3.17 \(\int \frac{(d+c d x)^2 (a+b \tanh ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=81 \[ -\frac{d^2 (c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{b c^2 d^2}{x}+\frac{4}{3} b c^3 d^2 \log (x)-\frac{4}{3} b c^3 d^2 \log (1-c x)-\frac{b c d^2}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0849651, antiderivative size = 81, 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} \[ -\frac{d^2 (c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac{b c^2 d^2}{x}+\frac{4}{3} b c^3 d^2 \log (x)-\frac{4}{3} b c^3 d^2 \log (1-c x)-\frac{b c d^2}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^4,x]

[Out]

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

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

Mathematica [A]  time = 0.0938123, size = 103, normalized size = 1.27 \[ -\frac{d^2 \left (6 a c^2 x^2+6 a c x+2 a+6 b c^2 x^2-8 b c^3 x^3 \log (x)+7 b c^3 x^3 \log (1-c x)+b c^3 x^3 \log (c x+1)+2 b \left (3 c^2 x^2+3 c x+1\right ) \tanh ^{-1}(c x)+b c x\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^4,x]

[Out]

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

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Maple [A]  time = 0.037, size = 141, normalized size = 1.7 \begin{align*} -{\frac{{c}^{2}{d}^{2}a}{x}}-{\frac{c{d}^{2}a}{{x}^{2}}}-{\frac{{d}^{2}a}{3\,{x}^{3}}}-{\frac{{c}^{2}{d}^{2}b{\it Artanh} \left ( cx \right ) }{x}}-{\frac{c{d}^{2}b{\it Artanh} \left ( cx \right ) }{{x}^{2}}}-{\frac{{d}^{2}b{\it Artanh} \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{7\,{c}^{3}{d}^{2}b\ln \left ( cx-1 \right ) }{6}}-{\frac{c{d}^{2}b}{6\,{x}^{2}}}-{\frac{{c}^{2}{d}^{2}b}{x}}+{\frac{4\,{c}^{3}{d}^{2}b\ln \left ( cx \right ) }{3}}-{\frac{{c}^{3}{d}^{2}b\ln \left ( cx+1 \right ) }{6}} \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^4,x)

[Out]

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

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Maxima [B]  time = 0.961842, size = 212, normalized size = 2.62 \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^{2} d^{2} + \frac{1}{2} \,{\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 d^{2} - \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 d^{2} - \frac{a c^{2} d^{2}}{x} - \frac{a c d^{2}}{x^{2}} - \frac{a d^{2}}{3 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.17284, size = 293, normalized size = 3.62 \begin{align*} -\frac{b c^{3} d^{2} x^{3} \log \left (c x + 1\right ) + 7 \, b c^{3} d^{2} x^{3} \log \left (c x - 1\right ) - 8 \, b c^{3} d^{2} x^{3} \log \left (x\right ) + 6 \,{\left (a + b\right )} c^{2} d^{2} x^{2} +{\left (6 \, a + b\right )} c d^{2} x + 2 \, a d^{2} +{\left (3 \, b c^{2} d^{2} x^{2} + 3 \, b c d^{2} x + b d^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{6 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.61552, size = 158, normalized size = 1.95 \begin{align*} \begin{cases} - \frac{a c^{2} d^{2}}{x} - \frac{a c d^{2}}{x^{2}} - \frac{a d^{2}}{3 x^{3}} + \frac{4 b c^{3} d^{2} \log{\left (x \right )}}{3} - \frac{4 b c^{3} d^{2} \log{\left (x - \frac{1}{c} \right )}}{3} - \frac{b c^{3} d^{2} \operatorname{atanh}{\left (c x \right )}}{3} - \frac{b c^{2} d^{2} \operatorname{atanh}{\left (c x \right )}}{x} - \frac{b c^{2} d^{2}}{x} - \frac{b c d^{2} \operatorname{atanh}{\left (c x \right )}}{x^{2}} - \frac{b c d^{2}}{6 x^{2}} - \frac{b d^{2} \operatorname{atanh}{\left (c x \right )}}{3 x^{3}} & \text{for}\: c \neq 0 \\- \frac{a d^{2}}{3 x^{3}} & \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**4,x)

[Out]

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

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Giac [A]  time = 1.30308, size = 188, normalized size = 2.32 \begin{align*} -\frac{1}{6} \, b c^{3} d^{2} \log \left (c x + 1\right ) - \frac{7}{6} \, b c^{3} d^{2} \log \left (c x - 1\right ) + \frac{4}{3} \, b c^{3} d^{2} \log \left (x\right ) - \frac{{\left (3 \, b c^{2} d^{2} x^{2} + 3 \, b c d^{2} x + b d^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{6 \, x^{3}} - \frac{6 \, a c^{2} d^{2} x^{2} + 6 \, b c^{2} d^{2} x^{2} + 6 \, a c d^{2} x + b c d^{2} x + 2 \, a d^{2}}{6 \, x^{3}} \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^4,x, algorithm="giac")

[Out]

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