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

Optimal. Leaf size=150 \[ -\frac{3}{2} b c d^3 \text{PolyLog}(2,-c x)+\frac{3}{2} b c d^3 \text{PolyLog}(2,c x)+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+3 a c^2 d^3 x+3 a c d^3 \log (x)+b c d^3 \log \left (1-c^2 x^2\right )+\frac{1}{2} b c^2 d^3 x+3 b c^2 d^3 x \tanh ^{-1}(c x)+b c d^3 \log (x)-\frac{1}{2} b c d^3 \tanh ^{-1}(c x) \]

[Out]

3*a*c^2*d^3*x + (b*c^2*d^3*x)/2 - (b*c*d^3*ArcTanh[c*x])/2 + 3*b*c^2*d^3*x*ArcTanh[c*x] - (d^3*(a + b*ArcTanh[
c*x]))/x + (c^3*d^3*x^2*(a + b*ArcTanh[c*x]))/2 + 3*a*c*d^3*Log[x] + b*c*d^3*Log[x] + b*c*d^3*Log[1 - c^2*x^2]
- (3*b*c*d^3*PolyLog[2, -(c*x)])/2 + (3*b*c*d^3*PolyLog[2, c*x])/2

________________________________________________________________________________________

Rubi [A]  time = 0.157628, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.55, Rules used = {5940, 5910, 260, 5916, 266, 36, 29, 31, 5912, 321, 206} \[ -\frac{3}{2} b c d^3 \text{PolyLog}(2,-c x)+\frac{3}{2} b c d^3 \text{PolyLog}(2,c x)+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+3 a c^2 d^3 x+3 a c d^3 \log (x)+b c d^3 \log \left (1-c^2 x^2\right )+\frac{1}{2} b c^2 d^3 x+3 b c^2 d^3 x \tanh ^{-1}(c x)+b c d^3 \log (x)-\frac{1}{2} b c d^3 \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

3*a*c^2*d^3*x + (b*c^2*d^3*x)/2 - (b*c*d^3*ArcTanh[c*x])/2 + 3*b*c^2*d^3*x*ArcTanh[c*x] - (d^3*(a + b*ArcTanh[
c*x]))/x + (c^3*d^3*x^2*(a + b*ArcTanh[c*x]))/2 + 3*a*c*d^3*Log[x] + b*c*d^3*Log[x] + b*c*d^3*Log[1 - c^2*x^2]
- (3*b*c*d^3*PolyLog[2, -(c*x)])/2 + (3*b*c*d^3*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 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 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 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 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]

Rule 321

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

Rubi steps

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

Mathematica [A]  time = 0.1409, size = 149, normalized size = 0.99 \[ \frac{d^3 \left (-6 b c x \text{PolyLog}(2,-c x)+6 b c x \text{PolyLog}(2,c x)+2 a c^3 x^3+12 a c^2 x^2+12 a c x \log (x)-4 a+2 b c^2 x^2+4 b c x \log \left (1-c^2 x^2\right )+2 b c^3 x^3 \tanh ^{-1}(c x)+12 b c^2 x^2 \tanh ^{-1}(c x)+4 b c x \log (c x)+b c x \log (1-c x)-b c x \log (c x+1)-4 b \tanh ^{-1}(c x)\right )}{4 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*(-4*a + 12*a*c^2*x^2 + 2*b*c^2*x^2 + 2*a*c^3*x^3 - 4*b*ArcTanh[c*x] + 12*b*c^2*x^2*ArcTanh[c*x] + 2*b*c^3
*x^3*ArcTanh[c*x] + 12*a*c*x*Log[x] + 4*b*c*x*Log[c*x] + b*c*x*Log[1 - c*x] - b*c*x*Log[1 + c*x] + 4*b*c*x*Log
[1 - c^2*x^2] - 6*b*c*x*PolyLog[2, -(c*x)] + 6*b*c*x*PolyLog[2, c*x]))/(4*x)

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 189, normalized size = 1.3 \begin{align*}{\frac{{d}^{3}a{c}^{3}{x}^{2}}{2}}+3\,a{c}^{2}{d}^{3}x-{\frac{{d}^{3}a}{x}}+3\,c{d}^{3}a\ln \left ( cx \right ) +{\frac{{d}^{3}b{\it Artanh} \left ( cx \right ){c}^{3}{x}^{2}}{2}}+3\,b{c}^{2}{d}^{3}x{\it Artanh} \left ( cx \right ) -{\frac{{d}^{3}b{\it Artanh} \left ( cx \right ) }{x}}+3\,c{d}^{3}b{\it Artanh} \left ( cx \right ) \ln \left ( cx \right ) -{\frac{3\,c{d}^{3}b{\it dilog} \left ( cx \right ) }{2}}-{\frac{3\,c{d}^{3}b{\it dilog} \left ( cx+1 \right ) }{2}}-{\frac{3\,c{d}^{3}b\ln \left ( cx \right ) \ln \left ( cx+1 \right ) }{2}}+{\frac{b{c}^{2}{d}^{3}x}{2}}+{\frac{5\,c{d}^{3}b\ln \left ( cx-1 \right ) }{4}}+c{d}^{3}b\ln \left ( cx \right ) +{\frac{3\,c{d}^{3}b\ln \left ( cx+1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.44768, size = 309, normalized size = 2.06 \begin{align*} \frac{1}{4} \, b c^{3} d^{3} x^{2} \log \left (c x + 1\right ) - \frac{1}{4} \, b c^{3} d^{3} x^{2} \log \left (-c x + 1\right ) + \frac{1}{2} \, a c^{3} d^{3} x^{2} + 3 \, a c^{2} d^{3} x + \frac{1}{2} \, b c^{2} d^{3} x + \frac{3}{2} \,{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b c d^{3} - \frac{3}{2} \,{\left (\log \left (c x\right ) \log \left (-c x + 1\right ) +{\rm Li}_2\left (-c x + 1\right )\right )} b c d^{3} + \frac{3}{2} \,{\left (\log \left (c x + 1\right ) \log \left (-c x\right ) +{\rm Li}_2\left (c x + 1\right )\right )} b c d^{3} - \frac{1}{4} \, b c d^{3} \log \left (c x + 1\right ) + \frac{1}{4} \, b c d^{3} \log \left (c x - 1\right ) + 3 \, a c d^{3} \log \left (x\right ) - \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^{3} - \frac{a d^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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