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

Optimal. Leaf size=77 \[ \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (c x+1)^2}-\frac{3 b}{8 c^2 d^3 (c x+1)}+\frac{b}{8 c^2 d^3 (c x+1)^2}-\frac{b \tanh ^{-1}(c x)}{8 c^2 d^3} \]

[Out]

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

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Rubi [A]  time = 0.0822645, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {37, 5936, 12, 88, 207} \[ \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (c x+1)^2}-\frac{3 b}{8 c^2 d^3 (c x+1)}+\frac{b}{8 c^2 d^3 (c x+1)^2}-\frac{b \tanh ^{-1}(c x)}{8 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.106985, size = 99, normalized size = 1.29 \[ -\frac{16 a c x+8 a-3 b c^2 x^2 \log (c x+1)+6 b c x-6 b c x \log (c x+1)+3 b (c x+1)^2 \log (1-c x)-3 b \log (c x+1)+8 (2 b c x+b) \tanh ^{-1}(c x)+4 b}{16 c^2 d^3 (c x+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(8*a + 4*b + 16*a*c*x + 6*b*c*x + 8*(b + 2*b*c*x)*ArcTanh[c*x] + 3*b*(1 + c*x)^2*Log[1 - c*x] - 3*b*Log[1 + c
*x] - 6*b*c*x*Log[1 + c*x] - 3*b*c^2*x^2*Log[1 + c*x])/(16*c^2*d^3*(1 + c*x)^2)

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Maple [A]  time = 0.037, size = 136, normalized size = 1.8 \begin{align*}{\frac{a}{2\,{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{a}{{c}^{2}{d}^{3} \left ( cx+1 \right ) }}+{\frac{b{\it Artanh} \left ( cx \right ) }{2\,{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{b{\it Artanh} \left ( cx \right ) }{{c}^{2}{d}^{3} \left ( cx+1 \right ) }}-{\frac{3\,b\ln \left ( cx-1 \right ) }{16\,{c}^{2}{d}^{3}}}+{\frac{b}{8\,{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{3\,b}{8\,{c}^{2}{d}^{3} \left ( cx+1 \right ) }}+{\frac{3\,b\ln \left ( cx+1 \right ) }{16\,{c}^{2}{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x))/(c*d*x+d)^3,x)

[Out]

1/2/c^2*a/d^3/(c*x+1)^2-1/c^2*a/d^3/(c*x+1)+1/2/c^2*b/d^3*arctanh(c*x)/(c*x+1)^2-1/c^2*b/d^3*arctanh(c*x)/(c*x
+1)-3/16/c^2*b/d^3*ln(c*x-1)+1/8*b/c^2/d^3/(c*x+1)^2-3/8*b/c^2/d^3/(c*x+1)+3/16/c^2*b/d^3*ln(c*x+1)

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Maxima [B]  time = 0.988733, size = 205, normalized size = 2.66 \begin{align*} -\frac{1}{16} \,{\left (c{\left (\frac{2 \,{\left (3 \, c x + 2\right )}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} - \frac{3 \, \log \left (c x + 1\right )}{c^{3} d^{3}} + \frac{3 \, \log \left (c x - 1\right )}{c^{3} d^{3}}\right )} + \frac{8 \,{\left (2 \, c x + 1\right )} \operatorname{artanh}\left (c x\right )}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}}\right )} b - \frac{{\left (2 \, c x + 1\right )} a}{2 \,{\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

-1/16*(c*(2*(3*c*x + 2)/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) - 3*log(c*x + 1)/(c^3*d^3) + 3*log(c*x - 1)/(c^3
*d^3)) + 8*(2*c*x + 1)*arctanh(c*x)/(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3))*b - 1/2*(2*c*x + 1)*a/(c^4*d^3*x^2
+ 2*c^3*d^3*x + c^2*d^3)

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Fricas [A]  time = 2.13021, size = 180, normalized size = 2.34 \begin{align*} -\frac{2 \,{\left (8 \, a + 3 \, b\right )} c x -{\left (3 \, b c^{2} x^{2} - 2 \, b c x - b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 8 \, a + 4 \, b}{16 \,{\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.76345, size = 389, normalized size = 5.05 \begin{align*} \begin{cases} \frac{8 a c^{2} x^{2}}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} - \frac{8 a c x}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} - \frac{4 a}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} + \frac{9 b c^{2} x^{2} \operatorname{atanh}{\left (c x \right )}}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} + \frac{3 b c^{2} x^{2}}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} - \frac{6 b c x \operatorname{atanh}{\left (c x \right )}}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} - \frac{3 b c x}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} - \frac{3 b \operatorname{atanh}{\left (c x \right )}}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} - \frac{3 b}{24 c^{4} d^{3} x^{2} + 48 c^{3} d^{3} x + 24 c^{2} d^{3}} & \text{for}\: d \neq 0 \\\tilde{\infty } \left (\frac{a x^{2}}{2} + \frac{b x^{2} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{b x}{2 c} - \frac{b \operatorname{atanh}{\left (c x \right )}}{2 c^{2}}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atanh(c*x))/(c*d*x+d)**3,x)

[Out]

Piecewise((8*a*c**2*x**2/(24*c**4*d**3*x**2 + 48*c**3*d**3*x + 24*c**2*d**3) - 8*a*c*x/(24*c**4*d**3*x**2 + 48
*c**3*d**3*x + 24*c**2*d**3) - 4*a/(24*c**4*d**3*x**2 + 48*c**3*d**3*x + 24*c**2*d**3) + 9*b*c**2*x**2*atanh(c
*x)/(24*c**4*d**3*x**2 + 48*c**3*d**3*x + 24*c**2*d**3) + 3*b*c**2*x**2/(24*c**4*d**3*x**2 + 48*c**3*d**3*x +
24*c**2*d**3) - 6*b*c*x*atanh(c*x)/(24*c**4*d**3*x**2 + 48*c**3*d**3*x + 24*c**2*d**3) - 3*b*c*x/(24*c**4*d**3
*x**2 + 48*c**3*d**3*x + 24*c**2*d**3) - 3*b*atanh(c*x)/(24*c**4*d**3*x**2 + 48*c**3*d**3*x + 24*c**2*d**3) -
3*b/(24*c**4*d**3*x**2 + 48*c**3*d**3*x + 24*c**2*d**3), Ne(d, 0)), (zoo*(a*x**2/2 + b*x**2*atanh(c*x)/2 + b*x
/(2*c) - b*atanh(c*x)/(2*c**2)), True))

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Giac [A]  time = 1.20421, size = 178, normalized size = 2.31 \begin{align*} -\frac{{\left (2 \, b c x + b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{4 \,{\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} - \frac{8 \, a c x + 3 \, b c x + 4 \, a + 2 \, b}{8 \,{\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} + \frac{3 \, b \log \left (c x + 1\right )}{16 \, c^{2} d^{3}} - \frac{3 \, b \log \left (c x - 1\right )}{16 \, c^{2} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="giac")

[Out]

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