3.37 \(\int \frac{(d+c d x)^4 (a+b \tanh ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=156 \[ -3 b c^2 d^4 \text{PolyLog}(2,-c x)+3 b c^2 d^4 \text{PolyLog}(2,c x)+\frac{1}{2} c^4 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+4 a c^3 d^4 x+6 a c^2 d^4 \log (x)+\frac{1}{2} b c^3 d^4 x+4 b c^2 d^4 \log (x)+4 b c^3 d^4 x \tanh ^{-1}(c x)-\frac{b c d^4}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.189288, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 12, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5940, 5910, 260, 5916, 325, 206, 266, 36, 29, 31, 5912, 321} \[ -3 b c^2 d^4 \text{PolyLog}(2,-c x)+3 b c^2 d^4 \text{PolyLog}(2,c x)+\frac{1}{2} c^4 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+4 a c^3 d^4 x+6 a c^2 d^4 \log (x)+\frac{1}{2} b c^3 d^4 x+4 b c^2 d^4 \log (x)+4 b c^3 d^4 x \tanh ^{-1}(c x)-\frac{b c d^4}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^3,x]

[Out]

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

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 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]

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]

Rubi steps

\begin{align*} \int \frac{(d+c d x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^3} \, dx &=\int \left (4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^3}+\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+\frac{6 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+c^4 d^4 x \left (a+b \tanh ^{-1}(c x)\right )\right ) \, dx\\ &=d^4 \int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx+\left (4 c d^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (6 c^2 d^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x} \, dx+\left (4 c^3 d^4\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\left (c^4 d^4\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx\\ &=4 a c^3 d^4 x-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{1}{2} c^4 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )+6 a c^2 d^4 \log (x)-3 b c^2 d^4 \text{Li}_2(-c x)+3 b c^2 d^4 \text{Li}_2(c x)+\frac{1}{2} \left (b c d^4\right ) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (4 b c^2 d^4\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx+\left (4 b c^3 d^4\right ) \int \tanh ^{-1}(c x) \, dx-\frac{1}{2} \left (b c^5 d^4\right ) \int \frac{x^2}{1-c^2 x^2} \, dx\\ &=-\frac{b c d^4}{2 x}+4 a c^3 d^4 x+\frac{1}{2} b c^3 d^4 x+4 b c^3 d^4 x \tanh ^{-1}(c x)-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{1}{2} c^4 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )+6 a c^2 d^4 \log (x)-3 b c^2 d^4 \text{Li}_2(-c x)+3 b c^2 d^4 \text{Li}_2(c x)+\left (2 b c^2 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )-\left (4 b c^4 d^4\right ) \int \frac{x}{1-c^2 x^2} \, dx\\ &=-\frac{b c d^4}{2 x}+4 a c^3 d^4 x+\frac{1}{2} b c^3 d^4 x+4 b c^3 d^4 x \tanh ^{-1}(c x)-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{1}{2} c^4 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )+6 a c^2 d^4 \log (x)+2 b c^2 d^4 \log \left (1-c^2 x^2\right )-3 b c^2 d^4 \text{Li}_2(-c x)+3 b c^2 d^4 \text{Li}_2(c x)+\left (2 b c^2 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\left (2 b c^4 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b c d^4}{2 x}+4 a c^3 d^4 x+\frac{1}{2} b c^3 d^4 x+4 b c^3 d^4 x \tanh ^{-1}(c x)-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{1}{2} c^4 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )+6 a c^2 d^4 \log (x)+4 b c^2 d^4 \log (x)-3 b c^2 d^4 \text{Li}_2(-c x)+3 b c^2 d^4 \text{Li}_2(c x)\\ \end{align*}

Mathematica [A]  time = 0.166052, size = 143, normalized size = 0.92 \[ \frac{d^4 \left (-6 b c^2 x^2 \text{PolyLog}(2,-c x)+6 b c^2 x^2 \text{PolyLog}(2,c x)+a c^4 x^4+8 a c^3 x^3+12 a c^2 x^2 \log (x)-8 a c x-a+b c^3 x^3+8 b c^2 x^2 \log (c x)+b c^4 x^4 \tanh ^{-1}(c x)+8 b c^3 x^3 \tanh ^{-1}(c x)-b c x-8 b c x \tanh ^{-1}(c x)-b \tanh ^{-1}(c x)\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^3,x]

[Out]

(d^4*(-a - 8*a*c*x - b*c*x + 8*a*c^3*x^3 + b*c^3*x^3 + a*c^4*x^4 - b*ArcTanh[c*x] - 8*b*c*x*ArcTanh[c*x] + 8*b
*c^3*x^3*ArcTanh[c*x] + b*c^4*x^4*ArcTanh[c*x] + 12*a*c^2*x^2*Log[x] + 8*b*c^2*x^2*Log[c*x] - 6*b*c^2*x^2*Poly
Log[2, -(c*x)] + 6*b*c^2*x^2*PolyLog[2, c*x]))/(2*x^2)

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Maple [A]  time = 0.045, size = 210, normalized size = 1.4 \begin{align*}{\frac{{c}^{4}{d}^{4}a{x}^{2}}{2}}+4\,a{c}^{3}{d}^{4}x-4\,{\frac{ca{d}^{4}}{x}}+6\,{c}^{2}{d}^{4}a\ln \left ( cx \right ) -{\frac{{d}^{4}a}{2\,{x}^{2}}}+{\frac{{c}^{4}{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{2}}{2}}+4\,b{c}^{3}{d}^{4}x{\it Artanh} \left ( cx \right ) -4\,{\frac{{d}^{4}bc{\it Artanh} \left ( cx \right ) }{x}}+6\,{c}^{2}{d}^{4}b{\it Artanh} \left ( cx \right ) \ln \left ( cx \right ) -{\frac{{d}^{4}b{\it Artanh} \left ( cx \right ) }{2\,{x}^{2}}}-3\,{c}^{2}{d}^{4}b{\it dilog} \left ( cx \right ) -3\,{c}^{2}{d}^{4}b{\it dilog} \left ( cx+1 \right ) -3\,{c}^{2}{d}^{4}b\ln \left ( cx \right ) \ln \left ( cx+1 \right ) +{\frac{b{c}^{3}{d}^{4}x}{2}}-{\frac{{d}^{4}bc}{2\,x}}+4\,{c}^{2}{d}^{4}b\ln \left ( cx \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c^4*d^4*a*x^2+4*a*c^3*d^4*x-4*c*d^4*a/x+6*c^2*d^4*a*ln(c*x)-1/2*d^4*a/x^2+1/2*c^4*d^4*b*arctanh(c*x)*x^2+4
*b*c^3*d^4*x*arctanh(c*x)-4*c*d^4*b*arctanh(c*x)/x+6*c^2*d^4*b*arctanh(c*x)*ln(c*x)-1/2*d^4*b*arctanh(c*x)/x^2
-3*c^2*d^4*b*dilog(c*x)-3*c^2*d^4*b*dilog(c*x+1)-3*c^2*d^4*b*ln(c*x)*ln(c*x+1)+1/2*b*c^3*d^4*x-1/2*b*c*d^4/x+4
*c^2*d^4*b*ln(c*x)

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Maxima [B]  time = 1.44967, size = 396, normalized size = 2.54 \begin{align*} \frac{1}{4} \, b c^{4} d^{4} x^{2} \log \left (c x + 1\right ) - \frac{1}{4} \, b c^{4} d^{4} x^{2} \log \left (-c x + 1\right ) + \frac{1}{2} \, a c^{4} d^{4} x^{2} + 4 \, a c^{3} d^{4} x + \frac{1}{2} \, b c^{3} d^{4} x + 2 \,{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b c^{2} d^{4} - 3 \,{\left (\log \left (c x\right ) \log \left (-c x + 1\right ) +{\rm Li}_2\left (-c x + 1\right )\right )} b c^{2} d^{4} + 3 \,{\left (\log \left (c x + 1\right ) \log \left (-c x\right ) +{\rm Li}_2\left (c x + 1\right )\right )} b c^{2} d^{4} - \frac{1}{4} \, b c^{2} d^{4} \log \left (c x + 1\right ) + \frac{1}{4} \, b c^{2} d^{4} \log \left (c x - 1\right ) + 6 \, a c^{2} d^{4} \log \left (x\right ) - 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 c d^{4} + \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^{4} - \frac{4 \, a c d^{4}}{x} - \frac{a d^{4}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^3,x, algorithm="maxima")

[Out]

1/4*b*c^4*d^4*x^2*log(c*x + 1) - 1/4*b*c^4*d^4*x^2*log(-c*x + 1) + 1/2*a*c^4*d^4*x^2 + 4*a*c^3*d^4*x + 1/2*b*c
^3*d^4*x + 2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*c^2*d^4 - 3*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))
*b*c^2*d^4 + 3*(log(c*x + 1)*log(-c*x) + dilog(c*x + 1))*b*c^2*d^4 - 1/4*b*c^2*d^4*log(c*x + 1) + 1/4*b*c^2*d^
4*log(c*x - 1) + 6*a*c^2*d^4*log(x) - 2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*b*c*d^4 + 1/4*((c
*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*d^4 - 4*a*c*d^4/x - 1/2*a*d^4/x^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a c^{4} d^{4} x^{4} + 4 \, a c^{3} d^{4} x^{3} + 6 \, a c^{2} d^{4} x^{2} + 4 \, a c d^{4} x + a d^{4} +{\left (b c^{4} d^{4} x^{4} + 4 \, b c^{3} d^{4} x^{3} + 6 \, b c^{2} d^{4} x^{2} + 4 \, b c d^{4} x + b d^{4}\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)^4*(a+b*arctanh(c*x))/x^3,x, algorithm="fricas")

[Out]

integral((a*c^4*d^4*x^4 + 4*a*c^3*d^4*x^3 + 6*a*c^2*d^4*x^2 + 4*a*c*d^4*x + a*d^4 + (b*c^4*d^4*x^4 + 4*b*c^3*d
^4*x^3 + 6*b*c^2*d^4*x^2 + 4*b*c*d^4*x + b*d^4)*arctanh(c*x))/x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{4} \left (\int 4 a c^{3}\, dx + \int \frac{a}{x^{3}}\, dx + \int \frac{4 a c}{x^{2}}\, dx + \int \frac{6 a c^{2}}{x}\, dx + \int a c^{4} x\, dx + \int 4 b c^{3} \operatorname{atanh}{\left (c x \right )}\, dx + \int \frac{b \operatorname{atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{4 b c \operatorname{atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{6 b c^{2} \operatorname{atanh}{\left (c x \right )}}{x}\, dx + \int b c^{4} 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)**4*(a+b*atanh(c*x))/x**3,x)

[Out]

d**4*(Integral(4*a*c**3, x) + Integral(a/x**3, x) + Integral(4*a*c/x**2, x) + Integral(6*a*c**2/x, x) + Integr
al(a*c**4*x, x) + Integral(4*b*c**3*atanh(c*x), x) + Integral(b*atanh(c*x)/x**3, x) + Integral(4*b*c*atanh(c*x
)/x**2, x) + Integral(6*b*c**2*atanh(c*x)/x, x) + Integral(b*c**4*x*atanh(c*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 )}^{4}{\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)^4*(a+b*arctanh(c*x))/x^3,x, algorithm="giac")

[Out]

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