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

Optimal. Leaf size=164 \[ -\frac{3 i b^3 e \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right )}{2 d}-\frac{3 b^2 e \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d}-\frac{3 i b e \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac{3 b e (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}+\frac{e (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d}+\frac{e \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.243152, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {5043, 12, 4852, 4916, 4846, 4920, 4854, 2402, 2315, 4884} \[ -\frac{3 i b^3 e \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right )}{2 d}-\frac{3 b^2 e \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d}-\frac{3 i b e \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-\frac{3 b e (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}+\frac{e (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d}+\frac{e \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.267458, size = 196, normalized size = 1.2 \[ \frac{e \left (3 i b^3 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )+3 b \tan ^{-1}(c+d x) \left (a \left (a \left (c^2+2 c d x+d^2 x^2+1\right )-2 b (c+d x)\right )-2 b^2 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )+a \left (a (c+d x) (a c+a d x-3 b)-6 b^2 \log \left (\frac{1}{\sqrt{(c+d x)^2+1}}\right )\right )+3 b^2 (c+d x-i) \tan ^{-1}(c+d x)^2 (-b+a (c+d x+i))+b^3 \left (c^2+2 c d x+d^2 x^2+1\right ) \tan ^{-1}(c+d x)^3\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.132, size = 567, normalized size = 3.5 \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)*(a+b*arctan(d*x+c))^3,x)

[Out]

3/4*I/d*e*b^3*ln(d*x+c-I)*ln(1+(d*x+c)^2)-3/4*I/d*e*b^3*ln(d*x+c+I)*ln(1+(d*x+c)^2)+3*arctan(d*x+c)*x*a^2*b*c*
e+3/4*I/d*e*b^3*ln(d*x+c+I)*ln(1/2*I*(d*x+c-I))-3/4*I/d*e*b^3*ln(d*x+c-I)*ln(-1/2*I*(d*x+c+I))-3/2/d*a^2*b*c*e
+3/2/d*e*b^3*arctan(d*x+c)*ln(1+(d*x+c)^2)+3/4*I/d*e*b^3*dilog(1/2*I*(d*x+c-I))-3*arctan(d*x+c)*x*a*b^2*e+1/2*
d*arctan(d*x+c)^3*x^2*b^3*e-3/2/d*arctan(d*x+c)^2*b^3*c*e+arctan(d*x+c)^3*x*b^3*c*e-3/8*I/d*e*b^3*ln(d*x+c-I)^
2-3/4*I/d*e*b^3*dilog(-1/2*I*(d*x+c+I))+3/8*I/d*e*b^3*ln(d*x+c+I)^2+1/2/d*arctan(d*x+c)^3*b^3*c^2*e+3/2/d*e*a*
b^2*ln(1+(d*x+c)^2)+3/2/d*e*a^2*b*arctan(d*x+c)+3/2/d*e*a*b^2*arctan(d*x+c)^2+3/2/d*arctan(d*x+c)^2*a*b^2*c^2*
e+3/2*d*arctan(d*x+c)^2*x^2*a*b^2*e+3*arctan(d*x+c)^2*x*a*b^2*c*e-3/d*arctan(d*x+c)*a*b^2*c*e+3/2/d*arctan(d*x
+c)*a^2*b*c^2*e+3/2*d*arctan(d*x+c)*x^2*a^2*b*e+1/2/d*e*b^3*arctan(d*x+c)^3+1/2/d*a^3*c^2*e-3/2*e*x*a^2*b+x*a^
3*c*e+1/2*d*x^2*a^3*e-3/2*arctan(d*x+c)^2*x*b^3*e

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

[Out]

1/2*a^3*d*e*x^2 + 3/2*(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^2*b*d*e + a^3*c*e*x + 3/2*(2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*a^
2*b*c*e/d + 1/32*(8*(b^3*d^2*e*x^2 + 2*b^3*c*d*e*x + (b^3*c^2 + b^3)*e)*arctan(d*x + c)^3 + 12*(a*b^2*d^2*e*x^
2 + (2*a*b^2*c - b^3)*d*e*x)*arctan(d*x + c)^2 - 3*(a*b^2*d^2*e*x^2 + (2*a*b^2*c - b^3)*d*e*x)*log(d^2*x^2 + 2
*c*d*x + c^2 + 1)^2 + 4*(4*b^3*c^3*e*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 18*a*b^2*c^3*e*arctan(d*x +
c)^2*arctan((d^2*x + c*d)/d)/d - 6*(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^3*e - (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^3*e - 3*b^3*c^2*e*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d + 4*b^
3*c*e*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 128*b^3*d^3*e*integrate(1/32*x^3*arctan(d*x + c)^3/(d^2*x^
2 + 2*c*d*x + c^2 + 1), x) + 576*a*b^2*d^3*e*integrate(1/32*x^3*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1
), x) + 384*b^3*c*d^2*e*integrate(1/32*x^2*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 48*a*b^2*d^3*
e*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) + 1728*a*b^2*c*d^2*e
*integrate(1/32*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 384*b^3*c^2*d*e*integrate(1/32*x*arc
tan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 96*a*b^2*d^3*e*integrate(1/32*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) + 144*a*b^2*c*d^2*e*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) + 1728*a*b^2*c^2*d*e*integrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c
*d*x + c^2 + 1), x) + 288*a*b^2*c*d^2*e*integrate(1/32*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) + 144*a*b^2*c^2*d*e*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) + 192*a*b^2*c^2*d*e*integrate(1/32*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) + 48*a*b^2*c^3*e*integrate(1/32*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)*b^3*c^2*e + 18*a*b^2*c*e*arctan(d
*x + c)^2*arctan((d^2*x + c*d)/d)/d - 96*b^3*d^2*e*integrate(1/32*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c
^2 + 1), x) - 24*b^3*d^2*e*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) - 192*a*b^2*d^2*e*integrate(1/32*x^2*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 192*b^3*c*d*e*in
tegrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 96*b^3*d^2*e*integrate(1/32*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) - 48*b^3*c*d*e*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) - 384*a*b^2*c*d*e*integrate(1/32*x*arctan(d*x + c)/(d^2*x^2 +
2*c*d*x + c^2 + 1), x) - 96*b^3*c*d*e*integrate(1/32*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) - 24*b^3*c^2*e*integrate(1/32*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
) - 6*(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*e - (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*e - 3*b^3*e*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d + 128*b^3*d*e*integrate(1/32*x*arctan(d*x +
c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 576*a*b^2*d*e*integrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x +
c^2 + 1), x) + 48*a*b^2*d*e*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) + 192*b^3*d*e*integrate(1/32*x*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 48*a*b^2*c*e*integrate
(1/32*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)*b^3*e - 24*b^3*e*integrate(1/32*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))*d)/d

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{3} d e x + a^{3} c e +{\left (b^{3} d e x + b^{3} c e\right )} \arctan \left (d x + c\right )^{3} + 3 \,{\left (a b^{2} d e x + a b^{2} c e\right )} \arctan \left (d x + c\right )^{2} + 3 \,{\left (a^{2} b d e x + a^{2} b c e\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)*(a+b*arctan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e \left (\int a^{3} c\, dx + \int a^{3} d x\, dx + \int b^{3} c \operatorname{atan}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} c \operatorname{atan}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b c \operatorname{atan}{\left (c + d x \right )}\, dx + \int b^{3} d x \operatorname{atan}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} d x \operatorname{atan}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b 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)*(a+b*atan(d*x+c))**3,x)

[Out]

e*(Integral(a**3*c, x) + Integral(a**3*d*x, x) + Integral(b**3*c*atan(c + d*x)**3, x) + Integral(3*a*b**2*c*at
an(c + d*x)**2, x) + Integral(3*a**2*b*c*atan(c + d*x), x) + Integral(b**3*d*x*atan(c + d*x)**3, x) + Integral
(3*a*b**2*d*x*atan(c + d*x)**2, x) + Integral(3*a**2*b*d*x*atan(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 )}{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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