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

Optimal. Leaf size=271 \[ -\frac {i b^2 e^2 \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d}+a b^2 e^2 x-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}+\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {b e^2 \log \left (\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d}-\frac {b^3 e^2 \text {Li}_3\left (1-\frac {2}{i (c+d x)+1}\right )}{2 d}-\frac {b^3 e^2 \log \left ((c+d x)^2+1\right )}{2 d}+\frac {b^3 e^2 (c+d x) \tan ^{-1}(c+d x)}{d} \]

[Out]

a*b^2*e^2*x+b^3*e^2*(d*x+c)*arctan(d*x+c)/d-1/2*b*e^2*(a+b*arctan(d*x+c))^2/d-1/2*b*e^2*(d*x+c)^2*(a+b*arctan(
d*x+c))^2/d-1/3*I*e^2*(a+b*arctan(d*x+c))^3/d+1/3*e^2*(d*x+c)^3*(a+b*arctan(d*x+c))^3/d-b*e^2*(a+b*arctan(d*x+
c))^2*ln(2/(1+I*(d*x+c)))/d-1/2*b^3*e^2*ln(1+(d*x+c)^2)/d-I*b^2*e^2*(a+b*arctan(d*x+c))*polylog(2,1-2/(1+I*(d*
x+c)))/d-1/2*b^3*e^2*polylog(3,1-2/(1+I*(d*x+c)))/d

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Rubi [A]  time = 0.44, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {5043, 12, 4852, 4916, 4846, 260, 4884, 4920, 4854, 4994, 6610} \[ -\frac {i b^2 e^2 \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d}-\frac {b^3 e^2 \text {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right )}{2 d}+a b^2 e^2 x-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}+\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {b e^2 \log \left (\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d}-\frac {b^3 e^2 \log \left ((c+d x)^2+1\right )}{2 d}+\frac {b^3 e^2 (c+d x) \tan ^{-1}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 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 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 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 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 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 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 4994

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

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 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int (c e+d e x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int e^2 x^2 \left (a+b \tan ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int x^2 \left (a+b \tan ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x^3 \left (a+b \tan ^{-1}(x)\right )^2}{1+x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int x \left (a+b \tan ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}+\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \tan ^{-1}(x)\right )^2}{1+x^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac {b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}+\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{i-x} \, dx,x,c+d x\right )}{d}+\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \tan ^{-1}(x)\right )}{1+x^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac {b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}+\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}+\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}-\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{d}+\frac {\left (2 b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d}\\ &=a b^2 e^2 x-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}+\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}-\frac {i b^2 e^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{d}+\frac {\left (i b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d}+\frac {\left (b^3 e^2\right ) \operatorname {Subst}\left (\int \tan ^{-1}(x) \, dx,x,c+d x\right )}{d}\\ &=a b^2 e^2 x+\frac {b^3 e^2 (c+d x) \tan ^{-1}(c+d x)}{d}-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}+\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}-\frac {i b^2 e^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{d}-\frac {b^3 e^2 \text {Li}_3\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d}-\frac {\left (b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,c+d x\right )}{d}\\ &=a b^2 e^2 x+\frac {b^3 e^2 (c+d x) \tan ^{-1}(c+d x)}{d}-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {b e^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac {i e^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}+\frac {e^2 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d}-\frac {b e^2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}-\frac {b^3 e^2 \log \left (1+(c+d x)^2\right )}{2 d}-\frac {i b^2 e^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{d}-\frac {b^3 e^2 \text {Li}_3\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.77, size = 349, normalized size = 1.29 \[ \frac {e^2 \left (2 a^3 (c+d x)^3-3 a^2 b (c+d x)^2+3 a^2 b \log \left ((c+d x)^2+1\right )+6 a^2 b (c+d x)^3 \tan ^{-1}(c+d x)+6 a b^2 \left (i \text {Li}_2\left (-e^{2 i \tan ^{-1}(c+d x)}\right )+(c+d x)^3 \tan ^{-1}(c+d x)^2-(c+d x)^2 \tan ^{-1}(c+d x)+i \tan ^{-1}(c+d x)^2-\tan ^{-1}(c+d x)-2 \tan ^{-1}(c+d x) \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )+c+d x\right )+b^3 \left (6 i \tan ^{-1}(c+d x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c+d x)}\right )-3 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c+d x)}\right )+6 \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )-2 (c+d x) \tan ^{-1}(c+d x)^3+2 (c+d x) \left ((c+d x)^2+1\right ) \tan ^{-1}(c+d x)^3+2 i \tan ^{-1}(c+d x)^3-3 \left ((c+d x)^2+1\right ) \tan ^{-1}(c+d x)^2+6 (c+d x) \tan ^{-1}(c+d x)-6 \tan ^{-1}(c+d x)^2 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )\right )}{6 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(e^2*(-3*a^2*b*(c + d*x)^2 + 2*a^3*(c + d*x)^3 + 6*a^2*b*(c + d*x)^3*ArcTan[c + d*x] + 3*a^2*b*Log[1 + (c + d*
x)^2] + 6*a*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*A
rcTan[c + d*x]^2 - 2*ArcTan[c + d*x]*Log[1 + E^((2*I)*ArcTan[c + d*x])] + I*PolyLog[2, -E^((2*I)*ArcTan[c + d*
x])]) + b^3*(6*(c + d*x)*ArcTan[c + d*x] - 3*(1 + (c + d*x)^2)*ArcTan[c + d*x]^2 + (2*I)*ArcTan[c + d*x]^3 - 2
*(c + d*x)*ArcTan[c + d*x]^3 + 2*(c + d*x)*(1 + (c + d*x)^2)*ArcTan[c + d*x]^3 - 6*ArcTan[c + d*x]^2*Log[1 + E
^((2*I)*ArcTan[c + d*x])] + 6*Log[1/Sqrt[1 + (c + d*x)^2]] + (6*I)*ArcTan[c + d*x]*PolyLog[2, -E^((2*I)*ArcTan
[c + d*x])] - 3*PolyLog[3, -E^((2*I)*ArcTan[c + d*x])])))/(6*d)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{3} d^{2} e^{2} x^{2} + 2 \, a^{3} c d e^{2} x + a^{3} c^{2} e^{2} + {\left (b^{3} d^{2} e^{2} x^{2} + 2 \, b^{3} c d e^{2} x + b^{3} c^{2} e^{2}\right )} \arctan \left (d x + c\right )^{3} + 3 \, {\left (a b^{2} d^{2} e^{2} x^{2} + 2 \, a b^{2} c d e^{2} x + a b^{2} c^{2} e^{2}\right )} \arctan \left (d x + c\right )^{2} + 3 \, {\left (a^{2} b d^{2} e^{2} x^{2} + 2 \, a^{2} b c d e^{2} x + a^{2} b c^{2} e^{2}\right )} \arctan \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [C]  time = 2.25, size = 3242, normalized size = 11.96 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^2*arctan(d*x+c)^2*
Pi*b^3*c*e^2+1/4*I/d*e^2*b^3*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^2*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^
2)+1))*arctan(d*x+c)^2*Pi-1/8*I/d*e^2*b^3*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)*csgn(I*((1+I*(d*x+c))^2/
(1+(d*x+c)^2)+1))^2*arctan(d*x+c)^2*Pi-1/4*I/d*e^2*b^3*csgn(I/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)*arctan(d*x+
c)^2*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^2*Pi-1/2*I/d*e^2*b^3*arctan(d*x
+c)^2*csgn(I*(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2*Pi+1/4*I/d*e^2*b^3*csg
n(I*(1+I*(d*x+c))^4/(1+(d*x+c)^2)^2+2*I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)^2*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2
)+I)*arctan(d*x+c)^2*Pi-1/4/d*csgn(I*(1+I*(d*x+c))^4/(1+(d*x+c)^2)^2+2*I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)^2*cs
gn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)*arctan(d*x+c)^2*Pi*b^3*c*e^2-1/4*I/d*e^2*b^3*arctan(d*x+c)^2*csgn(I*(1+I
*(d*x+c))^2/(1+(d*x+c)^2)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^2*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*Pi-1/8*
I/d*e^2*b^3*csgn(I*(1+I*(d*x+c))^4/(1+(d*x+c)^2)^2+2*I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)*csgn(I*(1+I*(d*x+c))^2
/(1+(d*x+c)^2)+I)^2*arctan(d*x+c)^2*Pi+1/4*I/d*e^2*b^3*arctan(d*x+c)^2*csgn(I*(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2
))^2*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*Pi-1/2*d*x^2*a^2*b*e^2+d*x^2*a^3*c*e^2+arctan(d*x+c)^3*x*b^3*c^2*e^
2-arctan(d*x+c)^2*x*b^3*c*e^2-1/d*e^2*b^3*arctan(d*x+c)^2*ln((1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+1/2/d*e^2*b^3*
arctan(d*x+c)^2*ln(1+(d*x+c)^2)+1/d*arctan(d*x+c)*b^3*c*e^2+1/3/d*arctan(d*x+c)^3*b^3*c^3*e^2-1/2/d*arctan(d*x
+c)^2*b^3*c^2*e^2-1/d*e^2*b^3*ln(2)*arctan(d*x+c)^2-1/d*e^2*a*b^2*arctan(d*x+c)+1/3*d^2*arctan(d*x+c)^3*x^3*b^
3*e^2-1/2*d*arctan(d*x+c)^2*x^2*b^3*e^2+1/2/d*e^2*a^2*b*ln(1+(d*x+c)^2)+1/3*I/d*e^2*b^3*arctan(d*x+c)^3-I/d*e^
2*b^3*arctan(d*x+c)-d*arctan(d*x+c)*x^2*a*b^2*e^2+d^2*arctan(d*x+c)*x^3*a^2*b*e^2+1/d*arctan(d*x+c)^2*a*b^2*c^
3*e^2-1/d*arctan(d*x+c)*a*b^2*c^2*e^2+d^2*arctan(d*x+c)^2*x^3*a*b^2*e^2+d*arctan(d*x+c)^3*x^2*b^3*c*e^2+1/3*d^
2*x^3*a^3*e^2+1/d*e^2*b^3*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)-1/2/d*e^2*b^3*arctan(d*x+c)^2-1/2/d*e^2*b^3*poly
log(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+arctan(d*x+c)*x*b^3*e^2+x*a^3*c^2*e^2+1/d*a*b^2*c*e^2-x*a^2*b*c*e^2+a*b^
2*e^2*x-1/8*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)*arctan(d*x
+c)^2*Pi*x*b^3*e^2-1/4*csgn(I*(1+I*(d*x+c))^4/(1+(d*x+c)^2)^2+2*I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)^2*csgn(I*(1
+I*(d*x+c))^2/(1+(d*x+c)^2)+I)*arctan(d*x+c)^2*Pi*x*b^3*e^2+1/8*csgn(I*(1+I*(d*x+c))^4/(1+(d*x+c)^2)^2+2*I*(1+
I*(d*x+c))^2/(1+(d*x+c)^2)+I)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)^2*arctan(d*x+c)^2*Pi*x*b^3*e^2+3*d*arcta
n(d*x+c)^2*x^2*a*b^2*c*e^2-1/8/d*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^3*arctan(d*x+c)^2*Pi*b^3*c*e^2+1/
8/d*csgn(I*(1+I*(d*x+c))^4/(1+(d*x+c)^2)^2+2*I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)^3*arctan(d*x+c)^2*Pi*b^3*c*e^2
-1/2*I/d*e^2*a*b^2*ln(I+d*x+c)*ln(1+(d*x+c)^2)+1/2*I/d*e^2*a*b^2*ln(I+d*x+c)*ln(1/2*I*(d*x+c-I))-1/8*I/d*e^2*b
^3*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^3*arctan(d*x+c)^2*Pi-1/8*I/d*e^2*b^3*csgn(I*(1+I*(d*x+c))^4/(1+
(d*x+c)^2)^2+2*I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)^3*arctan(d*x+c)^2*Pi+1/4*I/d*e^2*b^3*arctan(d*x+c)^2*csgn(I*
(1+I*(d*x+c))^2/(1+(d*x+c)^2)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^3*Pi+1/4*I/d*e^2*b^3*arctan(d*x+c)^2*csgn(I
*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^3*Pi+3*d*arctan(d*x+c)*x^2*a^2*b*c*e^2+1/2*I/d*e^2*a*b^2*ln(d*x+c-I)*ln(1+(d*x
+c)^2)-1/2*I/d*e^2*a*b^2*ln(d*x+c-I)*ln(-1/2*I*(I+d*x+c))+1/3/d*a^3*c^3*e^2+1/d*arctan(d*x+c)*a^2*b*c^3*e^2+1/
d*e^2*a*b^2*arctan(d*x+c)*ln(1+(d*x+c)^2)+I/d*e^2*b^3*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-
1/8*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^3*arctan(d*x+c)^2*Pi*x*b^3*e^2+3*arctan(d*x+c)^2*x*a*b^2*c^2*e
^2+3*arctan(d*x+c)*x*a^2*b*c^2*e^2+1/8*csgn(I*(1+I*(d*x+c))^4/(1+(d*x+c)^2)^2+2*I*(1+I*(d*x+c))^2/(1+(d*x+c)^2
)+I)^3*arctan(d*x+c)^2*Pi*x*b^3*e^2-2*arctan(d*x+c)*x*a*b^2*c*e^2+1/4*I/d*e^2*a*b^2*ln(I+d*x+c)^2+1/2*I/d*e^2*
a*b^2*dilog(1/2*I*(d*x+c-I))-1/4*I/d*e^2*a*b^2*ln(d*x+c-I)^2-1/2*I/d*e^2*a*b^2*dilog(-1/2*I*(I+d*x+c))+1/4*I/d
*e^2*b^3*csgn(I/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)*arctan(d*x+c)^2*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/((1+
I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*Pi-1/2/d*a^2*b*c^2*e^2+1/4*csgn(I*((1+I
*(d*x+c))^2/(1+(d*x+c)^2)+1))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^2*arctan(d*x+c)^2*Pi*x*b^3*e^2+1/8/d
*csgn(I*(1+I*(d*x+c))^4/(1+(d*x+c)^2)^2+2*I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+I)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)
^2)+I)^2*arctan(d*x+c)^2*Pi*b^3*c*e^2-1/8/d*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*csgn(I*((1+I*(d*x+c))^
2/(1+(d*x+c)^2)+1)^2)*arctan(d*x+c)^2*Pi*b^3*c*e^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/8*b^3*c^4*e^2*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 3*a*b^2*c^4*e^2*arctan(d*x + c)^2*arctan((d^2*x
+ c*d)/d)/d - (3*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2*c^4*e^2 - 7/
32*(6*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)^2/d - 4*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^3/d + arctan((
d^2*x + c*d)/d)^4/d)*b^3*c^4*e^2 + 1/3*a^3*d^2*e^2*x^3 + 7/8*b^3*c^2*e^2*arctan(d*x + c)^3*arctan((d^2*x + c*d
)/d)/d + 28*b^3*d^4*e^2*integrate(1/32*x^4*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*d^4*e^2
*integrate(1/32*x^4*arctan(d*x + c)*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) + 96*
a*b^2*d^4*e^2*integrate(1/32*x^4*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 112*b^3*c*d^3*e^2*integ
rate(1/32*x^3*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 4*b^3*d^4*e^2*integrate(1/32*x^4*arctan(d*
x + c)*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^3*c*d^3*e^2*integrate(1/32*x^
3*arctan(d*x + c)*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) + 384*a*b^2*c*d^3*e^2*i
ntegrate(1/32*x^3*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 168*b^3*c^2*d^2*e^2*integrate(1/32*x^2
*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 16*b^3*c*d^3*e^2*integrate(1/32*x^3*arctan(d*x + c)*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^3*c^2*d^2*e^2*integrate(1/32*x^2*arctan
(d*x + c)*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) + 576*a*b^2*c^2*d^2*e^2*integra
te(1/32*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 112*b^3*c^3*d*e^2*integrate(1/32*x*arctan(d*
x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 24*b^3*c^2*d^2*e^2*integrate(1/32*x^2*arctan(d*x + c)*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^3*c^3*d*e^2*integrate(1/32*x*arctan(d*x + c)*lo
g(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 384*a*b^2*c^3*d*e^2*integrate(1/32*x*arct
an(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3*c^3*d*e^2*integrate(1/32*x*arctan(d*x + c)*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^3*c^4*e^2*integrate(1/32*arctan(d*x + c)*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^3*c*d*e^2*x^2 + 3*a*b^2*c^2*e^2*arctan(d*x
+ c)^2*arctan((d^2*x + c*d)/d)/d - 4*b^3*d^3*e^2*integrate(1/32*x^3*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2
+ 1), x) + b^3*d^3*e^2*integrate(1/32*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
) - 12*b^3*c*d^2*e^2*integrate(1/32*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*c*d^2*e^2*
integrate(1/32*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) - 12*b^3*c^2*d*e^2*int
egrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*c^2*d*e^2*integrate(1/32*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*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d -
arctan((d^2*x + c*d)/d)^3/d)*a*b^2*c^2*e^2 - 7/32*(6*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)^2/d - 4*arctan
(d*x + c)*arctan((d^2*x + c*d)/d)^3/d + arctan((d^2*x + c*d)/d)^4/d)*b^3*c^2*e^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^2*b*c*d*e^2 + 1/2*
(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^2*b*d^2*e^2 + a^3*c^2*e^2*x + 28*b^3*d^2*e^2*integrate(1/32*x^2*arctan(d*
x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*d^2*e^2*integrate(1/32*x^2*arctan(d*x + c)*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) + 96*a*b^2*d^2*e^2*integrate(1/32*x^2*arctan(d*x + c)^2/(
d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 56*b^3*c*d*e^2*integrate(1/32*x*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2
+ 1), x) + 6*b^3*c*d*e^2*integrate(1/32*x*arctan(d*x + c)*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) + 192*a*b^2*c*d*e^2*integrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*
b^3*c^2*e^2*integrate(1/32*arctan(d*x + c)*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
) + 3/2*(2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*a^2*b*c^2*e^2/d + 1/24*(b^3*d^2*e^2*x^3 + 3*b^3*c
*d*e^2*x^2 + 3*b^3*c^2*e^2*x)*arctan(d*x + c)^3 - 1/32*(b^3*d^2*e^2*x^3 + 3*b^3*c*d*e^2*x^2 + 3*b^3*c^2*e^2*x)
*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,e+d\,e\,x\right )}^2\,{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{2} \left (\int a^{3} c^{2}\, dx + \int a^{3} d^{2} x^{2}\, dx + \int b^{3} c^{2} \operatorname {atan}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} c^{2} \operatorname {atan}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b c^{2} \operatorname {atan}{\left (c + d x \right )}\, dx + \int 2 a^{3} c d x\, dx + \int b^{3} d^{2} x^{2} \operatorname {atan}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} d^{2} x^{2} \operatorname {atan}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b d^{2} x^{2} \operatorname {atan}{\left (c + d x \right )}\, dx + \int 2 b^{3} c d x \operatorname {atan}^{3}{\left (c + d x \right )}\, dx + \int 6 a b^{2} c d x \operatorname {atan}^{2}{\left (c + d x \right )}\, dx + \int 6 a^{2} b c d x \operatorname {atan}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**2*(Integral(a**3*c**2, x) + Integral(a**3*d**2*x**2, x) + Integral(b**3*c**2*atan(c + d*x)**3, x) + Integra
l(3*a*b**2*c**2*atan(c + d*x)**2, x) + Integral(3*a**2*b*c**2*atan(c + d*x), x) + Integral(2*a**3*c*d*x, x) +
Integral(b**3*d**2*x**2*atan(c + d*x)**3, x) + Integral(3*a*b**2*d**2*x**2*atan(c + d*x)**2, x) + Integral(3*a
**2*b*d**2*x**2*atan(c + d*x), x) + Integral(2*b**3*c*d*x*atan(c + d*x)**3, x) + Integral(6*a*b**2*c*d*x*atan(
c + d*x)**2, x) + Integral(6*a**2*b*c*d*x*atan(c + d*x), x))

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