\(\int x^3 (a+b \arctan (c x)) (d+e \log (1+c^2 x^2)) \, dx\) [1287]

Optimal result

Integrand size = 26, antiderivative size = 221 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {b (2 d-3 e) x}{8 c^3}-\frac {2 b e x}{3 c^3}-\frac {b (2 d-e) x^3}{24 c}+\frac {b e x^3}{18 c}-\frac {b (2 d-3 e) \arctan (c x)}{8 c^4}+\frac {2 b e \arctan (c x)}{3 c^4}+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))+\frac {b e x \log \left (1+c^2 x^2\right )}{4 c^3}-\frac {b e x^3 \log \left (1+c^2 x^2\right )}{12 c}-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \]

[Out]

1/8*b*(2*d-3*e)*x/c^3-2/3*b*e*x/c^3-1/24*b*(2*d-e)*x^3/c+1/18*b*e*x^3/c-1/8*b*(2*d-3*e)*arctan(c*x)/c^4+2/3*b*
e*arctan(c*x)/c^4+1/4*e*x^2*(a+b*arctan(c*x))/c^2-1/8*e*x^4*(a+b*arctan(c*x))+1/4*b*e*x*ln(c^2*x^2+1)/c^3-1/12
*b*e*x^3*ln(c^2*x^2+1)/c-1/4*e*(a+b*arctan(c*x))*ln(c^2*x^2+1)/c^4+1/4*x^4*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1
))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {2504, 2442, 45, 5139, 470, 327, 209, 2521, 2498, 2505, 308} \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {1}{4} x^4 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {e \log \left (c^2 x^2+1\right ) (a+b \arctan (c x))}{4 c^4}-\frac {1}{8} e x^4 (a+b \arctan (c x))-\frac {b (2 d-3 e) \arctan (c x)}{8 c^4}+\frac {2 b e \arctan (c x)}{3 c^4}+\frac {b x (2 d-3 e)}{8 c^3}-\frac {2 b e x}{3 c^3}-\frac {b e x^3 \log \left (c^2 x^2+1\right )}{12 c}+\frac {b e x \log \left (c^2 x^2+1\right )}{4 c^3}-\frac {b x^3 (2 d-e)}{24 c}+\frac {b e x^3}{18 c} \]

[In]

Int[x^3*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]

[Out]

(b*(2*d - 3*e)*x)/(8*c^3) - (2*b*e*x)/(3*c^3) - (b*(2*d - e)*x^3)/(24*c) + (b*e*x^3)/(18*c) - (b*(2*d - 3*e)*A
rcTan[c*x])/(8*c^4) + (2*b*e*ArcTan[c*x])/(3*c^4) + (e*x^2*(a + b*ArcTan[c*x]))/(4*c^2) - (e*x^4*(a + b*ArcTan
[c*x]))/8 + (b*e*x*Log[1 + c^2*x^2])/(4*c^3) - (b*e*x^3*Log[1 + c^2*x^2])/(12*c) - (e*(a + b*ArcTan[c*x])*Log[
1 + c^2*x^2])/(4*c^4) + (x^4*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/4

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 327

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

Rule 470

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

Rule 2442

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

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2505

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

Rule 2521

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol]
:> With[{t = ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; Free
Q[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && IntegerQ[r] && IntegerQ[s] && (EqQ[
q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))

Rule 5139

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> With
[{u = IntHide[x^m*(d + e*Log[f + g*x^2]), x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[ExpandIntegrand[u
/(1 + c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[(m + 1)/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-(b c) \int \left (\frac {x^2 \left (2 e+c^2 (2 d-e) x^2\right )}{8 c^2 \left (1+c^2 x^2\right )}+\frac {e \left (-1+c^2 x^2\right ) \log \left (1+c^2 x^2\right )}{4 c^4}\right ) \, dx \\ & = \frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \int \frac {x^2 \left (2 e+c^2 (2 d-e) x^2\right )}{1+c^2 x^2} \, dx}{8 c}-\frac {(b e) \int \left (-1+c^2 x^2\right ) \log \left (1+c^2 x^2\right ) \, dx}{4 c^3} \\ & = -\frac {b (2 d-e) x^3}{24 c}+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {(b (2 d-3 e)) \int \frac {x^2}{1+c^2 x^2} \, dx}{8 c}-\frac {(b e) \int \left (-\log \left (1+c^2 x^2\right )+c^2 x^2 \log \left (1+c^2 x^2\right )\right ) \, dx}{4 c^3} \\ & = \frac {b (2 d-3 e) x}{8 c^3}-\frac {b (2 d-e) x^3}{24 c}+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(b (2 d-3 e)) \int \frac {1}{1+c^2 x^2} \, dx}{8 c^3}+\frac {(b e) \int \log \left (1+c^2 x^2\right ) \, dx}{4 c^3}-\frac {(b e) \int x^2 \log \left (1+c^2 x^2\right ) \, dx}{4 c} \\ & = \frac {b (2 d-3 e) x}{8 c^3}-\frac {b (2 d-e) x^3}{24 c}-\frac {b (2 d-3 e) \arctan (c x)}{8 c^4}+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))+\frac {b e x \log \left (1+c^2 x^2\right )}{4 c^3}-\frac {b e x^3 \log \left (1+c^2 x^2\right )}{12 c}-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(b e) \int \frac {x^2}{1+c^2 x^2} \, dx}{2 c}+\frac {1}{6} (b c e) \int \frac {x^4}{1+c^2 x^2} \, dx \\ & = \frac {b (2 d-3 e) x}{8 c^3}-\frac {b e x}{2 c^3}-\frac {b (2 d-e) x^3}{24 c}-\frac {b (2 d-3 e) \arctan (c x)}{8 c^4}+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))+\frac {b e x \log \left (1+c^2 x^2\right )}{4 c^3}-\frac {b e x^3 \log \left (1+c^2 x^2\right )}{12 c}-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {(b e) \int \frac {1}{1+c^2 x^2} \, dx}{2 c^3}+\frac {1}{6} (b c e) \int \left (-\frac {1}{c^4}+\frac {x^2}{c^2}+\frac {1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {b (2 d-3 e) x}{8 c^3}-\frac {2 b e x}{3 c^3}-\frac {b (2 d-e) x^3}{24 c}+\frac {b e x^3}{18 c}-\frac {b (2 d-3 e) \arctan (c x)}{8 c^4}+\frac {b e \arctan (c x)}{2 c^4}+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))+\frac {b e x \log \left (1+c^2 x^2\right )}{4 c^3}-\frac {b e x^3 \log \left (1+c^2 x^2\right )}{12 c}-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )+\frac {(b e) \int \frac {1}{1+c^2 x^2} \, dx}{6 c^3} \\ & = \frac {b (2 d-3 e) x}{8 c^3}-\frac {2 b e x}{3 c^3}-\frac {b (2 d-e) x^3}{24 c}+\frac {b e x^3}{18 c}-\frac {b (2 d-3 e) \arctan (c x)}{8 c^4}+\frac {2 b e \arctan (c x)}{3 c^4}+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))+\frac {b e x \log \left (1+c^2 x^2\right )}{4 c^3}-\frac {b e x^3 \log \left (1+c^2 x^2\right )}{12 c}-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.74 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {c x \left (18 a c^3 d x^3-6 b d \left (-3+c^2 x^2\right )-9 a c e x \left (-2+c^2 x^2\right )+b e \left (-75+7 c^2 x^2\right )\right )-6 e \left (b c x \left (-3+c^2 x^2\right )+a \left (3-3 c^4 x^4\right )\right ) \log \left (1+c^2 x^2\right )+3 b \arctan (c x) \left (e \left (25+6 c^2 x^2-3 c^4 x^4\right )+6 d \left (-1+c^4 x^4\right )+6 e \left (-1+c^4 x^4\right ) \log \left (1+c^2 x^2\right )\right )}{72 c^4} \]

[In]

Integrate[x^3*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]

[Out]

(c*x*(18*a*c^3*d*x^3 - 6*b*d*(-3 + c^2*x^2) - 9*a*c*e*x*(-2 + c^2*x^2) + b*e*(-75 + 7*c^2*x^2)) - 6*e*(b*c*x*(
-3 + c^2*x^2) + a*(3 - 3*c^4*x^4))*Log[1 + c^2*x^2] + 3*b*ArcTan[c*x]*(e*(25 + 6*c^2*x^2 - 3*c^4*x^4) + 6*d*(-
1 + c^4*x^4) + 6*e*(-1 + c^4*x^4)*Log[1 + c^2*x^2]))/(72*c^4)

Maple [A] (verified)

Time = 1.67 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.10

method	result	size
parallelrisch	\(\frac {18 e b \ln \left (c^{2} x^{2}+1\right ) \arctan \left (c x \right ) x^{4} c^{4}+18 x^{4} \arctan \left (c x \right ) b \,c^{4} d -9 x^{4} \arctan \left (c x \right ) b \,c^{4} e +18 e a \ln \left (c^{2} x^{2}+1\right ) x^{4} c^{4}+18 c^{4} a d \,x^{4}-9 x^{4} a \,c^{4} e -6 e b \ln \left (c^{2} x^{2}+1\right ) x^{3} c^{3}-6 b \,c^{3} d \,x^{3}+7 b \,c^{3} e \,x^{3}+18 \arctan \left (c x \right ) b \,c^{2} e \,x^{2}+18 a \,c^{2} e \,x^{2}+18 \ln \left (c^{2} x^{2}+1\right ) b c e x +18 b c d x -75 b c e x -18 \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right ) b e -18 \arctan \left (c x \right ) b d +75 e b \arctan \left (c x \right )-18 \ln \left (c^{2} x^{2}+1\right ) a e -18 e a}{72 c^{4}}\)	\(243\)
default	\(\text {Expression too large to display}\)	\(3752\)
parts	\(\text {Expression too large to display}\)	\(3752\)
risch	\(\text {Expression too large to display}\)	\(22188\)

[In]

int(x^3*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1)),x,method=_RETURNVERBOSE)

[Out]

1/72*(18*e*b*ln(c^2*x^2+1)*arctan(c*x)*x^4*c^4+18*x^4*arctan(c*x)*b*c^4*d-9*x^4*arctan(c*x)*b*c^4*e+18*e*a*ln(
c^2*x^2+1)*x^4*c^4+18*c^4*a*d*x^4-9*x^4*a*c^4*e-6*e*b*ln(c^2*x^2+1)*x^3*c^3-6*b*c^3*d*x^3+7*b*c^3*e*x^3+18*arc
tan(c*x)*b*c^2*e*x^2+18*a*c^2*e*x^2+18*ln(c^2*x^2+1)*b*c*e*x+18*b*c*d*x-75*b*c*e*x-18*arctan(c*x)*ln(c^2*x^2+1
)*b*e-18*arctan(c*x)*b*d+75*e*b*arctan(c*x)-18*ln(c^2*x^2+1)*a*e-18*e*a)/c^4

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.81 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {18 \, a c^{2} e x^{2} + 9 \, {\left (2 \, a c^{4} d - a c^{4} e\right )} x^{4} - {\left (6 \, b c^{3} d - 7 \, b c^{3} e\right )} x^{3} + 3 \, {\left (6 \, b c d - 25 \, b c e\right )} x + 3 \, {\left (6 \, b c^{2} e x^{2} + 3 \, {\left (2 \, b c^{4} d - b c^{4} e\right )} x^{4} - 6 \, b d + 25 \, b e\right )} \arctan \left (c x\right ) + 6 \, {\left (3 \, a c^{4} e x^{4} - b c^{3} e x^{3} + 3 \, b c e x - 3 \, a e + 3 \, {\left (b c^{4} e x^{4} - b e\right )} \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )}{72 \, c^{4}} \]

[In]

integrate(x^3*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="fricas")

[Out]

1/72*(18*a*c^2*e*x^2 + 9*(2*a*c^4*d - a*c^4*e)*x^4 - (6*b*c^3*d - 7*b*c^3*e)*x^3 + 3*(6*b*c*d - 25*b*c*e)*x +
3*(6*b*c^2*e*x^2 + 3*(2*b*c^4*d - b*c^4*e)*x^4 - 6*b*d + 25*b*e)*arctan(c*x) + 6*(3*a*c^4*e*x^4 - b*c^3*e*x^3
+ 3*b*c*e*x - 3*a*e + 3*(b*c^4*e*x^4 - b*e)*arctan(c*x))*log(c^2*x^2 + 1))/c^4

Sympy [A] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.26 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{4}}{4} + \frac {a e x^{4} \log {\left (c^{2} x^{2} + 1 \right )}}{4} - \frac {a e x^{4}}{8} + \frac {a e x^{2}}{4 c^{2}} - \frac {a e \log {\left (c^{2} x^{2} + 1 \right )}}{4 c^{4}} + \frac {b d x^{4} \operatorname {atan}{\left (c x \right )}}{4} + \frac {b e x^{4} \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )}}{4} - \frac {b e x^{4} \operatorname {atan}{\left (c x \right )}}{8} - \frac {b d x^{3}}{12 c} - \frac {b e x^{3} \log {\left (c^{2} x^{2} + 1 \right )}}{12 c} + \frac {7 b e x^{3}}{72 c} + \frac {b e x^{2} \operatorname {atan}{\left (c x \right )}}{4 c^{2}} + \frac {b d x}{4 c^{3}} + \frac {b e x \log {\left (c^{2} x^{2} + 1 \right )}}{4 c^{3}} - \frac {25 b e x}{24 c^{3}} - \frac {b d \operatorname {atan}{\left (c x \right )}}{4 c^{4}} - \frac {b e \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )}}{4 c^{4}} + \frac {25 b e \operatorname {atan}{\left (c x \right )}}{24 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(a+b*atan(c*x))*(d+e*ln(c**2*x**2+1)),x)

[Out]

Piecewise((a*d*x**4/4 + a*e*x**4*log(c**2*x**2 + 1)/4 - a*e*x**4/8 + a*e*x**2/(4*c**2) - a*e*log(c**2*x**2 + 1
)/(4*c**4) + b*d*x**4*atan(c*x)/4 + b*e*x**4*log(c**2*x**2 + 1)*atan(c*x)/4 - b*e*x**4*atan(c*x)/8 - b*d*x**3/
(12*c) - b*e*x**3*log(c**2*x**2 + 1)/(12*c) + 7*b*e*x**3/(72*c) + b*e*x**2*atan(c*x)/(4*c**2) + b*d*x/(4*c**3)
+ b*e*x*log(c**2*x**2 + 1)/(4*c**3) - 25*b*e*x/(24*c**3) - b*d*atan(c*x)/(4*c**4) - b*e*log(c**2*x**2 + 1)*at
an(c*x)/(4*c**4) + 25*b*e*atan(c*x)/(24*c**4), Ne(c, 0)), (a*d*x**4/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.01 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {1}{4} \, a d x^{4} + \frac {1}{72} \, b c e {\left (\frac {7 \, c^{2} x^{3} - 6 \, {\left (c^{2} x^{3} - 3 \, x\right )} \log \left (c^{2} x^{2} + 1\right ) - 75 \, x}{c^{4}} + \frac {75 \, \arctan \left (c x\right )}{c^{5}}\right )} + \frac {1}{8} \, {\left (2 \, x^{4} \log \left (c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e \arctan \left (c x\right ) + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d + \frac {1}{8} \, {\left (2 \, x^{4} \log \left (c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} a e \]

[In]

integrate(x^3*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="maxima")

[Out]

1/4*a*d*x^4 + 1/72*b*c*e*((7*c^2*x^3 - 6*(c^2*x^3 - 3*x)*log(c^2*x^2 + 1) - 75*x)/c^4 + 75*arctan(c*x)/c^5) +
1/8*(2*x^4*log(c^2*x^2 + 1) - c^2*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*e*arctan(c*x) + 1/12*(3*
x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*d + 1/8*(2*x^4*log(c^2*x^2 + 1) - c^2*((c^2*x
^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*a*e

Giac [F]

\[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )} x^{3} \,d x } \]

[In]

integrate(x^3*(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 = 1.94 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.34 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {a\,d\,x^4}{4}-\frac {a\,e\,x^4}{8}+\frac {b\,d\,x}{4\,c^3}-\frac {25\,b\,e\,x}{24\,c^3}+\frac {b\,d\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}-\frac {b\,e\,x^4\,\mathrm {atan}\left (c\,x\right )}{8}-\frac {a\,e\,\ln \left (c^2\,x^2+1\right )}{4\,c^4}+\frac {a\,e\,x^2}{4\,c^2}-\frac {b\,d\,x^3}{12\,c}-\frac {b\,d\,\mathrm {atan}\left (\frac {6\,b\,c\,d\,x}{6\,b\,d-25\,b\,e}-\frac {25\,b\,c\,e\,x}{6\,b\,d-25\,b\,e}\right )}{4\,c^4}+\frac {7\,b\,e\,x^3}{72\,c}+\frac {25\,b\,e\,\mathrm {atan}\left (\frac {6\,b\,c\,d\,x}{6\,b\,d-25\,b\,e}-\frac {25\,b\,c\,e\,x}{6\,b\,d-25\,b\,e}\right )}{24\,c^4}+\frac {a\,e\,x^4\,\ln \left (c^2\,x^2+1\right )}{4}+\frac {b\,e\,x\,\ln \left (c^2\,x^2+1\right )}{4\,c^3}-\frac {b\,e\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right )}{4\,c^4}+\frac {b\,e\,x^2\,\mathrm {atan}\left (c\,x\right )}{4\,c^2}+\frac {b\,e\,x^4\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right )}{4}-\frac {b\,e\,x^3\,\ln \left (c^2\,x^2+1\right )}{12\,c} \]

[In]

int(x^3*(a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)),x)

[Out]

(a*d*x^4)/4 - (a*e*x^4)/8 + (b*d*x)/(4*c^3) - (25*b*e*x)/(24*c^3) + (b*d*x^4*atan(c*x))/4 - (b*e*x^4*atan(c*x)
)/8 - (a*e*log(c^2*x^2 + 1))/(4*c^4) + (a*e*x^2)/(4*c^2) - (b*d*x^3)/(12*c) - (b*d*atan((6*b*c*d*x)/(6*b*d - 2
5*b*e) - (25*b*c*e*x)/(6*b*d - 25*b*e)))/(4*c^4) + (7*b*e*x^3)/(72*c) + (25*b*e*atan((6*b*c*d*x)/(6*b*d - 25*b
*e) - (25*b*c*e*x)/(6*b*d - 25*b*e)))/(24*c^4) + (a*e*x^4*log(c^2*x^2 + 1))/4 + (b*e*x*log(c^2*x^2 + 1))/(4*c^
3) - (b*e*atan(c*x)*log(c^2*x^2 + 1))/(4*c^4) + (b*e*x^2*atan(c*x))/(4*c^2) + (b*e*x^4*atan(c*x)*log(c^2*x^2 +
1))/4 - (b*e*x^3*log(c^2*x^2 + 1))/(12*c)