3.19 \(\int \frac {(d+c d x)^2 (a+b \tanh ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=161 \[ -\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^4}+\frac {8}{15} b c^5 d^2 \log (x)-\frac {31}{60} b c^5 d^2 \log (1-c x)-\frac {1}{60} b c^5 d^2 \log (c x+1)-\frac {b c^4 d^2}{2 x}-\frac {4 b c^3 d^2}{15 x^2}-\frac {b c^2 d^2}{6 x^3}-\frac {b c d^2}{20 x^4} \]

[Out]

-1/20*b*c*d^2/x^4-1/6*b*c^2*d^2/x^3-4/15*b*c^3*d^2/x^2-1/2*b*c^4*d^2/x-1/5*d^2*(a+b*arctanh(c*x))/x^5-1/2*c*d^
2*(a+b*arctanh(c*x))/x^4-1/3*c^2*d^2*(a+b*arctanh(c*x))/x^3+8/15*b*c^5*d^2*ln(x)-31/60*b*c^5*d^2*ln(-c*x+1)-1/
60*b*c^5*d^2*ln(c*x+1)

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Rubi [A]  time = 0.16, antiderivative size = 161, 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 = {43, 5936, 12, 1802} \[ -\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^4}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {4 b c^3 d^2}{15 x^2}-\frac {b c^2 d^2}{6 x^3}-\frac {b c^4 d^2}{2 x}+\frac {8}{15} b c^5 d^2 \log (x)-\frac {31}{60} b c^5 d^2 \log (1-c x)-\frac {1}{60} b c^5 d^2 \log (c x+1)-\frac {b c d^2}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^6,x]

[Out]

-(b*c*d^2)/(20*x^4) - (b*c^2*d^2)/(6*x^3) - (4*b*c^3*d^2)/(15*x^2) - (b*c^4*d^2)/(2*x) - (d^2*(a + b*ArcTanh[c
*x]))/(5*x^5) - (c*d^2*(a + b*ArcTanh[c*x]))/(2*x^4) - (c^2*d^2*(a + b*ArcTanh[c*x]))/(3*x^3) + (8*b*c^5*d^2*L
og[x])/15 - (31*b*c^5*d^2*Log[1 - c*x])/60 - (b*c^5*d^2*Log[1 + c*x])/60

Rule 12

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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)^2 \left (a+b \tanh ^{-1}(c x)\right )}{x^6} \, dx &=-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^4}-\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-(b c) \int \frac {d^2 \left (-6-15 c x-10 c^2 x^2\right )}{30 x^5 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^4}-\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {1}{30} \left (b c d^2\right ) \int \frac {-6-15 c x-10 c^2 x^2}{x^5 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^4}-\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {1}{30} \left (b c d^2\right ) \int \left (-\frac {6}{x^5}-\frac {15 c}{x^4}-\frac {16 c^2}{x^3}-\frac {15 c^3}{x^2}-\frac {16 c^4}{x}+\frac {31 c^5}{2 (-1+c x)}+\frac {c^5}{2 (1+c x)}\right ) \, dx\\ &=-\frac {b c d^2}{20 x^4}-\frac {b c^2 d^2}{6 x^3}-\frac {4 b c^3 d^2}{15 x^2}-\frac {b c^4 d^2}{2 x}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 x^5}-\frac {c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^4}-\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}+\frac {8}{15} b c^5 d^2 \log (x)-\frac {31}{60} b c^5 d^2 \log (1-c x)-\frac {1}{60} b c^5 d^2 \log (1+c x)\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 122, normalized size = 0.76 \[ -\frac {d^2 \left (20 a c^2 x^2+30 a c x+12 a-32 b c^5 x^5 \log (x)+31 b c^5 x^5 \log (1-c x)+b c^5 x^5 \log (c x+1)+30 b c^4 x^4+16 b c^3 x^3+10 b c^2 x^2+2 b \left (10 c^2 x^2+15 c x+6\right ) \tanh ^{-1}(c x)+3 b c x\right )}{60 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^6,x]

[Out]

-1/60*(d^2*(12*a + 30*a*c*x + 3*b*c*x + 20*a*c^2*x^2 + 10*b*c^2*x^2 + 16*b*c^3*x^3 + 30*b*c^4*x^4 + 2*b*(6 + 1
5*c*x + 10*c^2*x^2)*ArcTanh[c*x] - 32*b*c^5*x^5*Log[x] + 31*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 = 0.54, size = 156, normalized size = 0.97 \[ -\frac {b c^{5} d^{2} x^{5} \log \left (c x + 1\right ) + 31 \, b c^{5} d^{2} x^{5} \log \left (c x - 1\right ) - 32 \, b c^{5} d^{2} x^{5} \log \relax (x) + 30 \, b c^{4} d^{2} x^{4} + 16 \, b c^{3} d^{2} x^{3} + 10 \, {\left (2 \, a + b\right )} c^{2} d^{2} x^{2} + 3 \, {\left (10 \, a + b\right )} c d^{2} x + 12 \, a d^{2} + {\left (10 \, b c^{2} d^{2} x^{2} + 15 \, b c d^{2} x + 6 \, b d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{60 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/60*(b*c^5*d^2*x^5*log(c*x + 1) + 31*b*c^5*d^2*x^5*log(c*x - 1) - 32*b*c^5*d^2*x^5*log(x) + 30*b*c^4*d^2*x^4
+ 16*b*c^3*d^2*x^3 + 10*(2*a + b)*c^2*d^2*x^2 + 3*(10*a + b)*c*d^2*x + 12*a*d^2 + (10*b*c^2*d^2*x^2 + 15*b*c*
d^2*x + 6*b*d^2)*log(-(c*x + 1)/(c*x - 1)))/x^5

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giac [B]  time = 0.27, size = 532, normalized size = 3.30 \[ \frac {4}{15} \, {\left (2 \, b c^{4} d^{2} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - 2 \, b c^{4} d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {15 \, {\left (c x + 1\right )}^{4} b c^{4} d^{2}}{{\left (c x - 1\right )}^{4}} + \frac {15 \, {\left (c x + 1\right )}^{3} b c^{4} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {20 \, {\left (c x + 1\right )}^{2} b c^{4} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {10 \, {\left (c x + 1\right )} b c^{4} d^{2}}{c x - 1} + 2 \, b c^{4} d^{2}\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 {30 \, {\left (c x + 1\right )}^{4} a c^{4} d^{2}}{{\left (c x - 1\right )}^{4}} + \frac {30 \, {\left (c x + 1\right )}^{3} a c^{4} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {40 \, {\left (c x + 1\right )}^{2} a c^{4} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {20 \, {\left (c x + 1\right )} a c^{4} d^{2}}{c x - 1} + 4 \, a c^{4} d^{2} + \frac {13 \, {\left (c x + 1\right )}^{4} b c^{4} d^{2}}{{\left (c x - 1\right )}^{4}} + \frac {36 \, {\left (c x + 1\right )}^{3} b c^{4} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {41 \, {\left (c x + 1\right )}^{2} b c^{4} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {23 \, {\left (c x + 1\right )} b c^{4} d^{2}}{c x - 1} + 5 \, b c^{4} d^{2}}{\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)^2*(a+b*arctanh(c*x))/x^6,x, algorithm="giac")

[Out]

4/15*(2*b*c^4*d^2*log(-(c*x + 1)/(c*x - 1) - 1) - 2*b*c^4*d^2*log(-(c*x + 1)/(c*x - 1)) + (15*(c*x + 1)^4*b*c^
4*d^2/(c*x - 1)^4 + 15*(c*x + 1)^3*b*c^4*d^2/(c*x - 1)^3 + 20*(c*x + 1)^2*b*c^4*d^2/(c*x - 1)^2 + 10*(c*x + 1)
*b*c^4*d^2/(c*x - 1) + 2*b*c^4*d^2)*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) + (30*(c*x + 1)^4*
a*c^4*d^2/(c*x - 1)^4 + 30*(c*x + 1)^3*a*c^4*d^2/(c*x - 1)^3 + 40*(c*x + 1)^2*a*c^4*d^2/(c*x - 1)^2 + 20*(c*x
+ 1)*a*c^4*d^2/(c*x - 1) + 4*a*c^4*d^2 + 13*(c*x + 1)^4*b*c^4*d^2/(c*x - 1)^4 + 36*(c*x + 1)^3*b*c^4*d^2/(c*x
- 1)^3 + 41*(c*x + 1)^2*b*c^4*d^2/(c*x - 1)^2 + 23*(c*x + 1)*b*c^4*d^2/(c*x - 1) + 5*b*c^4*d^2)/((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 = 165, normalized size = 1.02 \[ -\frac {c^{2} d^{2} a}{3 x^{3}}-\frac {c \,d^{2} a}{2 x^{4}}-\frac {d^{2} a}{5 x^{5}}-\frac {c^{2} d^{2} b \arctanh \left (c x \right )}{3 x^{3}}-\frac {c \,d^{2} b \arctanh \left (c x \right )}{2 x^{4}}-\frac {d^{2} b \arctanh \left (c x \right )}{5 x^{5}}-\frac {b c \,d^{2}}{20 x^{4}}-\frac {b \,c^{2} d^{2}}{6 x^{3}}-\frac {4 b \,c^{3} d^{2}}{15 x^{2}}-\frac {b \,c^{4} d^{2}}{2 x}+\frac {8 c^{5} d^{2} b \ln \left (c x \right )}{15}-\frac {31 c^{5} d^{2} b \ln \left (c x -1\right )}{60}-\frac {b \,c^{5} d^{2} \ln \left (c x +1\right )}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^2*(a+b*arctanh(c*x))/x^6,x)

[Out]

-1/3*c^2*d^2*a/x^3-1/2*c*d^2*a/x^4-1/5*d^2*a/x^5-1/3*c^2*d^2*b*arctanh(c*x)/x^3-1/2*c*d^2*b*arctanh(c*x)/x^4-1
/5*d^2*b*arctanh(c*x)/x^5-1/20*b*c*d^2/x^4-1/6*b*c^2*d^2/x^3-4/15*b*c^3*d^2/x^2-1/2*b*c^4*d^2/x+8/15*c^5*d^2*b
*ln(c*x)-31/60*c^5*d^2*b*ln(c*x-1)-1/60*b*c^5*d^2*ln(c*x+1)

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maxima [A]  time = 0.32, size = 194, normalized size = 1.20 \[ -\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c^{2} d^{2} + \frac {1}{12} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b c d^{2} - \frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} - 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) + \frac {2 \, c^{2} x^{2} + 1}{x^{4}}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x\right )}{x^{5}}\right )} b d^{2} - \frac {a c^{2} d^{2}}{3 \, x^{3}} - \frac {a c d^{2}}{2 \, x^{4}} - \frac {a d^{2}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/6*((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^2 + 1/12*((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^2 - 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^2 - 1/3*a*c^2*d^2/x^3 - 1/2*a*c*d^2/x^
4 - 1/5*a*d^2/x^5

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mupad [B]  time = 0.95, size = 182, normalized size = 1.13 \[ -\frac {12\,a\,d^2+12\,b\,d^2\,\mathrm {atanh}\left (c\,x\right )+20\,a\,c^2\,d^2\,x^2+10\,b\,c^2\,d^2\,x^2+16\,b\,c^3\,d^2\,x^3+30\,b\,c^4\,d^2\,x^4+30\,a\,c\,d^2\,x+3\,b\,c\,d^2\,x-32\,b\,c^5\,d^2\,x^5\,\ln \relax (x)+20\,b\,c^2\,d^2\,x^2\,\mathrm {atanh}\left (c\,x\right )+16\,b\,c^5\,d^2\,x^5\,\ln \left (c^2\,x^2-1\right )+30\,b\,c\,d^2\,x\,\mathrm {atanh}\left (c\,x\right )-30\,b\,c^4\,d^2\,x^5\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )\,\sqrt {-c^2}}{60\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x)^2)/x^6,x)

[Out]

-(12*a*d^2 + 12*b*d^2*atanh(c*x) + 20*a*c^2*d^2*x^2 + 10*b*c^2*d^2*x^2 + 16*b*c^3*d^2*x^3 + 30*b*c^4*d^2*x^4 +
30*a*c*d^2*x + 3*b*c*d^2*x - 32*b*c^5*d^2*x^5*log(x) + 20*b*c^2*d^2*x^2*atanh(c*x) + 16*b*c^5*d^2*x^5*log(c^2
*x^2 - 1) + 30*b*c*d^2*x*atanh(c*x) - 30*b*c^4*d^2*x^5*atan((c^2*x)/(-c^2)^(1/2))*(-c^2)^(1/2))/(60*x^5)

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sympy [A]  time = 2.56, size = 199, normalized size = 1.24 \[ \begin {cases} - \frac {a c^{2} d^{2}}{3 x^{3}} - \frac {a c d^{2}}{2 x^{4}} - \frac {a d^{2}}{5 x^{5}} + \frac {8 b c^{5} d^{2} \log {\relax (x )}}{15} - \frac {8 b c^{5} d^{2} \log {\left (x - \frac {1}{c} \right )}}{15} - \frac {b c^{5} d^{2} \operatorname {atanh}{\left (c x \right )}}{30} - \frac {b c^{4} d^{2}}{2 x} - \frac {4 b c^{3} d^{2}}{15 x^{2}} - \frac {b c^{2} d^{2} \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} - \frac {b c^{2} d^{2}}{6 x^{3}} - \frac {b c d^{2} \operatorname {atanh}{\left (c x \right )}}{2 x^{4}} - \frac {b c d^{2}}{20 x^{4}} - \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a d^{2}}{5 x^{5}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**2*(a+b*atanh(c*x))/x**6,x)

[Out]

Piecewise((-a*c**2*d**2/(3*x**3) - a*c*d**2/(2*x**4) - a*d**2/(5*x**5) + 8*b*c**5*d**2*log(x)/15 - 8*b*c**5*d*
*2*log(x - 1/c)/15 - b*c**5*d**2*atanh(c*x)/30 - b*c**4*d**2/(2*x) - 4*b*c**3*d**2/(15*x**2) - b*c**2*d**2*ata
nh(c*x)/(3*x**3) - b*c**2*d**2/(6*x**3) - b*c*d**2*atanh(c*x)/(2*x**4) - b*c*d**2/(20*x**4) - b*d**2*atanh(c*x
)/(5*x**5), Ne(c, 0)), (-a*d**2/(5*x**5), True))

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