3.1.20 \(\int \frac {(a+b \text {ArcTan}(c+d x))^3}{(c e+d e x)^4} \, dx\) [20]

Optimal. Leaf size=287 \[ -\frac {b^2 (a+b \text {ArcTan}(c+d x))}{d e^4 (c+d x)}-\frac {b (a+b \text {ArcTan}(c+d x))^2}{2 d e^4}-\frac {b (a+b \text {ArcTan}(c+d x))^2}{2 d e^4 (c+d x)^2}+\frac {i (a+b \text {ArcTan}(c+d x))^3}{3 d e^4}-\frac {(a+b \text {ArcTan}(c+d x))^3}{3 d e^4 (c+d x)^3}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left (1+(c+d x)^2\right )}{2 d e^4}-\frac {b (a+b \text {ArcTan}(c+d x))^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \text {PolyLog}\left (3,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4} \]

[Out]

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

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Rubi [A]
time = 0.35, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {5151, 12, 4946, 5038, 272, 36, 29, 31, 5004, 5044, 4988, 5112, 6745} \begin {gather*} \frac {i b^2 \text {Li}_2\left (\frac {2}{1-i (c+d x)}-1\right ) (a+b \text {ArcTan}(c+d x))}{d e^4}-\frac {b^2 (a+b \text {ArcTan}(c+d x))}{d e^4 (c+d x)}-\frac {b (a+b \text {ArcTan}(c+d x))^2}{2 d e^4 (c+d x)^2}-\frac {b (a+b \text {ArcTan}(c+d x))^2}{2 d e^4}-\frac {(a+b \text {ArcTan}(c+d x))^3}{3 d e^4 (c+d x)^3}+\frac {i (a+b \text {ArcTan}(c+d x))^3}{3 d e^4}-\frac {b \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))^2}{d e^4}-\frac {b^3 \text {Li}_3\left (\frac {2}{1-i (c+d x)}-1\right )}{2 d e^4}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left ((c+d x)^2+1\right )}{2 d e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

-((b^2*(a + b*ArcTan[c + d*x]))/(d*e^4*(c + d*x))) - (b*(a + b*ArcTan[c + d*x])^2)/(2*d*e^4) - (b*(a + b*ArcTa
n[c + d*x])^2)/(2*d*e^4*(c + d*x)^2) + ((I/3)*(a + b*ArcTan[c + d*x])^3)/(d*e^4) - (a + b*ArcTan[c + d*x])^3/(
3*d*e^4*(c + d*x)^3) + (b^3*Log[c + d*x])/(d*e^4) - (b^3*Log[1 + (c + d*x)^2])/(2*d*e^4) - (b*(a + b*ArcTan[c
+ d*x])^2*Log[2 - 2/(1 - I*(c + d*x))])/(d*e^4) + (I*b^2*(a + b*ArcTan[c + d*x])*PolyLog[2, -1 + 2/(1 - I*(c +
d*x))])/(d*e^4) - (b^3*PolyLog[3, -1 + 2/(1 - I*(c + d*x))])/(2*d*e^4)

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 4988

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTan[c*x])
^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 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 5004

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 5038

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), 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] && LtQ[m, -1]

Rule 5044

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

Rule 5112

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[I*(a + b*ArcTa
n[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 5151

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 6745

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 \frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{(c e+d e x)^4} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x^3 \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x^3} \, dx,x,c+d x\right )}{d e^4}-\frac {b \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {(i b) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x (i+x)} \, dx,x,c+d x\right )}{d e^4}+\frac {b^2 \text {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {b^2 \text {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^4}-\frac {b^2 \text {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{d e^4}+\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \log \left (2-\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {\left (i b^3\right ) \text {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 e^4}+\frac {b^3 \text {Subst}\left (\int \frac {1}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4}+\frac {b^3 \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4}+\frac {b^3 \text {Subst}\left (\int \frac {1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^4}-\frac {b^3 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left (1+(c+d x)^2\right )}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4}\\ \end {align*}

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Mathematica [A]
time = 0.87, size = 360, normalized size = 1.25 \begin {gather*} \frac {-\frac {8 a^3}{(c+d x)^3}-\frac {12 a^2 b}{(c+d x)^2}-\frac {24 a^2 b \text {ArcTan}(c+d x)}{(c+d x)^3}-24 a^2 b \log (c+d x)+12 a^2 b \log \left (1+c^2+2 c d x+d^2 x^2\right )+24 a b^2 \left (-\frac {(c+d x)^2+\text {ArcTan}(c+d x)^2}{(c+d x)^3}+\text {ArcTan}(c+d x) \left (-1-\frac {1}{(c+d x)^2}+i \text {ArcTan}(c+d x)-2 \log \left (1-e^{2 i \text {ArcTan}(c+d x)}\right )\right )+i \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c+d x)}\right )\right )+b^3 \left (i \pi ^3-\frac {24 \text {ArcTan}(c+d x)}{c+d x}-12 \text {ArcTan}(c+d x)^2-\frac {12 \text {ArcTan}(c+d x)^2}{(c+d x)^2}-8 i \text {ArcTan}(c+d x)^3-\frac {8 \text {ArcTan}(c+d x)^3}{(c+d x)^3}-24 \text {ArcTan}(c+d x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c+d x)}\right )+24 \log \left (\frac {c+d x}{\sqrt {1+(c+d x)^2}}\right )-24 i \text {ArcTan}(c+d x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c+d x)}\right )-12 \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c+d x)}\right )\right )}{24 d e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

((-8*a^3)/(c + d*x)^3 - (12*a^2*b)/(c + d*x)^2 - (24*a^2*b*ArcTan[c + d*x])/(c + d*x)^3 - 24*a^2*b*Log[c + d*x
] + 12*a^2*b*Log[1 + c^2 + 2*c*d*x + d^2*x^2] + 24*a*b^2*(-(((c + d*x)^2 + ArcTan[c + d*x]^2)/(c + d*x)^3) + A
rcTan[c + d*x]*(-1 - (c + d*x)^(-2) + I*ArcTan[c + d*x] - 2*Log[1 - E^((2*I)*ArcTan[c + d*x])]) + I*PolyLog[2,
E^((2*I)*ArcTan[c + d*x])]) + b^3*(I*Pi^3 - (24*ArcTan[c + d*x])/(c + d*x) - 12*ArcTan[c + d*x]^2 - (12*ArcTa
n[c + d*x]^2)/(c + d*x)^2 - (8*I)*ArcTan[c + d*x]^3 - (8*ArcTan[c + d*x]^3)/(c + d*x)^3 - 24*ArcTan[c + d*x]^2
*Log[1 - E^((-2*I)*ArcTan[c + d*x])] + 24*Log[(c + d*x)/Sqrt[1 + (c + d*x)^2]] - (24*I)*ArcTan[c + d*x]*PolyLo
g[2, E^((-2*I)*ArcTan[c + d*x])] - 12*PolyLog[3, E^((-2*I)*ArcTan[c + d*x])]))/(24*d*e^4)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 4.13, size = 6805, normalized size = 23.71

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(6805\)
default	\(\text {Expression too large to display}\)	\(6805\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

1/2*(d*(e^(-4)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^2 - 2*e^(-4)*log(d*x + c)/d^2 - 1/(d^4*x^2*e^4 + 2*c*d^3*x*e
^4 + c^2*d^2*e^4)) - 2*arctan(d*x + c)/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4))*a^2*b -
1/3*a^3/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4) - 1/96*(4*b^3*arctan(d*x + c)^3 - 3*b^3*
arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 - 96*(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3
*d*e^4)*integrate(1/32*(28*(b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2 + b^3)*arctan(d*x + c)^3 + 4*(24*a*b^2*d^2*x^2
+ 24*a*b^2*c^2 + b^3*c + 24*a*b^2 + (48*a*b^2*c + b^3)*d*x)*arctan(d*x + c)^2 - 4*(b^3*d^2*x^2 + 2*b^3*c*d*x
+ b^3*c^2)*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - (b^3*d*x + b^3*c - 3*(b^3*d^2*x^2 + 2*b^3*c*d*x
+ b^3*c^2 + b^3)*arctan(d*x + c))*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2)/(d^6*x^6*e^4 + 6*c*d^5*x^5*e^4 + (15*c^2
*e^4 + e^4)*d^4*x^4 + 4*(5*c^3*e^4 + c*e^4)*d^3*x^3 + c^6*e^4 + 3*(5*c^4*e^4 + 2*c^2*e^4)*d^2*x^2 + c^4*e^4 +
2*(3*c^5*e^4 + 2*c^3*e^4)*d*x), x))/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

integral((b^3*arctan(d*x + c)^3 + 3*a*b^2*arctan(d*x + c)^2 + 3*a^2*b*arctan(d*x + c) + a^3)*e^(-4)/(d^4*x^4 +
4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(d*x+c))**3/(d*e*x+c*e)**4,x)

[Out]

(Integral(a**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**3*atan(c +
d*x)**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a*b**2*atan(c + d
*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a**2*b*atan(c + d*x
)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))^3/(c*e + d*e*x)^4,x)

[Out]

int((a + b*atan(c + d*x))^3/(c*e + d*e*x)^4, x)

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