3.1.19 \(\int \frac {(a+b \text {ArcTan}(c+d x))^3}{(c e+d e x)^3} \, dx\) [19]

Optimal. Leaf size=180 \[ -\frac {3 i b (a+b \text {ArcTan}(c+d x))^2}{2 d e^3}-\frac {3 b (a+b \text {ArcTan}(c+d x))^2}{2 d e^3 (c+d x)}-\frac {(a+b \text {ArcTan}(c+d x))^3}{2 d e^3}-\frac {(a+b \text {ArcTan}(c+d x))^3}{2 d e^3 (c+d x)^2}+\frac {3 b^2 (a+b \text {ArcTan}(c+d x)) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^3}-\frac {3 i b^3 \text {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^3} \]

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5151, 12, 4946, 5038, 5044, 4988, 2497, 5004} \begin {gather*} \frac {3 b^2 \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))}{d e^3}-\frac {3 b (a+b \text {ArcTan}(c+d x))^2}{2 d e^3 (c+d x)}-\frac {3 i b (a+b \text {ArcTan}(c+d x))^2}{2 d e^3}-\frac {(a+b \text {ArcTan}(c+d x))^3}{2 d e^3 (c+d x)^2}-\frac {(a+b \text {ArcTan}(c+d x))^3}{2 d e^3}-\frac {3 i b^3 \text {Li}_2\left (\frac {2}{1-i (c+d x)}-1\right )}{2 d e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 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]

Rubi steps

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

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Mathematica [A]
time = 0.20, size = 225, normalized size = 1.25 \begin {gather*} -\frac {a^3+b^3 \left (1+c^2+2 c d x+d^2 x^2\right ) \text {ArcTan}(c+d x)^3+3 a^2 b \left (c+d x+\left (1+(c+d x)^2\right ) \text {ArcTan}(c+d x)\right )+3 a b^2 \left (2 (c+d x) \text {ArcTan}(c+d x)+\left (1+(c+d x)^2\right ) \text {ArcTan}(c+d x)^2-2 (c+d x)^2 \log \left (\frac {c+d x}{\sqrt {1+(c+d x)^2}}\right )\right )+3 b^3 (c+d x) \left (\text {ArcTan}(c+d x)^2-2 (c+d x) \text {ArcTan}(c+d x) \log \left (1-e^{2 i \text {ArcTan}(c+d x)}\right )+i (c+d x) \left (\text {ArcTan}(c+d x)^2+\text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c+d x)}\right )\right )\right )}{2 d e^3 (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (166 ) = 332\).
time = 0.65, size = 557, normalized size = 3.09

method	result	size
derivativedivides	\(\frac {-\frac {a^{3}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {b^{3} \arctan \left (d x +c \right )^{3}}{2 e^{3}}-\frac {3 a^{2} b \arctan \left (d x +c \right )}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \arctan \left (d x +c \right )}{e^{3} \left (d x +c \right )}-\frac {3 a \,b^{2} \arctan \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 a^{2} b}{2 e^{3} \left (d x +c \right )}-\frac {3 a^{2} b \arctan \left (d x +c \right )}{2 e^{3}}-\frac {3 a \,b^{2} \arctan \left (d x +c \right )^{2}}{2 e^{3}}-\frac {3 a \,b^{2} \ln \left (1+\left (d x +c \right )^{2}\right )}{2 e^{3}}+\frac {3 a \,b^{2} \ln \left (d x +c \right )}{e^{3}}-\frac {3 b^{3} \arctan \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )}-\frac {3 b^{3} \arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2 e^{3}}+\frac {3 b^{3} \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{e^{3}}-\frac {b^{3} \arctan \left (d x +c \right )^{3}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 i b^{3} \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{2 e^{3}}+\frac {3 i b^{3} \ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{4 e^{3}}-\frac {3 i b^{3} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )}{4 e^{3}}+\frac {3 i b^{3} \ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{4 e^{3}}+\frac {3 i b^{3} \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{2 e^{3}}-\frac {3 i b^{3} \ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{4 e^{3}}+\frac {3 i b^{3} \ln \left (d x +c -i\right )^{2}}{8 e^{3}}-\frac {3 i b^{3} \ln \left (d x +c +i\right )^{2}}{8 e^{3}}+\frac {3 i b^{3} \dilog \left (1+i \left (d x +c \right )\right )}{2 e^{3}}+\frac {3 i b^{3} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{4 e^{3}}-\frac {3 i b^{3} \dilog \left (1-i \left (d x +c \right )\right )}{2 e^{3}}-\frac {3 i b^{3} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right )}{4 e^{3}}}{d}\)	\(557\)
default	\(\frac {-\frac {a^{3}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {b^{3} \arctan \left (d x +c \right )^{3}}{2 e^{3}}-\frac {3 a^{2} b \arctan \left (d x +c \right )}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \arctan \left (d x +c \right )}{e^{3} \left (d x +c \right )}-\frac {3 a \,b^{2} \arctan \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 a^{2} b}{2 e^{3} \left (d x +c \right )}-\frac {3 a^{2} b \arctan \left (d x +c \right )}{2 e^{3}}-\frac {3 a \,b^{2} \arctan \left (d x +c \right )^{2}}{2 e^{3}}-\frac {3 a \,b^{2} \ln \left (1+\left (d x +c \right )^{2}\right )}{2 e^{3}}+\frac {3 a \,b^{2} \ln \left (d x +c \right )}{e^{3}}-\frac {3 b^{3} \arctan \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )}-\frac {3 b^{3} \arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2 e^{3}}+\frac {3 b^{3} \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{e^{3}}-\frac {b^{3} \arctan \left (d x +c \right )^{3}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 i b^{3} \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{2 e^{3}}+\frac {3 i b^{3} \ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{4 e^{3}}-\frac {3 i b^{3} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )}{4 e^{3}}+\frac {3 i b^{3} \ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{4 e^{3}}+\frac {3 i b^{3} \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{2 e^{3}}-\frac {3 i b^{3} \ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{4 e^{3}}+\frac {3 i b^{3} \ln \left (d x +c -i\right )^{2}}{8 e^{3}}-\frac {3 i b^{3} \ln \left (d x +c +i\right )^{2}}{8 e^{3}}+\frac {3 i b^{3} \dilog \left (1+i \left (d x +c \right )\right )}{2 e^{3}}+\frac {3 i b^{3} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{4 e^{3}}-\frac {3 i b^{3} \dilog \left (1-i \left (d x +c \right )\right )}{2 e^{3}}-\frac {3 i b^{3} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right )}{4 e^{3}}}{d}\)	\(557\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^3,x, algorithm="maxima")

[Out]

-3/2*(d*(arctan((d^2*x + c*d)/d)*e^(-3)/d^2 + 1/(d^3*x*e^3 + c*d^2*e^3)) + arctan(d*x + c)/(d^3*x^2*e^3 + 2*c*
d^2*x*e^3 + c^2*d*e^3))*a^2*b - 3/2*(2*d*(arctan((d^2*x + c*d)/d)*e^(-3)/d^2 + 1/(d^3*x*e^3 + c*d^2*e^3))*arct
an(d*x + c) - (arctan(d*x + c)^2 - log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 2*log(d*x + c))*e^(-3)/d)*a*b^2 - 3/2*a*
b^2*arctan(d*x + c)^2/(d^3*x^2*e^3 + 2*c*d^2*x*e^3 + c^2*d*e^3) - 1/32*(8*(d^2*x^2 + 2*c*d*x + c^2 + 1)*arctan
(d*x + c)^3 + 12*(d*x + c)*arctan(d*x + c)^2 - 3*(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 - 32*(d^3*x^2*e^
3 + 2*c*d^2*x*e^3 + c^2*d*e^3)*integrate(1/32*(16*(d^2*x^2 + 2*c*d*x + c^2 + 1)*arctan(d*x + c)^3 + 12*(d^3*x^
3 + 3*c*d^2*x^2 + c^3 + (3*c^2 + 1)*d*x + c)*arctan(d*x + c)^2 + 3*(d^3*x^3 + 3*c*d^2*x^2 + c^3 + (3*c^2 + 1)*
d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 + 24*(d^2*x^2 + 2*c*d*x + c^2)*arctan(d*x + c) - 12*(d^3*x^3 + 3*c
*d^2*x^2 + 3*c^2*d*x + c^3)*log(d^2*x^2 + 2*c*d*x + c^2 + 1))/(d^5*x^5*e^3 + 5*c*d^4*x^4*e^3 + (10*c^2*e^3 + e
^3)*d^3*x^3 + c^5*e^3 + (10*c^3*e^3 + 3*c*e^3)*d^2*x^2 + c^3*e^3 + (5*c^4*e^3 + 3*c^2*e^3)*d*x), x))*b^3/(d^3*
x^2*e^3 + 2*c*d^2*x*e^3 + c^2*d*e^3) - 1/2*a^3/(d^3*x^2*e^3 + 2*c*d^2*x*e^3 + c^2*d*e^3)

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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)^3,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^(-3)/(d^3*x^3 +
3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

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

[Out]

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

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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)^3,x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^3} \,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)^3,x)

[Out]

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

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