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

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

[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

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Rubi [A]  time = 0.160266, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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 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]))

Rule 12

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

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]

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*}

Mathematica [A]  time = 0.0954493, size = 122, normalized size = 0.76 \[ -\frac{d^2 \left (20 a c^2 x^2+30 a c x+12 a+30 b c^4 x^4+16 b c^3 x^3+10 b c^2 x^2-32 b c^5 x^5 \log (x)+31 b c^5 x^5 \log (1-c x)+b c^5 x^5 \log (c x+1)+2 b \left (10 c^2 x^2+15 c x+6\right ) \tanh ^{-1}(c x)+3 b c x\right )}{60 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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 + 15*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]))/(60*x
^5)

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Maple [A]  time = 0.04, size = 165, normalized size = 1. \begin{align*} -{\frac{c{d}^{2}a}{2\,{x}^{4}}}-{\frac{{d}^{2}a}{5\,{x}^{5}}}-{\frac{{c}^{2}{d}^{2}a}{3\,{x}^{3}}}-{\frac{c{d}^{2}b{\it Artanh} \left ( cx \right ) }{2\,{x}^{4}}}-{\frac{{d}^{2}b{\it Artanh} \left ( cx \right ) }{5\,{x}^{5}}}-{\frac{{c}^{2}{d}^{2}b{\it Artanh} \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{31\,{c}^{5}{d}^{2}b\ln \left ( cx-1 \right ) }{60}}-{\frac{c{d}^{2}b}{20\,{x}^{4}}}-{\frac{{c}^{2}{d}^{2}b}{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 ( cx \right ) }{15}}-{\frac{b{c}^{5}{d}^{2}\ln \left ( cx+1 \right ) }{60}} \end{align*}

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]

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

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Maxima [A]  time = 0.969873, size = 262, normalized size = 1.63 \begin{align*} -\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}} \end{align*}

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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Fricas [A]  time = 2.36032, size = 363, normalized size = 2.25 \begin{align*} -\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 \left (x\right ) + 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}} \end{align*}

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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Sympy [A]  time = 4.64253, size = 199, normalized size = 1.24 \begin{align*} \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{\left (x \right )}}{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} \end{align*}

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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Giac [A]  time = 1.41537, size = 223, normalized size = 1.39 \begin{align*} -\frac{1}{60} \, b c^{5} d^{2} \log \left (c x + 1\right ) - \frac{31}{60} \, b c^{5} d^{2} \log \left (c x - 1\right ) + \frac{8}{15} \, b c^{5} d^{2} \log \left (x\right ) - \frac{{\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}} - \frac{30 \, b c^{4} d^{2} x^{4} + 16 \, b c^{3} d^{2} x^{3} + 20 \, a c^{2} d^{2} x^{2} + 10 \, b c^{2} d^{2} x^{2} + 30 \, a c d^{2} x + 3 \, b c d^{2} x + 12 \, a d^{2}}{60 \, x^{5}} \end{align*}

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]

-1/60*b*c^5*d^2*log(c*x + 1) - 31/60*b*c^5*d^2*log(c*x - 1) + 8/15*b*c^5*d^2*log(x) - 1/60*(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 - 1/60*(30*b*c^4*d^2*x^4 + 16*b*c^3*d^2*x^3 + 20*a*c^2*
d^2*x^2 + 10*b*c^2*d^2*x^2 + 30*a*c*d^2*x + 3*b*c*d^2*x + 12*a*d^2)/x^5