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

Optimal. Leaf size=143 \[ \frac{1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b d^2 x}{2 c^2}+\frac{31 b d^2 \log (1-c x)}{60 c^3}+\frac{b d^2 \log (c x+1)}{60 c^3}+\frac{1}{20} b c d^2 x^4+\frac{4 b d^2 x^2}{15 c}+\frac{1}{6} b d^2 x^3 \]

[Out]

(b*d^2*x)/(2*c^2) + (4*b*d^2*x^2)/(15*c) + (b*d^2*x^3)/6 + (b*c*d^2*x^4)/20 + (d^2*x^3*(a + b*ArcTanh[c*x]))/3
 + (c*d^2*x^4*(a + b*ArcTanh[c*x]))/2 + (c^2*d^2*x^5*(a + b*ArcTanh[c*x]))/5 + (31*b*d^2*Log[1 - c*x])/(60*c^3
) + (b*d^2*Log[1 + c*x])/(60*c^3)

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Rubi [A]  time = 0.154411, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {43, 5936, 12, 1802, 633, 31} \[ \frac{1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b d^2 x}{2 c^2}+\frac{31 b d^2 \log (1-c x)}{60 c^3}+\frac{b d^2 \log (c x+1)}{60 c^3}+\frac{1}{20} b c d^2 x^4+\frac{4 b d^2 x^2}{15 c}+\frac{1}{6} b d^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^2*x)/(2*c^2) + (4*b*d^2*x^2)/(15*c) + (b*d^2*x^3)/6 + (b*c*d^2*x^4)/20 + (d^2*x^3*(a + b*ArcTanh[c*x]))/3
 + (c*d^2*x^4*(a + b*ArcTanh[c*x]))/2 + (c^2*d^2*x^5*(a + b*ArcTanh[c*x]))/5 + (31*b*d^2*Log[1 - c*x])/(60*c^3
) + (b*d^2*Log[1 + c*x])/(60*c^3)

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]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int x^2 (d+c d x)^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )-(b c) \int \frac{d^2 x^3 \left (10+15 c x+6 c^2 x^2\right )}{30 \left (1-c^2 x^2\right )} \, dx\\ &=\frac{1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{30} \left (b c d^2\right ) \int \frac{x^3 \left (10+15 c x+6 c^2 x^2\right )}{1-c^2 x^2} \, dx\\ &=\frac{1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{30} \left (b c d^2\right ) \int \left (-\frac{15}{c^3}-\frac{16 x}{c^2}-\frac{15 x^2}{c}-6 x^3+\frac{15+16 c x}{c^3 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac{b d^2 x}{2 c^2}+\frac{4 b d^2 x^2}{15 c}+\frac{1}{6} b d^2 x^3+\frac{1}{20} b c d^2 x^4+\frac{1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac{\left (b d^2\right ) \int \frac{15+16 c x}{1-c^2 x^2} \, dx}{30 c^2}\\ &=\frac{b d^2 x}{2 c^2}+\frac{4 b d^2 x^2}{15 c}+\frac{1}{6} b d^2 x^3+\frac{1}{20} b c d^2 x^4+\frac{1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac{\left (b d^2\right ) \int \frac{1}{-c-c^2 x} \, dx}{60 c}-\frac{\left (31 b d^2\right ) \int \frac{1}{c-c^2 x} \, dx}{60 c}\\ &=\frac{b d^2 x}{2 c^2}+\frac{4 b d^2 x^2}{15 c}+\frac{1}{6} b d^2 x^3+\frac{1}{20} b c d^2 x^4+\frac{1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac{31 b d^2 \log (1-c x)}{60 c^3}+\frac{b d^2 \log (1+c x)}{60 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0936998, size = 115, normalized size = 0.8 \[ \frac{d^2 \left (12 a c^5 x^5+30 a c^4 x^4+20 a c^3 x^3+3 b c^4 x^4+10 b c^3 x^3+16 b c^2 x^2+2 b c^3 x^3 \left (6 c^2 x^2+15 c x+10\right ) \tanh ^{-1}(c x)+30 b c x+31 b \log (1-c x)+b \log (c x+1)\right )}{60 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(30*b*c*x + 16*b*c^2*x^2 + 20*a*c^3*x^3 + 10*b*c^3*x^3 + 30*a*c^4*x^4 + 3*b*c^4*x^4 + 12*a*c^5*x^5 + 2*b*
c^3*x^3*(10 + 15*c*x + 6*c^2*x^2)*ArcTanh[c*x] + 31*b*Log[1 - c*x] + b*Log[1 + c*x]))/(60*c^3)

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Maple [A]  time = 0.029, size = 147, normalized size = 1. \begin{align*}{\frac{{c}^{2}{d}^{2}a{x}^{5}}{5}}+{\frac{c{d}^{2}a{x}^{4}}{2}}+{\frac{{d}^{2}a{x}^{3}}{3}}+{\frac{{c}^{2}{d}^{2}b{\it Artanh} \left ( cx \right ){x}^{5}}{5}}+{\frac{c{d}^{2}b{\it Artanh} \left ( cx \right ){x}^{4}}{2}}+{\frac{{d}^{2}b{\it Artanh} \left ( cx \right ){x}^{3}}{3}}+{\frac{bc{d}^{2}{x}^{4}}{20}}+{\frac{b{d}^{2}{x}^{3}}{6}}+{\frac{4\,b{d}^{2}{x}^{2}}{15\,c}}+{\frac{b{d}^{2}x}{2\,{c}^{2}}}+{\frac{31\,{d}^{2}b\ln \left ( cx-1 \right ) }{60\,{c}^{3}}}+{\frac{{d}^{2}b\ln \left ( cx+1 \right ) }{60\,{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*d*x+d)^2*(a+b*arctanh(c*x)),x)

[Out]

1/5*c^2*d^2*a*x^5+1/2*c*d^2*a*x^4+1/3*d^2*a*x^3+1/5*c^2*d^2*b*arctanh(c*x)*x^5+1/2*c*d^2*b*arctanh(c*x)*x^4+1/
3*d^2*b*arctanh(c*x)*x^3+1/20*b*c*d^2*x^4+1/6*b*d^2*x^3+4/15*b*d^2*x^2/c+1/2*b*d^2*x/c^2+31/60/c^3*d^2*b*ln(c*
x-1)+1/60*b*d^2*ln(c*x+1)/c^3

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Maxima [A]  time = 0.975962, size = 248, normalized size = 1.73 \begin{align*} \frac{1}{5} \, a c^{2} d^{2} x^{5} + \frac{1}{2} \, a c d^{2} 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^{2} d^{2} + \frac{1}{3} \, a d^{2} x^{3} + \frac{1}{12} \,{\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 d^{2} + \frac{1}{6} \,{\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 d^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/5*a*c^2*d^2*x^5 + 1/2*a*c*d^2*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^2*d^2 + 1/3*a*d^2*x^3 + 1/12*(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*d^2 + 1/6*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*d^2

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Fricas [A]  time = 2.05371, size = 332, normalized size = 2.32 \begin{align*} \frac{12 \, a c^{5} d^{2} x^{5} + 3 \,{\left (10 \, a + b\right )} c^{4} d^{2} x^{4} + 10 \,{\left (2 \, a + b\right )} c^{3} d^{2} x^{3} + 16 \, b c^{2} d^{2} x^{2} + 30 \, b c d^{2} x + b d^{2} \log \left (c x + 1\right ) + 31 \, b d^{2} \log \left (c x - 1\right ) +{\left (6 \, b c^{5} d^{2} x^{5} + 15 \, b c^{4} d^{2} x^{4} + 10 \, b c^{3} d^{2} x^{3}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{60 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/60*(12*a*c^5*d^2*x^5 + 3*(10*a + b)*c^4*d^2*x^4 + 10*(2*a + b)*c^3*d^2*x^3 + 16*b*c^2*d^2*x^2 + 30*b*c*d^2*x
 + b*d^2*log(c*x + 1) + 31*b*d^2*log(c*x - 1) + (6*b*c^5*d^2*x^5 + 15*b*c^4*d^2*x^4 + 10*b*c^3*d^2*x^3)*log(-(
c*x + 1)/(c*x - 1)))/c^3

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Sympy [A]  time = 3.03088, size = 177, normalized size = 1.24 \begin{align*} \begin{cases} \frac{a c^{2} d^{2} x^{5}}{5} + \frac{a c d^{2} x^{4}}{2} + \frac{a d^{2} x^{3}}{3} + \frac{b c^{2} d^{2} x^{5} \operatorname{atanh}{\left (c x \right )}}{5} + \frac{b c d^{2} x^{4} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{b c d^{2} x^{4}}{20} + \frac{b d^{2} x^{3} \operatorname{atanh}{\left (c x \right )}}{3} + \frac{b d^{2} x^{3}}{6} + \frac{4 b d^{2} x^{2}}{15 c} + \frac{b d^{2} x}{2 c^{2}} + \frac{8 b d^{2} \log{\left (x - \frac{1}{c} \right )}}{15 c^{3}} + \frac{b d^{2} \operatorname{atanh}{\left (c x \right )}}{30 c^{3}} & \text{for}\: c \neq 0 \\\frac{a d^{2} x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(c*d*x+d)**2*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*c**2*d**2*x**5/5 + a*c*d**2*x**4/2 + a*d**2*x**3/3 + b*c**2*d**2*x**5*atanh(c*x)/5 + b*c*d**2*x**
4*atanh(c*x)/2 + b*c*d**2*x**4/20 + b*d**2*x**3*atanh(c*x)/3 + b*d**2*x**3/6 + 4*b*d**2*x**2/(15*c) + b*d**2*x
/(2*c**2) + 8*b*d**2*log(x - 1/c)/(15*c**3) + b*d**2*atanh(c*x)/(30*c**3), Ne(c, 0)), (a*d**2*x**3/3, True))

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Giac [A]  time = 1.32498, size = 203, normalized size = 1.42 \begin{align*} \frac{1}{5} \, a c^{2} d^{2} x^{5} + \frac{1}{20} \,{\left (10 \, a c d^{2} + b c d^{2}\right )} x^{4} + \frac{4 \, b d^{2} x^{2}}{15 \, c} + \frac{1}{6} \,{\left (2 \, a d^{2} + b d^{2}\right )} x^{3} + \frac{b d^{2} x}{2 \, c^{2}} + \frac{1}{60} \,{\left (6 \, b c^{2} d^{2} x^{5} + 15 \, b c d^{2} x^{4} + 10 \, b d^{2} x^{3}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) + \frac{b d^{2} \log \left (c x + 1\right )}{60 \, c^{3}} + \frac{31 \, b d^{2} \log \left (c x - 1\right )}{60 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

1/5*a*c^2*d^2*x^5 + 1/20*(10*a*c*d^2 + b*c*d^2)*x^4 + 4/15*b*d^2*x^2/c + 1/6*(2*a*d^2 + b*d^2)*x^3 + 1/2*b*d^2
*x/c^2 + 1/60*(6*b*c^2*d^2*x^5 + 15*b*c*d^2*x^4 + 10*b*d^2*x^3)*log(-(c*x + 1)/(c*x - 1)) + 1/60*b*d^2*log(c*x
 + 1)/c^3 + 31/60*b*d^2*log(c*x - 1)/c^3