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

Optimal. Leaf size=135 \[ \frac{d^3 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{b d^3 (c x+1)^4}{20 c^2}+\frac{b d^3 (c x+1)^3}{20 c^2}+\frac{3 b d^3 (c x+1)^2}{20 c^2}+\frac{6 b d^3 \log (1-c x)}{5 c^2}+\frac{3 b d^3 x}{5 c} \]

[Out]

(3*b*d^3*x)/(5*c) + (3*b*d^3*(1 + c*x)^2)/(20*c^2) + (b*d^3*(1 + c*x)^3)/(20*c^2) + (b*d^3*(1 + c*x)^4)/(20*c^
2) - (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(4*c^2) + (d^3*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c^2) + (6*b*d^
3*Log[1 - c*x])/(5*c^2)

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Rubi [A]  time = 0.10111, antiderivative size = 135, 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, 77} \[ \frac{d^3 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{b d^3 (c x+1)^4}{20 c^2}+\frac{b d^3 (c x+1)^3}{20 c^2}+\frac{3 b d^3 (c x+1)^2}{20 c^2}+\frac{6 b d^3 \log (1-c x)}{5 c^2}+\frac{3 b d^3 x}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*d^3*x)/(5*c) + (3*b*d^3*(1 + c*x)^2)/(20*c^2) + (b*d^3*(1 + c*x)^3)/(20*c^2) + (b*d^3*(1 + c*x)^4)/(20*c^
2) - (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(4*c^2) + (d^3*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c^2) + (6*b*d^
3*Log[1 - c*x])/(5*c^2)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int x (d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-(b c) \int \frac{(-1+4 c x) (d+c d x)^3}{20 c^2 (1-c x)} \, dx\\ &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-\frac{b \int \frac{(-1+4 c x) (d+c d x)^3}{1-c x} \, dx}{20 c}\\ &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-\frac{b \int \left (-12 d^3-\frac{24 d^3}{-1+c x}-6 d^2 (d+c d x)-3 d (d+c d x)^2-4 (d+c d x)^3\right ) \, dx}{20 c}\\ &=\frac{3 b d^3 x}{5 c}+\frac{3 b d^3 (1+c x)^2}{20 c^2}+\frac{b d^3 (1+c x)^3}{20 c^2}+\frac{b d^3 (1+c x)^4}{20 c^2}-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{6 b d^3 \log (1-c x)}{5 c^2}\\ \end{align*}

Mathematica [A]  time = 0.10686, size = 133, normalized size = 0.99 \[ \frac{d^3 \left (8 a c^5 x^5+30 a c^4 x^4+40 a c^3 x^3+20 a c^2 x^2+2 b c^4 x^4+10 b c^3 x^3+24 b c^2 x^2+2 b c^2 x^2 \left (4 c^3 x^3+15 c^2 x^2+20 c x+10\right ) \tanh ^{-1}(c x)+50 b c x+49 b \log (1-c x)-b \log (c x+1)\right )}{40 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(50*b*c*x + 20*a*c^2*x^2 + 24*b*c^2*x^2 + 40*a*c^3*x^3 + 10*b*c^3*x^3 + 30*a*c^4*x^4 + 2*b*c^4*x^4 + 8*a*
c^5*x^5 + 2*b*c^2*x^2*(10 + 20*c*x + 15*c^2*x^2 + 4*c^3*x^3)*ArcTanh[c*x] + 49*b*Log[1 - c*x] - b*Log[1 + c*x]
))/(40*c^2)

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Maple [A]  time = 0.029, size = 173, normalized size = 1.3 \begin{align*}{\frac{{c}^{3}{d}^{3}a{x}^{5}}{5}}+{\frac{3\,{c}^{2}{d}^{3}a{x}^{4}}{4}}+c{d}^{3}a{x}^{3}+{\frac{{d}^{3}a{x}^{2}}{2}}+{\frac{{c}^{3}{d}^{3}b{\it Artanh} \left ( cx \right ){x}^{5}}{5}}+{\frac{3\,{c}^{2}{d}^{3}b{\it Artanh} \left ( cx \right ){x}^{4}}{4}}+c{d}^{3}b{\it Artanh} \left ( cx \right ){x}^{3}+{\frac{{d}^{3}b{\it Artanh} \left ( cx \right ){x}^{2}}{2}}+{\frac{{c}^{2}{d}^{3}b{x}^{4}}{20}}+{\frac{c{d}^{3}b{x}^{3}}{4}}+{\frac{3\,{d}^{3}b{x}^{2}}{5}}+{\frac{5\,b{d}^{3}x}{4\,c}}+{\frac{49\,{d}^{3}b\ln \left ( cx-1 \right ) }{40\,{c}^{2}}}-{\frac{{d}^{3}b\ln \left ( cx+1 \right ) }{40\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*c^3*d^3*a*x^5+3/4*c^2*d^3*a*x^4+c*d^3*a*x^3+1/2*d^3*a*x^2+1/5*c^3*d^3*b*arctanh(c*x)*x^5+3/4*c^2*d^3*b*arc
tanh(c*x)*x^4+c*d^3*b*arctanh(c*x)*x^3+1/2*d^3*b*arctanh(c*x)*x^2+1/20*c^2*d^3*b*x^4+1/4*c*d^3*b*x^3+3/5*d^3*b
*x^2+5/4*b*d^3*x/c+49/40/c^2*d^3*b*ln(c*x-1)-1/40/c^2*d^3*b*ln(c*x+1)

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Maxima [B]  time = 0.968479, size = 329, normalized size = 2.44 \begin{align*} \frac{1}{5} \, a c^{3} d^{3} x^{5} + \frac{3}{4} \, a c^{2} d^{3} x^{4} + \frac{1}{20} \,{\left (4 \, x^{5} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{3} d^{3} + a c d^{3} x^{3} + \frac{1}{8} \,{\left (6 \, x^{4} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \,{\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac{3 \, \log \left (c x + 1\right )}{c^{5}} + \frac{3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{2} d^{3} + \frac{1}{2} \,{\left (2 \, x^{3} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{x^{2}}{c^{2}} + \frac{\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c d^{3} + \frac{1}{2} \, a d^{3} x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \, x}{c^{2}} - \frac{\log \left (c x + 1\right )}{c^{3}} + \frac{\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^3*d^3*x^5 + 3/4*a*c^2*d^3*x^4 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 -
1)/c^6))*b*c^3*d^3 + a*c*d^3*x^3 + 1/8*(6*x^4*arctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3
*log(c*x - 1)/c^5))*b*c^2*d^3 + 1/2*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*c*d^3 + 1/2*a*
d^3*x^2 + 1/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b*d^3

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Fricas [A]  time = 1.89864, size = 369, normalized size = 2.73 \begin{align*} \frac{8 \, a c^{5} d^{3} x^{5} + 2 \,{\left (15 \, a + b\right )} c^{4} d^{3} x^{4} + 10 \,{\left (4 \, a + b\right )} c^{3} d^{3} x^{3} + 4 \,{\left (5 \, a + 6 \, b\right )} c^{2} d^{3} x^{2} + 50 \, b c d^{3} x - b d^{3} \log \left (c x + 1\right ) + 49 \, b d^{3} \log \left (c x - 1\right ) +{\left (4 \, b c^{5} d^{3} x^{5} + 15 \, b c^{4} d^{3} x^{4} + 20 \, b c^{3} d^{3} x^{3} + 10 \, b c^{2} d^{3} x^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{40 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(8*a*c^5*d^3*x^5 + 2*(15*a + b)*c^4*d^3*x^4 + 10*(4*a + b)*c^3*d^3*x^3 + 4*(5*a + 6*b)*c^2*d^3*x^2 + 50*b
*c*d^3*x - b*d^3*log(c*x + 1) + 49*b*d^3*log(c*x - 1) + (4*b*c^5*d^3*x^5 + 15*b*c^4*d^3*x^4 + 20*b*c^3*d^3*x^3
+ 10*b*c^2*d^3*x^2)*log(-(c*x + 1)/(c*x - 1)))/c^2

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Sympy [A]  time = 4.20865, size = 211, normalized size = 1.56 \begin{align*} \begin{cases} \frac{a c^{3} d^{3} x^{5}}{5} + \frac{3 a c^{2} d^{3} x^{4}}{4} + a c d^{3} x^{3} + \frac{a d^{3} x^{2}}{2} + \frac{b c^{3} d^{3} x^{5} \operatorname{atanh}{\left (c x \right )}}{5} + \frac{3 b c^{2} d^{3} x^{4} \operatorname{atanh}{\left (c x \right )}}{4} + \frac{b c^{2} d^{3} x^{4}}{20} + b c d^{3} x^{3} \operatorname{atanh}{\left (c x \right )} + \frac{b c d^{3} x^{3}}{4} + \frac{b d^{3} x^{2} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{3 b d^{3} x^{2}}{5} + \frac{5 b d^{3} x}{4 c} + \frac{6 b d^{3} \log{\left (x - \frac{1}{c} \right )}}{5 c^{2}} - \frac{b d^{3} \operatorname{atanh}{\left (c x \right )}}{20 c^{2}} & \text{for}\: c \neq 0 \\\frac{a d^{3} x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**3*d**3*x**5/5 + 3*a*c**2*d**3*x**4/4 + a*c*d**3*x**3 + a*d**3*x**2/2 + b*c**3*d**3*x**5*atanh(
c*x)/5 + 3*b*c**2*d**3*x**4*atanh(c*x)/4 + b*c**2*d**3*x**4/20 + b*c*d**3*x**3*atanh(c*x) + b*c*d**3*x**3/4 +
b*d**3*x**2*atanh(c*x)/2 + 3*b*d**3*x**2/5 + 5*b*d**3*x/(4*c) + 6*b*d**3*log(x - 1/c)/(5*c**2) - b*d**3*atanh(
c*x)/(20*c**2), Ne(c, 0)), (a*d**3*x**2/2, True))

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Giac [A]  time = 1.21847, size = 235, normalized size = 1.74 \begin{align*} \frac{1}{5} \, a c^{3} d^{3} x^{5} + \frac{1}{20} \,{\left (15 \, a c^{2} d^{3} + b c^{2} d^{3}\right )} x^{4} + \frac{5 \, b d^{3} x}{4 \, c} + \frac{1}{4} \,{\left (4 \, a c d^{3} + b c d^{3}\right )} x^{3} + \frac{1}{10} \,{\left (5 \, a d^{3} + 6 \, b d^{3}\right )} x^{2} - \frac{b d^{3} \log \left (c x + 1\right )}{40 \, c^{2}} + \frac{49 \, b d^{3} \log \left (c x - 1\right )}{40 \, c^{2}} + \frac{1}{40} \,{\left (4 \, b c^{3} d^{3} x^{5} + 15 \, b c^{2} d^{3} x^{4} + 20 \, b c d^{3} x^{3} + 10 \, b d^{3} x^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^3*d^3*x^5 + 1/20*(15*a*c^2*d^3 + b*c^2*d^3)*x^4 + 5/4*b*d^3*x/c + 1/4*(4*a*c*d^3 + b*c*d^3)*x^3 + 1/10
*(5*a*d^3 + 6*b*d^3)*x^2 - 1/40*b*d^3*log(c*x + 1)/c^2 + 49/40*b*d^3*log(c*x - 1)/c^2 + 1/40*(4*b*c^3*d^3*x^5
+ 15*b*c^2*d^3*x^4 + 20*b*c*d^3*x^3 + 10*b*d^3*x^2)*log(-(c*x + 1)/(c*x - 1))