3.18 \(\int \frac{(d+c d x)^2 (a+b \tanh ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=147 \[ -\frac{c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{b c^2 d^2}{3 x^2}-\frac{3 b c^3 d^2}{4 x}+\frac{2}{3} b c^4 d^2 \log (x)-\frac{17}{24} b c^4 d^2 \log (1-c x)+\frac{1}{24} b c^4 d^2 \log (c x+1)-\frac{b c d^2}{12 x^3} \]

[Out]

-(b*c*d^2)/(12*x^3) - (b*c^2*d^2)/(3*x^2) - (3*b*c^3*d^2)/(4*x) - (d^2*(a + b*ArcTanh[c*x]))/(4*x^4) - (2*c*d^
2*(a + b*ArcTanh[c*x]))/(3*x^3) - (c^2*d^2*(a + b*ArcTanh[c*x]))/(2*x^2) + (2*b*c^4*d^2*Log[x])/3 - (17*b*c^4*
d^2*Log[1 - c*x])/24 + (b*c^4*d^2*Log[1 + c*x])/24

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Rubi [A]  time = 0.149283, antiderivative size = 147, 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} \[ -\frac{c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{b c^2 d^2}{3 x^2}-\frac{3 b c^3 d^2}{4 x}+\frac{2}{3} b c^4 d^2 \log (x)-\frac{17}{24} b c^4 d^2 \log (1-c x)+\frac{1}{24} b c^4 d^2 \log (c x+1)-\frac{b c d^2}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

-(b*c*d^2)/(12*x^3) - (b*c^2*d^2)/(3*x^2) - (3*b*c^3*d^2)/(4*x) - (d^2*(a + b*ArcTanh[c*x]))/(4*x^4) - (2*c*d^
2*(a + b*ArcTanh[c*x]))/(3*x^3) - (c^2*d^2*(a + b*ArcTanh[c*x]))/(2*x^2) + (2*b*c^4*d^2*Log[x])/3 - (17*b*c^4*
d^2*Log[1 - c*x])/24 + (b*c^4*d^2*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 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^5} \, dx &=-\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-(b c) \int \frac{d^2 \left (-3-8 c x-6 c^2 x^2\right )}{12 x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{1}{12} \left (b c d^2\right ) \int \frac{-3-8 c x-6 c^2 x^2}{x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac{1}{12} \left (b c d^2\right ) \int \left (-\frac{3}{x^4}-\frac{8 c}{x^3}-\frac{9 c^2}{x^2}-\frac{8 c^3}{x}+\frac{17 c^4}{2 (-1+c x)}-\frac{c^4}{2 (1+c x)}\right ) \, dx\\ &=-\frac{b c d^2}{12 x^3}-\frac{b c^2 d^2}{3 x^2}-\frac{3 b c^3 d^2}{4 x}-\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}+\frac{2}{3} b c^4 d^2 \log (x)-\frac{17}{24} b c^4 d^2 \log (1-c x)+\frac{1}{24} b c^4 d^2 \log (1+c x)\\ \end{align*}

Mathematica [A]  time = 0.0967844, size = 114, normalized size = 0.78 \[ -\frac{d^2 \left (12 a c^2 x^2+16 a c x+6 a+18 b c^3 x^3+8 b c^2 x^2-16 b c^4 x^4 \log (x)+17 b c^4 x^4 \log (1-c x)-b c^4 x^4 \log (c x+1)+2 b \left (6 c^2 x^2+8 c x+3\right ) \tanh ^{-1}(c x)+2 b c x\right )}{24 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

-(d^2*(6*a + 16*a*c*x + 2*b*c*x + 12*a*c^2*x^2 + 8*b*c^2*x^2 + 18*b*c^3*x^3 + 2*b*(3 + 8*c*x + 6*c^2*x^2)*ArcT
anh[c*x] - 16*b*c^4*x^4*Log[x] + 17*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 = 153, normalized size = 1. \begin{align*} -{\frac{{d}^{2}a}{4\,{x}^{4}}}-{\frac{{c}^{2}{d}^{2}a}{2\,{x}^{2}}}-{\frac{2\,c{d}^{2}a}{3\,{x}^{3}}}-{\frac{{d}^{2}b{\it Artanh} \left ( cx \right ) }{4\,{x}^{4}}}-{\frac{{c}^{2}{d}^{2}b{\it Artanh} \left ( cx \right ) }{2\,{x}^{2}}}-{\frac{2\,c{d}^{2}b{\it Artanh} \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{17\,{c}^{4}{d}^{2}b\ln \left ( cx-1 \right ) }{24}}-{\frac{c{d}^{2}b}{12\,{x}^{3}}}-{\frac{{c}^{2}{d}^{2}b}{3\,{x}^{2}}}-{\frac{3\,b{c}^{3}{d}^{2}}{4\,x}}+{\frac{2\,{c}^{4}{d}^{2}b\ln \left ( cx \right ) }{3}}+{\frac{b{c}^{4}{d}^{2}\ln \left ( cx+1 \right ) }{24}} \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^5,x)

[Out]

-1/4*d^2*a/x^4-1/2*c^2*d^2*a/x^2-2/3*c*d^2*a/x^3-1/4*d^2*b*arctanh(c*x)/x^4-1/2*c^2*d^2*b*arctanh(c*x)/x^2-2/3
*c*d^2*b*arctanh(c*x)/x^3-17/24*c^4*d^2*b*ln(c*x-1)-1/12*b*c*d^2/x^3-1/3*b*c^2*d^2/x^2-3/4*b*c^3*d^2/x+2/3*c^4
*d^2*b*ln(c*x)+1/24*b*c^4*d^2*ln(c*x+1)

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Maxima [A]  time = 0.975368, size = 240, normalized size = 1.63 \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^{2} d^{2} - \frac{1}{3} \,{\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^{2} + \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^{2} - \frac{a c^{2} d^{2}}{2 \, x^{2}} - \frac{2 \, a c d^{2}}{3 \, x^{3}} - \frac{a d^{2}}{4 \, x^{4}} \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^5,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^2*d^2 - 1/3*((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^2 + 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^2 - 1/2*a*c^2*d^2/x^2 - 2/3*a*c*d^2/x^3 - 1/4*a*d^2/x^4

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Fricas [A]  time = 2.09943, size = 332, normalized size = 2.26 \begin{align*} \frac{b c^{4} d^{2} x^{4} \log \left (c x + 1\right ) - 17 \, b c^{4} d^{2} x^{4} \log \left (c x - 1\right ) + 16 \, b c^{4} d^{2} x^{4} \log \left (x\right ) - 18 \, b c^{3} d^{2} x^{3} - 4 \,{\left (3 \, a + 2 \, b\right )} c^{2} d^{2} x^{2} - 2 \,{\left (8 \, a + b\right )} c d^{2} x - 6 \, a d^{2} -{\left (6 \, b c^{2} d^{2} x^{2} + 8 \, b c d^{2} x + 3 \, b d^{2}\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)^2*(a+b*arctanh(c*x))/x^5,x, algorithm="fricas")

[Out]

1/24*(b*c^4*d^2*x^4*log(c*x + 1) - 17*b*c^4*d^2*x^4*log(c*x - 1) + 16*b*c^4*d^2*x^4*log(x) - 18*b*c^3*d^2*x^3
- 4*(3*a + 2*b)*c^2*d^2*x^2 - 2*(8*a + b)*c*d^2*x - 6*a*d^2 - (6*b*c^2*d^2*x^2 + 8*b*c*d^2*x + 3*b*d^2)*log(-(
c*x + 1)/(c*x - 1)))/x^4

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Sympy [A]  time = 5.26977, size = 189, normalized size = 1.29 \begin{align*} \begin{cases} - \frac{a c^{2} d^{2}}{2 x^{2}} - \frac{2 a c d^{2}}{3 x^{3}} - \frac{a d^{2}}{4 x^{4}} + \frac{2 b c^{4} d^{2} \log{\left (x \right )}}{3} - \frac{2 b c^{4} d^{2} \log{\left (x - \frac{1}{c} \right )}}{3} + \frac{b c^{4} d^{2} \operatorname{atanh}{\left (c x \right )}}{12} - \frac{3 b c^{3} d^{2}}{4 x} - \frac{b c^{2} d^{2} \operatorname{atanh}{\left (c x \right )}}{2 x^{2}} - \frac{b c^{2} d^{2}}{3 x^{2}} - \frac{2 b c d^{2} \operatorname{atanh}{\left (c x \right )}}{3 x^{3}} - \frac{b c d^{2}}{12 x^{3}} - \frac{b d^{2} \operatorname{atanh}{\left (c x \right )}}{4 x^{4}} & \text{for}\: c \neq 0 \\- \frac{a d^{2}}{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)**2*(a+b*atanh(c*x))/x**5,x)

[Out]

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

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Giac [A]  time = 1.25085, size = 205, normalized size = 1.39 \begin{align*} \frac{1}{24} \, b c^{4} d^{2} \log \left (c x + 1\right ) - \frac{17}{24} \, b c^{4} d^{2} \log \left (c x - 1\right ) + \frac{2}{3} \, b c^{4} d^{2} \log \left (x\right ) - \frac{{\left (6 \, b c^{2} d^{2} x^{2} + 8 \, b c d^{2} x + 3 \, b d^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{24 \, x^{4}} - \frac{9 \, b c^{3} d^{2} x^{3} + 6 \, a c^{2} d^{2} x^{2} + 4 \, b c^{2} d^{2} x^{2} + 8 \, a c d^{2} x + b c d^{2} x + 3 \, a d^{2}}{12 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

1/24*b*c^4*d^2*log(c*x + 1) - 17/24*b*c^4*d^2*log(c*x - 1) + 2/3*b*c^4*d^2*log(x) - 1/24*(6*b*c^2*d^2*x^2 + 8*
b*c*d^2*x + 3*b*d^2)*log(-(c*x + 1)/(c*x - 1))/x^4 - 1/12*(9*b*c^3*d^2*x^3 + 6*a*c^2*d^2*x^2 + 4*b*c^2*d^2*x^2
+ 8*a*c*d^2*x + b*c*d^2*x + 3*a*d^2)/x^4