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

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

[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

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Rubi [A]  time = 0.118844, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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 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 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 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]))

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*}

Mathematica [A]  time = 0.123228, size = 140, normalized size = 1.02 \[ -\frac{d^3 \left (20 a c^3 x^3+40 a c^2 x^2+30 a c x+8 a+50 b c^4 x^4+24 b c^3 x^3+10 b c^2 x^2-48 b c^5 x^5 \log (x)+49 b c^5 x^5 \log (1-c x)-b c^5 x^5 \log (c x+1)+2 b \left (10 c^3 x^3+20 c^2 x^2+15 c x+4\right ) \tanh ^{-1}(c x)+2 b c x\right )}{40 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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]))/(40*x^5)

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Maple [A]  time = 0.039, size = 193, normalized size = 1.4 \begin{align*} -{\frac{3\,c{d}^{3}a}{4\,{x}^{4}}}-{\frac{{d}^{3}a}{5\,{x}^{5}}}-{\frac{{c}^{3}{d}^{3}a}{2\,{x}^{2}}}-{\frac{{c}^{2}{d}^{3}a}{{x}^{3}}}-{\frac{3\,c{d}^{3}b{\it Artanh} \left ( cx \right ) }{4\,{x}^{4}}}-{\frac{{d}^{3}b{\it Artanh} \left ( cx \right ) }{5\,{x}^{5}}}-{\frac{{c}^{3}{d}^{3}b{\it Artanh} \left ( cx \right ) }{2\,{x}^{2}}}-{\frac{{d}^{3}b{c}^{2}{\it Artanh} \left ( cx \right ) }{{x}^{3}}}-{\frac{49\,{c}^{5}{d}^{3}b\ln \left ( cx-1 \right ) }{40}}-{\frac{c{d}^{3}b}{20\,{x}^{4}}}-{\frac{{d}^{3}b{c}^{2}}{4\,{x}^{3}}}-{\frac{3\,{c}^{3}{d}^{3}b}{5\,{x}^{2}}}-{\frac{5\,b{c}^{4}{d}^{3}}{4\,x}}+{\frac{6\,{c}^{5}{d}^{3}b\ln \left ( cx \right ) }{5}}+{\frac{{c}^{5}{d}^{3}b\ln \left ( cx+1 \right ) }{40}} \end{align*}

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]

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

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Maxima [B]  time = 0.96778, size = 338, normalized size = 2.47 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 2.07202, size = 398, normalized size = 2.91 \begin{align*} \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 \left (x\right ) - 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}} \end{align*}

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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Sympy [A]  time = 4.86091, size = 233, normalized size = 1.7 \begin{align*} \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{\left (x \right )}}{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} \end{align*}

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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Giac [A]  time = 1.53538, size = 254, normalized size = 1.85 \begin{align*} \frac{1}{40} \, b c^{5} d^{3} \log \left (c x + 1\right ) - \frac{49}{40} \, b c^{5} d^{3} \log \left (c x - 1\right ) + \frac{6}{5} \, b c^{5} d^{3} \log \left (x\right ) - \frac{{\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}} - \frac{25 \, b c^{4} d^{3} x^{4} + 10 \, a c^{3} d^{3} x^{3} + 12 \, b c^{3} d^{3} x^{3} + 20 \, a c^{2} d^{3} x^{2} + 5 \, b c^{2} d^{3} x^{2} + 15 \, a c d^{3} x + b c d^{3} x + 4 \, a d^{3}}{20 \, x^{5}} \end{align*}

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]

1/40*b*c^5*d^3*log(c*x + 1) - 49/40*b*c^5*d^3*log(c*x - 1) + 6/5*b*c^5*d^3*log(x) - 1/40*(10*b*c^3*d^3*x^3 + 2
0*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 - 1/20*(25*b*c^4*d^3*x^4 + 10*a*c^3*d^
3*x^3 + 12*b*c^3*d^3*x^3 + 20*a*c^2*d^3*x^2 + 5*b*c^2*d^3*x^2 + 15*a*c*d^3*x + b*c*d^3*x + 4*a*d^3)/x^5