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

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

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

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

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 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 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 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]))

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*}

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Mathematica [A]  time = 0.09, size = 115, normalized size = 0.80 \[ \frac {d^2 \left (12 a c^5 x^5+30 a c^4 x^4+20 a c^3 x^3+3 b c^4 x^4+10 b c^3 x^3+16 b c^2 x^2+2 b c^3 x^3 \left (6 c^2 x^2+15 c x+10\right ) \tanh ^{-1}(c x)+30 b c x+31 b \log (1-c x)+b \log (c x+1)\right )}{60 c^3} \]

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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fricas [A]  time = 0.50, size = 146, normalized size = 1.02 \[ \frac {12 \, a c^{5} d^{2} x^{5} + 3 \, {\left (10 \, a + b\right )} c^{4} d^{2} x^{4} + 10 \, {\left (2 \, a + b\right )} c^{3} d^{2} x^{3} + 16 \, b c^{2} d^{2} x^{2} + 30 \, b c d^{2} x + b d^{2} \log \left (c x + 1\right ) + 31 \, b d^{2} \log \left (c x - 1\right ) + {\left (6 \, b c^{5} d^{2} x^{5} + 15 \, b c^{4} d^{2} x^{4} + 10 \, b c^{3} d^{2} x^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{60 \, c^{3}} \]

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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giac [B]  time = 0.23, size = 525, normalized size = 3.67 \[ \frac {4}{15} \, c {\left (\frac {{\left (\frac {15 \, {\left (c x + 1\right )}^{4} b d^{2}}{{\left (c x - 1\right )}^{4}} - \frac {15 \, {\left (c x + 1\right )}^{3} b d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {20 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} - \frac {10 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} + 2 \, b d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5} c^{4}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{4}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{4}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{4}}{c x - 1} - c^{4}} + \frac {\frac {30 \, {\left (c x + 1\right )}^{4} a d^{2}}{{\left (c x - 1\right )}^{4}} - \frac {30 \, {\left (c x + 1\right )}^{3} a d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {40 \, {\left (c x + 1\right )}^{2} a d^{2}}{{\left (c x - 1\right )}^{2}} - \frac {20 \, {\left (c x + 1\right )} a d^{2}}{c x - 1} + 4 \, a d^{2} + \frac {13 \, {\left (c x + 1\right )}^{4} b d^{2}}{{\left (c x - 1\right )}^{4}} - \frac {36 \, {\left (c x + 1\right )}^{3} b d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {41 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} - \frac {23 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} + 5 \, b d^{2}}{\frac {{\left (c x + 1\right )}^{5} c^{4}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{4}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{4}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{4}}{c x - 1} - c^{4}} - \frac {2 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{4}} + \frac {2 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{4}}\right )} \]

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]

4/15*c*((15*(c*x + 1)^4*b*d^2/(c*x - 1)^4 - 15*(c*x + 1)^3*b*d^2/(c*x - 1)^3 + 20*(c*x + 1)^2*b*d^2/(c*x - 1)^
2 - 10*(c*x + 1)*b*d^2/(c*x - 1) + 2*b*d^2)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^4/(c*x - 1)^5 - 5*(c*x +
1)^4*c^4/(c*x - 1)^4 + 10*(c*x + 1)^3*c^4/(c*x - 1)^3 - 10*(c*x + 1)^2*c^4/(c*x - 1)^2 + 5*(c*x + 1)*c^4/(c*x
- 1) - c^4) + (30*(c*x + 1)^4*a*d^2/(c*x - 1)^4 - 30*(c*x + 1)^3*a*d^2/(c*x - 1)^3 + 40*(c*x + 1)^2*a*d^2/(c*x
- 1)^2 - 20*(c*x + 1)*a*d^2/(c*x - 1) + 4*a*d^2 + 13*(c*x + 1)^4*b*d^2/(c*x - 1)^4 - 36*(c*x + 1)^3*b*d^2/(c*
x - 1)^3 + 41*(c*x + 1)^2*b*d^2/(c*x - 1)^2 - 23*(c*x + 1)*b*d^2/(c*x - 1) + 5*b*d^2)/((c*x + 1)^5*c^4/(c*x -
1)^5 - 5*(c*x + 1)^4*c^4/(c*x - 1)^4 + 10*(c*x + 1)^3*c^4/(c*x - 1)^3 - 10*(c*x + 1)^2*c^4/(c*x - 1)^2 + 5*(c*
x + 1)*c^4/(c*x - 1) - c^4) - 2*b*d^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^4 + 2*b*d^2*log(-(c*x + 1)/(c*x - 1))/c^
4)

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maple [A]  time = 0.03, size = 147, normalized size = 1.03 \[ \frac {c^{2} d^{2} a \,x^{5}}{5}+\frac {c \,d^{2} a \,x^{4}}{2}+\frac {d^{2} a \,x^{3}}{3}+\frac {c^{2} d^{2} b \arctanh \left (c x \right ) x^{5}}{5}+\frac {c \,d^{2} b \arctanh \left (c x \right ) x^{4}}{2}+\frac {d^{2} b \arctanh \left (c x \right ) x^{3}}{3}+\frac {b c \,d^{2} x^{4}}{20}+\frac {b \,d^{2} x^{3}}{6}+\frac {4 b \,d^{2} x^{2}}{15 c}+\frac {b \,d^{2} x}{2 c^{2}}+\frac {31 d^{2} b \ln \left (c x -1\right )}{60 c^{3}}+\frac {b \,d^{2} \ln \left (c x +1\right )}{60 c^{3}} \]

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 = 0.31, size = 184, normalized size = 1.29 \[ \frac {1}{5} \, a c^{2} d^{2} x^{5} + \frac {1}{2} \, a c d^{2} x^{4} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{2} d^{2} + \frac {1}{3} \, a d^{2} x^{3} + \frac {1}{12} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c d^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d^{2} \]

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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mupad [B]  time = 1.03, size = 134, normalized size = 0.94 \[ \frac {\frac {4\,b\,c^2\,d^2\,x^2}{15}-\frac {d^2\,\left (30\,b\,\mathrm {atanh}\left (c\,x\right )-16\,b\,\ln \left (c^2\,x^2-1\right )\right )}{60}+\frac {b\,c\,d^2\,x}{2}}{c^3}+\frac {d^2\,\left (20\,a\,x^3+10\,b\,x^3+20\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c^2\,d^2\,\left (12\,a\,x^5+12\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c\,d^2\,\left (30\,a\,x^4+3\,b\,x^4+30\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*b*c^2*d^2*x^2)/15 - (d^2*(30*b*atanh(c*x) - 16*b*log(c^2*x^2 - 1)))/60 + (b*c*d^2*x)/2)/c^3 + (d^2*(20*a*x
^3 + 10*b*x^3 + 20*b*x^3*atanh(c*x)))/60 + (c^2*d^2*(12*a*x^5 + 12*b*x^5*atanh(c*x)))/60 + (c*d^2*(30*a*x^4 +
3*b*x^4 + 30*b*x^4*atanh(c*x)))/60

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sympy [A]  time = 1.71, size = 177, normalized size = 1.24 \[ \begin {cases} \frac {a c^{2} d^{2} x^{5}}{5} + \frac {a c d^{2} x^{4}}{2} + \frac {a d^{2} x^{3}}{3} + \frac {b c^{2} d^{2} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b c d^{2} x^{4} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b c d^{2} x^{4}}{20} + \frac {b d^{2} x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {b d^{2} x^{3}}{6} + \frac {4 b d^{2} x^{2}}{15 c} + \frac {b d^{2} x}{2 c^{2}} + \frac {8 b d^{2} \log {\left (x - \frac {1}{c} \right )}}{15 c^{3}} + \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{30 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{3}}{3} & \text {otherwise} \end {cases} \]

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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