3.29 \(\int \frac {(d+c d x)^3 (a+b \tanh ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=137 \[ -\frac {d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}+\frac {c d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 x^4}+\frac {6}{5} b c^5 d^3 \log (x)-\frac {6}{5} b c^5 d^3 \log (1-c x)-\frac {5 b c^4 d^3}{4 x}-\frac {3 b c^3 d^3}{5 x^2}-\frac {b c^2 d^3}{4 x^3}-\frac {b c d^3}{20 x^4} \]

[Out]

-1/20*b*c*d^3/x^4-1/4*b*c^2*d^3/x^3-3/5*b*c^3*d^3/x^2-5/4*b*c^4*d^3/x-1/5*d^3*(c*x+1)^4*(a+b*arctanh(c*x))/x^5
+1/20*c*d^3*(c*x+1)^4*(a+b*arctanh(c*x))/x^4+6/5*b*c^5*d^3*ln(x)-6/5*b*c^5*d^3*ln(-c*x+1)

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Rubi [A]  time = 0.12, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {45, 37, 5936, 12, 148} \[ \frac {c d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 x^4}-\frac {d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {3 b c^3 d^3}{5 x^2}-\frac {b c^2 d^3}{4 x^3}-\frac {5 b c^4 d^3}{4 x}+\frac {6}{5} b c^5 d^3 \log (x)-\frac {6}{5} b c^5 d^3 \log (1-c x)-\frac {b c d^3}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^3*(a + b*ArcTanh[c*x]))/x^6,x]

[Out]

-(b*c*d^3)/(20*x^4) - (b*c^2*d^3)/(4*x^3) - (3*b*c^3*d^3)/(5*x^2) - (5*b*c^4*d^3)/(4*x) - (d^3*(1 + c*x)^4*(a
+ b*ArcTanh[c*x]))/(5*x^5) + (c*d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(20*x^4) + (6*b*c^5*d^3*Log[x])/5 - (6*b
*c^5*d^3*Log[1 - c*x])/5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 148

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g
, h, m}, x] && (IntegersQ[m, n, p] || (IGtQ[n, 0] && IGtQ[p, 0]))

Rule 5936

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[a + b*ArcTanh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 - c^2*
x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && IntegerQ[2*m] && ((IGtQ[m, 0] && IGtQ[q,
0]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0]))

Rubi steps

\begin {align*} \int \frac {(d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^6} \, dx &=-\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}+\frac {c d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 x^4}-(b c) \int \frac {(-4+c x) (d+c d x)^3}{20 x^5 (1-c x)} \, dx\\ &=-\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}+\frac {c d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 x^4}-\frac {1}{20} (b c) \int \frac {(-4+c x) (d+c d x)^3}{x^5 (1-c x)} \, dx\\ &=-\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}+\frac {c d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 x^4}-\frac {1}{20} (b c) \int \left (-\frac {4 d^3}{x^5}-\frac {15 c d^3}{x^4}-\frac {24 c^2 d^3}{x^3}-\frac {25 c^3 d^3}{x^2}-\frac {24 c^4 d^3}{x}+\frac {24 c^5 d^3}{-1+c x}\right ) \, dx\\ &=-\frac {b c d^3}{20 x^4}-\frac {b c^2 d^3}{4 x^3}-\frac {3 b c^3 d^3}{5 x^2}-\frac {5 b c^4 d^3}{4 x}-\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}+\frac {c d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 x^4}+\frac {6}{5} b c^5 d^3 \log (x)-\frac {6}{5} b c^5 d^3 \log (1-c x)\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 140, normalized size = 1.02 \[ -\frac {d^3 \left (20 a c^3 x^3+40 a c^2 x^2+30 a c x+8 a-48 b c^5 x^5 \log (x)+49 b c^5 x^5 \log (1-c x)-b c^5 x^5 \log (c x+1)+50 b c^4 x^4+24 b c^3 x^3+10 b c^2 x^2+2 b \left (10 c^3 x^3+20 c^2 x^2+15 c x+4\right ) \tanh ^{-1}(c x)+2 b c x\right )}{40 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^3*(a + b*ArcTanh[c*x]))/x^6,x]

[Out]

-1/40*(d^3*(8*a + 30*a*c*x + 2*b*c*x + 40*a*c^2*x^2 + 10*b*c^2*x^2 + 20*a*c^3*x^3 + 24*b*c^3*x^3 + 50*b*c^4*x^
4 + 2*b*(4 + 15*c*x + 20*c^2*x^2 + 10*c^3*x^3)*ArcTanh[c*x] - 48*b*c^5*x^5*Log[x] + 49*b*c^5*x^5*Log[1 - c*x]
- b*c^5*x^5*Log[1 + c*x]))/x^5

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fricas [A]  time = 0.45, size = 175, normalized size = 1.28 \[ \frac {b c^{5} d^{3} x^{5} \log \left (c x + 1\right ) - 49 \, b c^{5} d^{3} x^{5} \log \left (c x - 1\right ) + 48 \, b c^{5} d^{3} x^{5} \log \relax (x) - 50 \, b c^{4} d^{3} x^{4} - 4 \, {\left (5 \, a + 6 \, b\right )} c^{3} d^{3} x^{3} - 10 \, {\left (4 \, a + b\right )} c^{2} d^{3} x^{2} - 2 \, {\left (15 \, a + b\right )} c d^{3} x - 8 \, a d^{3} - {\left (10 \, b c^{3} d^{3} x^{3} + 20 \, b c^{2} d^{3} x^{2} + 15 \, b c d^{3} x + 4 \, b d^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{40 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^6,x, algorithm="fricas")

[Out]

1/40*(b*c^5*d^3*x^5*log(c*x + 1) - 49*b*c^5*d^3*x^5*log(c*x - 1) + 48*b*c^5*d^3*x^5*log(x) - 50*b*c^4*d^3*x^4
- 4*(5*a + 6*b)*c^3*d^3*x^3 - 10*(4*a + b)*c^2*d^3*x^2 - 2*(15*a + b)*c*d^3*x - 8*a*d^3 - (10*b*c^3*d^3*x^3 +
20*b*c^2*d^3*x^2 + 15*b*c*d^3*x + 4*b*d^3)*log(-(c*x + 1)/(c*x - 1)))/x^5

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giac [B]  time = 0.20, size = 533, normalized size = 3.89 \[ \frac {1}{5} \, {\left (6 \, b c^{4} d^{3} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - 6 \, b c^{4} d^{3} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {2 \, {\left (\frac {20 \, {\left (c x + 1\right )}^{4} b c^{4} d^{3}}{{\left (c x - 1\right )}^{4}} + \frac {30 \, {\left (c x + 1\right )}^{3} b c^{4} d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {30 \, {\left (c x + 1\right )}^{2} b c^{4} d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {15 \, {\left (c x + 1\right )} b c^{4} d^{3}}{c x - 1} + 3 \, b c^{4} d^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5}}{{\left (c x - 1\right )}^{5}} + \frac {5 \, {\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {80 \, {\left (c x + 1\right )}^{4} a c^{4} d^{3}}{{\left (c x - 1\right )}^{4}} + \frac {120 \, {\left (c x + 1\right )}^{3} a c^{4} d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {120 \, {\left (c x + 1\right )}^{2} a c^{4} d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {60 \, {\left (c x + 1\right )} a c^{4} d^{3}}{c x - 1} + 12 \, a c^{4} d^{3} + \frac {34 \, {\left (c x + 1\right )}^{4} b c^{4} d^{3}}{{\left (c x - 1\right )}^{4}} + \frac {103 \, {\left (c x + 1\right )}^{3} b c^{4} d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {123 \, {\left (c x + 1\right )}^{2} b c^{4} d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {69 \, {\left (c x + 1\right )} b c^{4} d^{3}}{c x - 1} + 15 \, b c^{4} d^{3}}{\frac {{\left (c x + 1\right )}^{5}}{{\left (c x - 1\right )}^{5}} + \frac {5 \, {\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^6,x, algorithm="giac")

[Out]

1/5*(6*b*c^4*d^3*log(-(c*x + 1)/(c*x - 1) - 1) - 6*b*c^4*d^3*log(-(c*x + 1)/(c*x - 1)) + 2*(20*(c*x + 1)^4*b*c
^4*d^3/(c*x - 1)^4 + 30*(c*x + 1)^3*b*c^4*d^3/(c*x - 1)^3 + 30*(c*x + 1)^2*b*c^4*d^3/(c*x - 1)^2 + 15*(c*x + 1
)*b*c^4*d^3/(c*x - 1) + 3*b*c^4*d^3)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5/(c*x - 1)^5 + 5*(c*x + 1)^4/(c*x -
1)^4 + 10*(c*x + 1)^3/(c*x - 1)^3 + 10*(c*x + 1)^2/(c*x - 1)^2 + 5*(c*x + 1)/(c*x - 1) + 1) + (80*(c*x + 1)^4
*a*c^4*d^3/(c*x - 1)^4 + 120*(c*x + 1)^3*a*c^4*d^3/(c*x - 1)^3 + 120*(c*x + 1)^2*a*c^4*d^3/(c*x - 1)^2 + 60*(c
*x + 1)*a*c^4*d^3/(c*x - 1) + 12*a*c^4*d^3 + 34*(c*x + 1)^4*b*c^4*d^3/(c*x - 1)^4 + 103*(c*x + 1)^3*b*c^4*d^3/
(c*x - 1)^3 + 123*(c*x + 1)^2*b*c^4*d^3/(c*x - 1)^2 + 69*(c*x + 1)*b*c^4*d^3/(c*x - 1) + 15*b*c^4*d^3)/((c*x +
1)^5/(c*x - 1)^5 + 5*(c*x + 1)^4/(c*x - 1)^4 + 10*(c*x + 1)^3/(c*x - 1)^3 + 10*(c*x + 1)^2/(c*x - 1)^2 + 5*(c
*x + 1)/(c*x - 1) + 1))*c

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maple [A]  time = 0.04, size = 193, normalized size = 1.41 \[ -\frac {c^{2} d^{3} a}{x^{3}}-\frac {c^{3} d^{3} a}{2 x^{2}}-\frac {3 c \,d^{3} a}{4 x^{4}}-\frac {d^{3} a}{5 x^{5}}-\frac {c^{2} d^{3} b \arctanh \left (c x \right )}{x^{3}}-\frac {c^{3} d^{3} b \arctanh \left (c x \right )}{2 x^{2}}-\frac {3 c \,d^{3} b \arctanh \left (c x \right )}{4 x^{4}}-\frac {d^{3} b \arctanh \left (c x \right )}{5 x^{5}}-\frac {b c \,d^{3}}{20 x^{4}}-\frac {b \,c^{2} d^{3}}{4 x^{3}}-\frac {3 b \,c^{3} d^{3}}{5 x^{2}}-\frac {5 b \,c^{4} d^{3}}{4 x}+\frac {6 c^{5} d^{3} b \ln \left (c x \right )}{5}-\frac {49 c^{5} d^{3} b \ln \left (c x -1\right )}{40}+\frac {c^{5} d^{3} b \ln \left (c x +1\right )}{40} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^3*(a+b*arctanh(c*x))/x^6,x)

[Out]

-c^2*d^3*a/x^3-1/2*c^3*d^3*a/x^2-3/4*c*d^3*a/x^4-1/5*d^3*a/x^5-c^2*d^3*b*arctanh(c*x)/x^3-1/2*c^3*d^3*b*arctan
h(c*x)/x^2-3/4*c*d^3*b*arctanh(c*x)/x^4-1/5*d^3*b*arctanh(c*x)/x^5-1/20*b*c*d^3/x^4-1/4*b*c^2*d^3/x^3-3/5*b*c^
3*d^3/x^2-5/4*b*c^4*d^3/x+6/5*c^5*d^3*b*ln(c*x)-49/40*c^5*d^3*b*ln(c*x-1)+1/40*c^5*d^3*b*ln(c*x+1)

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maxima [B]  time = 0.33, size = 250, normalized size = 1.82 \[ \frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b c^{3} d^{3} - \frac {1}{2} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c^{2} d^{3} + \frac {1}{8} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b c d^{3} - \frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} - 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) + \frac {2 \, c^{2} x^{2} + 1}{x^{4}}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x\right )}{x^{5}}\right )} b d^{3} - \frac {a c^{3} d^{3}}{2 \, x^{2}} - \frac {a c^{2} d^{3}}{x^{3}} - \frac {3 \, a c d^{3}}{4 \, x^{4}} - \frac {a d^{3}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^6,x, algorithm="maxima")

[Out]

1/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*c^3*d^3 - 1/2*((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^2*d^3 + 1/8*((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*c*d^3 - 1/20*((2*c^4*log(c^2*x^2 - 1) - 2*c^4*log(x^2) + (2*c^2*x
^2 + 1)/x^4)*c + 4*arctanh(c*x)/x^5)*b*d^3 - 1/2*a*c^3*d^3/x^2 - a*c^2*d^3/x^3 - 3/4*a*c*d^3/x^4 - 1/5*a*d^3/x
^5

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mupad [B]  time = 0.98, size = 233, normalized size = 1.70 \[ -\frac {4\,a\,d^3+4\,b\,d^3\,\mathrm {atanh}\left (c\,x\right )+20\,a\,c^2\,d^3\,x^2+10\,a\,c^3\,d^3\,x^3+10\,a\,c^5\,d^3\,x^5+5\,b\,c^2\,d^3\,x^2+12\,b\,c^3\,d^3\,x^3+25\,b\,c^4\,d^3\,x^4+12\,b\,c^5\,d^3\,x^5+15\,a\,c\,d^3\,x+b\,c\,d^3\,x-24\,b\,c^5\,d^3\,x^5\,\ln \relax (x)+20\,b\,c^2\,d^3\,x^2\,\mathrm {atanh}\left (c\,x\right )+10\,b\,c^3\,d^3\,x^3\,\mathrm {atanh}\left (c\,x\right )+12\,b\,c^5\,d^3\,x^5\,\ln \left (c^2\,x^2-1\right )+15\,b\,c\,d^3\,x\,\mathrm {atanh}\left (c\,x\right )-25\,b\,c^4\,d^3\,x^5\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )\,\sqrt {-c^2}}{20\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x)^3)/x^6,x)

[Out]

-(4*a*d^3 + 4*b*d^3*atanh(c*x) + 20*a*c^2*d^3*x^2 + 10*a*c^3*d^3*x^3 + 10*a*c^5*d^3*x^5 + 5*b*c^2*d^3*x^2 + 12
*b*c^3*d^3*x^3 + 25*b*c^4*d^3*x^4 + 12*b*c^5*d^3*x^5 + 15*a*c*d^3*x + b*c*d^3*x - 24*b*c^5*d^3*x^5*log(x) + 20
*b*c^2*d^3*x^2*atanh(c*x) + 10*b*c^3*d^3*x^3*atanh(c*x) + 12*b*c^5*d^3*x^5*log(c^2*x^2 - 1) + 15*b*c*d^3*x*ata
nh(c*x) - 25*b*c^4*d^3*x^5*atan((c^2*x)/(-c^2)^(1/2))*(-c^2)^(1/2))/(20*x^5)

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sympy [A]  time = 2.62, size = 233, normalized size = 1.70 \[ \begin {cases} - \frac {a c^{3} d^{3}}{2 x^{2}} - \frac {a c^{2} d^{3}}{x^{3}} - \frac {3 a c d^{3}}{4 x^{4}} - \frac {a d^{3}}{5 x^{5}} + \frac {6 b c^{5} d^{3} \log {\relax (x )}}{5} - \frac {6 b c^{5} d^{3} \log {\left (x - \frac {1}{c} \right )}}{5} + \frac {b c^{5} d^{3} \operatorname {atanh}{\left (c x \right )}}{20} - \frac {5 b c^{4} d^{3}}{4 x} - \frac {b c^{3} d^{3} \operatorname {atanh}{\left (c x \right )}}{2 x^{2}} - \frac {3 b c^{3} d^{3}}{5 x^{2}} - \frac {b c^{2} d^{3} \operatorname {atanh}{\left (c x \right )}}{x^{3}} - \frac {b c^{2} d^{3}}{4 x^{3}} - \frac {3 b c d^{3} \operatorname {atanh}{\left (c x \right )}}{4 x^{4}} - \frac {b c d^{3}}{20 x^{4}} - \frac {b d^{3} \operatorname {atanh}{\left (c x \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a d^{3}}{5 x^{5}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**3*(a+b*atanh(c*x))/x**6,x)

[Out]

Piecewise((-a*c**3*d**3/(2*x**2) - a*c**2*d**3/x**3 - 3*a*c*d**3/(4*x**4) - a*d**3/(5*x**5) + 6*b*c**5*d**3*lo
g(x)/5 - 6*b*c**5*d**3*log(x - 1/c)/5 + b*c**5*d**3*atanh(c*x)/20 - 5*b*c**4*d**3/(4*x) - b*c**3*d**3*atanh(c*
x)/(2*x**2) - 3*b*c**3*d**3/(5*x**2) - b*c**2*d**3*atanh(c*x)/x**3 - b*c**2*d**3/(4*x**3) - 3*b*c*d**3*atanh(c
*x)/(4*x**4) - b*c*d**3/(20*x**4) - b*d**3*atanh(c*x)/(5*x**5), Ne(c, 0)), (-a*d**3/(5*x**5), True))

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