3.10 \(\int x^3 (d+c d x)^2 (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=157 \[ \frac {1}{6} c^2 d^2 x^6 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{5} c d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {49 b d^2 \log (1-c x)}{120 c^4}-\frac {b d^2 \log (c x+1)}{120 c^4}+\frac {5 b d^2 x}{12 c^3}+\frac {b d^2 x^2}{5 c^2}+\frac {1}{30} b c d^2 x^5+\frac {5 b d^2 x^3}{36 c}+\frac {1}{10} b d^2 x^4 \]

[Out]

5/12*b*d^2*x/c^3+1/5*b*d^2*x^2/c^2+5/36*b*d^2*x^3/c+1/10*b*d^2*x^4+1/30*b*c*d^2*x^5+1/4*d^2*x^4*(a+b*arctanh(c
*x))+2/5*c*d^2*x^5*(a+b*arctanh(c*x))+1/6*c^2*d^2*x^6*(a+b*arctanh(c*x))+49/120*b*d^2*ln(-c*x+1)/c^4-1/120*b*d
^2*ln(c*x+1)/c^4

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Rubi [A]  time = 0.17, antiderivative size = 157, 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} \[ \frac {1}{6} c^2 d^2 x^6 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{5} c d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b d^2 x^2}{5 c^2}+\frac {5 b d^2 x}{12 c^3}+\frac {49 b d^2 \log (1-c x)}{120 c^4}-\frac {b d^2 \log (c x+1)}{120 c^4}+\frac {1}{30} b c d^2 x^5+\frac {5 b d^2 x^3}{36 c}+\frac {1}{10} b d^2 x^4 \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + c*d*x)^2*(a + b*ArcTanh[c*x]),x]

[Out]

(5*b*d^2*x)/(12*c^3) + (b*d^2*x^2)/(5*c^2) + (5*b*d^2*x^3)/(36*c) + (b*d^2*x^4)/10 + (b*c*d^2*x^5)/30 + (d^2*x
^4*(a + b*ArcTanh[c*x]))/4 + (2*c*d^2*x^5*(a + b*ArcTanh[c*x]))/5 + (c^2*d^2*x^6*(a + b*ArcTanh[c*x]))/6 + (49
*b*d^2*Log[1 - c*x])/(120*c^4) - (b*d^2*Log[1 + c*x])/(120*c^4)

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^3 (d+c d x)^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {1}{4} d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{5} c d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{6} c^2 d^2 x^6 \left (a+b \tanh ^{-1}(c x)\right )-(b c) \int \frac {d^2 x^4 \left (15+24 c x+10 c^2 x^2\right )}{60 \left (1-c^2 x^2\right )} \, dx\\ &=\frac {1}{4} d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{5} c d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{6} c^2 d^2 x^6 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{60} \left (b c d^2\right ) \int \frac {x^4 \left (15+24 c x+10 c^2 x^2\right )}{1-c^2 x^2} \, dx\\ &=\frac {1}{4} d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{5} c d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{6} c^2 d^2 x^6 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{60} \left (b c d^2\right ) \int \left (-\frac {25}{c^4}-\frac {24 x}{c^3}-\frac {25 x^2}{c^2}-\frac {24 x^3}{c}-10 x^4+\frac {25+24 c x}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac {5 b d^2 x}{12 c^3}+\frac {b d^2 x^2}{5 c^2}+\frac {5 b d^2 x^3}{36 c}+\frac {1}{10} b d^2 x^4+\frac {1}{30} b c d^2 x^5+\frac {1}{4} d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{5} c d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{6} c^2 d^2 x^6 \left (a+b \tanh ^{-1}(c x)\right )-\frac {\left (b d^2\right ) \int \frac {25+24 c x}{1-c^2 x^2} \, dx}{60 c^3}\\ &=\frac {5 b d^2 x}{12 c^3}+\frac {b d^2 x^2}{5 c^2}+\frac {5 b d^2 x^3}{36 c}+\frac {1}{10} b d^2 x^4+\frac {1}{30} b c d^2 x^5+\frac {1}{4} d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{5} c d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{6} c^2 d^2 x^6 \left (a+b \tanh ^{-1}(c x)\right )+\frac {\left (b d^2\right ) \int \frac {1}{-c-c^2 x} \, dx}{120 c^2}-\frac {\left (49 b d^2\right ) \int \frac {1}{c-c^2 x} \, dx}{120 c^2}\\ &=\frac {5 b d^2 x}{12 c^3}+\frac {b d^2 x^2}{5 c^2}+\frac {5 b d^2 x^3}{36 c}+\frac {1}{10} b d^2 x^4+\frac {1}{30} b c d^2 x^5+\frac {1}{4} d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{5} c d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{6} c^2 d^2 x^6 \left (a+b \tanh ^{-1}(c x)\right )+\frac {49 b d^2 \log (1-c x)}{120 c^4}-\frac {b d^2 \log (1+c x)}{120 c^4}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 125, normalized size = 0.80 \[ \frac {d^2 \left (60 a c^6 x^6+144 a c^5 x^5+90 a c^4 x^4+12 b c^5 x^5+36 b c^4 x^4+50 b c^3 x^3+72 b c^2 x^2+6 b c^4 x^4 \left (10 c^2 x^2+24 c x+15\right ) \tanh ^{-1}(c x)+150 b c x+147 b \log (1-c x)-3 b \log (c x+1)\right )}{360 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + c*d*x)^2*(a + b*ArcTanh[c*x]),x]

[Out]

(d^2*(150*b*c*x + 72*b*c^2*x^2 + 50*b*c^3*x^3 + 90*a*c^4*x^4 + 36*b*c^4*x^4 + 144*a*c^5*x^5 + 12*b*c^5*x^5 + 6
0*a*c^6*x^6 + 6*b*c^4*x^4*(15 + 24*c*x + 10*c^2*x^2)*ArcTanh[c*x] + 147*b*Log[1 - c*x] - 3*b*Log[1 + c*x]))/(3
60*c^4)

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fricas [A]  time = 0.45, size = 162, normalized size = 1.03 \[ \frac {60 \, a c^{6} d^{2} x^{6} + 12 \, {\left (12 \, a + b\right )} c^{5} d^{2} x^{5} + 18 \, {\left (5 \, a + 2 \, b\right )} c^{4} d^{2} x^{4} + 50 \, b c^{3} d^{2} x^{3} + 72 \, b c^{2} d^{2} x^{2} + 150 \, b c d^{2} x - 3 \, b d^{2} \log \left (c x + 1\right ) + 147 \, b d^{2} \log \left (c x - 1\right ) + 3 \, {\left (10 \, b c^{6} d^{2} x^{6} + 24 \, b c^{5} d^{2} x^{5} + 15 \, b c^{4} d^{2} x^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{360 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/360*(60*a*c^6*d^2*x^6 + 12*(12*a + b)*c^5*d^2*x^5 + 18*(5*a + 2*b)*c^4*d^2*x^4 + 50*b*c^3*d^2*x^3 + 72*b*c^2
*d^2*x^2 + 150*b*c*d^2*x - 3*b*d^2*log(c*x + 1) + 147*b*d^2*log(c*x - 1) + 3*(10*b*c^6*d^2*x^6 + 24*b*c^5*d^2*
x^5 + 15*b*c^4*d^2*x^4)*log(-(c*x + 1)/(c*x - 1)))/c^4

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giac [B]  time = 0.28, size = 620, normalized size = 3.95 \[ \frac {1}{45} \, c {\left (\frac {6 \, {\left (\frac {30 \, {\left (c x + 1\right )}^{5} b d^{2}}{{\left (c x - 1\right )}^{5}} - \frac {30 \, {\left (c x + 1\right )}^{4} b d^{2}}{{\left (c x - 1\right )}^{4}} + \frac {70 \, {\left (c x + 1\right )}^{3} b d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {45 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {18 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} - 3 \, b d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{6} c^{5}}{{\left (c x - 1\right )}^{6}} - \frac {6 \, {\left (c x + 1\right )}^{5} c^{5}}{{\left (c x - 1\right )}^{5}} + \frac {15 \, {\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {20 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {15 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {6 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}} + \frac {\frac {360 \, {\left (c x + 1\right )}^{5} a d^{2}}{{\left (c x - 1\right )}^{5}} - \frac {360 \, {\left (c x + 1\right )}^{4} a d^{2}}{{\left (c x - 1\right )}^{4}} + \frac {840 \, {\left (c x + 1\right )}^{3} a d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {540 \, {\left (c x + 1\right )}^{2} a d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {216 \, {\left (c x + 1\right )} a d^{2}}{c x - 1} - 36 \, a d^{2} + \frac {162 \, {\left (c x + 1\right )}^{5} b d^{2}}{{\left (c x - 1\right )}^{5}} - \frac {531 \, {\left (c x + 1\right )}^{4} b d^{2}}{{\left (c x - 1\right )}^{4}} + \frac {818 \, {\left (c x + 1\right )}^{3} b d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {696 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {300 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} - 53 \, b d^{2}}{\frac {{\left (c x + 1\right )}^{6} c^{5}}{{\left (c x - 1\right )}^{6}} - \frac {6 \, {\left (c x + 1\right )}^{5} c^{5}}{{\left (c x - 1\right )}^{5}} + \frac {15 \, {\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {20 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {15 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {6 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}} - \frac {18 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{5}} + \frac {18 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

1/45*c*(6*(30*(c*x + 1)^5*b*d^2/(c*x - 1)^5 - 30*(c*x + 1)^4*b*d^2/(c*x - 1)^4 + 70*(c*x + 1)^3*b*d^2/(c*x - 1
)^3 - 45*(c*x + 1)^2*b*d^2/(c*x - 1)^2 + 18*(c*x + 1)*b*d^2/(c*x - 1) - 3*b*d^2)*log(-(c*x + 1)/(c*x - 1))/((c
*x + 1)^6*c^5/(c*x - 1)^6 - 6*(c*x + 1)^5*c^5/(c*x - 1)^5 + 15*(c*x + 1)^4*c^5/(c*x - 1)^4 - 20*(c*x + 1)^3*c^
5/(c*x - 1)^3 + 15*(c*x + 1)^2*c^5/(c*x - 1)^2 - 6*(c*x + 1)*c^5/(c*x - 1) + c^5) + (360*(c*x + 1)^5*a*d^2/(c*
x - 1)^5 - 360*(c*x + 1)^4*a*d^2/(c*x - 1)^4 + 840*(c*x + 1)^3*a*d^2/(c*x - 1)^3 - 540*(c*x + 1)^2*a*d^2/(c*x
- 1)^2 + 216*(c*x + 1)*a*d^2/(c*x - 1) - 36*a*d^2 + 162*(c*x + 1)^5*b*d^2/(c*x - 1)^5 - 531*(c*x + 1)^4*b*d^2/
(c*x - 1)^4 + 818*(c*x + 1)^3*b*d^2/(c*x - 1)^3 - 696*(c*x + 1)^2*b*d^2/(c*x - 1)^2 + 300*(c*x + 1)*b*d^2/(c*x
- 1) - 53*b*d^2)/((c*x + 1)^6*c^5/(c*x - 1)^6 - 6*(c*x + 1)^5*c^5/(c*x - 1)^5 + 15*(c*x + 1)^4*c^5/(c*x - 1)^
4 - 20*(c*x + 1)^3*c^5/(c*x - 1)^3 + 15*(c*x + 1)^2*c^5/(c*x - 1)^2 - 6*(c*x + 1)*c^5/(c*x - 1) + c^5) - 18*b*
d^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^5 + 18*b*d^2*log(-(c*x + 1)/(c*x - 1))/c^5)

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maple [A]  time = 0.03, size = 159, normalized size = 1.01 \[ \frac {c^{2} d^{2} a \,x^{6}}{6}+\frac {2 c \,d^{2} a \,x^{5}}{5}+\frac {a \,x^{4} d^{2}}{4}+\frac {c^{2} d^{2} b \arctanh \left (c x \right ) x^{6}}{6}+\frac {2 c \,d^{2} b \arctanh \left (c x \right ) x^{5}}{5}+\frac {d^{2} b \arctanh \left (c x \right ) x^{4}}{4}+\frac {b c \,d^{2} x^{5}}{30}+\frac {b \,d^{2} x^{4}}{10}+\frac {5 b \,d^{2} x^{3}}{36 c}+\frac {b \,d^{2} x^{2}}{5 c^{2}}+\frac {5 b \,d^{2} x}{12 c^{3}}+\frac {49 d^{2} b \ln \left (c x -1\right )}{120 c^{4}}-\frac {b \,d^{2} \ln \left (c x +1\right )}{120 c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*d*x+d)^2*(a+b*arctanh(c*x)),x)

[Out]

1/6*c^2*d^2*a*x^6+2/5*c*d^2*a*x^5+1/4*a*x^4*d^2+1/6*c^2*d^2*b*arctanh(c*x)*x^6+2/5*c*d^2*b*arctanh(c*x)*x^5+1/
4*d^2*b*arctanh(c*x)*x^4+1/30*b*c*d^2*x^5+1/10*b*d^2*x^4+5/36*b*d^2*x^3/c+1/5*b*d^2*x^2/c^2+5/12*b*d^2*x/c^3+4
9/120/c^4*d^2*b*ln(c*x-1)-1/120*b*d^2*ln(c*x+1)/c^4

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maxima [A]  time = 0.33, size = 210, normalized size = 1.34 \[ \frac {1}{6} \, a c^{2} d^{2} x^{6} + \frac {2}{5} \, a c d^{2} x^{5} + \frac {1}{4} \, a d^{2} x^{4} + \frac {1}{180} \, {\left (30 \, x^{6} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac {15 \, \log \left (c x + 1\right )}{c^{7}} + \frac {15 \, \log \left (c x - 1\right )}{c^{7}}\right )}\right )} b c^{2} d^{2} + \frac {1}{10} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c d^{2} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b d^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/6*a*c^2*d^2*x^6 + 2/5*a*c*d^2*x^5 + 1/4*a*d^2*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^2*d^2 + 1/10*(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*d^2 + 1/24*(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*d^2

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mupad [B]  time = 1.04, size = 146, normalized size = 0.93 \[ \frac {\frac {b\,c^2\,d^2\,x^2}{5}-\frac {d^2\,\left (75\,b\,\mathrm {atanh}\left (c\,x\right )-36\,b\,\ln \left (c^2\,x^2-1\right )\right )}{180}+\frac {5\,b\,c^3\,d^2\,x^3}{36}+\frac {5\,b\,c\,d^2\,x}{12}}{c^4}+\frac {d^2\,\left (45\,a\,x^4+18\,b\,x^4+45\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{180}+\frac {c^2\,d^2\,\left (30\,a\,x^6+30\,b\,x^6\,\mathrm {atanh}\left (c\,x\right )\right )}{180}+\frac {c\,d^2\,\left (72\,a\,x^5+6\,b\,x^5+72\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{180} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x))*(d + c*d*x)^2,x)

[Out]

((b*c^2*d^2*x^2)/5 - (d^2*(75*b*atanh(c*x) - 36*b*log(c^2*x^2 - 1)))/180 + (5*b*c^3*d^2*x^3)/36 + (5*b*c*d^2*x
)/12)/c^4 + (d^2*(45*a*x^4 + 18*b*x^4 + 45*b*x^4*atanh(c*x)))/180 + (c^2*d^2*(30*a*x^6 + 30*b*x^6*atanh(c*x)))
/180 + (c*d^2*(72*a*x^5 + 6*b*x^5 + 72*b*x^5*atanh(c*x)))/180

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sympy [A]  time = 2.16, size = 196, normalized size = 1.25 \[ \begin {cases} \frac {a c^{2} d^{2} x^{6}}{6} + \frac {2 a c d^{2} x^{5}}{5} + \frac {a d^{2} x^{4}}{4} + \frac {b c^{2} d^{2} x^{6} \operatorname {atanh}{\left (c x \right )}}{6} + \frac {2 b c d^{2} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b c d^{2} x^{5}}{30} + \frac {b d^{2} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b d^{2} x^{4}}{10} + \frac {5 b d^{2} x^{3}}{36 c} + \frac {b d^{2} x^{2}}{5 c^{2}} + \frac {5 b d^{2} x}{12 c^{3}} + \frac {2 b d^{2} \log {\left (x - \frac {1}{c} \right )}}{5 c^{4}} - \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{60 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{4}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c*d*x+d)**2*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*c**2*d**2*x**6/6 + 2*a*c*d**2*x**5/5 + a*d**2*x**4/4 + b*c**2*d**2*x**6*atanh(c*x)/6 + 2*b*c*d**2
*x**5*atanh(c*x)/5 + b*c*d**2*x**5/30 + b*d**2*x**4*atanh(c*x)/4 + b*d**2*x**4/10 + 5*b*d**2*x**3/(36*c) + b*d
**2*x**2/(5*c**2) + 5*b*d**2*x/(12*c**3) + 2*b*d**2*log(x - 1/c)/(5*c**4) - b*d**2*atanh(c*x)/(60*c**4), Ne(c,
0)), (a*d**2*x**4/4, True))

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