3.1249 \(\int x (d+e x^2) (a+b \tan ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=199 \[ \frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac{a b e x}{2 c^3}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{a b d x}{c}+\frac{1}{4} e x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{b^2 d \log \left (c^2 x^2+1\right )}{2 c^2}+\frac{b^2 e x^2}{12 c^2}-\frac{b^2 e \log \left (c^2 x^2+1\right )}{3 c^4}+\frac{b^2 e x \tan ^{-1}(c x)}{2 c^3}-\frac{b^2 d x \tan ^{-1}(c x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.398327, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {4980, 4852, 4916, 4846, 260, 4884, 266, 43} \[ \frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac{a b e x}{2 c^3}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{a b d x}{c}+\frac{1}{4} e x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{b^2 d \log \left (c^2 x^2+1\right )}{2 c^2}+\frac{b^2 e x^2}{12 c^2}-\frac{b^2 e \log \left (c^2 x^2+1\right )}{3 c^4}+\frac{b^2 e x \tan ^{-1}(c x)}{2 c^3}-\frac{b^2 d x \tan ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

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 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 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 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 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 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])

Rubi steps

\begin{align*} \int x \left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (d x \left (a+b \tan ^{-1}(c x)\right )^2+e x^3 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+e \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e x^4 \left (a+b \tan ^{-1}(c x)\right )^2-(b c d) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac{1}{2} (b c e) \int \frac{x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{(b d) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c}+\frac{(b d) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c}-\frac{(b e) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}+\frac{(b e) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c}\\ &=-\frac{a b d x}{c}-\frac{b e x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{\left (b^2 d\right ) \int \tan ^{-1}(c x) \, dx}{c}+\frac{1}{6} \left (b^2 e\right ) \int \frac{x^3}{1+c^2 x^2} \, dx+\frac{(b e) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac{(b e) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^3}\\ &=-\frac{a b d x}{c}+\frac{a b e x}{2 c^3}-\frac{b^2 d x \tan ^{-1}(c x)}{c}-\frac{b e x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\left (b^2 d\right ) \int \frac{x}{1+c^2 x^2} \, dx+\frac{1}{12} \left (b^2 e\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )+\frac{\left (b^2 e\right ) \int \tan ^{-1}(c x) \, dx}{2 c^3}\\ &=-\frac{a b d x}{c}+\frac{a b e x}{2 c^3}-\frac{b^2 d x \tan ^{-1}(c x)}{c}+\frac{b^2 e x \tan ^{-1}(c x)}{2 c^3}-\frac{b e x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{b^2 d \log \left (1+c^2 x^2\right )}{2 c^2}+\frac{1}{12} \left (b^2 e\right ) \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 )-\frac{\left (b^2 e\right ) \int \frac{x}{1+c^2 x^2} \, dx}{2 c^2}\\ &=-\frac{a b d x}{c}+\frac{a b e x}{2 c^3}+\frac{b^2 e x^2}{12 c^2}-\frac{b^2 d x \tan ^{-1}(c x)}{c}+\frac{b^2 e x \tan ^{-1}(c x)}{2 c^3}-\frac{b e x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{b^2 d \log \left (1+c^2 x^2\right )}{2 c^2}-\frac{b^2 e \log \left (1+c^2 x^2\right )}{3 c^4}\\ \end{align*}

Mathematica [A]  time = 0.173684, size = 179, normalized size = 0.9 \[ \frac{c x \left (3 a^2 c^3 x \left (2 d+e x^2\right )-2 a b c^2 \left (6 d+e x^2\right )+6 a b e+b^2 c e x\right )+2 b \tan ^{-1}(c x) \left (3 a c^4 \left (2 d x^2+e x^4\right )+6 a c^2 d-3 a e-b c^3 x \left (6 d+e x^2\right )+3 b c e x\right )+2 b^2 \left (3 c^2 d-2 e\right ) \log \left (c^2 x^2+1\right )+3 b^2 \tan ^{-1}(c x)^2 \left (c^4 \left (2 d x^2+e x^4\right )+2 c^2 d-e\right )}{12 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 249, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}e{x}^{4}}{4}}+{\frac{{a}^{2}{x}^{2}d}{2}}+{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}e{x}^{4}}{4}}+{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}d{x}^{2}}{2}}-{\frac{{b}^{2}\arctan \left ( cx \right ){x}^{3}e}{6\,c}}-{\frac{{b}^{2}dx\arctan \left ( cx \right ) }{c}}+{\frac{{b}^{2}ex\arctan \left ( cx \right ) }{2\,{c}^{3}}}+{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}d}{2\,{c}^{2}}}-{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}e}{4\,{c}^{4}}}+{\frac{{b}^{2}e{x}^{2}}{12\,{c}^{2}}}+{\frac{{b}^{2}d\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{2}}}-{\frac{{b}^{2}e\ln \left ({c}^{2}{x}^{2}+1 \right ) }{3\,{c}^{4}}}+{\frac{ab\arctan \left ( cx \right ) e{x}^{4}}{2}}+ab\arctan \left ( cx \right ) d{x}^{2}-{\frac{ab{x}^{3}e}{6\,c}}-{\frac{abdx}{c}}+{\frac{abex}{2\,{c}^{3}}}+{\frac{ab\arctan \left ( cx \right ) d}{{c}^{2}}}-{\frac{ab\arctan \left ( cx \right ) e}{2\,{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.61354, size = 333, normalized size = 1.67 \begin{align*} \frac{1}{4} \, b^{2} e x^{4} \arctan \left (c x\right )^{2} + \frac{1}{4} \, a^{2} e x^{4} + \frac{1}{2} \, b^{2} d x^{2} \arctan \left (c x\right )^{2} + \frac{1}{2} \, a^{2} d x^{2} +{\left (x^{2} \arctan \left (c x\right ) - c{\left (\frac{x}{c^{2}} - \frac{\arctan \left (c x\right )}{c^{3}}\right )}\right )} a b d - \frac{1}{2} \,{\left (2 \, c{\left (\frac{x}{c^{2}} - \frac{\arctan \left (c x\right )}{c^{3}}\right )} \arctan \left (c x\right ) + \frac{\arctan \left (c x\right )^{2} - \log \left (c^{2} x^{2} + 1\right )}{c^{2}}\right )} b^{2} d + \frac{1}{6} \,{\left (3 \, x^{4} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{3} - 3 \, x}{c^{4}} + \frac{3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} a b e - \frac{1}{12} \,{\left (2 \, c{\left (\frac{c^{2} x^{3} - 3 \, x}{c^{4}} + \frac{3 \, \arctan \left (c x\right )}{c^{5}}\right )} \arctan \left (c x\right ) - \frac{c^{2} x^{2} + 3 \, \arctan \left (c x\right )^{2} - 4 \, \log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )} b^{2} e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.04188, size = 471, normalized size = 2.37 \begin{align*} \frac{3 \, a^{2} c^{4} e x^{4} - 2 \, a b c^{3} e x^{3} +{\left (6 \, a^{2} c^{4} d + b^{2} c^{2} e\right )} x^{2} + 3 \,{\left (b^{2} c^{4} e x^{4} + 2 \, b^{2} c^{4} d x^{2} + 2 \, b^{2} c^{2} d - b^{2} e\right )} \arctan \left (c x\right )^{2} - 6 \,{\left (2 \, a b c^{3} d - a b c e\right )} x + 2 \,{\left (3 \, a b c^{4} e x^{4} + 6 \, a b c^{4} d x^{2} - b^{2} c^{3} e x^{3} + 6 \, a b c^{2} d - 3 \, a b e - 3 \,{\left (2 \, b^{2} c^{3} d - b^{2} c e\right )} x\right )} \arctan \left (c x\right ) + 2 \,{\left (3 \, b^{2} c^{2} d - 2 \, b^{2} e\right )} \log \left (c^{2} x^{2} + 1\right )}{12 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.61472, size = 296, normalized size = 1.49 \begin{align*} \begin{cases} \frac{a^{2} d x^{2}}{2} + \frac{a^{2} e x^{4}}{4} + a b d x^{2} \operatorname{atan}{\left (c x \right )} + \frac{a b e x^{4} \operatorname{atan}{\left (c x \right )}}{2} - \frac{a b d x}{c} - \frac{a b e x^{3}}{6 c} + \frac{a b d \operatorname{atan}{\left (c x \right )}}{c^{2}} + \frac{a b e x}{2 c^{3}} - \frac{a b e \operatorname{atan}{\left (c x \right )}}{2 c^{4}} + \frac{b^{2} d x^{2} \operatorname{atan}^{2}{\left (c x \right )}}{2} + \frac{b^{2} e x^{4} \operatorname{atan}^{2}{\left (c x \right )}}{4} - \frac{b^{2} d x \operatorname{atan}{\left (c x \right )}}{c} - \frac{b^{2} e x^{3} \operatorname{atan}{\left (c x \right )}}{6 c} + \frac{b^{2} d \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c^{2}} + \frac{b^{2} d \operatorname{atan}^{2}{\left (c x \right )}}{2 c^{2}} + \frac{b^{2} e x^{2}}{12 c^{2}} + \frac{b^{2} e x \operatorname{atan}{\left (c x \right )}}{2 c^{3}} - \frac{b^{2} e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{3 c^{4}} - \frac{b^{2} e \operatorname{atan}^{2}{\left (c x \right )}}{4 c^{4}} & \text{for}\: c \neq 0 \\a^{2} \left (\frac{d x^{2}}{2} + \frac{e x^{4}}{4}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*d*x**2/2 + a**2*e*x**4/4 + a*b*d*x**2*atan(c*x) + a*b*e*x**4*atan(c*x)/2 - a*b*d*x/c - a*b*e*x
**3/(6*c) + a*b*d*atan(c*x)/c**2 + a*b*e*x/(2*c**3) - a*b*e*atan(c*x)/(2*c**4) + b**2*d*x**2*atan(c*x)**2/2 +
b**2*e*x**4*atan(c*x)**2/4 - b**2*d*x*atan(c*x)/c - b**2*e*x**3*atan(c*x)/(6*c) + b**2*d*log(x**2 + c**(-2))/(
2*c**2) + b**2*d*atan(c*x)**2/(2*c**2) + b**2*e*x**2/(12*c**2) + b**2*e*x*atan(c*x)/(2*c**3) - b**2*e*log(x**2
+ c**(-2))/(3*c**4) - b**2*e*atan(c*x)**2/(4*c**4), Ne(c, 0)), (a**2*(d*x**2/2 + e*x**4/4), True))

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Giac [A]  time = 1.28589, size = 382, normalized size = 1.92 \begin{align*} \frac{3 \, b^{2} c^{4} x^{4} \arctan \left (c x\right )^{2} e + 6 \, a b c^{4} x^{4} \arctan \left (c x\right ) e + 6 \, b^{2} c^{4} d x^{2} \arctan \left (c x\right )^{2} + 3 \, a^{2} c^{4} x^{4} e + 12 \, a b c^{4} d x^{2} \arctan \left (c x\right ) - 2 \, b^{2} c^{3} x^{3} \arctan \left (c x\right ) e + 6 \, a^{2} c^{4} d x^{2} - 2 \, a b c^{3} x^{3} e - 12 \, b^{2} c^{3} d x \arctan \left (c x\right ) - 12 \, \pi a b c^{2} d \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - 12 \, a b c^{3} d x + 6 \, b^{2} c^{2} d \arctan \left (c x\right )^{2} + b^{2} c^{2} x^{2} e + 12 \, a b c^{2} d \arctan \left (c x\right ) + 6 \, b^{2} c x \arctan \left (c x\right ) e + 6 \, b^{2} c^{2} d \log \left (c^{2} x^{2} + 1\right ) + 6 \, a b c x e - 3 \, b^{2} \arctan \left (c x\right )^{2} e - 6 \, a b \arctan \left (c x\right ) e - 4 \, b^{2} e \log \left (c^{2} x^{2} + 1\right )}{12 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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