh ^{-1}(c x)-\frac{b c d^2}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.139936, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {5940, 5916, 325, 206, 266, 36, 29, 31, 5912} \[ -\frac{1}{2} b c^2 d^2 \text{PolyLog}(2,-c x)+\frac{1}{2} b c^2 d^2 \text{PolyLog}(2,c x)-\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^2 d^2 \log (x)-b c^2 d^2 \log \left (1-c^2 x^2\right )+2 b c^2 d^2 \log (x)+\frac{1}{2} b c^2 d^2 \tanh ^{-1}(c x)-\frac{b c d^2}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^3,x]

[Out]

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

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])

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 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]

Rubi steps

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

Mathematica [A]  time = 0.100833, size = 143, normalized size = 1.04 \[ \frac{d^2 \left (-2 b c^2 x^2 \text{PolyLog}(2,-c x)+2 b c^2 x^2 \text{PolyLog}(2,c x)+4 a c^2 x^2 \log (x)-8 a c x-2 a+8 b c^2 x^2 \log (c x)-b c^2 x^2 \log (1-c x)+b c^2 x^2 \log (c x+1)-4 b c^2 x^2 \log \left (1-c^2 x^2\right )-2 b c x-8 b c x \tanh ^{-1}(c x)-2 b \tanh ^{-1}(c x)\right )}{4 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^3,x]

[Out]

(d^2*(-2*a - 8*a*c*x - 2*b*c*x - 2*b*ArcTanh[c*x] - 8*b*c*x*ArcTanh[c*x] + 4*a*c^2*x^2*Log[x] + 8*b*c^2*x^2*Lo
g[c*x] - b*c^2*x^2*Log[1 - c*x] + b*c^2*x^2*Log[1 + c*x] - 4*b*c^2*x^2*Log[1 - c^2*x^2] - 2*b*c^2*x^2*PolyLog[
2, -(c*x)] + 2*b*c^2*x^2*PolyLog[2, c*x]))/(4*x^2)

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Maple [A]  time = 0.046, size = 176, normalized size = 1.3 \begin{align*} -2\,{\frac{c{d}^{2}a}{x}}+{c}^{2}{d}^{2}a\ln \left ( cx \right ) -{\frac{{d}^{2}a}{2\,{x}^{2}}}-2\,{\frac{c{d}^{2}b{\it Artanh} \left ( cx \right ) }{x}}+{c}^{2}{d}^{2}b{\it Artanh} \left ( cx \right ) \ln \left ( cx \right ) -{\frac{{d}^{2}b{\it Artanh} \left ( cx \right ) }{2\,{x}^{2}}}-{\frac{5\,{c}^{2}{d}^{2}b\ln \left ( cx-1 \right ) }{4}}-{\frac{c{d}^{2}b}{2\,x}}+2\,{c}^{2}{d}^{2}b\ln \left ( cx \right ) -{\frac{3\,{c}^{2}{d}^{2}b\ln \left ( cx+1 \right ) }{4}}-{\frac{{c}^{2}{d}^{2}b{\it dilog} \left ( cx \right ) }{2}}-{\frac{{c}^{2}{d}^{2}b{\it dilog} \left ( cx+1 \right ) }{2}}-{\frac{{c}^{2}{d}^{2}b\ln \left ( cx \right ) \ln \left ( cx+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c*d^2*a/x+c^2*d^2*a*ln(c*x)-1/2*d^2*a/x^2-2*c*d^2*b*arctanh(c*x)/x+c^2*d^2*b*arctanh(c*x)*ln(c*x)-1/2*d^2*b
*arctanh(c*x)/x^2-5/4*c^2*d^2*b*ln(c*x-1)-1/2*b*c*d^2/x+2*c^2*d^2*b*ln(c*x)-3/4*c^2*d^2*b*ln(c*x+1)-1/2*c^2*d^
2*b*dilog(c*x)-1/2*c^2*d^2*b*dilog(c*x+1)-1/2*c^2*d^2*b*ln(c*x)*ln(c*x+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a c^{2} d^{2} x^{2} + 2 \, a c d^{2} x + a d^{2} +{\left (b c^{2} d^{2} x^{2} + 2 \, b c d^{2} x + b d^{2}\right )} \operatorname{artanh}\left (c x\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} \left (\int \frac{a}{x^{3}}\, dx + \int \frac{2 a c}{x^{2}}\, dx + \int \frac{a c^{2}}{x}\, dx + \int \frac{b \operatorname{atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{2 b c \operatorname{atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{b c^{2} \operatorname{atanh}{\left (c x \right )}}{x}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^3,x, algorithm="giac")

[Out]

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