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

Optimal. Leaf size=209 \[ -\frac {4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}+a c^4 d^4 \log (x)-\frac {1}{2} b c^4 d^4 \text {Li}_2(-c x)+\frac {1}{2} b c^4 d^4 \text {Li}_2(c x)+\frac {16}{3} b c^4 d^4 \log (x)+\frac {13}{4} b c^4 d^4 \tanh ^{-1}(c x)-\frac {13 b c^3 d^4}{4 x}-\frac {2 b c^2 d^4}{3 x^2}-\frac {8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac {b c d^4}{12 x^3} \]

[Out]

-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+13/4*b*c^4*d^4*arctanh(c*x)-1/4*d^4*(a+b*arctanh(c*x))/x^
4-4/3*c*d^4*(a+b*arctanh(c*x))/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*l
n(x)+16/3*b*c^4*d^4*ln(x)-8/3*b*c^4*d^4*ln(-c^2*x^2+1)-1/2*b*c^4*d^4*polylog(2,-c*x)+1/2*b*c^4*d^4*polylog(2,c
*x)

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Rubi [A]  time = 0.23, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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 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 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 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.16, size = 206, normalized size = 0.99 \[ \frac {d^4 \left (24 a c^4 x^4 \log (x)-96 a c^3 x^3-72 a c^2 x^2-32 a c x-6 a-12 b c^4 x^4 \text {Li}_2(-c x)+12 b c^4 x^4 \text {Li}_2(c x)+128 b c^4 x^4 \log (c x)-39 b c^4 x^4 \log (1-c x)+39 b c^4 x^4 \log (c x+1)-78 b c^3 x^3-96 b c^3 x^3 \tanh ^{-1}(c x)-16 b c^2 x^2-72 b c^2 x^2 \tanh ^{-1}(c x)-64 b c^4 x^4 \log \left (1-c^2 x^2\right )-2 b c x-32 b c x \tanh ^{-1}(c x)-6 b \tanh ^{-1}(c x)\right )}{24 x^4} \]

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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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a c^{4} d^{4} x^{4} + 4 \, a c^{3} d^{4} x^{3} + 6 \, a c^{2} d^{4} x^{2} + 4 \, a c d^{4} x + a d^{4} + {\left (b c^{4} d^{4} x^{4} + 4 \, b c^{3} d^{4} x^{3} + 6 \, b c^{2} d^{4} x^{2} + 4 \, b c d^{4} x + b d^{4}\right )} \operatorname {artanh}\left (c x\right )}{x^{5}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d x + d\right )}^{4} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x^{5}}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 256, normalized size = 1.22 \[ c^{4} d^{4} a \ln \left (c x \right )-\frac {4 c^{3} d^{4} a}{x}-\frac {4 c \,d^{4} a}{3 x^{3}}-\frac {3 c^{2} d^{4} a}{x^{2}}-\frac {d^{4} a}{4 x^{4}}+c^{4} d^{4} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {4 c^{3} d^{4} b \arctanh \left (c x \right )}{x}-\frac {4 c \,d^{4} b \arctanh \left (c x \right )}{3 x^{3}}-\frac {3 c^{2} d^{4} b \arctanh \left (c x \right )}{x^{2}}-\frac {d^{4} b \arctanh \left (c x \right )}{4 x^{4}}-\frac {b c \,d^{4}}{12 x^{3}}-\frac {2 b \,c^{2} d^{4}}{3 x^{2}}-\frac {13 b \,c^{3} d^{4}}{4 x}+\frac {16 c^{4} d^{4} b \ln \left (c x \right )}{3}-\frac {103 c^{4} d^{4} b \ln \left (c x -1\right )}{24}-\frac {25 c^{4} d^{4} b \ln \left (c x +1\right )}{24}-\frac {c^{4} d^{4} b \dilog \left (c x \right )}{2}-\frac {c^{4} d^{4} b \dilog \left (c x +1\right )}{2}-\frac {c^{4} d^{4} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2} \]

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]

c^4*d^4*a*ln(c*x)-4*c^3*d^4*a/x-4/3*c*d^4*a/x^3-3*c^2*d^4*a/x^2-1/4*d^4*a/x^4+c^4*d^4*b*arctanh(c*x)*ln(c*x)-4
*c^3*d^4*b*arctanh(c*x)/x-4/3*c*d^4*b*arctanh(c*x)/x^3-3*c^2*d^4*b*arctanh(c*x)/x^2-1/4*d^4*b*arctanh(c*x)/x^4
-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)-103/24*c^4*d^4*b*ln(c*x-1)-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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, b c^{4} d^{4} \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{x}\,{d x} + a c^{4} d^{4} \log \relax (x) - 2 \, {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} b c^{3} d^{4} + \frac {3}{2} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b c^{2} d^{4} - \frac {2}{3} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c d^{4} - \frac {4 \, a c^{3} d^{4}}{x} + \frac {1}{24} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b d^{4} - \frac {3 \, a c^{2} d^{4}}{x^{2}} - \frac {4 \, a c d^{4}}{3 \, x^{3}} - \frac {a d^{4}}{4 \, x^{4}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^4}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x)^4)/x^5,x)

[Out]

int(((a + b*atanh(c*x))*(d + c*d*x)^4)/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{4} \left (\int \frac {a}{x^{5}}\, dx + \int \frac {4 a c}{x^{4}}\, dx + \int \frac {6 a c^{2}}{x^{3}}\, dx + \int \frac {4 a c^{3}}{x^{2}}\, dx + \int \frac {a c^{4}}{x}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x^{5}}\, dx + \int \frac {4 b c \operatorname {atanh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {6 b c^{2} \operatorname {atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {4 b c^{3} \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {b c^{4} \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \]

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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