3.7 \(\int (c e+d e x)^3 (a+b \tan ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=157 \[ \frac{e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}-\frac{b e^3 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )}{6 d}-\frac{e^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}+\frac{1}{2} a b e^3 x+\frac{b^2 e^3 (c+d x)^2}{12 d}-\frac{b^2 e^3 \log \left ((c+d x)^2+1\right )}{3 d}+\frac{b^2 e^3 (c+d x) \tan ^{-1}(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.22059, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5043, 12, 4852, 4916, 266, 43, 4846, 260, 4884} \[ \frac{e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}-\frac{b e^3 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )}{6 d}-\frac{e^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}+\frac{1}{2} a b e^3 x+\frac{b^2 e^3 (c+d x)^2}{12 d}-\frac{b^2 e^3 \log \left ((c+d x)^2+1\right )}{3 d}+\frac{b^2 e^3 (c+d x) \tan ^{-1}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3*(a + b*ArcTan[c + d*x])^2,x]

[Out]

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0980365, size = 216, normalized size = 1.38 \[ \frac{e^3 \left ((c+d x) \left (3 a^2 (c+d x)^3-2 a b \left (c^2+2 c d x+d^2 x^2-3\right )+b^2 (c+d x)\right )+2 b \tan ^{-1}(c+d x) \left (3 a \left (6 c^2 d^2 x^2+4 c^3 d x+c^4+4 c d^3 x^3+d^4 x^4-1\right )-b \left (3 c^2 d x+c^3+3 c d^2 x^2-3 c+d^3 x^3-3 d x\right )\right )+3 b^2 \left (6 c^2 d^2 x^2+4 c^3 d x+c^4+4 c d^3 x^3+d^4 x^4-1\right ) \tan ^{-1}(c+d x)^2-4 b^2 \log \left ((c+d x)^2+1\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^3*(a + b*ArcTan[c + d*x])^2,x]

[Out]

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

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Maple [B]  time = 0.046, size = 543, normalized size = 3.5 \begin{align*}{\frac{{b}^{2}{c}^{2}{e}^{3}}{12\,d}}+{\frac{{a}^{2}{c}^{4}{e}^{3}}{4\,d}}-{\frac{d{x}^{2}abc{e}^{3}}{2}}-{\frac{d\arctan \left ( dx+c \right ){x}^{2}{b}^{2}c{e}^{3}}{2}}+{\frac{{d}^{3}\arctan \left ( dx+c \right ){x}^{4}ab{e}^{3}}{2}}+{\frac{\arctan \left ( dx+c \right ) ab{c}^{4}{e}^{3}}{2\,d}}+{d}^{2} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}{x}^{3}{b}^{2}c{e}^{3}+2\,{d}^{2}\arctan \left ( dx+c \right ){x}^{3}abc{e}^{3}+3\,d\arctan \left ( dx+c \right ){x}^{2}ab{c}^{2}{e}^{3}-{\frac{xab{c}^{2}{e}^{3}}{2}}+{\frac{{d}^{3}{x}^{4}{a}^{2}{e}^{3}}{4}}-{\frac{{e}^{3}{b}^{2} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}}{4\,d}}+{\frac{\arctan \left ( dx+c \right ) x{b}^{2}{e}^{3}}{2}}+{\frac{d{x}^{2}{b}^{2}{e}^{3}}{12}}+{\frac{x{b}^{2}c{e}^{3}}{6}}+x{a}^{2}{c}^{3}{e}^{3}-{\frac{\arctan \left ( dx+c \right ) x{b}^{2}{c}^{2}{e}^{3}}{2}}-{\frac{{d}^{2}{x}^{3}ab{e}^{3}}{6}}+{d}^{2}{x}^{3}{a}^{2}c{e}^{3}+{\frac{3\,d{x}^{2}{a}^{2}{c}^{2}{e}^{3}}{2}}+{\frac{{d}^{3} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}{x}^{4}{b}^{2}{e}^{3}}{4}}-{\frac{{d}^{2}\arctan \left ( dx+c \right ){x}^{3}{b}^{2}{e}^{3}}{6}}-{\frac{\arctan \left ( dx+c \right ){b}^{2}{c}^{3}{e}^{3}}{6\,d}}-{\frac{{e}^{3}ab\arctan \left ( dx+c \right ) }{2\,d}}+{\frac{ \left ( \arctan \left ( dx+c \right ) \right ) ^{2}{b}^{2}{c}^{4}{e}^{3}}{4\,d}}+{\frac{\arctan \left ( dx+c \right ){b}^{2}c{e}^{3}}{2\,d}}+{\frac{abc{e}^{3}}{2\,d}}-{\frac{{e}^{3}{b}^{2}\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{3\,d}}+ \left ( \arctan \left ( dx+c \right ) \right ) ^{2}x{b}^{2}{c}^{3}{e}^{3}+{\frac{3\,d \left ( \arctan \left ( dx+c \right ) \right ) ^{2}{x}^{2}{b}^{2}{c}^{2}{e}^{3}}{2}}+2\,\arctan \left ( dx+c \right ) xab{c}^{3}{e}^{3}+{\frac{ab{e}^{3}x}{2}}-{\frac{ab{c}^{3}{e}^{3}}{6\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3*(a+b*arctan(d*x+c))^2,x)

[Out]

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

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Maxima [B]  time = 5.51433, size = 806, normalized size = 5.13 \begin{align*} \frac{1}{4} \, a^{2} d^{3} e^{3} x^{4} + a^{2} c d^{2} e^{3} x^{3} + \frac{3}{2} \, a^{2} c^{2} d e^{3} x^{2} + 3 \,{\left (x^{2} \arctan \left (d x + c\right ) - d{\left (\frac{x}{d^{2}} + \frac{{\left (c^{2} - 1\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{3}} - \frac{c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} a b c^{2} d e^{3} +{\left (2 \, x^{3} \arctan \left (d x + c\right ) - d{\left (\frac{d x^{2} - 4 \, c x}{d^{3}} - \frac{2 \,{\left (c^{3} - 3 \, c\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{4}} + \frac{{\left (3 \, c^{2} - 1\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4}}\right )}\right )} a b c d^{2} e^{3} + \frac{1}{6} \,{\left (3 \, x^{4} \arctan \left (d x + c\right ) - d{\left (\frac{d^{2} x^{3} - 3 \, c d x^{2} + 3 \,{\left (3 \, c^{2} - 1\right )} x}{d^{4}} + \frac{3 \,{\left (c^{4} - 6 \, c^{2} + 1\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{5}} - \frac{6 \,{\left (c^{3} - c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{5}}\right )}\right )} a b d^{3} e^{3} + a^{2} c^{3} e^{3} x + \frac{{\left (2 \,{\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} a b c^{3} e^{3}}{d} + \frac{b^{2} d^{2} e^{3} x^{2} + 2 \, b^{2} c d e^{3} x - 4 \, b^{2} e^{3} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 3 \,{\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x +{\left (b^{2} c^{4} - b^{2}\right )} e^{3}\right )} \arctan \left (d x + c\right )^{2} - 2 \,{\left (b^{2} d^{3} e^{3} x^{3} + 3 \, b^{2} c d^{2} e^{3} x^{2} + 3 \,{\left (b^{2} c^{2} - b^{2}\right )} d e^{3} x +{\left (b^{2} c^{3} - 3 \, b^{2} c\right )} e^{3}\right )} \arctan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3*(a+b*arctan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.79675, size = 703, normalized size = 4.48 \begin{align*} \frac{3 \, a^{2} d^{4} e^{3} x^{4} + 2 \,{\left (6 \, a^{2} c - a b\right )} d^{3} e^{3} x^{3} +{\left (18 \, a^{2} c^{2} - 6 \, a b c + b^{2}\right )} d^{2} e^{3} x^{2} + 2 \,{\left (6 \, a^{2} c^{3} - 3 \, a b c^{2} + b^{2} c + 3 \, a b\right )} d e^{3} x - 4 \, b^{2} e^{3} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 3 \,{\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x +{\left (b^{2} c^{4} - b^{2}\right )} e^{3}\right )} \arctan \left (d x + c\right )^{2} + 2 \,{\left (3 \, a b d^{4} e^{3} x^{4} +{\left (12 \, a b c - b^{2}\right )} d^{3} e^{3} x^{3} + 3 \,{\left (6 \, a b c^{2} - b^{2} c\right )} d^{2} e^{3} x^{2} + 3 \,{\left (4 \, a b c^{3} - b^{2} c^{2} + b^{2}\right )} d e^{3} x +{\left (3 \, a b c^{4} - b^{2} c^{3} + 3 \, b^{2} c - 3 \, a b\right )} e^{3}\right )} \arctan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3*(a+b*arctan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(3*a^2*d^4*e^3*x^4 + 2*(6*a^2*c - a*b)*d^3*e^3*x^3 + (18*a^2*c^2 - 6*a*b*c + b^2)*d^2*e^3*x^2 + 2*(6*a^2*
c^3 - 3*a*b*c^2 + b^2*c + 3*a*b)*d*e^3*x - 4*b^2*e^3*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 3*(b^2*d^4*e^3*x^4 + 4
*b^2*c*d^3*e^3*x^3 + 6*b^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (b^2*c^4 - b^2)*e^3)*arctan(d*x + c)^2 + 2*(3
*a*b*d^4*e^3*x^4 + (12*a*b*c - b^2)*d^3*e^3*x^3 + 3*(6*a*b*c^2 - b^2*c)*d^2*e^3*x^2 + 3*(4*a*b*c^3 - b^2*c^2 +
b^2)*d*e^3*x + (3*a*b*c^4 - b^2*c^3 + 3*b^2*c - 3*a*b)*e^3)*arctan(d*x + c))/d

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Sympy [A]  time = 11.7642, size = 575, normalized size = 3.66 \begin{align*} \begin{cases} a^{2} c^{3} e^{3} x + \frac{3 a^{2} c^{2} d e^{3} x^{2}}{2} + a^{2} c d^{2} e^{3} x^{3} + \frac{a^{2} d^{3} e^{3} x^{4}}{4} + \frac{a b c^{4} e^{3} \operatorname{atan}{\left (c + d x \right )}}{2 d} + 2 a b c^{3} e^{3} x \operatorname{atan}{\left (c + d x \right )} + 3 a b c^{2} d e^{3} x^{2} \operatorname{atan}{\left (c + d x \right )} - \frac{a b c^{2} e^{3} x}{2} + 2 a b c d^{2} e^{3} x^{3} \operatorname{atan}{\left (c + d x \right )} - \frac{a b c d e^{3} x^{2}}{2} + \frac{a b d^{3} e^{3} x^{4} \operatorname{atan}{\left (c + d x \right )}}{2} - \frac{a b d^{2} e^{3} x^{3}}{6} + \frac{a b e^{3} x}{2} - \frac{a b e^{3} \operatorname{atan}{\left (c + d x \right )}}{2 d} + \frac{b^{2} c^{4} e^{3} \operatorname{atan}^{2}{\left (c + d x \right )}}{4 d} + b^{2} c^{3} e^{3} x \operatorname{atan}^{2}{\left (c + d x \right )} - \frac{b^{2} c^{3} e^{3} \operatorname{atan}{\left (c + d x \right )}}{6 d} + \frac{3 b^{2} c^{2} d e^{3} x^{2} \operatorname{atan}^{2}{\left (c + d x \right )}}{2} - \frac{b^{2} c^{2} e^{3} x \operatorname{atan}{\left (c + d x \right )}}{2} + b^{2} c d^{2} e^{3} x^{3} \operatorname{atan}^{2}{\left (c + d x \right )} - \frac{b^{2} c d e^{3} x^{2} \operatorname{atan}{\left (c + d x \right )}}{2} + \frac{b^{2} c e^{3} x}{6} + \frac{b^{2} c e^{3} \operatorname{atan}{\left (c + d x \right )}}{2 d} + \frac{b^{2} d^{3} e^{3} x^{4} \operatorname{atan}^{2}{\left (c + d x \right )}}{4} - \frac{b^{2} d^{2} e^{3} x^{3} \operatorname{atan}{\left (c + d x \right )}}{6} + \frac{b^{2} d e^{3} x^{2}}{12} + \frac{b^{2} e^{3} x \operatorname{atan}{\left (c + d x \right )}}{2} - \frac{b^{2} e^{3} \log{\left (\frac{c^{2}}{d^{2}} + \frac{2 c x}{d} + x^{2} + \frac{1}{d^{2}} \right )}}{3 d} - \frac{b^{2} e^{3} \operatorname{atan}^{2}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname{atan}{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3*(a+b*atan(d*x+c))**2,x)

[Out]

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

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Giac [B]  time = 1.37464, size = 784, normalized size = 4.99 \begin{align*} \frac{3 \, b^{2} d^{4} x^{4} \arctan \left (d x + c\right )^{2} e^{3} + 6 \, a b d^{4} x^{4} \arctan \left (d x + c\right ) e^{3} + 12 \, b^{2} c d^{3} x^{3} \arctan \left (d x + c\right )^{2} e^{3} + 3 \, a^{2} d^{4} x^{4} e^{3} + 24 \, a b c d^{3} x^{3} \arctan \left (d x + c\right ) e^{3} + 18 \, b^{2} c^{2} d^{2} x^{2} \arctan \left (d x + c\right )^{2} e^{3} + 12 \, a^{2} c d^{3} x^{3} e^{3} + 36 \, a b c^{2} d^{2} x^{2} \arctan \left (d x + c\right ) e^{3} - 2 \, b^{2} d^{3} x^{3} \arctan \left (d x + c\right ) e^{3} + 12 \, b^{2} c^{3} d x \arctan \left (d x + c\right )^{2} e^{3} + 18 \, a^{2} c^{2} d^{2} x^{2} e^{3} - 2 \, a b d^{3} x^{3} e^{3} + 24 \, a b c^{3} d x \arctan \left (d x + c\right ) e^{3} - 6 \, b^{2} c d^{2} x^{2} \arctan \left (d x + c\right ) e^{3} + 3 \, b^{2} c^{4} \arctan \left (d x + c\right )^{2} e^{3} + 3 \, \pi a b c^{4} e^{3} \mathrm{sgn}\left (d x + c\right ) - 3 \, \pi a b c^{4} e^{3} + 12 \, a^{2} c^{3} d x e^{3} - 6 \, a b c d^{2} x^{2} e^{3} - 6 \, b^{2} c^{2} d x \arctan \left (d x + c\right ) e^{3} - 6 \, a b c^{4} \arctan \left (\frac{1}{d x + c}\right ) e^{3} - \pi b^{2} c^{3} e^{3} \mathrm{sgn}\left (d x + c\right ) + \pi b^{2} c^{3} e^{3} - 6 \, a b c^{2} d x e^{3} + b^{2} d^{2} x^{2} e^{3} + 2 \, b^{2} c^{3} \arctan \left (\frac{1}{d x + c}\right ) e^{3} + 2 \, b^{2} c d x e^{3} + 6 \, b^{2} d x \arctan \left (d x + c\right ) e^{3} + 3 \, \pi b^{2} c e^{3} \mathrm{sgn}\left (d x + c\right ) - 3 \, \pi b^{2} c e^{3} + 6 \, a b d x e^{3} - 3 \, b^{2} \arctan \left (d x + c\right )^{2} e^{3} - 6 \, b^{2} c \arctan \left (\frac{1}{d x + c}\right ) e^{3} - 3 \, \pi a b e^{3} \mathrm{sgn}\left (d x + c\right ) + 3 \, \pi a b e^{3} + 6 \, a b \arctan \left (\frac{1}{d x + c}\right ) e^{3} - 4 \, b^{2} e^{3} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3*(a+b*arctan(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(3*b^2*d^4*x^4*arctan(d*x + c)^2*e^3 + 6*a*b*d^4*x^4*arctan(d*x + c)*e^3 + 12*b^2*c*d^3*x^3*arctan(d*x +
c)^2*e^3 + 3*a^2*d^4*x^4*e^3 + 24*a*b*c*d^3*x^3*arctan(d*x + c)*e^3 + 18*b^2*c^2*d^2*x^2*arctan(d*x + c)^2*e^3
+ 12*a^2*c*d^3*x^3*e^3 + 36*a*b*c^2*d^2*x^2*arctan(d*x + c)*e^3 - 2*b^2*d^3*x^3*arctan(d*x + c)*e^3 + 12*b^2*
c^3*d*x*arctan(d*x + c)^2*e^3 + 18*a^2*c^2*d^2*x^2*e^3 - 2*a*b*d^3*x^3*e^3 + 24*a*b*c^3*d*x*arctan(d*x + c)*e^
3 - 6*b^2*c*d^2*x^2*arctan(d*x + c)*e^3 + 3*b^2*c^4*arctan(d*x + c)^2*e^3 + 3*pi*a*b*c^4*e^3*sgn(d*x + c) - 3*
pi*a*b*c^4*e^3 + 12*a^2*c^3*d*x*e^3 - 6*a*b*c*d^2*x^2*e^3 - 6*b^2*c^2*d*x*arctan(d*x + c)*e^3 - 6*a*b*c^4*arct
an(1/(d*x + c))*e^3 - pi*b^2*c^3*e^3*sgn(d*x + c) + pi*b^2*c^3*e^3 - 6*a*b*c^2*d*x*e^3 + b^2*d^2*x^2*e^3 + 2*b
^2*c^3*arctan(1/(d*x + c))*e^3 + 2*b^2*c*d*x*e^3 + 6*b^2*d*x*arctan(d*x + c)*e^3 + 3*pi*b^2*c*e^3*sgn(d*x + c)
- 3*pi*b^2*c*e^3 + 6*a*b*d*x*e^3 - 3*b^2*arctan(d*x + c)^2*e^3 - 6*b^2*c*arctan(1/(d*x + c))*e^3 - 3*pi*a*b*e
^3*sgn(d*x + c) + 3*pi*a*b*e^3 + 6*a*b*arctan(1/(d*x + c))*e^3 - 4*b^2*e^3*log(d^2*x^2 + 2*c*d*x + c^2 + 1))/d