3.1250 \(\int (d+e x^2) (a+b \tan ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=231 \[ -\frac{i b^2 e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c}-\frac{i e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}-\frac{2 b e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}+d x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{i d \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{1}{3} e x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{b^2 e x}{3 c^2}-\frac{b^2 e \tan ^{-1}(c x)}{3 c^3} \]

[Out]

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

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Rubi [A]  time = 0.35754, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {4914, 4846, 4920, 4854, 2402, 2315, 4852, 4916, 321, 203} \[ -\frac{i b^2 e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c}-\frac{i e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}-\frac{2 b e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}+d x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{i d \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{1}{3} e x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{b^2 e x}{3 c^2}-\frac{b^2 e \tan ^{-1}(c x)}{3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4914

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcTan[c*x])^p, (d + e*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[q] && IGtQ[p, 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 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]

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])

Rubi steps

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

Mathematica [A]  time = 0.433747, size = 208, normalized size = 0.9 \[ \frac{-i b^2 \left (3 c^2 d-e\right ) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+3 a^2 c^3 d x+a^2 c^3 e x^3-b \tan ^{-1}(c x) \left (-2 a c^3 x \left (3 d+e x^2\right )+2 b \left (e-3 c^2 d\right ) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+b \left (c^2 e x^2+e\right )\right )-3 a b c^2 d \log \left (c^2 x^2+1\right )-a b c^2 e x^2+a b e \log \left (c^2 x^2+1\right )+b^2 \tan ^{-1}(c x)^2 \left (c^3 \left (3 d x+e x^3\right )-3 i c^2 d+i e\right )+b^2 c e x}{3 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.128, size = 570, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, a^{2} e x^{3} + 36 \, b^{2} c^{2} e \int \frac{x^{4} \arctan \left (c x\right )^{2}}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 3 \, b^{2} c^{2} e \int \frac{x^{4} \log \left (c^{2} x^{2} + 1\right )^{2}}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 4 \, b^{2} c^{2} e \int \frac{x^{4} \log \left (c^{2} x^{2} + 1\right )}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 36 \, b^{2} c^{2} d \int \frac{x^{2} \arctan \left (c x\right )^{2}}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 3 \, b^{2} c^{2} d \int \frac{x^{2} \log \left (c^{2} x^{2} + 1\right )^{2}}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 12 \, b^{2} c^{2} d \int \frac{x^{2} \log \left (c^{2} x^{2} + 1\right )}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac{b^{2} d \arctan \left (c x\right )^{3}}{4 \, c} - 8 \, b^{2} c e \int \frac{x^{3} \arctan \left (c x\right )}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} - 24 \, b^{2} c d \int \frac{x \arctan \left (c x\right )}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac{1}{3} \,{\left (2 \, x^{3} \arctan \left (c x\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} a b e + a^{2} d x + 36 \, b^{2} e \int \frac{x^{2} \arctan \left (c x\right )^{2}}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 3 \, b^{2} e \int \frac{x^{2} \log \left (c^{2} x^{2} + 1\right )^{2}}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 3 \, b^{2} d \int \frac{\log \left (c^{2} x^{2} + 1\right )^{2}}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} a b d}{c} + \frac{1}{12} \,{\left (b^{2} e x^{3} + 3 \, b^{2} d x\right )} \arctan \left (c x\right )^{2} - \frac{1}{48} \,{\left (b^{2} e x^{3} + 3 \, b^{2} d x\right )} \log \left (c^{2} x^{2} + 1\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*e*x^3 + 36*b^2*c^2*e*integrate(1/48*x^4*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^2*c^2*e*integrate(1/48*x
^4*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 4*b^2*c^2*e*integrate(1/48*x^4*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) +
36*b^2*c^2*d*integrate(1/48*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^2*c^2*d*integrate(1/48*x^2*log(c^2*x^2 +
1)^2/(c^2*x^2 + 1), x) + 12*b^2*c^2*d*integrate(1/48*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 1/4*b^2*d*arcta
n(c*x)^3/c - 8*b^2*c*e*integrate(1/48*x^3*arctan(c*x)/(c^2*x^2 + 1), x) - 24*b^2*c*d*integrate(1/48*x*arctan(c
*x)/(c^2*x^2 + 1), x) + 1/3*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*e + a^2*d*x + 36*b^2*
e*integrate(1/48*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^2*e*integrate(1/48*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2
+ 1), x) + 3*b^2*d*integrate(1/48*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + (2*c*x*arctan(c*x) - log(c^2*x^2 + 1)
)*a*b*d/c + 1/12*(b^2*e*x^3 + 3*b^2*d*x)*arctan(c*x)^2 - 1/48*(b^2*e*x^3 + 3*b^2*d*x)*log(c^2*x^2 + 1)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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