3.1286 \(\int x^4 (a+b \tan ^{-1}(c x)) (d+e \log (1+c^2 x^2)) \, dx\)

Optimal. Leaf size=278 \[ \frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac {2 a e \tan ^{-1}(c x)}{5 c^5}-\frac {2 a e x}{5 c^4}+\frac {2 a e x^3}{15 c^2}-\frac {2}{25} a e x^5+\frac {b e \tan ^{-1}(c x)^2}{5 c^5}-\frac {2 b e x \tan ^{-1}(c x)}{5 c^4}-\frac {77 b e x^2}{300 c^3}-\frac {b x^4 \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 c}+\frac {2 b e x^3 \tan ^{-1}(c x)}{15 c^2}-\frac {b \log \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^5}+\frac {b e \log ^2\left (c^2 x^2+1\right )}{20 c^5}+\frac {137 b e \log \left (c^2 x^2+1\right )}{300 c^5}+\frac {b x^2 \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^3}-\frac {2}{25} b e x^5 \tan ^{-1}(c x)+\frac {9 b e x^4}{200 c} \]

[Out]

-2/5*a*e*x/c^4-77/300*b*e*x^2/c^3+2/15*a*e*x^3/c^2+9/200*b*e*x^4/c-2/25*a*e*x^5+2/5*a*e*arctan(c*x)/c^5-2/5*b*
e*x*arctan(c*x)/c^4+2/15*b*e*x^3*arctan(c*x)/c^2-2/25*b*e*x^5*arctan(c*x)+1/5*b*e*arctan(c*x)^2/c^5+137/300*b*
e*ln(c^2*x^2+1)/c^5+1/20*b*e*ln(c^2*x^2+1)^2/c^5+1/10*b*x^2*(d+e*ln(c^2*x^2+1))/c^3-1/20*b*x^4*(d+e*ln(c^2*x^2
+1))/c+1/5*x^5*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))-1/10*b*ln(c^2*x^2+1)*(d+e*ln(c^2*x^2+1))/c^5

________________________________________________________________________________________

Rubi [A]  time = 0.69, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 15, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {4852, 266, 43, 5021, 6725, 1802, 635, 203, 260, 4916, 4846, 4884, 2475, 2390, 2301} \[ \frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac {2 a e x^3}{15 c^2}-\frac {2 a e x}{5 c^4}+\frac {2 a e \tan ^{-1}(c x)}{5 c^5}-\frac {2}{25} a e x^5-\frac {b x^4 \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 c}+\frac {b x^2 \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^3}-\frac {b \log \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^5}-\frac {77 b e x^2}{300 c^3}+\frac {b e \log ^2\left (c^2 x^2+1\right )}{20 c^5}+\frac {137 b e \log \left (c^2 x^2+1\right )}{300 c^5}+\frac {2 b e x^3 \tan ^{-1}(c x)}{15 c^2}-\frac {2 b e x \tan ^{-1}(c x)}{5 c^4}+\frac {b e \tan ^{-1}(c x)^2}{5 c^5}+\frac {9 b e x^4}{200 c}-\frac {2}{25} b e x^5 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]

[Out]

(-2*a*e*x)/(5*c^4) - (77*b*e*x^2)/(300*c^3) + (2*a*e*x^3)/(15*c^2) + (9*b*e*x^4)/(200*c) - (2*a*e*x^5)/25 + (2
*a*e*ArcTan[c*x])/(5*c^5) - (2*b*e*x*ArcTan[c*x])/(5*c^4) + (2*b*e*x^3*ArcTan[c*x])/(15*c^2) - (2*b*e*x^5*ArcT
an[c*x])/25 + (b*e*ArcTan[c*x]^2)/(5*c^5) + (137*b*e*Log[1 + c^2*x^2])/(300*c^5) + (b*e*Log[1 + c^2*x^2]^2)/(2
0*c^5) + (b*x^2*(d + e*Log[1 + c^2*x^2]))/(10*c^3) - (b*x^4*(d + e*Log[1 + c^2*x^2]))/(20*c) + (x^5*(a + b*Arc
Tan[c*x])*(d + e*Log[1 + c^2*x^2]))/5 - (b*Log[1 + c^2*x^2]*(d + e*Log[1 + c^2*x^2]))/(10*c^5)

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 203

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

Rule 260

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 266

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 635

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 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 2301

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 2390

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 2475

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 4846

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

Rule 4852

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

Rule 4884

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 4916

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 5021

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 6725

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*} \int x^4 \left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \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 )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}-\left (2 c^2 e\right ) \int \left (\frac {2 b x^3-b c^2 x^5+4 a c^3 x^6+4 b c^3 x^6 \tan ^{-1}(c x)}{20 c^3 \left (1+c^2 x^2\right )}-\frac {b x \log \left (1+c^2 x^2\right )}{10 c^5 \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 )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}+\frac {(b e) \int \frac {x \log \left (1+c^2 x^2\right )}{1+c^2 x^2} \, dx}{5 c^3}-\frac {e \int \frac {2 b x^3-b c^2 x^5+4 a c^3 x^6+4 b c^3 x^6 \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{10 c}\\ &=\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}+\frac {(b e) \operatorname {Subst}\left (\int \frac {\log \left (1+c^2 x\right )}{1+c^2 x} \, dx,x,x^2\right )}{10 c^3}-\frac {e \int \left (\frac {x^3 \left (2 b-b c^2 x^2+4 a c^3 x^3\right )}{1+c^2 x^2}+\frac {4 b c^3 x^6 \tan ^{-1}(c x)}{1+c^2 x^2}\right ) \, dx}{10 c}\\ &=\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}+\frac {(b e) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+c^2 x^2\right )}{10 c^5}-\frac {e \int \frac {x^3 \left (2 b-b c^2 x^2+4 a c^3 x^3\right )}{1+c^2 x^2} \, dx}{10 c}-\frac {1}{5} \left (2 b c^2 e\right ) \int \frac {x^6 \tan ^{-1}(c x)}{1+c^2 x^2} \, dx\\ &=\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}-\frac {1}{5} (2 b e) \int x^4 \tan ^{-1}(c x) \, dx+\frac {1}{5} (2 b e) \int \frac {x^4 \tan ^{-1}(c x)}{1+c^2 x^2} \, dx-\frac {e \int \left (\frac {4 a}{c^3}+\frac {3 b x}{c^2}-\frac {4 a x^2}{c}-b x^3+4 a c x^4-\frac {4 a+3 b c x}{c^3 \left (1+c^2 x^2\right )}\right ) \, dx}{10 c}\\ &=-\frac {2 a e x}{5 c^4}-\frac {3 b e x^2}{20 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {b e x^4}{40 c}-\frac {2}{25} a e x^5-\frac {2}{25} b e x^5 \tan ^{-1}(c x)+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}+\frac {e \int \frac {4 a+3 b c x}{1+c^2 x^2} \, dx}{10 c^4}+\frac {(2 b e) \int x^2 \tan ^{-1}(c x) \, dx}{5 c^2}-\frac {(2 b e) \int \frac {x^2 \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{5 c^2}+\frac {1}{25} (2 b c e) \int \frac {x^5}{1+c^2 x^2} \, dx\\ &=-\frac {2 a e x}{5 c^4}-\frac {3 b e x^2}{20 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {b e x^4}{40 c}-\frac {2}{25} a e x^5+\frac {2 b e x^3 \tan ^{-1}(c x)}{15 c^2}-\frac {2}{25} b e x^5 \tan ^{-1}(c x)+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}+\frac {(2 a e) \int \frac {1}{1+c^2 x^2} \, dx}{5 c^4}-\frac {(2 b e) \int \tan ^{-1}(c x) \, dx}{5 c^4}+\frac {(2 b e) \int \frac {\tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{5 c^4}+\frac {(3 b e) \int \frac {x}{1+c^2 x^2} \, dx}{10 c^3}-\frac {(2 b e) \int \frac {x^3}{1+c^2 x^2} \, dx}{15 c}+\frac {1}{25} (b c e) \operatorname {Subst}\left (\int \frac {x^2}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {2 a e x}{5 c^4}-\frac {3 b e x^2}{20 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {b e x^4}{40 c}-\frac {2}{25} a e x^5+\frac {2 a e \tan ^{-1}(c x)}{5 c^5}-\frac {2 b e x \tan ^{-1}(c x)}{5 c^4}+\frac {2 b e x^3 \tan ^{-1}(c x)}{15 c^2}-\frac {2}{25} b e x^5 \tan ^{-1}(c x)+\frac {b e \tan ^{-1}(c x)^2}{5 c^5}+\frac {3 b e \log \left (1+c^2 x^2\right )}{20 c^5}+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}+\frac {(2 b e) \int \frac {x}{1+c^2 x^2} \, dx}{5 c^3}-\frac {(b e) \operatorname {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )}{15 c}+\frac {1}{25} (b c e) \operatorname {Subst}\left (\int \left (-\frac {1}{c^4}+\frac {x}{c^2}+\frac {1}{c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {2 a e x}{5 c^4}-\frac {19 b e x^2}{100 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {9 b e x^4}{200 c}-\frac {2}{25} a e x^5+\frac {2 a e \tan ^{-1}(c x)}{5 c^5}-\frac {2 b e x \tan ^{-1}(c x)}{5 c^4}+\frac {2 b e x^3 \tan ^{-1}(c x)}{15 c^2}-\frac {2}{25} b e x^5 \tan ^{-1}(c x)+\frac {b e \tan ^{-1}(c x)^2}{5 c^5}+\frac {39 b e \log \left (1+c^2 x^2\right )}{100 c^5}+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}-\frac {(b e) \operatorname {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{15 c}\\ &=-\frac {2 a e x}{5 c^4}-\frac {77 b e x^2}{300 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {9 b e x^4}{200 c}-\frac {2}{25} a e x^5+\frac {2 a e \tan ^{-1}(c x)}{5 c^5}-\frac {2 b e x \tan ^{-1}(c x)}{5 c^4}+\frac {2 b e x^3 \tan ^{-1}(c x)}{15 c^2}-\frac {2}{25} b e x^5 \tan ^{-1}(c x)+\frac {b e \tan ^{-1}(c x)^2}{5 c^5}+\frac {137 b e \log \left (1+c^2 x^2\right )}{300 c^5}+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right ) \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 )}{10 c^5}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 214, normalized size = 0.77 \[ \frac {c x \left (8 a \left (15 c^4 d x^4-2 e \left (3 c^4 x^4-5 c^2 x^2+15\right )\right )+b c x \left (e \left (27 c^2 x^2-154\right )-30 d \left (c^2 x^2-2\right )\right )\right )+\log \left (c^2 x^2+1\right ) \left (120 a c^5 e x^5+2 b e \left (-15 c^4 x^4+30 c^2 x^2+137\right )-60 b d\right )+8 \tan ^{-1}(c x) \left (30 a e+15 b c^5 d x^5+15 b c^5 e x^5 \log \left (c^2 x^2+1\right )-2 b c e x \left (3 c^4 x^4-5 c^2 x^2+15\right )\right )-30 b e \log ^2\left (c^2 x^2+1\right )+120 b e \tan ^{-1}(c x)^2}{600 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]

[Out]

(c*x*(b*c*x*(-30*d*(-2 + c^2*x^2) + e*(-154 + 27*c^2*x^2)) + 8*a*(15*c^4*d*x^4 - 2*e*(15 - 5*c^2*x^2 + 3*c^4*x
^4))) + 120*b*e*ArcTan[c*x]^2 + (-60*b*d + 120*a*c^5*e*x^5 + 2*b*e*(137 + 30*c^2*x^2 - 15*c^4*x^4))*Log[1 + c^
2*x^2] - 30*b*e*Log[1 + c^2*x^2]^2 + 8*ArcTan[c*x]*(30*a*e + 15*b*c^5*d*x^5 - 2*b*c*e*x*(15 - 5*c^2*x^2 + 3*c^
4*x^4) + 15*b*c^5*e*x^5*Log[1 + c^2*x^2]))/(600*c^5)

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fricas [A]  time = 0.56, size = 220, normalized size = 0.79 \[ \frac {80 \, a c^{3} e x^{3} + 24 \, {\left (5 \, a c^{5} d - 2 \, a c^{5} e\right )} x^{5} - 3 \, {\left (10 \, b c^{4} d - 9 \, b c^{4} e\right )} x^{4} - 240 \, a c e x + 120 \, b e \arctan \left (c x\right )^{2} - 30 \, b e \log \left (c^{2} x^{2} + 1\right )^{2} + 2 \, {\left (30 \, b c^{2} d - 77 \, b c^{2} e\right )} x^{2} + 8 \, {\left (10 \, b c^{3} e x^{3} + 3 \, {\left (5 \, b c^{5} d - 2 \, b c^{5} e\right )} x^{5} - 30 \, b c e x + 30 \, a e\right )} \arctan \left (c x\right ) + 2 \, {\left (60 \, b c^{5} e x^{5} \arctan \left (c x\right ) + 60 \, a c^{5} e x^{5} - 15 \, b c^{4} e x^{4} + 30 \, b c^{2} e x^{2} - 30 \, b d + 137 \, b e\right )} \log \left (c^{2} x^{2} + 1\right )}{600 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="fricas")

[Out]

1/600*(80*a*c^3*e*x^3 + 24*(5*a*c^5*d - 2*a*c^5*e)*x^5 - 3*(10*b*c^4*d - 9*b*c^4*e)*x^4 - 240*a*c*e*x + 120*b*
e*arctan(c*x)^2 - 30*b*e*log(c^2*x^2 + 1)^2 + 2*(30*b*c^2*d - 77*b*c^2*e)*x^2 + 8*(10*b*c^3*e*x^3 + 3*(5*b*c^5
*d - 2*b*c^5*e)*x^5 - 30*b*c*e*x + 30*a*e)*arctan(c*x) + 2*(60*b*c^5*e*x^5*arctan(c*x) + 60*a*c^5*e*x^5 - 15*b
*c^4*e*x^4 + 30*b*c^2*e*x^2 - 30*b*d + 137*b*e)*log(c^2*x^2 + 1))/c^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 3.80, size = 4941, normalized size = 17.77 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1)),x)

[Out]

-181/600*e/c^5*b-1/10*I*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2
+1)+1)^2)*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*arctan(c*x)*Pi*x^5*e+1/40*I/c*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)
)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*Pi*x^4*e
-1/20*I/c^3*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*cs
gn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*Pi*x^2*e-1/10*I/c^5*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2
+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*e*Pi*ln((1+I*c*x)^2/(c^2*x^2+1
)+1)+3/40*I/c^5*b*Pi*e*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*
x^2+1)+1)^2)^2+3/20*I/c^5*b*Pi*e*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2-1/10*I/
c^5*b*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*e*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3+1/10*I/c^5*b*ln((1+I*c*x)^2/(c^2*x^
2+1)+1)*e*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3-1/10*I/c^5*b*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*e*Pi*csgn(I*(1
+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3-3/20*I/c^5*b*Pi*e*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*c
sgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2-2/25*a*e*x^5-2/5*b*e*x*arctan(c*x)/c^4+2/15*b*e*x^3*arctan(c*x)/c^2+3/1
0/c^5*b*ln(2)*e-2/5*a*e*x/c^4-77/300*b*e*x^2/c^3+2/15*a*e*x^3/c^2+2/5*a*e*arctan(c*x)/c^5-2/25*b*e*x^5*arctan(
c*x)-1/10/c^5*b*arctan(c*x)*Pi*e*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3-1/10/c^5*b*arctan(c*x)*Pi*e*csgn(I*(1+I*c*x
)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3+1/10/c^5*b*arctan(c*x)*Pi*e*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1
)^2)^3+3/40*I/c^5*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*Pi*e-3/40*I/c^5*b*Pi*e*csgn(I*(1+I*c*x)^2/(c^2*x^2
+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3-3/40*I/c^5*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*Pi*e-2/5*I/c^5*b*arctan(c*
x)*ln(2)*e+1/5*x^5*a*d+1/10/c^5*b*e*(4*arctan(c*x)*x^5*c^5-c^4*x^4-4*I*arctan(c*x)+2*c^2*x^2+4*ln((1+I*c*x)^2/
(c^2*x^2+1)+1)+3)*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))+1/10*I*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/
(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*arctan(c*x)*Pi*x^5*e-1/5*I*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*
csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*arctan(c*x)*Pi*x^5*e-1/20/c*b*x^4*d+1/10/c^3*b*d*x^2+1/5*a*e*x^5*ln(c^
2*x^2+1)+1/5/c^5*b*arctan(c*x)*Pi*e*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2+1/10
/c^5*b*arctan(c*x)*Pi*e*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2
*x^2+1)+1)^2)^2-137/150/c^5*b*e*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+1/5/c^5*b*d*ln((1+I*c*x)^2/(c^2*x^2+1)+1)-1/5/c^
5*b*e*ln((1+I*c*x)^2/(c^2*x^2+1)+1)^2+1/5*b*arctan(c*x)*x^5*d-1/5/c^5*b*arctan(c*x)*Pi*e*csgn(I*((1+I*c*x)^2/(
c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2+1/10/c^5*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/
(c^2*x^2+1)+1)^2)^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*arctan(c*x)*Pi*e+1/10/c*b*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*x^
4*e-1/5/c^3*b*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*x^2*e-1/10/c*b*ln(2)*x^4*e+1/5/c^3*b*ln(2)*x^2*e+2/5/c^5*b*e*ln(2)
*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+2/5*b*ln(2)*arctan(c*x)*x^5*e-2/5*b*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*arctan(c*x)*x
^5*e-1/5*I/c^5*b*arctan(c*x)*d+46/75*I/c^5*b*e*arctan(c*x)+1/5*I*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*csgn(I*(1
+I*c*x)/(c^2*x^2+1)^(1/2))*arctan(c*x)*Pi*x^5*e-1/10*I*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)/(c^2
*x^2+1)^(1/2))^2*arctan(c*x)*Pi*x^5*e+1/10*I*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2
*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*arctan(c*x)*Pi*x^5*e+1/10*I*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csg
n(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*arctan(c*x)*Pi*x^5*e-3/40*I/c^5*b*Pi*e*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2
)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)+1/40*I/c*b*csg
n(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))^2*Pi*x^4*e+1/20*I/c*b*csgn(I*((1+I*c*x)^2/(c^
2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*Pi*x^4*e-1/20*I/c*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*csg
n(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*Pi*x^4*e-1/40*I/c*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x
^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*Pi*x^4*e-1/10*I/c^3*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*
c*x)^2/(c^2*x^2+1)+1)^2)^2*Pi*x^2*e+1/10*I/c^3*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*csgn(I*(1+I*c*x)/(c^2*x^2+1
)^(1/2))*Pi*x^2*e+1/20*I/c^3*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*Pi*
x^2*e+1/20*I/c^3*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^
2)^2*Pi*x^2*e+1/20*I/c^3*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I/((1+I*c*x)^2
/(c^2*x^2+1)+1)^2)*Pi*x^2*e-1/20*I/c^3*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))^2
*Pi*x^2*e-1/40*I/c*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I/((1+I*c*x)^2/(c^2*
x^2+1)+1)^2)*Pi*x^4*e-1/40*I/c*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*P
i*x^4*e-1/5*I/c^5*b*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*e*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2
/(c^2*x^2+1)+1)^2)^2+1/10*I/c^5*b*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*e*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1
+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2-1/10*I/c^5*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+
I*c*x)/(c^2*x^2+1)^(1/2))^2*e*Pi*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+1/5*I/c^5*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*c
sgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*e*Pi*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+1/10*I/c^5*b*csgn(I*((1+I*c*x)^2/(c^2*x^
2+1)+1)^2)*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*Pi*e*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+1/10*I/c^5*b*ln((1+I*c*x)^
2/(c^2*x^2+1)+1)*e*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x
^2+1)+1)^2)^2-1/10/c^5*b*arctan(c*x)*Pi*e*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)
)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)-3/40*I/c^5*b*Pi*e*csgn(I*(1+I*c*x)/(c^2*x^2+1)
^(1/2))^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))-1/20*I/c^3*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*Pi*x^2*e+1/40*I/c*b*c
sgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*Pi*x^4*e+1/40*I/c*b*csgn(I*(1+I*c*x)^2/(c^2*x^2
+1))^3*Pi*x^4*e-1/40*I/c*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*Pi*x^4*e+1/20*I/c^3*b*csgn(I*((1+I*c*x)^2/(
c^2*x^2+1)+1)^2)^3*Pi*x^2*e-1/10*I*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*arctan(c*
x)*Pi*x^5*e+1/10*I*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*arctan(c*x)*Pi*x^5*e-1/10*I*b*csgn(I*(1+I*c*x)^2/
(c^2*x^2+1))^3*arctan(c*x)*Pi*x^5*e+1/10/c^5*b*arctan(c*x)*Pi*e*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*(
(1+I*c*x)^2/(c^2*x^2+1)+1)^2)+3/20/c^5*b*d-1/10/c^5*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)/(c^2*x^
2+1)^(1/2))^2*arctan(c*x)*Pi*e-1/20*I/c^3*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*Pi
*x^2*e+3/40*I/c^5*b*Pi*e*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+
1)+1)^2)^2+3/40*I/c^5*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*Pi*e+9/200
*b*e*x^4/c

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maxima [A]  time = 0.43, size = 256, normalized size = 0.92 \[ \frac {1}{5} \, a d x^{5} + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{2} {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b e \arctan \left (c x\right ) + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b d + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{2} {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} a e + \frac {{\left (27 \, c^{4} x^{4} - 154 \, c^{2} x^{2} - 120 \, \arctan \left (c x\right )^{2} - 2 \, {\left (15 \, c^{4} x^{4} - 30 \, c^{2} x^{2} - 137\right )} \log \left (c^{2} x^{2} + 1\right ) - 30 \, \log \left (c^{2} x^{2} + 1\right )^{2}\right )} b e}{600 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="maxima")

[Out]

1/5*a*d*x^5 + 1/75*(15*x^5*log(c^2*x^2 + 1) - 2*c^2*((3*c^4*x^5 - 5*c^2*x^3 + 15*x)/c^6 - 15*arctan(c*x)/c^7))
*b*e*arctan(c*x) + 1/20*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*d + 1/75*(1
5*x^5*log(c^2*x^2 + 1) - 2*c^2*((3*c^4*x^5 - 5*c^2*x^3 + 15*x)/c^6 - 15*arctan(c*x)/c^7))*a*e + 1/600*(27*c^4*
x^4 - 154*c^2*x^2 - 120*arctan(c*x)^2 - 2*(15*c^4*x^4 - 30*c^2*x^2 - 137)*log(c^2*x^2 + 1) - 30*log(c^2*x^2 +
1)^2)*b*e/c^5

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mupad [B]  time = 3.35, size = 276, normalized size = 0.99 \[ \frac {a\,d\,x^5}{5}-\frac {2\,a\,e\,x^5}{25}-\frac {b\,e\,{\ln \left (c^2\,x^2+1\right )}^2}{20\,c^5}-\frac {2\,a\,e\,x}{5\,c^4}+\frac {2\,a\,e\,\mathrm {atan}\left (c\,x\right )}{5\,c^5}+\frac {b\,d\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}-\frac {2\,b\,e\,x^5\,\mathrm {atan}\left (c\,x\right )}{25}-\frac {b\,d\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}+\frac {137\,b\,e\,\ln \left (c^2\,x^2+1\right )}{300\,c^5}+\frac {2\,a\,e\,x^3}{15\,c^2}-\frac {b\,d\,x^4}{20\,c}+\frac {b\,d\,x^2}{10\,c^3}+\frac {9\,b\,e\,x^4}{200\,c}-\frac {77\,b\,e\,x^2}{300\,c^3}+\frac {a\,e\,x^5\,\ln \left (c^2\,x^2+1\right )}{5}+\frac {b\,e\,{\mathrm {atan}\left (c\,x\right )}^2}{5\,c^5}+\frac {2\,b\,e\,x^3\,\mathrm {atan}\left (c\,x\right )}{15\,c^2}+\frac {b\,e\,x^5\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right )}{5}-\frac {b\,e\,x^4\,\ln \left (c^2\,x^2+1\right )}{20\,c}+\frac {b\,e\,x^2\,\ln \left (c^2\,x^2+1\right )}{10\,c^3}-\frac {2\,b\,e\,x\,\mathrm {atan}\left (c\,x\right )}{5\,c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)),x)

[Out]

(a*d*x^5)/5 - (2*a*e*x^5)/25 - (b*e*log(c^2*x^2 + 1)^2)/(20*c^5) - (2*a*e*x)/(5*c^4) + (2*a*e*atan(c*x))/(5*c^
5) + (b*d*x^5*atan(c*x))/5 - (2*b*e*x^5*atan(c*x))/25 - (b*d*log(c^2*x^2 + 1))/(10*c^5) + (137*b*e*log(c^2*x^2
+ 1))/(300*c^5) + (2*a*e*x^3)/(15*c^2) - (b*d*x^4)/(20*c) + (b*d*x^2)/(10*c^3) + (9*b*e*x^4)/(200*c) - (77*b*
e*x^2)/(300*c^3) + (a*e*x^5*log(c^2*x^2 + 1))/5 + (b*e*atan(c*x)^2)/(5*c^5) + (2*b*e*x^3*atan(c*x))/(15*c^2) +
(b*e*x^5*atan(c*x)*log(c^2*x^2 + 1))/5 - (b*e*x^4*log(c^2*x^2 + 1))/(20*c) + (b*e*x^2*log(c^2*x^2 + 1))/(10*c
^3) - (2*b*e*x*atan(c*x))/(5*c^4)

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sympy [A]  time = 13.49, size = 338, normalized size = 1.22 \[ \begin {cases} \frac {a d x^{5}}{5} + \frac {a e x^{5} \log {\left (c^{2} x^{2} + 1 \right )}}{5} - \frac {2 a e x^{5}}{25} + \frac {2 a e x^{3}}{15 c^{2}} - \frac {2 a e x}{5 c^{4}} + \frac {2 a e \operatorname {atan}{\left (c x \right )}}{5 c^{5}} + \frac {b d x^{5} \operatorname {atan}{\left (c x \right )}}{5} + \frac {b e x^{5} \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )}}{5} - \frac {2 b e x^{5} \operatorname {atan}{\left (c x \right )}}{25} - \frac {b d x^{4}}{20 c} - \frac {b e x^{4} \log {\left (c^{2} x^{2} + 1 \right )}}{20 c} + \frac {9 b e x^{4}}{200 c} + \frac {2 b e x^{3} \operatorname {atan}{\left (c x \right )}}{15 c^{2}} + \frac {b d x^{2}}{10 c^{3}} + \frac {b e x^{2} \log {\left (c^{2} x^{2} + 1 \right )}}{10 c^{3}} - \frac {77 b e x^{2}}{300 c^{3}} - \frac {2 b e x \operatorname {atan}{\left (c x \right )}}{5 c^{4}} - \frac {b d \log {\left (c^{2} x^{2} + 1 \right )}}{10 c^{5}} - \frac {b e \log {\left (c^{2} x^{2} + 1 \right )}^{2}}{20 c^{5}} + \frac {137 b e \log {\left (c^{2} x^{2} + 1 \right )}}{300 c^{5}} + \frac {b e \operatorname {atan}^{2}{\left (c x \right )}}{5 c^{5}} & \text {for}\: c \neq 0 \\\frac {a d x^{5}}{5} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atan(c*x))*(d+e*ln(c**2*x**2+1)),x)

[Out]

Piecewise((a*d*x**5/5 + a*e*x**5*log(c**2*x**2 + 1)/5 - 2*a*e*x**5/25 + 2*a*e*x**3/(15*c**2) - 2*a*e*x/(5*c**4
) + 2*a*e*atan(c*x)/(5*c**5) + b*d*x**5*atan(c*x)/5 + b*e*x**5*log(c**2*x**2 + 1)*atan(c*x)/5 - 2*b*e*x**5*ata
n(c*x)/25 - b*d*x**4/(20*c) - b*e*x**4*log(c**2*x**2 + 1)/(20*c) + 9*b*e*x**4/(200*c) + 2*b*e*x**3*atan(c*x)/(
15*c**2) + b*d*x**2/(10*c**3) + b*e*x**2*log(c**2*x**2 + 1)/(10*c**3) - 77*b*e*x**2/(300*c**3) - 2*b*e*x*atan(
c*x)/(5*c**4) - b*d*log(c**2*x**2 + 1)/(10*c**5) - b*e*log(c**2*x**2 + 1)**2/(20*c**5) + 137*b*e*log(c**2*x**2
+ 1)/(300*c**5) + b*e*atan(c*x)**2/(5*c**5), Ne(c, 0)), (a*d*x**5/5, True))

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