\(\int x^2 (a+b \arctan (c x)) (d+e \log (1+c^2 x^2)) \, dx\) [1288]

Optimal result

Integrand size = 26, antiderivative size = 213 \[ \int x^2 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {2 a e x}{3 c^2}+\frac {5 b e x^2}{18 c}-\frac {2}{9} a e x^3-\frac {2 a e \arctan (c x)}{3 c^3}+\frac {2 b e x \arctan (c x)}{3 c^2}-\frac {2}{9} b e x^3 \arctan (c x)-\frac {b e \arctan (c x)^2}{3 c^3}-\frac {11 b e \log \left (1+c^2 x^2\right )}{18 c^3}-\frac {b e \log ^2\left (1+c^2 x^2\right )}{12 c^3}-\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3} \]

[Out]

2/3*a*e*x/c^2+5/18*b*e*x^2/c-2/9*a*e*x^3-2/3*a*e*arctan(c*x)/c^3+2/3*b*e*x*arctan(c*x)/c^2-2/9*b*e*x^3*arctan(
c*x)-1/3*b*e*arctan(c*x)^2/c^3-11/18*b*e*ln(c^2*x^2+1)/c^3-1/12*b*e*ln(c^2*x^2+1)^2/c^3-1/6*b*x^2*(d+e*ln(c^2*
x^2+1))/c+1/3*x^3*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))+1/6*b*ln(c^2*x^2+1)*(d+e*ln(c^2*x^2+1))/c^3

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {4946, 272, 45, 5141, 6857, 815, 649, 209, 266, 5036, 4930, 5004, 2525, 2437, 2338} \[ \int x^2 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {1}{3} x^3 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {2 a e \arctan (c x)}{3 c^3}+\frac {2 a e x}{3 c^2}-\frac {2}{9} a e x^3-\frac {b e \arctan (c x)^2}{3 c^3}+\frac {2 b e x \arctan (c x)}{3 c^2}-\frac {2}{9} b e x^3 \arctan (c x)-\frac {b x^2 \left (e \log \left (c^2 x^2+1\right )+d\right )}{6 c}+\frac {b \log \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{6 c^3}-\frac {b e \log ^2\left (c^2 x^2+1\right )}{12 c^3}-\frac {11 b e \log \left (c^2 x^2+1\right )}{18 c^3}+\frac {5 b e x^2}{18 c} \]

[In]

Int[x^2*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]

[Out]

(2*a*e*x)/(3*c^2) + (5*b*e*x^2)/(18*c) - (2*a*e*x^3)/9 - (2*a*e*ArcTan[c*x])/(3*c^3) + (2*b*e*x*ArcTan[c*x])/(
3*c^2) - (2*b*e*x^3*ArcTan[c*x])/9 - (b*e*ArcTan[c*x]^2)/(3*c^3) - (11*b*e*Log[1 + c^2*x^2])/(18*c^3) - (b*e*L
og[1 + c^2*x^2]^2)/(12*c^3) - (b*x^2*(d + e*Log[1 + c^2*x^2]))/(6*c) + (x^3*(a + b*ArcTan[c*x])*(d + e*Log[1 +
c^2*x^2]))/3 + (b*Log[1 + c^2*x^2]*(d + e*Log[1 + c^2*x^2]))/(6*c^3)

Rule 45

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 649

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

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 5004

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 5036

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d*(f^2/e), Int[(f*x)^(m - 2)*((a + b*ArcTan[c*x])^p/
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 5141

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> With
[{u = IntHide[x^m*(a + b*ArcTan[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegrand
[x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3}-\left (2 c^2 e\right ) \int \left (\frac {x^3 (-b+2 a c x+2 b c x \arctan (c x))}{6 c \left (1+c^2 x^2\right )}+\frac {b x \log \left (1+c^2 x^2\right )}{6 c^3 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = -\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3}-\frac {(b e) \int \frac {x \log \left (1+c^2 x^2\right )}{1+c^2 x^2} \, dx}{3 c}-\frac {1}{3} (c e) \int \frac {x^3 (-b+2 a c x+2 b c x \arctan (c x))}{1+c^2 x^2} \, dx \\ & = -\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3}-\frac {(b e) \text {Subst}\left (\int \frac {\log \left (1+c^2 x\right )}{1+c^2 x} \, dx,x,x^2\right )}{6 c}-\frac {1}{3} (c e) \int \left (\frac {x^3 (-b+2 a c x)}{1+c^2 x^2}+\frac {2 b c x^4 \arctan (c x)}{1+c^2 x^2}\right ) \, dx \\ & = -\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3}-\frac {(b e) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+c^2 x^2\right )}{6 c^3}-\frac {1}{3} (c e) \int \frac {x^3 (-b+2 a c x)}{1+c^2 x^2} \, dx-\frac {1}{3} \left (2 b c^2 e\right ) \int \frac {x^4 \arctan (c x)}{1+c^2 x^2} \, dx \\ & = -\frac {b e \log ^2\left (1+c^2 x^2\right )}{12 c^3}-\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3}-\frac {1}{3} (2 b e) \int x^2 \arctan (c x) \, dx+\frac {1}{3} (2 b e) \int \frac {x^2 \arctan (c x)}{1+c^2 x^2} \, dx-\frac {1}{3} (c e) \int \left (-\frac {2 a}{c^3}-\frac {b x}{c^2}+\frac {2 a x^2}{c}+\frac {2 a+b c x}{c^3 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {2 a e x}{3 c^2}+\frac {b e x^2}{6 c}-\frac {2}{9} a e x^3-\frac {2}{9} b e x^3 \arctan (c x)-\frac {b e \log ^2\left (1+c^2 x^2\right )}{12 c^3}-\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3}-\frac {e \int \frac {2 a+b c x}{1+c^2 x^2} \, dx}{3 c^2}+\frac {(2 b e) \int \arctan (c x) \, dx}{3 c^2}-\frac {(2 b e) \int \frac {\arctan (c x)}{1+c^2 x^2} \, dx}{3 c^2}+\frac {1}{9} (2 b c e) \int \frac {x^3}{1+c^2 x^2} \, dx \\ & = \frac {2 a e x}{3 c^2}+\frac {b e x^2}{6 c}-\frac {2}{9} a e x^3+\frac {2 b e x \arctan (c x)}{3 c^2}-\frac {2}{9} b e x^3 \arctan (c x)-\frac {b e \arctan (c x)^2}{3 c^3}-\frac {b e \log ^2\left (1+c^2 x^2\right )}{12 c^3}-\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3}-\frac {(2 a e) \int \frac {1}{1+c^2 x^2} \, dx}{3 c^2}-\frac {(b e) \int \frac {x}{1+c^2 x^2} \, dx}{3 c}-\frac {(2 b e) \int \frac {x}{1+c^2 x^2} \, dx}{3 c}+\frac {1}{9} (b c e) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right ) \\ & = \frac {2 a e x}{3 c^2}+\frac {b e x^2}{6 c}-\frac {2}{9} a e x^3-\frac {2 a e \arctan (c x)}{3 c^3}+\frac {2 b e x \arctan (c x)}{3 c^2}-\frac {2}{9} b e x^3 \arctan (c x)-\frac {b e \arctan (c x)^2}{3 c^3}-\frac {b e \log \left (1+c^2 x^2\right )}{2 c^3}-\frac {b e \log ^2\left (1+c^2 x^2\right )}{12 c^3}-\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3}+\frac {1}{9} (b c e) \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right ) \\ & = \frac {2 a e x}{3 c^2}+\frac {5 b e x^2}{18 c}-\frac {2}{9} a e x^3-\frac {2 a e \arctan (c x)}{3 c^3}+\frac {2 b e x \arctan (c x)}{3 c^2}-\frac {2}{9} b e x^3 \arctan (c x)-\frac {b e \arctan (c x)^2}{3 c^3}-\frac {11 b e \log \left (1+c^2 x^2\right )}{18 c^3}-\frac {b e \log ^2\left (1+c^2 x^2\right )}{12 c^3}-\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.80 \[ \int x^2 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {2 c x \left (b c (-3 d+5 e) x+6 a c^2 d x^2-4 a e \left (-3+c^2 x^2\right )\right )-12 b e \arctan (c x)^2+2 \left (3 b d+6 a c^3 e x^3-b e \left (11+3 c^2 x^2\right )\right ) \log \left (1+c^2 x^2\right )+3 b e \log ^2\left (1+c^2 x^2\right )-4 \arctan (c x) \left (6 a e+b c x \left (-6 e-3 c^2 d x^2+2 c^2 e x^2\right )-3 b c^3 e x^3 \log \left (1+c^2 x^2\right )\right )}{36 c^3} \]

[In]

Integrate[x^2*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]

[Out]

(2*c*x*(b*c*(-3*d + 5*e)*x + 6*a*c^2*d*x^2 - 4*a*e*(-3 + c^2*x^2)) - 12*b*e*ArcTan[c*x]^2 + 2*(3*b*d + 6*a*c^3
*e*x^3 - b*e*(11 + 3*c^2*x^2))*Log[1 + c^2*x^2] + 3*b*e*Log[1 + c^2*x^2]^2 - 4*ArcTan[c*x]*(6*a*e + b*c*x*(-6*
e - 3*c^2*d*x^2 + 2*c^2*e*x^2) - 3*b*c^3*e*x^3*Log[1 + c^2*x^2]))/(36*c^3)

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.02

method	result	size
parallelrisch	\(\frac {12 e b \ln \left (c^{2} x^{2}+1\right ) \arctan \left (c x \right ) x^{3} c^{3}+12 b \arctan \left (c x \right ) x^{3} c^{3} d -8 x^{3} \arctan \left (c x \right ) b \,c^{3} e +12 e a \ln \left (c^{2} x^{2}+1\right ) x^{3} c^{3}+12 a \,c^{3} d \,x^{3}-8 a \,c^{3} e \,x^{3}-6 x^{2} \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} e -6 c^{2} x^{2} b d +10 b \,c^{2} e \,x^{2}+24 e b \arctan \left (c x \right ) x c +24 x a c e -12 e b \arctan \left (c x \right )^{2}+3 e b \ln \left (c^{2} x^{2}+1\right )^{2}-24 e a \arctan \left (c x \right )+6 \ln \left (c^{2} x^{2}+1\right ) b d -22 \ln \left (c^{2} x^{2}+1\right ) b e}{36 c^{3}}\)	\(217\)
default	\(\text {Expression too large to display}\)	\(4010\)
parts	\(\text {Expression too large to display}\)	\(4010\)
risch	\(\text {Expression too large to display}\)	\(22991\)

[In]

int(x^2*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1)),x,method=_RETURNVERBOSE)

[Out]

1/36*(12*e*b*ln(c^2*x^2+1)*arctan(c*x)*x^3*c^3+12*b*arctan(c*x)*x^3*c^3*d-8*x^3*arctan(c*x)*b*c^3*e+12*e*a*ln(
c^2*x^2+1)*x^3*c^3+12*a*c^3*d*x^3-8*a*c^3*e*x^3-6*x^2*ln(c^2*x^2+1)*b*c^2*e-6*c^2*x^2*b*d+10*b*c^2*e*x^2+24*e*
b*arctan(c*x)*x*c+24*x*a*c*e-12*e*b*arctan(c*x)^2+3*e*b*ln(c^2*x^2+1)^2-24*e*a*arctan(c*x)+6*ln(c^2*x^2+1)*b*d
-22*ln(c^2*x^2+1)*b*e)/c^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.79 \[ \int x^2 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {24 \, a c e x + 4 \, {\left (3 \, a c^{3} d - 2 \, a c^{3} e\right )} x^{3} - 12 \, b e \arctan \left (c x\right )^{2} + 3 \, b e \log \left (c^{2} x^{2} + 1\right )^{2} - 2 \, {\left (3 \, b c^{2} d - 5 \, b c^{2} e\right )} x^{2} + 4 \, {\left (6 \, b c e x + {\left (3 \, b c^{3} d - 2 \, b c^{3} e\right )} x^{3} - 6 \, a e\right )} \arctan \left (c x\right ) + 2 \, {\left (6 \, b c^{3} e x^{3} \arctan \left (c x\right ) + 6 \, a c^{3} e x^{3} - 3 \, b c^{2} e x^{2} + 3 \, b d - 11 \, b e\right )} \log \left (c^{2} x^{2} + 1\right )}{36 \, c^{3}} \]

[In]

integrate(x^2*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="fricas")

[Out]

1/36*(24*a*c*e*x + 4*(3*a*c^3*d - 2*a*c^3*e)*x^3 - 12*b*e*arctan(c*x)^2 + 3*b*e*log(c^2*x^2 + 1)^2 - 2*(3*b*c^
2*d - 5*b*c^2*e)*x^2 + 4*(6*b*c*e*x + (3*b*c^3*d - 2*b*c^3*e)*x^3 - 6*a*e)*arctan(c*x) + 2*(6*b*c^3*e*x^3*arct
an(c*x) + 6*a*c^3*e*x^3 - 3*b*c^2*e*x^2 + 3*b*d - 11*b*e)*log(c^2*x^2 + 1))/c^3

Sympy [A] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.21 \[ \int x^2 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{3}}{3} + \frac {a e x^{3} \log {\left (c^{2} x^{2} + 1 \right )}}{3} - \frac {2 a e x^{3}}{9} + \frac {2 a e x}{3 c^{2}} - \frac {2 a e \operatorname {atan}{\left (c x \right )}}{3 c^{3}} + \frac {b d x^{3} \operatorname {atan}{\left (c x \right )}}{3} + \frac {b e x^{3} \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )}}{3} - \frac {2 b e x^{3} \operatorname {atan}{\left (c x \right )}}{9} - \frac {b d x^{2}}{6 c} - \frac {b e x^{2} \log {\left (c^{2} x^{2} + 1 \right )}}{6 c} + \frac {5 b e x^{2}}{18 c} + \frac {2 b e x \operatorname {atan}{\left (c x \right )}}{3 c^{2}} + \frac {b d \log {\left (c^{2} x^{2} + 1 \right )}}{6 c^{3}} + \frac {b e \log {\left (c^{2} x^{2} + 1 \right )}^{2}}{12 c^{3}} - \frac {11 b e \log {\left (c^{2} x^{2} + 1 \right )}}{18 c^{3}} - \frac {b e \operatorname {atan}^{2}{\left (c x \right )}}{3 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d x^{3}}{3} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(a+b*atan(c*x))*(d+e*ln(c**2*x**2+1)),x)

[Out]

Piecewise((a*d*x**3/3 + a*e*x**3*log(c**2*x**2 + 1)/3 - 2*a*e*x**3/9 + 2*a*e*x/(3*c**2) - 2*a*e*atan(c*x)/(3*c
**3) + b*d*x**3*atan(c*x)/3 + b*e*x**3*log(c**2*x**2 + 1)*atan(c*x)/3 - 2*b*e*x**3*atan(c*x)/9 - b*d*x**2/(6*c
) - b*e*x**2*log(c**2*x**2 + 1)/(6*c) + 5*b*e*x**2/(18*c) + 2*b*e*x*atan(c*x)/(3*c**2) + b*d*log(c**2*x**2 + 1
)/(6*c**3) + b*e*log(c**2*x**2 + 1)**2/(12*c**3) - 11*b*e*log(c**2*x**2 + 1)/(18*c**3) - b*e*atan(c*x)**2/(3*c
**3), Ne(c, 0)), (a*d*x**3/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00 \[ \int x^2 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {1}{3} \, a d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{2} {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b e \arctan \left (c x\right ) + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{2} {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} a e + \frac {{\left (10 \, c^{2} x^{2} + 12 \, \arctan \left (c x\right )^{2} - 2 \, {\left (3 \, c^{2} x^{2} + 11\right )} \log \left (c^{2} x^{2} + 1\right ) + 3 \, \log \left (c^{2} x^{2} + 1\right )^{2}\right )} b e}{36 \, c^{3}} \]

[In]

integrate(x^2*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="maxima")

[Out]

1/3*a*d*x^3 + 1/9*(3*x^3*log(c^2*x^2 + 1) - 2*c^2*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*e*arctan(c*x) +
1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d + 1/9*(3*x^3*log(c^2*x^2 + 1) - 2*c^2*((c^2*
x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*a*e + 1/36*(10*c^2*x^2 + 12*arctan(c*x)^2 - 2*(3*c^2*x^2 + 11)*log(c^2*x^
2 + 1) + 3*log(c^2*x^2 + 1)^2)*b*e/c^3

Giac [F]

\[ \int x^2 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )} x^{2} \,d x } \]

[In]

integrate(x^2*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 2.79 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00 \[ \int x^2 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {a\,d\,x^3}{3}-\frac {2\,a\,e\,x^3}{9}+\frac {b\,e\,{\ln \left (c^2\,x^2+1\right )}^2}{12\,c^3}+\frac {2\,a\,e\,x}{3\,c^2}-\frac {2\,a\,e\,\mathrm {atan}\left (c\,x\right )}{3\,c^3}+\frac {b\,d\,x^3\,\mathrm {atan}\left (c\,x\right )}{3}-\frac {2\,b\,e\,x^3\,\mathrm {atan}\left (c\,x\right )}{9}+\frac {b\,d\,\ln \left (c^2\,x^2+1\right )}{6\,c^3}-\frac {11\,b\,e\,\ln \left (c^2\,x^2+1\right )}{18\,c^3}-\frac {b\,d\,x^2}{6\,c}+\frac {5\,b\,e\,x^2}{18\,c}+\frac {a\,e\,x^3\,\ln \left (c^2\,x^2+1\right )}{3}-\frac {b\,e\,{\mathrm {atan}\left (c\,x\right )}^2}{3\,c^3}+\frac {b\,e\,x^3\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right )}{3}-\frac {b\,e\,x^2\,\ln \left (c^2\,x^2+1\right )}{6\,c}+\frac {2\,b\,e\,x\,\mathrm {atan}\left (c\,x\right )}{3\,c^2} \]

[In]

int(x^2*(a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)),x)

[Out]

(a*d*x^3)/3 - (2*a*e*x^3)/9 + (b*e*log(c^2*x^2 + 1)^2)/(12*c^3) + (2*a*e*x)/(3*c^2) - (2*a*e*atan(c*x))/(3*c^3
) + (b*d*x^3*atan(c*x))/3 - (2*b*e*x^3*atan(c*x))/9 + (b*d*log(c^2*x^2 + 1))/(6*c^3) - (11*b*e*log(c^2*x^2 + 1
))/(18*c^3) - (b*d*x^2)/(6*c) + (5*b*e*x^2)/(18*c) + (a*e*x^3*log(c^2*x^2 + 1))/3 - (b*e*atan(c*x)^2)/(3*c^3)
+ (b*e*x^3*atan(c*x)*log(c^2*x^2 + 1))/3 - (b*e*x^2*log(c^2*x^2 + 1))/(6*c) + (2*b*e*x*atan(c*x))/(3*c^2)