3.28 \(\int \frac{(d+c d x)^3 (a+b \tanh ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=93 \[ -\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{b c^2 d^3}{2 x^2}-\frac{7 b c^3 d^3}{4 x}+2 b c^4 d^3 \log (x)-2 b c^4 d^3 \log (1-c x)-\frac{b c d^3}{12 x^3} \]

[Out]

-(b*c*d^3)/(12*x^3) - (b*c^2*d^3)/(2*x^2) - (7*b*c^3*d^3)/(4*x) - (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(4*x^
4) + 2*b*c^4*d^3*Log[x] - 2*b*c^4*d^3*Log[1 - c*x]

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Rubi [A]  time = 0.0985638, antiderivative size = 93, 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 = {37, 5936, 12, 88} \[ -\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{b c^2 d^3}{2 x^2}-\frac{7 b c^3 d^3}{4 x}+2 b c^4 d^3 \log (x)-2 b c^4 d^3 \log (1-c x)-\frac{b c d^3}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^3*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

-(b*c*d^3)/(12*x^3) - (b*c^2*d^3)/(2*x^2) - (7*b*c^3*d^3)/(4*x) - (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(4*x^
4) + 2*b*c^4*d^3*Log[x] - 2*b*c^4*d^3*Log[1 - c*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 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 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]))

Rubi steps

\begin{align*} \int \frac{(d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^5} \, dx &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-(b c) \int \frac{(d+c d x)^3}{4 x^4 (-1+c x)} \, dx\\ &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{1}{4} (b c) \int \frac{(d+c d x)^3}{x^4 (-1+c x)} \, dx\\ &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{1}{4} (b c) \int \left (-\frac{d^3}{x^4}-\frac{4 c d^3}{x^3}-\frac{7 c^2 d^3}{x^2}-\frac{8 c^3 d^3}{x}+\frac{8 c^4 d^3}{-1+c x}\right ) \, dx\\ &=-\frac{b c d^3}{12 x^3}-\frac{b c^2 d^3}{2 x^2}-\frac{7 b c^3 d^3}{4 x}-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}+2 b c^4 d^3 \log (x)-2 b c^4 d^3 \log (1-c x)\\ \end{align*}

Mathematica [A]  time = 0.119112, size = 131, normalized size = 1.41 \[ -\frac{d^3 \left (24 a c^3 x^3+36 a c^2 x^2+24 a c x+6 a+42 b c^3 x^3+12 b c^2 x^2-48 b c^4 x^4 \log (x)+45 b c^4 x^4 \log (1-c x)+3 b c^4 x^4 \log (c x+1)+6 b \left (4 c^3 x^3+6 c^2 x^2+4 c x+1\right ) \tanh ^{-1}(c x)+2 b c x\right )}{24 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^3*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

-(d^3*(6*a + 24*a*c*x + 2*b*c*x + 36*a*c^2*x^2 + 12*b*c^2*x^2 + 24*a*c^3*x^3 + 42*b*c^3*x^3 + 6*b*(1 + 4*c*x +
6*c^2*x^2 + 4*c^3*x^3)*ArcTanh[c*x] - 48*b*c^4*x^4*Log[x] + 45*b*c^4*x^4*Log[1 - c*x] + 3*b*c^4*x^4*Log[1 + c
*x]))/(24*x^4)

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Maple [B]  time = 0.04, size = 181, normalized size = 2. \begin{align*} -{\frac{{d}^{3}a}{4\,{x}^{4}}}-{\frac{{c}^{3}{d}^{3}a}{x}}-{\frac{3\,{c}^{2}{d}^{3}a}{2\,{x}^{2}}}-{\frac{c{d}^{3}a}{{x}^{3}}}-{\frac{{d}^{3}b{\it Artanh} \left ( cx \right ) }{4\,{x}^{4}}}-{\frac{{c}^{3}{d}^{3}b{\it Artanh} \left ( cx \right ) }{x}}-{\frac{3\,{d}^{3}b{c}^{2}{\it Artanh} \left ( cx \right ) }{2\,{x}^{2}}}-{\frac{c{d}^{3}b{\it Artanh} \left ( cx \right ) }{{x}^{3}}}-{\frac{15\,{c}^{4}{d}^{3}b\ln \left ( cx-1 \right ) }{8}}-{\frac{c{d}^{3}b}{12\,{x}^{3}}}-{\frac{{d}^{3}b{c}^{2}}{2\,{x}^{2}}}-{\frac{7\,{c}^{3}{d}^{3}b}{4\,x}}+2\,{c}^{4}{d}^{3}b\ln \left ( cx \right ) -{\frac{{c}^{4}{d}^{3}b\ln \left ( cx+1 \right ) }{8}} \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^5,x)

[Out]

-1/4*d^3*a/x^4-c^3*d^3*a/x-3/2*c^2*d^3*a/x^2-c*d^3*a/x^3-1/4*d^3*b*arctanh(c*x)/x^4-c^3*d^3*b*arctanh(c*x)/x-3
/2*c^2*d^3*b*arctanh(c*x)/x^2-c*d^3*b*arctanh(c*x)/x^3-15/8*c^4*d^3*b*ln(c*x-1)-1/12*b*c*d^3/x^3-1/2*b*c^2*d^3
/x^2-7/4*b*c^3*d^3/x+2*c^4*d^3*b*ln(c*x)-1/8*c^4*d^3*b*ln(c*x+1)

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Maxima [B]  time = 0.960071, size = 308, normalized size = 3.31 \begin{align*} -\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^{3} d^{3} + \frac{3}{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^{2} 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 d^{3} - \frac{a c^{3} d^{3}}{x} + \frac{1}{24} \,{\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 d^{3} - \frac{3 \, a c^{2} d^{3}}{2 \, x^{2}} - \frac{a c d^{3}}{x^{3}} - \frac{a d^{3}}{4 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

-1/2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*b*c^3*d^3 + 3/4*((c*log(c*x + 1) - c*log(c*x - 1) -
2/x)*c - 2*arctanh(c*x)/x^2)*b*c^2*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*d^3 - a*c^3*d^3/x + 1/24*((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*ar
ctanh(c*x)/x^4)*b*d^3 - 3/2*a*c^2*d^3/x^2 - a*c*d^3/x^3 - 1/4*a*d^3/x^4

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Fricas [A]  time = 2.15089, size = 373, normalized size = 4.01 \begin{align*} -\frac{3 \, b c^{4} d^{3} x^{4} \log \left (c x + 1\right ) + 45 \, b c^{4} d^{3} x^{4} \log \left (c x - 1\right ) - 48 \, b c^{4} d^{3} x^{4} \log \left (x\right ) + 6 \,{\left (4 \, a + 7 \, b\right )} c^{3} d^{3} x^{3} + 12 \,{\left (3 \, a + b\right )} c^{2} d^{3} x^{2} + 2 \,{\left (12 \, a + b\right )} c d^{3} x + 6 \, a d^{3} + 3 \,{\left (4 \, b c^{3} d^{3} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c d^{3} x + b d^{3}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{24 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

-1/24*(3*b*c^4*d^3*x^4*log(c*x + 1) + 45*b*c^4*d^3*x^4*log(c*x - 1) - 48*b*c^4*d^3*x^4*log(x) + 6*(4*a + 7*b)*
c^3*d^3*x^3 + 12*(3*a + b)*c^2*d^3*x^2 + 2*(12*a + b)*c*d^3*x + 6*a*d^3 + 3*(4*b*c^3*d^3*x^3 + 6*b*c^2*d^3*x^2
+ 4*b*c*d^3*x + b*d^3)*log(-(c*x + 1)/(c*x - 1)))/x^4

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Sympy [A]  time = 3.91854, size = 207, normalized size = 2.23 \begin{align*} \begin{cases} - \frac{a c^{3} d^{3}}{x} - \frac{3 a c^{2} d^{3}}{2 x^{2}} - \frac{a c d^{3}}{x^{3}} - \frac{a d^{3}}{4 x^{4}} + 2 b c^{4} d^{3} \log{\left (x \right )} - 2 b c^{4} d^{3} \log{\left (x - \frac{1}{c} \right )} - \frac{b c^{4} d^{3} \operatorname{atanh}{\left (c x \right )}}{4} - \frac{b c^{3} d^{3} \operatorname{atanh}{\left (c x \right )}}{x} - \frac{7 b c^{3} d^{3}}{4 x} - \frac{3 b c^{2} d^{3} \operatorname{atanh}{\left (c x \right )}}{2 x^{2}} - \frac{b c^{2} d^{3}}{2 x^{2}} - \frac{b c d^{3} \operatorname{atanh}{\left (c x \right )}}{x^{3}} - \frac{b c d^{3}}{12 x^{3}} - \frac{b d^{3} \operatorname{atanh}{\left (c x \right )}}{4 x^{4}} & \text{for}\: c \neq 0 \\- \frac{a d^{3}}{4 x^{4}} & \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**5,x)

[Out]

Piecewise((-a*c**3*d**3/x - 3*a*c**2*d**3/(2*x**2) - a*c*d**3/x**3 - a*d**3/(4*x**4) + 2*b*c**4*d**3*log(x) -
2*b*c**4*d**3*log(x - 1/c) - b*c**4*d**3*atanh(c*x)/4 - b*c**3*d**3*atanh(c*x)/x - 7*b*c**3*d**3/(4*x) - 3*b*c
**2*d**3*atanh(c*x)/(2*x**2) - b*c**2*d**3/(2*x**2) - b*c*d**3*atanh(c*x)/x**3 - b*c*d**3/(12*x**3) - b*d**3*a
tanh(c*x)/(4*x**4), Ne(c, 0)), (-a*d**3/(4*x**4), True))

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Giac [B]  time = 1.40483, size = 236, normalized size = 2.54 \begin{align*} -\frac{1}{8} \, b c^{4} d^{3} \log \left (c x + 1\right ) - \frac{15}{8} \, b c^{4} d^{3} \log \left (c x - 1\right ) + 2 \, b c^{4} d^{3} \log \left (x\right ) - \frac{{\left (4 \, b c^{3} d^{3} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c d^{3} x + b d^{3}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{8 \, x^{4}} - \frac{12 \, a c^{3} d^{3} x^{3} + 21 \, b c^{3} d^{3} x^{3} + 18 \, a c^{2} d^{3} x^{2} + 6 \, b c^{2} d^{3} x^{2} + 12 \, a c d^{3} x + b c d^{3} x + 3 \, a d^{3}}{12 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

-1/8*b*c^4*d^3*log(c*x + 1) - 15/8*b*c^4*d^3*log(c*x - 1) + 2*b*c^4*d^3*log(x) - 1/8*(4*b*c^3*d^3*x^3 + 6*b*c^
2*d^3*x^2 + 4*b*c*d^3*x + b*d^3)*log(-(c*x + 1)/(c*x - 1))/x^4 - 1/12*(12*a*c^3*d^3*x^3 + 21*b*c^3*d^3*x^3 + 1
8*a*c^2*d^3*x^2 + 6*b*c^2*d^3*x^2 + 12*a*c*d^3*x + b*c*d^3*x + 3*a*d^3)/x^4