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

Optimal. Leaf size=250 \[ \frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {b \left (3 c d^2+e^2\right ) \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \left (3 c d^2+e^2\right ) \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}-\frac {b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{3 \sqrt {2} c^{3/2}}-\frac {b d e \log \left (c^2 x^4+1\right )}{2 c}-\frac {b d^3 \tan ^{-1}\left (c x^2\right )}{3 e}-\frac {2 b e^2 x}{3 c} \]

[Out]

-2/3*b*e^2*x/c-1/3*b*d^3*arctan(c*x^2)/e+1/3*(e*x+d)^3*(a+b*arctan(c*x^2))/e-1/2*b*d*e*ln(c^2*x^4+1)/c-1/6*b*(
3*c*d^2-e^2)*arctan(-1+x*2^(1/2)*c^(1/2))/c^(3/2)*2^(1/2)-1/6*b*(3*c*d^2-e^2)*arctan(1+x*2^(1/2)*c^(1/2))/c^(3
/2)*2^(1/2)-1/12*b*(3*c*d^2+e^2)*ln(1+c*x^2-x*2^(1/2)*c^(1/2))/c^(3/2)*2^(1/2)+1/12*b*(3*c*d^2+e^2)*ln(1+c*x^2
+x*2^(1/2)*c^(1/2))/c^(3/2)*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 14, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {5205, 12, 1831, 1248, 635, 203, 260, 1280, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {b \left (3 c d^2+e^2\right ) \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \left (3 c d^2+e^2\right ) \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}-\frac {b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{3 \sqrt {2} c^{3/2}}-\frac {b d e \log \left (c^2 x^4+1\right )}{2 c}-\frac {b d^3 \tan ^{-1}\left (c x^2\right )}{3 e}-\frac {2 b e^2 x}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1280

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f*(f*x)^(m - 1)*
(a + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m - 2)*(a + c*x^4)^p*(a*e*
(m - 1) - c*d*(m + 4*p + 3)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] &
& IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1831

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[((c*x)^(m + ii)*(Coeff[Pq,
x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2)))/(c^ii*(a + b*x^n)), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; Fr
eeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n

Rule 5205

Int[((a_.) + ArcTan[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 + u^2), x], x]
, x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m +
1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]

Rubi steps

\begin {align*} \int (d+e x)^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \, dx &=\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {b \int \frac {2 c x (d+e x)^3}{1+c^2 x^4} \, dx}{3 e}\\ &=\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {(2 b c) \int \frac {x (d+e x)^3}{1+c^2 x^4} \, dx}{3 e}\\ &=\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {(2 b c) \int \left (\frac {x \left (d^3+3 d e^2 x^2\right )}{1+c^2 x^4}+\frac {x^2 \left (3 d^2 e+e^3 x^2\right )}{1+c^2 x^4}\right ) \, dx}{3 e}\\ &=\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {(2 b c) \int \frac {x \left (d^3+3 d e^2 x^2\right )}{1+c^2 x^4} \, dx}{3 e}-\frac {(2 b c) \int \frac {x^2 \left (3 d^2 e+e^3 x^2\right )}{1+c^2 x^4} \, dx}{3 e}\\ &=-\frac {2 b e^2 x}{3 c}+\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}+\frac {(2 b) \int \frac {e^3-3 c^2 d^2 e x^2}{1+c^2 x^4} \, dx}{3 c e}-\frac {(b c) \operatorname {Subst}\left (\int \frac {d^3+3 d e^2 x}{1+c^2 x^2} \, dx,x,x^2\right )}{3 e}\\ &=-\frac {2 b e^2 x}{3 c}+\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {\left (b c d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x^2} \, dx,x,x^2\right )}{3 e}-(b c d e) \operatorname {Subst}\left (\int \frac {x}{1+c^2 x^2} \, dx,x,x^2\right )-\frac {\left (b \left (3 c d^2-e^2\right )\right ) \int \frac {c+c^2 x^2}{1+c^2 x^4} \, dx}{3 c^2}+\frac {\left (b \left (3 c d^2+e^2\right )\right ) \int \frac {c-c^2 x^2}{1+c^2 x^4} \, dx}{3 c^2}\\ &=-\frac {2 b e^2 x}{3 c}-\frac {b d^3 \tan ^{-1}\left (c x^2\right )}{3 e}+\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {b d e \log \left (1+c^2 x^4\right )}{2 c}-\frac {\left (b \left (3 c d^2-e^2\right )\right ) \int \frac {1}{\frac {1}{c}-\frac {\sqrt {2} x}{\sqrt {c}}+x^2} \, dx}{6 c^2}-\frac {\left (b \left (3 c d^2-e^2\right )\right ) \int \frac {1}{\frac {1}{c}+\frac {\sqrt {2} x}{\sqrt {c}}+x^2} \, dx}{6 c^2}-\frac {\left (b \left (3 c d^2+e^2\right )\right ) \int \frac {\frac {\sqrt {2}}{\sqrt {c}}+2 x}{-\frac {1}{c}-\frac {\sqrt {2} x}{\sqrt {c}}-x^2} \, dx}{6 \sqrt {2} c^{3/2}}-\frac {\left (b \left (3 c d^2+e^2\right )\right ) \int \frac {\frac {\sqrt {2}}{\sqrt {c}}-2 x}{-\frac {1}{c}+\frac {\sqrt {2} x}{\sqrt {c}}-x^2} \, dx}{6 \sqrt {2} c^{3/2}}\\ &=-\frac {2 b e^2 x}{3 c}-\frac {b d^3 \tan ^{-1}\left (c x^2\right )}{3 e}+\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}-\frac {b \left (3 c d^2+e^2\right ) \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \left (3 c d^2+e^2\right ) \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}}-\frac {b d e \log \left (1+c^2 x^4\right )}{2 c}-\frac {\left (b \left (3 c d^2-e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}+\frac {\left (b \left (3 c d^2-e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}\\ &=-\frac {2 b e^2 x}{3 c}-\frac {b d^3 \tan ^{-1}\left (c x^2\right )}{3 e}+\frac {(d+e x)^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{3 e}+\frac {b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}-\frac {b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2} c^{3/2}}-\frac {b \left (3 c d^2+e^2\right ) \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}}+\frac {b \left (3 c d^2+e^2\right ) \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2} c^{3/2}}-\frac {b d e \log \left (1+c^2 x^4\right )}{2 c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.48, size = 252, normalized size = 1.01 \[ \frac {1}{12} \left (12 a d^2 x+12 a d e x^2+4 a e^2 x^3-\frac {\sqrt {2} b \left (3 c d^2+e^2\right ) \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{c^{3/2}}+\frac {\sqrt {2} b \left (3 c d^2+e^2\right ) \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{c^{3/2}}+\frac {2 \sqrt {2} b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{c^{3/2}}-\frac {2 \sqrt {2} b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{c^{3/2}}-\frac {6 b d e \log \left (c^2 x^4+1\right )}{c}+4 b x \tan ^{-1}\left (c x^2\right ) \left (3 d^2+3 d e x+e^2 x^2\right )-\frac {8 b e^2 x}{c}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*d^2*x - (8*b*e^2*x)/c + 12*a*d*e*x^2 + 4*a*e^2*x^3 + 4*b*x*(3*d^2 + 3*d*e*x + e^2*x^2)*ArcTan[c*x^2] + (
2*Sqrt[2]*b*(3*c*d^2 - e^2)*ArcTan[1 - Sqrt[2]*Sqrt[c]*x])/c^(3/2) - (2*Sqrt[2]*b*(3*c*d^2 - e^2)*ArcTan[1 + S
qrt[2]*Sqrt[c]*x])/c^(3/2) - (Sqrt[2]*b*(3*c*d^2 + e^2)*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2])/c^(3/2) + (Sqrt[2]
*b*(3*c*d^2 + e^2)*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/c^(3/2) - (6*b*d*e*Log[1 + c^2*x^4])/c)/12

________________________________________________________________________________________

fricas [B]  time = 0.65, size = 4861, normalized size = 19.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*sqrt(2)*c^7*sqrt((81*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8
+ 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b
^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(3/4)*sqrt((81*b^4*c^4*d^8 - 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)*arctan((sqrt(2)
*(c^11*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)*sqrt((81*b^4*c^4*d^8 - 18*b^4*c^2*d^4*e^4
+ b^4*e^8)/c^6) + 3*(9*b^2*c^11*d^6 + b^2*c^9*d^2*e^4)*sqrt((81*b^4*c^4*d^8 - 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c
^6))*sqrt((6561*b^6*c^8*d^16 - 162*b^6*c^4*d^8*e^8 + b^6*e^16)*x^2 + sqrt(2)*(3*(81*b^3*c^9*d^10 - 18*b^3*c^7*
d^6*e^4 + b^3*c^5*d^2*e^8)*x*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6) + (729*b^5*c^7*d^12*e^2
- 81*b^5*c^5*d^8*e^6 - 9*b^5*c^3*d^4*e^10 + b^5*c*e^14)*x)*sqrt((81*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^
8 - 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*
e^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(1/4) + (729*b^4*c^8*d^12 - 81*b^4*c^6*d
^8*e^4 - 9*b^4*c^4*d^4*e^8 + b^4*c^2*e^12)*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))*sqrt((81
*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^
8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6
)^(3/4) + sqrt(2)*((81*b^3*c^15*d^8*e^2 - b^3*c^11*e^10)*x*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8
)/c^6)*sqrt((81*b^4*c^4*d^8 - 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6) + 3*(729*b^5*c^15*d^14 + 81*b^5*c^13*d^10*e^4
- 9*b^5*c^11*d^6*e^8 - b^5*c^9*d^2*e^12)*x*sqrt((81*b^4*c^4*d^8 - 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))*sqrt((8
1*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e
^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^
6)^(3/4) + (6561*b^6*c^14*d^16 + 1458*b^6*c^12*d^12*e^4 - 18*b^6*c^8*d^4*e^12 - b^6*c^6*e^16)*sqrt((81*b^4*c^4
*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)*sqrt((81*b^4*c^4*d^8 - 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(531441*b
^10*c^12*d^24 + 118098*b^10*c^10*d^20*e^4 - 6561*b^10*c^8*d^16*e^8 - 2916*b^10*c^6*d^12*e^12 - 81*b^10*c^4*d^8
*e^16 + 18*b^10*c^2*d^4*e^20 + b^10*e^24)) + 4*sqrt(2)*c^7*sqrt((81*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8
- 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e
^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(3/4)*sqrt((81*b^4*c^4*d^8 - 18*b^4*c^2*d
^4*e^4 + b^4*e^8)/c^6)*arctan((sqrt(2)*(c^11*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)*sqr
t((81*b^4*c^4*d^8 - 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6) + 3*(9*b^2*c^11*d^6 + b^2*c^9*d^2*e^4)*sqrt((81*b^4*c^4
*d^8 - 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))*sqrt((6561*b^6*c^8*d^16 - 162*b^6*c^4*d^8*e^8 + b^6*e^16)*x^2 - sqr
t(2)*(3*(81*b^3*c^9*d^10 - 18*b^3*c^7*d^6*e^4 + b^3*c^5*d^2*e^8)*x*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 +
b^4*e^8)/c^6) + (729*b^5*c^7*d^12*e^2 - 81*b^5*c^5*d^8*e^6 - 9*b^5*c^3*d^4*e^10 + b^5*c*e^14)*x)*sqrt((81*b^2
*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c
^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(1
/4) + (729*b^4*c^8*d^12 - 81*b^4*c^6*d^8*e^4 - 9*b^4*c^4*d^4*e^8 + b^4*c^2*e^12)*sqrt((81*b^4*c^4*d^8 + 18*b^4
*c^2*d^4*e^4 + b^4*e^8)/c^6))*sqrt((81*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2*sqrt((81*b^4
*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*((81*b^4*c^4*d
^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(3/4) + sqrt(2)*((81*b^3*c^15*d^8*e^2 - b^3*c^11*e^10)*x*sqrt((81*b^4*
c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)*sqrt((81*b^4*c^4*d^8 - 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6) + 3*(72
9*b^5*c^15*d^14 + 81*b^5*c^13*d^10*e^4 - 9*b^5*c^11*d^6*e^8 - b^5*c^9*d^2*e^12)*x*sqrt((81*b^4*c^4*d^8 - 18*b^
4*c^2*d^4*e^4 + b^4*e^8)/c^6))*sqrt((81*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2*sqrt((81*b^
4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*((81*b^4*c^4*
d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(3/4) - (6561*b^6*c^14*d^16 + 1458*b^6*c^12*d^12*e^4 - 18*b^6*c^8*d^4
*e^12 - b^6*c^6*e^16)*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)*sqrt((81*b^4*c^4*d^8 - 18*b^4*
c^2*d^4*e^4 + b^4*e^8)/c^6))/(531441*b^10*c^12*d^24 + 118098*b^10*c^10*d^20*e^4 - 6561*b^10*c^8*d^16*e^8 - 291
6*b^10*c^6*d^12*e^12 - 81*b^10*c^4*d^8*e^16 + 18*b^10*c^2*d^4*e^20 + b^10*e^24)) - 4*(81*a*b^4*c^5*d^8*e^2 + 1
8*a*b^4*c^3*d^4*e^6 + a*b^4*c*e^10)*x^3 - 12*(81*a*b^4*c^5*d^9*e + 18*a*b^4*c^3*d^5*e^5 + a*b^4*c*d*e^9)*x^2 -
4*(243*a*b^4*c^5*d^10 - 162*b^5*c^4*d^8*e^2 + 54*a*b^4*c^3*d^6*e^4 - 36*b^5*c^2*d^4*e^6 + 3*a*b^4*c*d^2*e^8 -
2*b^5*e^10)*x - 4*((81*b^5*c^5*d^8*e^2 + 18*b^5*c^3*d^4*e^6 + b^5*c*e^10)*x^3 + 3*(81*b^5*c^5*d^9*e + 18*b^5*
c^3*d^5*e^5 + b^5*c*d*e^9)*x^2 + 3*(81*b^5*c^5*d^10 + 18*b^5*c^3*d^6*e^4 + b^5*c*d^2*e^8)*x)*arctan(c*x^2) + (
486*b^5*c^4*d^9*e + 108*b^5*c^2*d^5*e^5 + 6*b^5*d*e^9 - sqrt(2)*(81*b^4*c^5*d^8 + 18*b^4*c^3*d^4*e^4 + b^4*c*e
^8 + 6*b^2*c^5*d^2*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))*sqrt((81*b^2*c^4*d^8 + 18*b^
2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4
*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(1/4))*log((6561*b
^6*c^8*d^16 - 162*b^6*c^4*d^8*e^8 + b^6*e^16)*x^2 + sqrt(2)*(3*(81*b^3*c^9*d^10 - 18*b^3*c^7*d^6*e^4 + b^3*c^5
*d^2*e^8)*x*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6) + (729*b^5*c^7*d^12*e^2 - 81*b^5*c^5*d^8
*e^6 - 9*b^5*c^3*d^4*e^10 + b^5*c*e^14)*x)*sqrt((81*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2
*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*(
(81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(1/4) + (729*b^4*c^8*d^12 - 81*b^4*c^6*d^8*e^4 - 9*b^4*c^
4*d^4*e^8 + b^4*c^2*e^12)*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)) + (486*b^5*c^4*d^9*e + 10
8*b^5*c^2*d^5*e^5 + 6*b^5*d*e^9 + sqrt(2)*(81*b^4*c^5*d^8 + 18*b^4*c^3*d^4*e^4 + b^4*c*e^8 + 6*b^2*c^5*d^2*e^2
*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))*sqrt((81*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^
8 - 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*
e^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(1/4))*log((6561*b^6*c^8*d^16 - 162*b^6*
c^4*d^8*e^8 + b^6*e^16)*x^2 - sqrt(2)*(3*(81*b^3*c^9*d^10 - 18*b^3*c^7*d^6*e^4 + b^3*c^5*d^2*e^8)*x*sqrt((81*b
^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6) + (729*b^5*c^7*d^12*e^2 - 81*b^5*c^5*d^8*e^6 - 9*b^5*c^3*d^4*e
^10 + b^5*c*e^14)*x)*sqrt((81*b^2*c^4*d^8 + 18*b^2*c^2*d^4*e^4 + b^2*e^8 - 6*c^4*d^2*e^2*sqrt((81*b^4*c^4*d^8
+ 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6))/(81*b^2*c^4*d^8 - 18*b^2*c^2*d^4*e^4 + b^2*e^8))*((81*b^4*c^4*d^8 + 18*b
^4*c^2*d^4*e^4 + b^4*e^8)/c^6)^(1/4) + (729*b^4*c^8*d^12 - 81*b^4*c^6*d^8*e^4 - 9*b^4*c^4*d^4*e^8 + b^4*c^2*e^
12)*sqrt((81*b^4*c^4*d^8 + 18*b^4*c^2*d^4*e^4 + b^4*e^8)/c^6)))/(81*b^4*c^5*d^8 + 18*b^4*c^3*d^4*e^4 + b^4*c*e
^8)

________________________________________________________________________________________

giac [A]  time = 34.88, size = 303, normalized size = 1.21 \[ \frac {2 \, b c x^{3} \arctan \left (c x^{2}\right ) e^{2} + 6 \, b c d x^{2} \arctan \left (c x^{2}\right ) e + 6 \, b c d^{2} x \arctan \left (c x^{2}\right ) + 2 \, a c x^{3} e^{2} + 6 \, a c d x^{2} e + 6 \, a c d^{2} x - 3 \, b d e \log \left (c^{2} x^{4} + 1\right ) - 4 \, b x e^{2}}{6 \, c} - \frac {\sqrt {2} {\left (3 \, b c^{2} d^{2} + b {\left | c \right |} e^{2}\right )} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{12 \, c {\left | c \right |}^{\frac {3}{2}}} - \frac {\sqrt {2} {\left (3 \, b c^{2} d^{2} \sqrt {{\left | c \right |}} - b {\left | c \right |}^{\frac {3}{2}} e^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{6 \, c^{3}} - \frac {\sqrt {2} {\left (3 \, b c^{2} d^{2} \sqrt {{\left | c \right |}} - b {\left | c \right |}^{\frac {3}{2}} e^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{6 \, c^{3}} + \frac {\sqrt {2} {\left (3 \, b c^{2} d^{2} \sqrt {{\left | c \right |}} + b {\left | c \right |}^{\frac {3}{2}} e^{2}\right )} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{12 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b*c*x^3*arctan(c*x^2)*e^2 + 6*b*c*d*x^2*arctan(c*x^2)*e + 6*b*c*d^2*x*arctan(c*x^2) + 2*a*c*x^3*e^2 + 6
*a*c*d*x^2*e + 6*a*c*d^2*x - 3*b*d*e*log(c^2*x^4 + 1) - 4*b*x*e^2)/c - 1/12*sqrt(2)*(3*b*c^2*d^2 + b*abs(c)*e^
2)*log(x^2 - sqrt(2)*x/sqrt(abs(c)) + 1/abs(c))/(c*abs(c)^(3/2)) - 1/6*sqrt(2)*(3*b*c^2*d^2*sqrt(abs(c)) - b*a
bs(c)^(3/2)*e^2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/c^3 - 1/6*sqrt(2)*(3*b*c^2*d^2*
sqrt(abs(c)) - b*abs(c)^(3/2)*e^2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/c^3 + 1/12*sq
rt(2)*(3*b*c^2*d^2*sqrt(abs(c)) + b*abs(c)^(3/2)*e^2)*log(x^2 + sqrt(2)*x/sqrt(abs(c)) + 1/abs(c))/c^3

________________________________________________________________________________________

maple [A]  time = 0.05, size = 381, normalized size = 1.52 \[ \frac {a \,e^{2} x^{3}}{3}+a e d \,x^{2}+a x \,d^{2}+\frac {a \,d^{3}}{3 e}+\frac {b \,e^{2} \arctan \left (c \,x^{2}\right ) x^{3}}{3}+b e \arctan \left (c \,x^{2}\right ) x^{2} d +b \arctan \left (c \,x^{2}\right ) x \,d^{2}+\frac {b \,d^{3} \arctan \left (c \,x^{2}\right )}{3 e}-\frac {2 b \,e^{2} x}{3 c}+\frac {b \,e^{2} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )}{6 c}+\frac {b \,e^{2} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )}{12 c}+\frac {b \,e^{2} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )}{6 c}-\frac {b c \,d^{3} \arctan \left (x^{2} \sqrt {c^{2}}\right )}{3 e \sqrt {c^{2}}}-\frac {b \,d^{2} \sqrt {2}\, \ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b \,d^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b \,d^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b d e \ln \left (c^{2} x^{4}+1\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*e^2*x^3+a*e*d*x^2+a*x*d^2+1/3*a/e*d^3+1/3*b*e^2*arctan(c*x^2)*x^3+b*e*arctan(c*x^2)*x^2*d+b*arctan(c*x^2
)*x*d^2+1/3*b*d^3*arctan(c*x^2)/e-2/3*b*e^2*x/c+1/6*b*e^2/c*(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c^2)^(1/4)
*x-1)+1/12*b*e^2/c*(1/c^2)^(1/4)*2^(1/2)*ln((x^2+(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2))/(x^2-(1/c^2)^(1/4)*x*2
^(1/2)+(1/c^2)^(1/2)))+1/6*b*e^2/c*(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c^2)^(1/4)*x+1)-1/3*b/e*c*d^3/(c^2)
^(1/2)*arctan(x^2*(c^2)^(1/2))-1/4*b/c*d^2/(1/c^2)^(1/4)*2^(1/2)*ln((x^2-(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2)
)/(x^2+(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2)))-1/2*b/c*d^2/(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c^2)^(1/4)*
x+1)-1/2*b/c*d^2/(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c^2)^(1/4)*x-1)-1/2*b*d*e*ln(c^2*x^4+1)/c

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 323, normalized size = 1.29 \[ \frac {1}{3} \, a e^{2} x^{3} + a d e x^{2} - \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}}\right )} - 4 \, x \arctan \left (c x^{2}\right )\right )} b d^{2} + \frac {1}{12} \, {\left (4 \, x^{3} \arctan \left (c x^{2}\right ) - c {\left (\frac {8 \, x}{c^{2}} - \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}}}{c^{2}}\right )}\right )} b e^{2} + a d^{2} x + \frac {{\left (2 \, c x^{2} \arctan \left (c x^{2}\right ) - \log \left (c^{2} x^{4} + 1\right )\right )} b d e}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.37, size = 419, normalized size = 1.68 \[ \frac {a\,e^2\,x^3}{3}+a\,d^2\,x+\frac {b\,e^2\,x^3\,\mathrm {atan}\left (c\,x^2\right )}{3}+a\,d\,e\,x^2-\frac {3\,b\,d^2\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}-1\right )\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}}{2}+\frac {3\,b\,d^2\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}+1\right )\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}}{2}-\frac {b\,d^2\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}+1{}\mathrm {i}\right )\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}\,3{}\mathrm {i}}{2}+\frac {b\,d^2\,\ln \left (1+c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}\,3{}\mathrm {i}\right )\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}\,3{}\mathrm {i}}{2}-\frac {2\,b\,e^2\,x}{3\,c}+b\,d^2\,x\,\mathrm {atan}\left (c\,x^2\right )+\frac {b\,e^2\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}-1\right )\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}\,1{}\mathrm {i}}{2\,c}-\frac {b\,e^2\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}+1\right )\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}\,1{}\mathrm {i}}{2\,c}+\frac {b\,e^2\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}+1{}\mathrm {i}\right )\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}}{2\,c}-\frac {b\,e^2\,\ln \left (1+c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}\,3{}\mathrm {i}\right )\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}}{2\,c}+b\,d\,e\,x^2\,\mathrm {atan}\left (c\,x^2\right )-\frac {b\,d\,e\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}-1\right )}{2\,c}-\frac {b\,d\,e\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}+1\right )}{2\,c}-\frac {b\,d\,e\,\ln \left (3\,c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}+1{}\mathrm {i}\right )}{2\,c}-\frac {b\,d\,e\,\ln \left (1+c\,x\,\sqrt {\frac {1{}\mathrm {i}}{9\,c}}\,3{}\mathrm {i}\right )}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*e^2*x^3)/3 + a*d^2*x + (b*e^2*x^3*atan(c*x^2))/3 + a*d*e*x^2 - (3*b*d^2*log(3*c*x*(1i/(9*c))^(1/2) - 1)*(1i
/(9*c))^(1/2))/2 + (3*b*d^2*log(3*c*x*(1i/(9*c))^(1/2) + 1)*(1i/(9*c))^(1/2))/2 - (b*d^2*log(3*c*x*(1i/(9*c))^
(1/2) + 1i)*(1i/(9*c))^(1/2)*3i)/2 + (b*d^2*log(c*x*(1i/(9*c))^(1/2)*3i + 1)*(1i/(9*c))^(1/2)*3i)/2 - (2*b*e^2
*x)/(3*c) + b*d^2*x*atan(c*x^2) + (b*e^2*log(3*c*x*(1i/(9*c))^(1/2) - 1)*(1i/(9*c))^(1/2)*1i)/(2*c) - (b*e^2*l
og(3*c*x*(1i/(9*c))^(1/2) + 1)*(1i/(9*c))^(1/2)*1i)/(2*c) + (b*e^2*log(3*c*x*(1i/(9*c))^(1/2) + 1i)*(1i/(9*c))
^(1/2))/(2*c) - (b*e^2*log(c*x*(1i/(9*c))^(1/2)*3i + 1)*(1i/(9*c))^(1/2))/(2*c) + b*d*e*x^2*atan(c*x^2) - (b*d
*e*log(3*c*x*(1i/(9*c))^(1/2) - 1))/(2*c) - (b*d*e*log(3*c*x*(1i/(9*c))^(1/2) + 1))/(2*c) - (b*d*e*log(3*c*x*(
1i/(9*c))^(1/2) + 1i))/(2*c) - (b*d*e*log(c*x*(1i/(9*c))^(1/2)*3i + 1))/(2*c)

________________________________________________________________________________________

sympy [A]  time = 26.00, size = 464, normalized size = 1.86 \[ \begin {cases} a d^{2} x + a d e x^{2} + \frac {a e^{2} x^{3}}{3} + b d^{2} x \operatorname {atan}{\left (c x^{2} \right )} + b d e x^{2} \operatorname {atan}{\left (c x^{2} \right )} + \frac {b e^{2} x^{3} \operatorname {atan}{\left (c x^{2} \right )}}{3} + \frac {\left (-1\right )^{\frac {3}{4}} b d^{2} \log {\left (x - \sqrt [4]{-1} \sqrt [4]{\frac {1}{c^{2}}} \right )}}{c \sqrt [4]{\frac {1}{c^{2}}}} - \frac {\left (-1\right )^{\frac {3}{4}} b d^{2} \log {\left (x^{2} + i \sqrt {\frac {1}{c^{2}}} \right )}}{2 c \sqrt [4]{\frac {1}{c^{2}}}} - \frac {\left (-1\right )^{\frac {3}{4}} b d^{2} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} x}{\sqrt [4]{\frac {1}{c^{2}}}} \right )}}{c \sqrt [4]{\frac {1}{c^{2}}}} - \frac {b d e \log {\left (x^{2} + i \sqrt {\frac {1}{c^{2}}} \right )}}{c} - \frac {2 b e^{2} x}{3 c} + \frac {\sqrt [4]{-1} b d^{2} \operatorname {atan}{\left (c x^{2} \right )}}{c^{2} \left (\frac {1}{c^{2}}\right )^{\frac {3}{4}}} - \frac {i b d e \operatorname {atan}{\left (c x^{2} \right )}}{c^{2} \sqrt {\frac {1}{c^{2}}}} + \frac {\left (-1\right )^{\frac {3}{4}} b e^{2} \operatorname {atan}{\left (c x^{2} \right )}}{3 c^{2} \sqrt [4]{\frac {1}{c^{2}}}} - \frac {\sqrt [4]{-1} b e^{2} \log {\left (x - \sqrt [4]{-1} \sqrt [4]{\frac {1}{c^{2}}} \right )}}{3 c^{3} \left (\frac {1}{c^{2}}\right )^{\frac {3}{4}}} + \frac {\sqrt [4]{-1} b e^{2} \log {\left (x^{2} + i \sqrt {\frac {1}{c^{2}}} \right )}}{6 c^{3} \left (\frac {1}{c^{2}}\right )^{\frac {3}{4}}} - \frac {\sqrt [4]{-1} b e^{2} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} x}{\sqrt [4]{\frac {1}{c^{2}}}} \right )}}{3 c^{3} \left (\frac {1}{c^{2}}\right )^{\frac {3}{4}}} & \text {for}\: c \neq 0 \\a \left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**2*x + a*d*e*x**2 + a*e**2*x**3/3 + b*d**2*x*atan(c*x**2) + b*d*e*x**2*atan(c*x**2) + b*e**2*x*
*3*atan(c*x**2)/3 + (-1)**(3/4)*b*d**2*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(c*(c**(-2))**(1/4)) - (-1)**(3/4
)*b*d**2*log(x**2 + I*sqrt(c**(-2)))/(2*c*(c**(-2))**(1/4)) - (-1)**(3/4)*b*d**2*atan((-1)**(3/4)*x/(c**(-2))*
*(1/4))/(c*(c**(-2))**(1/4)) - b*d*e*log(x**2 + I*sqrt(c**(-2)))/c - 2*b*e**2*x/(3*c) + (-1)**(1/4)*b*d**2*ata
n(c*x**2)/(c**2*(c**(-2))**(3/4)) - I*b*d*e*atan(c*x**2)/(c**2*sqrt(c**(-2))) + (-1)**(3/4)*b*e**2*atan(c*x**2
)/(3*c**2*(c**(-2))**(1/4)) - (-1)**(1/4)*b*e**2*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(3*c**3*(c**(-2))**(3/4
)) + (-1)**(1/4)*b*e**2*log(x**2 + I*sqrt(c**(-2)))/(6*c**3*(c**(-2))**(3/4)) - (-1)**(1/4)*b*e**2*atan((-1)**
(3/4)*x/(c**(-2))**(1/4))/(3*c**3*(c**(-2))**(3/4)), Ne(c, 0)), (a*(d**2*x + d*e*x**2 + e**2*x**3/3), True))

________________________________________________________________________________________