3.39 \(\int \frac{(d+c d x)^4 (a+b \tanh ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=209 \[ -\frac{1}{2} b c^4 d^4 \text{PolyLog}(2,-c x)+\frac{1}{2} b c^4 d^4 \text{PolyLog}(2,c x)-\frac{3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}+a c^4 d^4 \log (x)-\frac{2 b c^2 d^4}{3 x^2}-\frac{8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac{13 b c^3 d^4}{4 x}+\frac{16}{3} b c^4 d^4 \log (x)+\frac{13}{4} b c^4 d^4 \tanh ^{-1}(c x)-\frac{b c d^4}{12 x^3} \]

[Out]

-(b*c*d^4)/(12*x^3) - (2*b*c^2*d^4)/(3*x^2) - (13*b*c^3*d^4)/(4*x) + (13*b*c^4*d^4*ArcTanh[c*x])/4 - (d^4*(a +
b*ArcTanh[c*x]))/(4*x^4) - (4*c*d^4*(a + b*ArcTanh[c*x]))/(3*x^3) - (3*c^2*d^4*(a + b*ArcTanh[c*x]))/x^2 - (4
*c^3*d^4*(a + b*ArcTanh[c*x]))/x + a*c^4*d^4*Log[x] + (16*b*c^4*d^4*Log[x])/3 - (8*b*c^4*d^4*Log[1 - c^2*x^2])
/3 - (b*c^4*d^4*PolyLog[2, -(c*x)])/2 + (b*c^4*d^4*PolyLog[2, c*x])/2

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Rubi [A]  time = 0.229174, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5940, 5916, 325, 206, 266, 44, 36, 29, 31, 5912} \[ -\frac{1}{2} b c^4 d^4 \text{PolyLog}(2,-c x)+\frac{1}{2} b c^4 d^4 \text{PolyLog}(2,c x)-\frac{3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}+a c^4 d^4 \log (x)-\frac{2 b c^2 d^4}{3 x^2}-\frac{8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac{13 b c^3 d^4}{4 x}+\frac{16}{3} b c^4 d^4 \log (x)+\frac{13}{4} b c^4 d^4 \tanh ^{-1}(c x)-\frac{b c d^4}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

-(b*c*d^4)/(12*x^3) - (2*b*c^2*d^4)/(3*x^2) - (13*b*c^3*d^4)/(4*x) + (13*b*c^4*d^4*ArcTanh[c*x])/4 - (d^4*(a +
b*ArcTanh[c*x]))/(4*x^4) - (4*c*d^4*(a + b*ArcTanh[c*x]))/(3*x^3) - (3*c^2*d^4*(a + b*ArcTanh[c*x]))/x^2 - (4
*c^3*d^4*(a + b*ArcTanh[c*x]))/x + a*c^4*d^4*Log[x] + (16*b*c^4*d^4*Log[x])/3 - (8*b*c^4*d^4*Log[1 - c^2*x^2])
/3 - (b*c^4*d^4*PolyLog[2, -(c*x)])/2 + (b*c^4*d^4*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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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)^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^5} \, dx &=\int \left (\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^5}+\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^4}+\frac{6 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^3}+\frac{4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+\frac{c^4 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^4 \int \frac{a+b \tanh ^{-1}(c x)}{x^5} \, dx+\left (4 c d^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^4} \, dx+\left (6 c^2 d^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx+\left (4 c^3 d^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (c^4 d^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x} \, dx\\ &=-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)-\frac{1}{2} b c^4 d^4 \text{Li}_2(-c x)+\frac{1}{2} b c^4 d^4 \text{Li}_2(c x)+\frac{1}{4} \left (b c d^4\right ) \int \frac{1}{x^4 \left (1-c^2 x^2\right )} \, dx+\frac{1}{3} \left (4 b c^2 d^4\right ) \int \frac{1}{x^3 \left (1-c^2 x^2\right )} \, dx+\left (3 b c^3 d^4\right ) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (4 b c^4 d^4\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b c d^4}{12 x^3}-\frac{3 b c^3 d^4}{x}-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)-\frac{1}{2} b c^4 d^4 \text{Li}_2(-c x)+\frac{1}{2} b c^4 d^4 \text{Li}_2(c x)+\frac{1}{3} \left (2 b c^2 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac{1}{4} \left (b c^3 d^4\right ) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (2 b c^4 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )+\left (3 b c^5 d^4\right ) \int \frac{1}{1-c^2 x^2} \, dx\\ &=-\frac{b c d^4}{12 x^3}-\frac{13 b c^3 d^4}{4 x}+3 b c^4 d^4 \tanh ^{-1}(c x)-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)-\frac{1}{2} b c^4 d^4 \text{Li}_2(-c x)+\frac{1}{2} b c^4 d^4 \text{Li}_2(c x)+\frac{1}{3} \left (2 b c^2 d^4\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{c^2}{x}-\frac{c^4}{-1+c^2 x}\right ) \, dx,x,x^2\right )+\left (2 b c^4 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (b c^5 d^4\right ) \int \frac{1}{1-c^2 x^2} \, dx+\left (2 b c^6 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b c d^4}{12 x^3}-\frac{2 b c^2 d^4}{3 x^2}-\frac{13 b c^3 d^4}{4 x}+\frac{13}{4} b c^4 d^4 \tanh ^{-1}(c x)-\frac{d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)+\frac{16}{3} b c^4 d^4 \log (x)-\frac{8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac{1}{2} b c^4 d^4 \text{Li}_2(-c x)+\frac{1}{2} b c^4 d^4 \text{Li}_2(c x)\\ \end{align*}

Mathematica [A]  time = 0.170063, size = 206, normalized size = 0.99 \[ \frac{d^4 \left (-12 b c^4 x^4 \text{PolyLog}(2,-c x)+12 b c^4 x^4 \text{PolyLog}(2,c x)-96 a c^3 x^3-72 a c^2 x^2+24 a c^4 x^4 \log (x)-32 a c x-6 a-78 b c^3 x^3-16 b c^2 x^2+128 b c^4 x^4 \log (c x)-39 b c^4 x^4 \log (1-c x)+39 b c^4 x^4 \log (c x+1)-64 b c^4 x^4 \log \left (1-c^2 x^2\right )-96 b c^3 x^3 \tanh ^{-1}(c x)-72 b c^2 x^2 \tanh ^{-1}(c x)-2 b c x-32 b c x \tanh ^{-1}(c x)-6 b \tanh ^{-1}(c x)\right )}{24 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

(d^4*(-6*a - 32*a*c*x - 2*b*c*x - 72*a*c^2*x^2 - 16*b*c^2*x^2 - 96*a*c^3*x^3 - 78*b*c^3*x^3 - 6*b*ArcTanh[c*x]
- 32*b*c*x*ArcTanh[c*x] - 72*b*c^2*x^2*ArcTanh[c*x] - 96*b*c^3*x^3*ArcTanh[c*x] + 24*a*c^4*x^4*Log[x] + 128*b
*c^4*x^4*Log[c*x] - 39*b*c^4*x^4*Log[1 - c*x] + 39*b*c^4*x^4*Log[1 + c*x] - 64*b*c^4*x^4*Log[1 - c^2*x^2] - 12
*b*c^4*x^4*PolyLog[2, -(c*x)] + 12*b*c^4*x^4*PolyLog[2, c*x]))/(24*x^4)

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Maple [A]  time = 0.049, size = 256, normalized size = 1.2 \begin{align*} -{\frac{{d}^{4}a}{4\,{x}^{4}}}-4\,{\frac{{c}^{3}{d}^{4}a}{x}}+{c}^{4}{d}^{4}a\ln \left ( cx \right ) -3\,{\frac{{c}^{2}{d}^{4}a}{{x}^{2}}}-{\frac{4\,c{d}^{4}a}{3\,{x}^{3}}}-{\frac{{d}^{4}b{\it Artanh} \left ( cx \right ) }{4\,{x}^{4}}}-4\,{\frac{{c}^{3}{d}^{4}b{\it Artanh} \left ( cx \right ) }{x}}+{c}^{4}{d}^{4}b{\it Artanh} \left ( cx \right ) \ln \left ( cx \right ) -3\,{\frac{{c}^{2}{d}^{4}b{\it Artanh} \left ( cx \right ) }{{x}^{2}}}-{\frac{4\,c{d}^{4}b{\it Artanh} \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{103\,{c}^{4}{d}^{4}b\ln \left ( cx-1 \right ) }{24}}-{\frac{c{d}^{4}b}{12\,{x}^{3}}}-{\frac{2\,{c}^{2}{d}^{4}b}{3\,{x}^{2}}}-{\frac{13\,{c}^{3}{d}^{4}b}{4\,x}}+{\frac{16\,{c}^{4}{d}^{4}b\ln \left ( cx \right ) }{3}}-{\frac{25\,{c}^{4}{d}^{4}b\ln \left ( cx+1 \right ) }{24}}-{\frac{{c}^{4}{d}^{4}b{\it dilog} \left ( cx \right ) }{2}}-{\frac{{c}^{4}{d}^{4}b{\it dilog} \left ( cx+1 \right ) }{2}}-{\frac{{c}^{4}{d}^{4}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)^4*(a+b*arctanh(c*x))/x^5,x)

[Out]

-1/4*d^4*a/x^4-4*c^3*d^4*a/x+c^4*d^4*a*ln(c*x)-3*c^2*d^4*a/x^2-4/3*c*d^4*a/x^3-1/4*d^4*b*arctanh(c*x)/x^4-4*c^
3*d^4*b*arctanh(c*x)/x+c^4*d^4*b*arctanh(c*x)*ln(c*x)-3*c^2*d^4*b*arctanh(c*x)/x^2-4/3*c*d^4*b*arctanh(c*x)/x^
3-103/24*c^4*d^4*b*ln(c*x-1)-1/12*b*c*d^4/x^3-2/3*b*c^2*d^4/x^2-13/4*b*c^3*d^4/x+16/3*c^4*d^4*b*ln(c*x)-25/24*
c^4*d^4*b*ln(c*x+1)-1/2*c^4*d^4*b*dilog(c*x)-1/2*c^4*d^4*b*dilog(c*x+1)-1/2*c^4*d^4*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^{4} d^{4} \int \frac{\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{x}\,{d x} + a c^{4} 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^{3} d^{4} + \frac{3}{2} \,{\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 c^{2} d^{4} - \frac{2}{3} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac{1}{x^{2}}\right )} c + \frac{2 \, \operatorname{artanh}\left (c x\right )}{x^{3}}\right )} b c d^{4} - \frac{4 \, a c^{3} d^{4}}{x} + \frac{1}{24} \,{\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac{2 \,{\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac{6 \, \operatorname{artanh}\left (c x\right )}{x^{4}}\right )} b d^{4} - \frac{3 \, a c^{2} d^{4}}{x^{2}} - \frac{4 \, a c d^{4}}{3 \, x^{3}} - \frac{a d^{4}}{4 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

1/2*b*c^4*d^4*integrate((log(c*x + 1) - log(-c*x + 1))/x, x) + a*c^4*d^4*log(x) - 2*(c*(log(c^2*x^2 - 1) - log
(x^2)) + 2*arctanh(c*x)/x)*b*c^3*d^4 + 3/2*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*
c^2*d^4 - 2/3*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3)*b*c*d^4 - 4*a*c^3*d^4/x +
1/24*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*b*d^4 - 3*a*c
^2*d^4/x^2 - 4/3*a*c*d^4/x^3 - 1/4*a*d^4/x^4

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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^{5}}, 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^5,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^5, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{4} \left (\int \frac{a}{x^{5}}\, dx + \int \frac{4 a c}{x^{4}}\, dx + \int \frac{6 a c^{2}}{x^{3}}\, dx + \int \frac{4 a c^{3}}{x^{2}}\, dx + \int \frac{a c^{4}}{x}\, dx + \int \frac{b \operatorname{atanh}{\left (c x \right )}}{x^{5}}\, dx + \int \frac{4 b c \operatorname{atanh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac{6 b c^{2} \operatorname{atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{4 b c^{3} \operatorname{atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{b c^{4} \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)**4*(a+b*atanh(c*x))/x**5,x)

[Out]

d**4*(Integral(a/x**5, x) + Integral(4*a*c/x**4, x) + Integral(6*a*c**2/x**3, x) + Integral(4*a*c**3/x**2, x)
+ Integral(a*c**4/x, x) + Integral(b*atanh(c*x)/x**5, x) + Integral(4*b*c*atanh(c*x)/x**4, x) + Integral(6*b*c
**2*atanh(c*x)/x**3, x) + Integral(4*b*c**3*atanh(c*x)/x**2, x) + Integral(b*c**4*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 )}^{4}{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}}{x^{5}}\,{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^5,x, algorithm="giac")

[Out]

integrate((c*d*x + d)^4*(b*arctanh(c*x) + a)/x^5, x)