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

Optimal. Leaf size=183 \[ -\frac{i b^2 e^2 \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right )}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{3 d}-\frac{i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{2 b e^2 \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{3 d}-\frac{b^2 e^2 \tan ^{-1}(c+d x)}{3 d}+\frac{1}{3} b^2 e^2 x \]

[Out]

(b^2*e^2*x)/3 - (b^2*e^2*ArcTan[c + d*x])/(3*d) - (b*e^2*(c + d*x)^2*(a + b*ArcTan[c + d*x]))/(3*d) - ((I/3)*e
^2*(a + b*ArcTan[c + d*x])^2)/d + (e^2*(c + d*x)^3*(a + b*ArcTan[c + d*x])^2)/(3*d) - (2*b*e^2*(a + b*ArcTan[c
+ d*x])*Log[2/(1 + I*(c + d*x))])/(3*d) - ((I/3)*b^2*e^2*PolyLog[2, 1 - 2/(1 + I*(c + d*x))])/d

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Rubi [A]  time = 0.221134, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {5043, 12, 4852, 4916, 321, 203, 4920, 4854, 2402, 2315} \[ -\frac{i b^2 e^2 \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right )}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{3 d}-\frac{i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{2 b e^2 \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{3 d}-\frac{b^2 e^2 \tan ^{-1}(c+d x)}{3 d}+\frac{1}{3} b^2 e^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*e^2*x)/3 - (b^2*e^2*ArcTan[c + d*x])/(3*d) - (b*e^2*(c + d*x)^2*(a + b*ArcTan[c + d*x]))/(3*d) - ((I/3)*e
^2*(a + b*ArcTan[c + d*x])^2)/d + (e^2*(c + d*x)^3*(a + b*ArcTan[c + d*x])^2)/(3*d) - (2*b*e^2*(a + b*ArcTan[c
+ d*x])*Log[2/(1 + I*(c + d*x))])/(3*d) - ((I/3)*b^2*e^2*PolyLog[2, 1 - 2/(1 + I*(c + d*x))])/d

Rule 5043

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

Rule 12

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

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 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 321

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 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 4920

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

Rule 4854

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

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

\begin{align*} \int (c e+d e x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int e^2 x^2 \left (a+b \tan ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int x^2 \left (a+b \tan ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{\left (2 b e^2\right ) \operatorname{Subst}\left (\int \frac{x^3 \left (a+b \tan ^{-1}(x)\right )}{1+x^2} \, dx,x,c+d x\right )}{3 d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{\left (2 b e^2\right ) \operatorname{Subst}\left (\int x \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{3 d}+\frac{\left (2 b e^2\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \tan ^{-1}(x)\right )}{1+x^2} \, dx,x,c+d x\right )}{3 d}\\ &=-\frac{b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{3 d}-\frac{i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{\left (2 b e^2\right ) \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{i-x} \, dx,x,c+d x\right )}{3 d}+\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,c+d x\right )}{3 d}\\ &=\frac{1}{3} b^2 e^2 x-\frac{b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{3 d}-\frac{i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{2 b e^2 \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2}{1+i (c+d x)}\right )}{3 d}-\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,c+d x\right )}{3 d}+\frac{\left (2 b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{3 d}\\ &=\frac{1}{3} b^2 e^2 x-\frac{b^2 e^2 \tan ^{-1}(c+d x)}{3 d}-\frac{b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{3 d}-\frac{i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{2 b e^2 \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2}{1+i (c+d x)}\right )}{3 d}-\frac{\left (2 i b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i (c+d x)}\right )}{3 d}\\ &=\frac{1}{3} b^2 e^2 x-\frac{b^2 e^2 \tan ^{-1}(c+d x)}{3 d}-\frac{b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{3 d}-\frac{i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d}-\frac{2 b e^2 \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2}{1+i (c+d x)}\right )}{3 d}-\frac{i b^2 e^2 \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.36222, size = 163, normalized size = 0.89 \[ \frac{e^2 \left (b^2 \left (i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )+(c+d x)^3 \tan ^{-1}(c+d x)^2-(c+d x)^2 \tan ^{-1}(c+d x)+i \tan ^{-1}(c+d x)^2-\tan ^{-1}(c+d x)-2 \tan ^{-1}(c+d x) \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )+c+d x\right )+a^2 (c+d x)^3+a b \left (-(c+d x)^2+\log \left ((c+d x)^2+1\right )+2 (c+d x)^3 \tan ^{-1}(c+d x)\right )\right )}{3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(e^2*(a^2*(c + d*x)^3 + a*b*(-(c + d*x)^2 + 2*(c + d*x)^3*ArcTan[c + d*x] + Log[1 + (c + d*x)^2]) + b^2*(c + d
*x - ArcTan[c + d*x] - (c + d*x)^2*ArcTan[c + d*x] + I*ArcTan[c + d*x]^2 + (c + d*x)^3*ArcTan[c + d*x]^2 - 2*A
rcTan[c + d*x]*Log[1 + E^((2*I)*ArcTan[c + d*x])] + I*PolyLog[2, -E^((2*I)*ArcTan[c + d*x])])))/(3*d)

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Maple [B]  time = 0.123, size = 593, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*b^2*c^4*e^2*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 1/4*(3*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2
/d - arctan((d^2*x + c*d)/d)^3/d)*b^2*c^4*e^2 + 1/3*a^2*d^2*e^2*x^3 + 36*b^2*d^4*e^2*integrate(1/48*x^4*arctan
(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^2*d^4*e^2*integrate(1/48*x^4*log(d^2*x^2 + 2*c*d*x + c^2 +
1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 144*b^2*c*d^3*e^2*integrate(1/48*x^3*arctan(d*x + c)^2/(d^2*x^2 + 2*
c*d*x + c^2 + 1), x) + 4*b^2*d^4*e^2*integrate(1/48*x^4*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x +
c^2 + 1), x) + 12*b^2*c*d^3*e^2*integrate(1/48*x^3*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2
+ 1), x) + 216*b^2*c^2*d^2*e^2*integrate(1/48*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 16*b^
2*c*d^3*e^2*integrate(1/48*x^3*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 18*b^2*c^2
*d^2*e^2*integrate(1/48*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 144*b^2*c^3
*d*e^2*integrate(1/48*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 24*b^2*c^2*d^2*e^2*integrate(1/4
8*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^2*c^3*d*e^2*integrate(1/48*x*l
og(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^2*c^3*d*e^2*integrate(1/48*x*log(d^
2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^2*c^4*e^2*integrate(1/48*log(d^2*x^2 + 2*c*
d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + a^2*c*d*e^2*x^2 + 3/4*b^2*c^2*e^2*arctan(d*x + c)^2*arcta
n((d^2*x + c*d)/d)/d - 8*b^2*d^3*e^2*integrate(1/48*x^3*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 24
*b^2*c*d^2*e^2*integrate(1/48*x^2*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 24*b^2*c^2*d*e^2*integra
te(1/48*x*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 1/4*(3*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2
/d - arctan((d^2*x + c*d)/d)^3/d)*b^2*c^2*e^2 + 2*(x^2*arctan(d*x + c) - d*(x/d^2 + (c^2 - 1)*arctan((d^2*x +
c*d)/d)/d^3 - c*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^3))*a*b*c*d*e^2 + 1/3*(2*x^3*arctan(d*x + c) - d*((d*x^2 -
4*c*x)/d^3 - 2*(c^3 - 3*c)*arctan((d^2*x + c*d)/d)/d^4 + (3*c^2 - 1)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^4))*a*
b*d^2*e^2 + a^2*c^2*e^2*x + 36*b^2*d^2*e^2*integrate(1/48*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1),
x) + 3*b^2*d^2*e^2*integrate(1/48*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) +
72*b^2*c*d*e^2*integrate(1/48*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 6*b^2*c*d*e^2*integrate(
1/48*x*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^2*c^2*e^2*integrate(1/48*log
(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + (2*(d*x + c)*arctan(d*x + c) - log((d*x +
c)^2 + 1))*a*b*c^2*e^2/d + 1/12*(b^2*d^2*e^2*x^3 + 3*b^2*c*d*e^2*x^2 + 3*b^2*c^2*e^2*x)*arctan(d*x + c)^2 - 1/
48*(b^2*d^2*e^2*x^3 + 3*b^2*c*d*e^2*x^2 + 3*b^2*c^2*e^2*x)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{2} d^{2} e^{2} x^{2} + 2 \, a^{2} c d e^{2} x + a^{2} c^{2} e^{2} +{\left (b^{2} d^{2} e^{2} x^{2} + 2 \, b^{2} c d e^{2} x + b^{2} c^{2} e^{2}\right )} \arctan \left (d x + c\right )^{2} + 2 \,{\left (a b d^{2} e^{2} x^{2} + 2 \, a b c d e^{2} x + a b c^{2} e^{2}\right )} \arctan \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{2} \left (\int a^{2} c^{2}\, dx + \int a^{2} d^{2} x^{2}\, dx + \int b^{2} c^{2} \operatorname{atan}^{2}{\left (c + d x \right )}\, dx + \int 2 a b c^{2} \operatorname{atan}{\left (c + d x \right )}\, dx + \int 2 a^{2} c d x\, dx + \int b^{2} d^{2} x^{2} \operatorname{atan}^{2}{\left (c + d x \right )}\, dx + \int 2 a b d^{2} x^{2} \operatorname{atan}{\left (c + d x \right )}\, dx + \int 2 b^{2} c d x \operatorname{atan}^{2}{\left (c + d x \right )}\, dx + \int 4 a b c d x \operatorname{atan}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**2*(Integral(a**2*c**2, x) + Integral(a**2*d**2*x**2, x) + Integral(b**2*c**2*atan(c + d*x)**2, x) + Integra
l(2*a*b*c**2*atan(c + d*x), x) + Integral(2*a**2*c*d*x, x) + Integral(b**2*d**2*x**2*atan(c + d*x)**2, x) + In
tegral(2*a*b*d**2*x**2*atan(c + d*x), x) + Integral(2*b**2*c*d*x*atan(c + d*x)**2, x) + Integral(4*a*b*c*d*x*a
tan(c + d*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{2}{\left (b \arctan \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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