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

Optimal. Leaf size=153 \[ \frac{d^4 (c x+1)^6 \left (a+b \tanh ^{-1}(c x)\right )}{6 c^2}-\frac{d^4 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{b d^4 (c x+1)^5}{30 c^2}+\frac{b d^4 (c x+1)^4}{30 c^2}+\frac{4 b d^4 (c x+1)^3}{45 c^2}+\frac{4 b d^4 (c x+1)^2}{15 c^2}+\frac{32 b d^4 \log (1-c x)}{15 c^2}+\frac{16 b d^4 x}{15 c} \]

[Out]

(16*b*d^4*x)/(15*c) + (4*b*d^4*(1 + c*x)^2)/(15*c^2) + (4*b*d^4*(1 + c*x)^3)/(45*c^2) + (b*d^4*(1 + c*x)^4)/(3
0*c^2) + (b*d^4*(1 + c*x)^5)/(30*c^2) - (d^4*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c^2) + (d^4*(1 + c*x)^6*(a +
b*ArcTanh[c*x]))/(6*c^2) + (32*b*d^4*Log[1 - c*x])/(15*c^2)

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Rubi [A]  time = 0.116179, antiderivative size = 153, 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} \[ \frac{d^4 (c x+1)^6 \left (a+b \tanh ^{-1}(c x)\right )}{6 c^2}-\frac{d^4 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{b d^4 (c x+1)^5}{30 c^2}+\frac{b d^4 (c x+1)^4}{30 c^2}+\frac{4 b d^4 (c x+1)^3}{45 c^2}+\frac{4 b d^4 (c x+1)^2}{15 c^2}+\frac{32 b d^4 \log (1-c x)}{15 c^2}+\frac{16 b d^4 x}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b*d^4*x)/(15*c) + (4*b*d^4*(1 + c*x)^2)/(15*c^2) + (4*b*d^4*(1 + c*x)^3)/(45*c^2) + (b*d^4*(1 + c*x)^4)/(3
0*c^2) + (b*d^4*(1 + c*x)^5)/(30*c^2) - (d^4*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c^2) + (d^4*(1 + c*x)^6*(a +
b*ArcTanh[c*x]))/(6*c^2) + (32*b*d^4*Log[1 - c*x])/(15*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)^4 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=-\frac{d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{d^4 (1+c x)^6 \left (a+b \tanh ^{-1}(c x)\right )}{6 c^2}-(b c) \int \frac{(-1+5 c x) (d+c d x)^4}{30 c^2 (1-c x)} \, dx\\ &=-\frac{d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{d^4 (1+c x)^6 \left (a+b \tanh ^{-1}(c x)\right )}{6 c^2}-\frac{b \int \frac{(-1+5 c x) (d+c d x)^4}{1-c x} \, dx}{30 c}\\ &=-\frac{d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{d^4 (1+c x)^6 \left (a+b \tanh ^{-1}(c x)\right )}{6 c^2}-\frac{b \int \left (-32 d^4-\frac{64 d^4}{-1+c x}-16 d^3 (d+c d x)-8 d^2 (d+c d x)^2-4 d (d+c d x)^3-5 (d+c d x)^4\right ) \, dx}{30 c}\\ &=\frac{16 b d^4 x}{15 c}+\frac{4 b d^4 (1+c x)^2}{15 c^2}+\frac{4 b d^4 (1+c x)^3}{45 c^2}+\frac{b d^4 (1+c x)^4}{30 c^2}+\frac{b d^4 (1+c x)^5}{30 c^2}-\frac{d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{d^4 (1+c x)^6 \left (a+b \tanh ^{-1}(c x)\right )}{6 c^2}+\frac{32 b d^4 \log (1-c x)}{15 c^2}\\ \end{align*}

Mathematica [A]  time = 0.141229, size = 159, normalized size = 1.04 \[ \frac{d^4 \left (30 a c^6 x^6+144 a c^5 x^5+270 a c^4 x^4+240 a c^3 x^3+90 a c^2 x^2+6 b c^5 x^5+36 b c^4 x^4+100 b c^3 x^3+192 b c^2 x^2+6 b c^2 x^2 \left (5 c^4 x^4+24 c^3 x^3+45 c^2 x^2+40 c x+15\right ) \tanh ^{-1}(c x)+390 b c x+387 b \log (1-c x)-3 b \log (c x+1)\right )}{180 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^4*(390*b*c*x + 90*a*c^2*x^2 + 192*b*c^2*x^2 + 240*a*c^3*x^3 + 100*b*c^3*x^3 + 270*a*c^4*x^4 + 36*b*c^4*x^4
+ 144*a*c^5*x^5 + 6*b*c^5*x^5 + 30*a*c^6*x^6 + 6*b*c^2*x^2*(15 + 40*c*x + 45*c^2*x^2 + 24*c^3*x^3 + 5*c^4*x^4)
*ArcTanh[c*x] + 387*b*Log[1 - c*x] - 3*b*Log[1 + c*x]))/(180*c^2)

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Maple [A]  time = 0.03, size = 215, normalized size = 1.4 \begin{align*}{\frac{{c}^{4}{d}^{4}a{x}^{6}}{6}}+{\frac{4\,{c}^{3}{d}^{4}a{x}^{5}}{5}}+{\frac{3\,{c}^{2}{d}^{4}a{x}^{4}}{2}}+{\frac{4\,c{d}^{4}a{x}^{3}}{3}}+{\frac{{d}^{4}a{x}^{2}}{2}}+{\frac{{c}^{4}{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{6}}{6}}+{\frac{4\,{c}^{3}{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{5}}{5}}+{\frac{3\,{c}^{2}{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{4}}{2}}+{\frac{4\,c{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{3}}{3}}+{\frac{{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{2}}{2}}+{\frac{{c}^{3}{d}^{4}b{x}^{5}}{30}}+{\frac{{c}^{2}{d}^{4}b{x}^{4}}{5}}+{\frac{5\,c{d}^{4}b{x}^{3}}{9}}+{\frac{16\,{d}^{4}b{x}^{2}}{15}}+{\frac{13\,b{d}^{4}x}{6\,c}}+{\frac{43\,{d}^{4}b\ln \left ( cx-1 \right ) }{20\,{c}^{2}}}-{\frac{{d}^{4}b\ln \left ( cx+1 \right ) }{60\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*d*x+d)^4*(a+b*arctanh(c*x)),x)

[Out]

1/6*c^4*d^4*a*x^6+4/5*c^3*d^4*a*x^5+3/2*c^2*d^4*a*x^4+4/3*c*d^4*a*x^3+1/2*d^4*a*x^2+1/6*c^4*d^4*b*arctanh(c*x)
*x^6+4/5*c^3*d^4*b*arctanh(c*x)*x^5+3/2*c^2*d^4*b*arctanh(c*x)*x^4+4/3*c*d^4*b*arctanh(c*x)*x^3+1/2*d^4*b*arct
anh(c*x)*x^2+1/30*c^3*d^4*b*x^5+1/5*c^2*d^4*b*x^4+5/9*c*d^4*b*x^3+16/15*d^4*b*x^2+13/6*b*d^4*x/c+43/20/c^2*d^4
*b*ln(c*x-1)-1/60/c^2*d^4*b*ln(c*x+1)

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Maxima [B]  time = 0.978578, size = 440, normalized size = 2.88 \begin{align*} \frac{1}{6} \, a c^{4} d^{4} x^{6} + \frac{4}{5} \, a c^{3} d^{4} x^{5} + \frac{3}{2} \, a c^{2} d^{4} x^{4} + \frac{1}{180} \,{\left (30 \, x^{6} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \,{\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac{15 \, \log \left (c x + 1\right )}{c^{7}} + \frac{15 \, \log \left (c x - 1\right )}{c^{7}}\right )}\right )} b c^{4} d^{4} + \frac{1}{5} \,{\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^{4} + \frac{4}{3} \, a c d^{4} x^{3} + \frac{1}{4} \,{\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^{4} + \frac{2}{3} \,{\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^{4} + \frac{1}{2} \, a d^{4} 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^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/6*a*c^4*d^4*x^6 + 4/5*a*c^3*d^4*x^5 + 3/2*a*c^2*d^4*x^4 + 1/180*(30*x^6*arctanh(c*x) + c*(2*(3*c^4*x^5 + 5*c
^2*x^3 + 15*x)/c^6 - 15*log(c*x + 1)/c^7 + 15*log(c*x - 1)/c^7))*b*c^4*d^4 + 1/5*(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^4 + 4/3*a*c*d^4*x^3 + 1/4*(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^4 + 2/3*(2*x^3*arctanh(c*x) + c*(x^2/c^2 +
log(c^2*x^2 - 1)/c^4))*b*c*d^4 + 1/2*a*d^4*x^2 + 1/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + lo
g(c*x - 1)/c^3))*b*d^4

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Fricas [A]  time = 2.12382, size = 452, normalized size = 2.95 \begin{align*} \frac{30 \, a c^{6} d^{4} x^{6} + 6 \,{\left (24 \, a + b\right )} c^{5} d^{4} x^{5} + 18 \,{\left (15 \, a + 2 \, b\right )} c^{4} d^{4} x^{4} + 20 \,{\left (12 \, a + 5 \, b\right )} c^{3} d^{4} x^{3} + 6 \,{\left (15 \, a + 32 \, b\right )} c^{2} d^{4} x^{2} + 390 \, b c d^{4} x - 3 \, b d^{4} \log \left (c x + 1\right ) + 387 \, b d^{4} \log \left (c x - 1\right ) + 3 \,{\left (5 \, b c^{6} d^{4} x^{6} + 24 \, b c^{5} d^{4} x^{5} + 45 \, b c^{4} d^{4} x^{4} + 40 \, b c^{3} d^{4} x^{3} + 15 \, b c^{2} d^{4} x^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{180 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/180*(30*a*c^6*d^4*x^6 + 6*(24*a + b)*c^5*d^4*x^5 + 18*(15*a + 2*b)*c^4*d^4*x^4 + 20*(12*a + 5*b)*c^3*d^4*x^3
+ 6*(15*a + 32*b)*c^2*d^4*x^2 + 390*b*c*d^4*x - 3*b*d^4*log(c*x + 1) + 387*b*d^4*log(c*x - 1) + 3*(5*b*c^6*d^
4*x^6 + 24*b*c^5*d^4*x^5 + 45*b*c^4*d^4*x^4 + 40*b*c^3*d^4*x^3 + 15*b*c^2*d^4*x^2)*log(-(c*x + 1)/(c*x - 1)))/
c^2

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Sympy [A]  time = 4.79549, size = 269, normalized size = 1.76 \begin{align*} \begin{cases} \frac{a c^{4} d^{4} x^{6}}{6} + \frac{4 a c^{3} d^{4} x^{5}}{5} + \frac{3 a c^{2} d^{4} x^{4}}{2} + \frac{4 a c d^{4} x^{3}}{3} + \frac{a d^{4} x^{2}}{2} + \frac{b c^{4} d^{4} x^{6} \operatorname{atanh}{\left (c x \right )}}{6} + \frac{4 b c^{3} d^{4} x^{5} \operatorname{atanh}{\left (c x \right )}}{5} + \frac{b c^{3} d^{4} x^{5}}{30} + \frac{3 b c^{2} d^{4} x^{4} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{b c^{2} d^{4} x^{4}}{5} + \frac{4 b c d^{4} x^{3} \operatorname{atanh}{\left (c x \right )}}{3} + \frac{5 b c d^{4} x^{3}}{9} + \frac{b d^{4} x^{2} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{16 b d^{4} x^{2}}{15} + \frac{13 b d^{4} x}{6 c} + \frac{32 b d^{4} \log{\left (x - \frac{1}{c} \right )}}{15 c^{2}} - \frac{b d^{4} \operatorname{atanh}{\left (c x \right )}}{30 c^{2}} & \text{for}\: c \neq 0 \\\frac{a d^{4} 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)**4*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*c**4*d**4*x**6/6 + 4*a*c**3*d**4*x**5/5 + 3*a*c**2*d**4*x**4/2 + 4*a*c*d**4*x**3/3 + a*d**4*x**2/
2 + b*c**4*d**4*x**6*atanh(c*x)/6 + 4*b*c**3*d**4*x**5*atanh(c*x)/5 + b*c**3*d**4*x**5/30 + 3*b*c**2*d**4*x**4
*atanh(c*x)/2 + b*c**2*d**4*x**4/5 + 4*b*c*d**4*x**3*atanh(c*x)/3 + 5*b*c*d**4*x**3/9 + b*d**4*x**2*atanh(c*x)
/2 + 16*b*d**4*x**2/15 + 13*b*d**4*x/(6*c) + 32*b*d**4*log(x - 1/c)/(15*c**2) - b*d**4*atanh(c*x)/(30*c**2), N
e(c, 0)), (a*d**4*x**2/2, True))

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Giac [A]  time = 1.29912, size = 285, normalized size = 1.86 \begin{align*} \frac{1}{6} \, a c^{4} d^{4} x^{6} + \frac{1}{30} \,{\left (24 \, a c^{3} d^{4} + b c^{3} d^{4}\right )} x^{5} + \frac{13 \, b d^{4} x}{6 \, c} + \frac{1}{10} \,{\left (15 \, a c^{2} d^{4} + 2 \, b c^{2} d^{4}\right )} x^{4} + \frac{1}{9} \,{\left (12 \, a c d^{4} + 5 \, b c d^{4}\right )} x^{3} - \frac{b d^{4} \log \left (c x + 1\right )}{60 \, c^{2}} + \frac{43 \, b d^{4} \log \left (c x - 1\right )}{20 \, c^{2}} + \frac{1}{30} \,{\left (15 \, a d^{4} + 32 \, b d^{4}\right )} x^{2} + \frac{1}{60} \,{\left (5 \, b c^{4} d^{4} x^{6} + 24 \, b c^{3} d^{4} x^{5} + 45 \, b c^{2} d^{4} x^{4} + 40 \, b c d^{4} x^{3} + 15 \, b d^{4} 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)^4*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

1/6*a*c^4*d^4*x^6 + 1/30*(24*a*c^3*d^4 + b*c^3*d^4)*x^5 + 13/6*b*d^4*x/c + 1/10*(15*a*c^2*d^4 + 2*b*c^2*d^4)*x
^4 + 1/9*(12*a*c*d^4 + 5*b*c*d^4)*x^3 - 1/60*b*d^4*log(c*x + 1)/c^2 + 43/20*b*d^4*log(c*x - 1)/c^2 + 1/30*(15*
a*d^4 + 32*b*d^4)*x^2 + 1/60*(5*b*c^4*d^4*x^6 + 24*b*c^3*d^4*x^5 + 45*b*c^2*d^4*x^4 + 40*b*c*d^4*x^3 + 15*b*d^
4*x^2)*log(-(c*x + 1)/(c*x - 1))