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

Optimal. Leaf size=110 \[ -\frac{c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{b c^2 d}{6 x^2}-\frac{b c^3 d}{4 x}+\frac{1}{3} b c^4 d \log (x)-\frac{7}{24} b c^4 d \log (1-c x)-\frac{1}{24} b c^4 d \log (c x+1)-\frac{b c d}{12 x^3} \]

[Out]

-(b*c*d)/(12*x^3) - (b*c^2*d)/(6*x^2) - (b*c^3*d)/(4*x) - (d*(a + b*ArcTanh[c*x]))/(4*x^4) - (c*d*(a + b*ArcTa
nh[c*x]))/(3*x^3) + (b*c^4*d*Log[x])/3 - (7*b*c^4*d*Log[1 - c*x])/24 - (b*c^4*d*Log[1 + c*x])/24

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Rubi [A]  time = 0.0924434, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {43, 5936, 12, 801} \[ -\frac{c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{b c^2 d}{6 x^2}-\frac{b c^3 d}{4 x}+\frac{1}{3} b c^4 d \log (x)-\frac{7}{24} b c^4 d \log (1-c x)-\frac{1}{24} b c^4 d \log (c x+1)-\frac{b c d}{12 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d)/(12*x^3) - (b*c^2*d)/(6*x^2) - (b*c^3*d)/(4*x) - (d*(a + b*ArcTanh[c*x]))/(4*x^4) - (c*d*(a + b*ArcTa
nh[c*x]))/(3*x^3) + (b*c^4*d*Log[x])/3 - (7*b*c^4*d*Log[1 - c*x])/24 - (b*c^4*d*Log[1 + c*x])/24

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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{(d+c d x) \left (a+b \tanh ^{-1}(c x)\right )}{x^5} \, dx &=-\frac{d \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-(b c) \int \frac{d (-3-4 c x)}{12 x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{d \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{1}{12} (b c d) \int \frac{-3-4 c x}{x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{d \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{1}{12} (b c d) \int \left (-\frac{3}{x^4}-\frac{4 c}{x^3}-\frac{3 c^2}{x^2}-\frac{4 c^3}{x}+\frac{7 c^4}{2 (-1+c x)}+\frac{c^4}{2 (1+c x)}\right ) \, dx\\ &=-\frac{b c d}{12 x^3}-\frac{b c^2 d}{6 x^2}-\frac{b c^3 d}{4 x}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}+\frac{1}{3} b c^4 d \log (x)-\frac{7}{24} b c^4 d \log (1-c x)-\frac{1}{24} b c^4 d \log (1+c x)\\ \end{align*}

Mathematica [A]  time = 0.0649337, size = 94, normalized size = 0.85 \[ -\frac{d \left (8 a c x+6 a+6 b c^3 x^3+4 b c^2 x^2-8 b c^4 x^4 \log (x)+7 b c^4 x^4 \log (1-c x)+b c^4 x^4 \log (c x+1)+2 b c x+2 b (4 c x+3) \tanh ^{-1}(c x)\right )}{24 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(6*a + 8*a*c*x + 2*b*c*x + 4*b*c^2*x^2 + 6*b*c^3*x^3 + 2*b*(3 + 4*c*x)*ArcTanh[c*x] - 8*b*c^4*x^4*Log[x] +
7*b*c^4*x^4*Log[1 - c*x] + b*c^4*x^4*Log[1 + c*x]))/(24*x^4)

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Maple [A]  time = 0.04, size = 105, normalized size = 1. \begin{align*} -{\frac{da}{4\,{x}^{4}}}-{\frac{cda}{3\,{x}^{3}}}-{\frac{db{\it Artanh} \left ( cx \right ) }{4\,{x}^{4}}}-{\frac{cdb{\it Artanh} \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{7\,{c}^{4}db\ln \left ( cx-1 \right ) }{24}}-{\frac{cdb}{12\,{x}^{3}}}-{\frac{b{c}^{2}d}{6\,{x}^{2}}}-{\frac{b{c}^{3}d}{4\,x}}+{\frac{{c}^{4}db\ln \left ( cx \right ) }{3}}-{\frac{b{c}^{4}d\ln \left ( cx+1 \right ) }{24}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)*(a+b*arctanh(c*x))/x^5,x)

[Out]

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

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Maxima [A]  time = 0.965634, size = 154, normalized size = 1.4 \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 d + \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 - \frac{a c d}{3 \, x^{3}} - \frac{a d}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^5,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*d + 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*arctanh(c*x)/x^4)*b*d - 1/3*a*c*d/x^3 - 1/4*a*d/x^4

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Fricas [A]  time = 2.10247, size = 267, normalized size = 2.43 \begin{align*} -\frac{b c^{4} d x^{4} \log \left (c x + 1\right ) + 7 \, b c^{4} d x^{4} \log \left (c x - 1\right ) - 8 \, b c^{4} d x^{4} \log \left (x\right ) + 6 \, b c^{3} d x^{3} + 4 \, b c^{2} d x^{2} + 2 \,{\left (4 \, a + b\right )} c d x + 6 \, a d +{\left (4 \, b c d x + 3 \, b d\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)*(a+b*arctanh(c*x))/x^5,x, algorithm="fricas")

[Out]

-1/24*(b*c^4*d*x^4*log(c*x + 1) + 7*b*c^4*d*x^4*log(c*x - 1) - 8*b*c^4*d*x^4*log(x) + 6*b*c^3*d*x^3 + 4*b*c^2*
d*x^2 + 2*(4*a + b)*c*d*x + 6*a*d + (4*b*c*d*x + 3*b*d)*log(-(c*x + 1)/(c*x - 1)))/x^4

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Sympy [A]  time = 3.09623, size = 129, normalized size = 1.17 \begin{align*} \begin{cases} - \frac{a c d}{3 x^{3}} - \frac{a d}{4 x^{4}} + \frac{b c^{4} d \log{\left (x \right )}}{3} - \frac{b c^{4} d \log{\left (x - \frac{1}{c} \right )}}{3} - \frac{b c^{4} d \operatorname{atanh}{\left (c x \right )}}{12} - \frac{b c^{3} d}{4 x} - \frac{b c^{2} d}{6 x^{2}} - \frac{b c d \operatorname{atanh}{\left (c x \right )}}{3 x^{3}} - \frac{b c d}{12 x^{3}} - \frac{b d \operatorname{atanh}{\left (c x \right )}}{4 x^{4}} & \text{for}\: c \neq 0 \\- \frac{a d}{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)*(a+b*atanh(c*x))/x**5,x)

[Out]

Piecewise((-a*c*d/(3*x**3) - a*d/(4*x**4) + b*c**4*d*log(x)/3 - b*c**4*d*log(x - 1/c)/3 - b*c**4*d*atanh(c*x)/
12 - b*c**3*d/(4*x) - b*c**2*d/(6*x**2) - b*c*d*atanh(c*x)/(3*x**3) - b*c*d/(12*x**3) - b*d*atanh(c*x)/(4*x**4
), Ne(c, 0)), (-a*d/(4*x**4), True))

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Giac [A]  time = 1.25656, size = 146, normalized size = 1.33 \begin{align*} -\frac{1}{24} \, b c^{4} d \log \left (c x + 1\right ) - \frac{7}{24} \, b c^{4} d \log \left (c x - 1\right ) + \frac{1}{3} \, b c^{4} d \log \left (x\right ) - \frac{{\left (4 \, b c d x + 3 \, b d\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{24 \, x^{4}} - \frac{3 \, b c^{3} d x^{3} + 2 \, b c^{2} d x^{2} + 4 \, a c d x + b c d x + 3 \, a d}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^5,x, algorithm="giac")

[Out]

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