3.40 \(\int \frac {(d+c d x)^4 (a+b \tanh ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=109 \[ -\frac {d^4 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}+\frac {16}{5} b c^5 d^4 \log (x)-\frac {16}{5} b c^5 d^4 \log (1-c x)-\frac {3 b c^4 d^4}{x}-\frac {11 b c^3 d^4}{10 x^2}-\frac {b c^2 d^4}{3 x^3}-\frac {b c d^4}{20 x^4} \]

[Out]

-1/20*b*c*d^4/x^4-1/3*b*c^2*d^4/x^3-11/10*b*c^3*d^4/x^2-3*b*c^4*d^4/x-1/5*d^4*(c*x+1)^5*(a+b*arctanh(c*x))/x^5
+16/5*b*c^5*d^4*ln(x)-16/5*b*c^5*d^4*ln(-c*x+1)

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Rubi [A]  time = 0.10, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {37, 5936, 12, 88} \[ -\frac {d^4 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {11 b c^3 d^4}{10 x^2}-\frac {b c^2 d^4}{3 x^3}-\frac {3 b c^4 d^4}{x}+\frac {16}{5} b c^5 d^4 \log (x)-\frac {16}{5} b c^5 d^4 \log (1-c x)-\frac {b c d^4}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^6,x]

[Out]

-(b*c*d^4)/(20*x^4) - (b*c^2*d^4)/(3*x^3) - (11*b*c^3*d^4)/(10*x^2) - (3*b*c^4*d^4)/x - (d^4*(1 + c*x)^5*(a +
b*ArcTanh[c*x]))/(5*x^5) + (16*b*c^5*d^4*Log[x])/5 - (16*b*c^5*d^4*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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)^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^6} \, dx &=-\frac {d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-(b c) \int \frac {(d+c d x)^4}{5 x^5 (-1+c x)} \, dx\\ &=-\frac {d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {1}{5} (b c) \int \frac {(d+c d x)^4}{x^5 (-1+c x)} \, dx\\ &=-\frac {d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {1}{5} (b c) \int \left (-\frac {d^4}{x^5}-\frac {5 c d^4}{x^4}-\frac {11 c^2 d^4}{x^3}-\frac {15 c^3 d^4}{x^2}-\frac {16 c^4 d^4}{x}+\frac {16 c^5 d^4}{-1+c x}\right ) \, dx\\ &=-\frac {b c d^4}{20 x^4}-\frac {b c^2 d^4}{3 x^3}-\frac {11 b c^3 d^4}{10 x^2}-\frac {3 b c^4 d^4}{x}-\frac {d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}+\frac {16}{5} b c^5 d^4 \log (x)-\frac {16}{5} b c^5 d^4 \log (1-c x)\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 157, normalized size = 1.44 \[ -\frac {d^4 \left (60 a c^4 x^4+120 a c^3 x^3+120 a c^2 x^2+60 a c x+12 a-192 b c^5 x^5 \log (x)+186 b c^5 x^5 \log (1-c x)+6 b c^5 x^5 \log (c x+1)+180 b c^4 x^4+66 b c^3 x^3+20 b c^2 x^2+12 b \left (5 c^4 x^4+10 c^3 x^3+10 c^2 x^2+5 c x+1\right ) \tanh ^{-1}(c x)+3 b c x\right )}{60 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^6,x]

[Out]

-1/60*(d^4*(12*a + 60*a*c*x + 3*b*c*x + 120*a*c^2*x^2 + 20*b*c^2*x^2 + 120*a*c^3*x^3 + 66*b*c^3*x^3 + 60*a*c^4
*x^4 + 180*b*c^4*x^4 + 12*b*(1 + 5*c*x + 10*c^2*x^2 + 10*c^3*x^3 + 5*c^4*x^4)*ArcTanh[c*x] - 192*b*c^5*x^5*Log
[x] + 186*b*c^5*x^5*Log[1 - c*x] + 6*b*c^5*x^5*Log[1 + c*x]))/x^5

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fricas [A]  time = 0.48, size = 191, normalized size = 1.75 \[ -\frac {6 \, b c^{5} d^{4} x^{5} \log \left (c x + 1\right ) + 186 \, b c^{5} d^{4} x^{5} \log \left (c x - 1\right ) - 192 \, b c^{5} d^{4} x^{5} \log \relax (x) + 60 \, {\left (a + 3 \, b\right )} c^{4} d^{4} x^{4} + 6 \, {\left (20 \, a + 11 \, b\right )} c^{3} d^{4} x^{3} + 20 \, {\left (6 \, a + b\right )} c^{2} d^{4} x^{2} + 3 \, {\left (20 \, a + b\right )} c d^{4} x + 12 \, a d^{4} + 6 \, {\left (5 \, b c^{4} d^{4} x^{4} + 10 \, b c^{3} d^{4} x^{3} + 10 \, b c^{2} d^{4} x^{2} + 5 \, b c d^{4} x + b d^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{60 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/60*(6*b*c^5*d^4*x^5*log(c*x + 1) + 186*b*c^5*d^4*x^5*log(c*x - 1) - 192*b*c^5*d^4*x^5*log(x) + 60*(a + 3*b)
*c^4*d^4*x^4 + 6*(20*a + 11*b)*c^3*d^4*x^3 + 20*(6*a + b)*c^2*d^4*x^2 + 3*(20*a + b)*c*d^4*x + 12*a*d^4 + 6*(5
*b*c^4*d^4*x^4 + 10*b*c^3*d^4*x^3 + 10*b*c^2*d^4*x^2 + 5*b*c*d^4*x + b*d^4)*log(-(c*x + 1)/(c*x - 1)))/x^5

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giac [B]  time = 0.23, size = 532, normalized size = 4.88 \[ \frac {4}{15} \, {\left (12 \, b c^{4} d^{4} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - 12 \, b c^{4} d^{4} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {12 \, {\left (\frac {5 \, {\left (c x + 1\right )}^{4} b c^{4} d^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} b c^{4} d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2} b c^{4} d^{4}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} b c^{4} d^{4}}{c x - 1} + b c^{4} d^{4}\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 {120 \, {\left (c x + 1\right )}^{4} a c^{4} d^{4}}{{\left (c x - 1\right )}^{4}} + \frac {240 \, {\left (c x + 1\right )}^{3} a c^{4} d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {240 \, {\left (c x + 1\right )}^{2} a c^{4} d^{4}}{{\left (c x - 1\right )}^{2}} + \frac {120 \, {\left (c x + 1\right )} a c^{4} d^{4}}{c x - 1} + 24 \, a c^{4} d^{4} + \frac {48 \, {\left (c x + 1\right )}^{4} b c^{4} d^{4}}{{\left (c x - 1\right )}^{4}} + \frac {156 \, {\left (c x + 1\right )}^{3} b c^{4} d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {196 \, {\left (c x + 1\right )}^{2} b c^{4} d^{4}}{{\left (c x - 1\right )}^{2}} + \frac {113 \, {\left (c x + 1\right )} b c^{4} d^{4}}{c x - 1} + 25 \, b c^{4} d^{4}}{\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)^4*(a+b*arctanh(c*x))/x^6,x, algorithm="giac")

[Out]

4/15*(12*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1) - 1) - 12*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1)) + 12*(5*(c*x + 1)^4*
b*c^4*d^4/(c*x - 1)^4 + 10*(c*x + 1)^3*b*c^4*d^4/(c*x - 1)^3 + 10*(c*x + 1)^2*b*c^4*d^4/(c*x - 1)^2 + 5*(c*x +
1)*b*c^4*d^4/(c*x - 1) + b*c^4*d^4)*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) + (120*(c*x + 1)^
4*a*c^4*d^4/(c*x - 1)^4 + 240*(c*x + 1)^3*a*c^4*d^4/(c*x - 1)^3 + 240*(c*x + 1)^2*a*c^4*d^4/(c*x - 1)^2 + 120*
(c*x + 1)*a*c^4*d^4/(c*x - 1) + 24*a*c^4*d^4 + 48*(c*x + 1)^4*b*c^4*d^4/(c*x - 1)^4 + 156*(c*x + 1)^3*b*c^4*d^
4/(c*x - 1)^3 + 196*(c*x + 1)^2*b*c^4*d^4/(c*x - 1)^2 + 113*(c*x + 1)*b*c^4*d^4/(c*x - 1) + 25*b*c^4*d^4)/((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 [B]  time = 0.04, size = 221, normalized size = 2.03 \[ -\frac {c^{4} d^{4} a}{x}-\frac {2 c^{2} d^{4} a}{x^{3}}-\frac {2 c^{3} d^{4} a}{x^{2}}-\frac {c \,d^{4} a}{x^{4}}-\frac {d^{4} a}{5 x^{5}}-\frac {c^{4} d^{4} b \arctanh \left (c x \right )}{x}-\frac {2 c^{2} d^{4} b \arctanh \left (c x \right )}{x^{3}}-\frac {2 c^{3} d^{4} b \arctanh \left (c x \right )}{x^{2}}-\frac {c \,d^{4} b \arctanh \left (c x \right )}{x^{4}}-\frac {d^{4} b \arctanh \left (c x \right )}{5 x^{5}}-\frac {b c \,d^{4}}{20 x^{4}}-\frac {b \,c^{2} d^{4}}{3 x^{3}}-\frac {11 b \,c^{3} d^{4}}{10 x^{2}}-\frac {3 b \,c^{4} d^{4}}{x}+\frac {16 c^{5} d^{4} b \ln \left (c x \right )}{5}-\frac {31 c^{5} d^{4} b \ln \left (c x -1\right )}{10}-\frac {c^{5} d^{4} b \ln \left (c x +1\right )}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^4*(a+b*arctanh(c*x))/x^6,x)

[Out]

-c^4*d^4*a/x-2*c^2*d^4*a/x^3-2*c^3*d^4*a/x^2-c*d^4*a/x^4-1/5*d^4*a/x^5-c^4*d^4*b*arctanh(c*x)/x-2*c^2*d^4*b*ar
ctanh(c*x)/x^3-2*c^3*d^4*b*arctanh(c*x)/x^2-c*d^4*b*arctanh(c*x)/x^4-1/5*d^4*b*arctanh(c*x)/x^5-1/20*b*c*d^4/x
^4-1/3*b*c^2*d^4/x^3-11/10*b*c^3*d^4/x^2-3*b*c^4*d^4/x+16/5*c^5*d^4*b*ln(c*x)-31/10*c^5*d^4*b*ln(c*x-1)-1/10*c
^5*d^4*b*ln(c*x+1)

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maxima [B]  time = 0.33, size = 299, normalized size = 2.74 \[ -\frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} b c^{4} d^{4} + {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b c^{3} d^{4} - {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c^{2} d^{4} - \frac {a c^{4} d^{4}}{x} + \frac {1}{6} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b c d^{4} - \frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} - 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) + \frac {2 \, c^{2} x^{2} + 1}{x^{4}}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x\right )}{x^{5}}\right )} b d^{4} - \frac {2 \, a c^{3} d^{4}}{x^{2}} - \frac {2 \, a c^{2} d^{4}}{x^{3}} - \frac {a c d^{4}}{x^{4}} - \frac {a d^{4}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^6,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.96, size = 179, normalized size = 1.64 \[ \frac {d^4\,\left (180\,b\,c^5\,\mathrm {atanh}\left (c\,x\right )-96\,b\,c^5\,\ln \left (c^2\,x^2-1\right )+192\,b\,c^5\,\ln \relax (x)\right )}{60}-\frac {\frac {d^4\,\left (12\,a+12\,b\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {d^4\,x\,\left (60\,a\,c+3\,b\,c+60\,b\,c\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {d^4\,x^2\,\left (120\,a\,c^2+20\,b\,c^2+120\,b\,c^2\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {d^4\,x^4\,\left (60\,a\,c^4+180\,b\,c^4+60\,b\,c^4\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {d^4\,x^3\,\left (120\,a\,c^3+66\,b\,c^3+120\,b\,c^3\,\mathrm {atanh}\left (c\,x\right )\right )}{60}}{x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x)^4)/x^6,x)

[Out]

(d^4*(180*b*c^5*atanh(c*x) - 96*b*c^5*log(c^2*x^2 - 1) + 192*b*c^5*log(x)))/60 - ((d^4*(12*a + 12*b*atanh(c*x)
))/60 + (d^4*x*(60*a*c + 3*b*c + 60*b*c*atanh(c*x)))/60 + (d^4*x^2*(120*a*c^2 + 20*b*c^2 + 120*b*c^2*atanh(c*x
)))/60 + (d^4*x^4*(60*a*c^4 + 180*b*c^4 + 60*b*c^4*atanh(c*x)))/60 + (d^4*x^3*(120*a*c^3 + 66*b*c^3 + 120*b*c^
3*atanh(c*x)))/60)/x^5

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sympy [A]  time = 2.70, size = 253, normalized size = 2.32 \[ \begin {cases} - \frac {a c^{4} d^{4}}{x} - \frac {2 a c^{3} d^{4}}{x^{2}} - \frac {2 a c^{2} d^{4}}{x^{3}} - \frac {a c d^{4}}{x^{4}} - \frac {a d^{4}}{5 x^{5}} + \frac {16 b c^{5} d^{4} \log {\relax (x )}}{5} - \frac {16 b c^{5} d^{4} \log {\left (x - \frac {1}{c} \right )}}{5} - \frac {b c^{5} d^{4} \operatorname {atanh}{\left (c x \right )}}{5} - \frac {b c^{4} d^{4} \operatorname {atanh}{\left (c x \right )}}{x} - \frac {3 b c^{4} d^{4}}{x} - \frac {2 b c^{3} d^{4} \operatorname {atanh}{\left (c x \right )}}{x^{2}} - \frac {11 b c^{3} d^{4}}{10 x^{2}} - \frac {2 b c^{2} d^{4} \operatorname {atanh}{\left (c x \right )}}{x^{3}} - \frac {b c^{2} d^{4}}{3 x^{3}} - \frac {b c d^{4} \operatorname {atanh}{\left (c x \right )}}{x^{4}} - \frac {b c d^{4}}{20 x^{4}} - \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a d^{4}}{5 x^{5}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**4*(a+b*atanh(c*x))/x**6,x)

[Out]

Piecewise((-a*c**4*d**4/x - 2*a*c**3*d**4/x**2 - 2*a*c**2*d**4/x**3 - a*c*d**4/x**4 - a*d**4/(5*x**5) + 16*b*c
**5*d**4*log(x)/5 - 16*b*c**5*d**4*log(x - 1/c)/5 - b*c**5*d**4*atanh(c*x)/5 - b*c**4*d**4*atanh(c*x)/x - 3*b*
c**4*d**4/x - 2*b*c**3*d**4*atanh(c*x)/x**2 - 11*b*c**3*d**4/(10*x**2) - 2*b*c**2*d**4*atanh(c*x)/x**3 - b*c**
2*d**4/(3*x**3) - b*c*d**4*atanh(c*x)/x**4 - b*c*d**4/(20*x**4) - b*d**4*atanh(c*x)/(5*x**5), Ne(c, 0)), (-a*d
**4/(5*x**5), True))

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