3.4.88 \(\int \frac {x^3 \text {ArcTan}(a x)^3}{c+a^2 c x^2} \, dx\) [388]

Optimal. Leaf size=260 \[ -\frac {3 i \text {ArcTan}(a x)^2}{2 a^4 c}-\frac {3 x \text {ArcTan}(a x)^2}{2 a^3 c}+\frac {\text {ArcTan}(a x)^3}{2 a^4 c}+\frac {x^2 \text {ArcTan}(a x)^3}{2 a^2 c}+\frac {i \text {ArcTan}(a x)^4}{4 a^4 c}-\frac {3 \text {ArcTan}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^4 c}+\frac {\text {ArcTan}(a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c}-\frac {3 i \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^4 c}+\frac {3 i \text {ArcTan}(a x)^2 \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^4 c}+\frac {3 \text {ArcTan}(a x) \text {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )}{2 a^4 c}-\frac {3 i \text {PolyLog}\left (4,1-\frac {2}{1+i a x}\right )}{4 a^4 c} \]

[Out]

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

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Rubi [A]
time = 0.33, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5036, 4946, 4930, 5040, 4964, 2449, 2352, 5004, 5114, 5118, 6745} \begin {gather*} \frac {3 i \text {ArcTan}(a x)^2 \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{2 a^4 c}+\frac {3 \text {ArcTan}(a x) \text {Li}_3\left (1-\frac {2}{i a x+1}\right )}{2 a^4 c}+\frac {i \text {ArcTan}(a x)^4}{4 a^4 c}+\frac {\text {ArcTan}(a x)^3}{2 a^4 c}-\frac {3 i \text {ArcTan}(a x)^2}{2 a^4 c}+\frac {\text {ArcTan}(a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c}-\frac {3 \text {ArcTan}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^4 c}-\frac {3 i \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{2 a^4 c}-\frac {3 i \text {Li}_4\left (1-\frac {2}{i a x+1}\right )}{4 a^4 c}-\frac {3 x \text {ArcTan}(a x)^2}{2 a^3 c}+\frac {x^2 \text {ArcTan}(a x)^3}{2 a^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[a*x]^3)/(c + a^2*c*x^2),x]

[Out]

(((-3*I)/2)*ArcTan[a*x]^2)/(a^4*c) - (3*x*ArcTan[a*x]^2)/(2*a^3*c) + ArcTan[a*x]^3/(2*a^4*c) + (x^2*ArcTan[a*x
]^3)/(2*a^2*c) + ((I/4)*ArcTan[a*x]^4)/(a^4*c) - (3*ArcTan[a*x]*Log[2/(1 + I*a*x)])/(a^4*c) + (ArcTan[a*x]^3*L
og[2/(1 + I*a*x)])/(a^4*c) - (((3*I)/2)*PolyLog[2, 1 - 2/(1 + I*a*x)])/(a^4*c) + (((3*I)/2)*ArcTan[a*x]^2*Poly
Log[2, 1 - 2/(1 + I*a*x)])/(a^4*c) + (3*ArcTan[a*x]*PolyLog[3, 1 - 2/(1 + I*a*x)])/(2*a^4*c) - (((3*I)/4)*Poly
Log[4, 1 - 2/(1 + I*a*x)])/(a^4*c)

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

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 4930

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

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 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 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 5036

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 5040

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*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 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 5118

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[I*(a +
b*ArcTan[c*x])^p*(PolyLog[k + 1, u]/(2*c*d)), x] - Dist[b*p*(I/2), Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[k
+ 1, u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 -
2*(I/(I - c*x)))^2, 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 {x^3 \tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx &=-\frac {\int \frac {x \tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx}{a^2}+\frac {\int x \tan ^{-1}(a x)^3 \, dx}{a^2 c}\\ &=\frac {x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac {i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac {\int \frac {\tan ^{-1}(a x)^3}{i-a x} \, dx}{a^3 c}-\frac {3 \int \frac {x^2 \tan ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{2 a c}\\ &=\frac {x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac {i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c}-\frac {3 \int \tan ^{-1}(a x)^2 \, dx}{2 a^3 c}+\frac {3 \int \frac {\tan ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{2 a^3 c}-\frac {3 \int \frac {\tan ^{-1}(a x)^2 \log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c}\\ &=-\frac {3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac {\tan ^{-1}(a x)^3}{2 a^4 c}+\frac {x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac {i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c}+\frac {3 i \tan ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}-\frac {(3 i) \int \frac {\tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c}+\frac {3 \int \frac {x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{a^2 c}\\ &=-\frac {3 i \tan ^{-1}(a x)^2}{2 a^4 c}-\frac {3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac {\tan ^{-1}(a x)^3}{2 a^4 c}+\frac {x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac {i \tan ^{-1}(a x)^4}{4 a^4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c}+\frac {3 i \tan ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}+\frac {3 \tan ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}-\frac {3 \int \frac {\text {Li}_3\left (1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{2 a^3 c}-\frac {3 \int \frac {\tan ^{-1}(a x)}{i-a x} \, dx}{a^3 c}\\ &=-\frac {3 i \tan ^{-1}(a x)^2}{2 a^4 c}-\frac {3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac {\tan ^{-1}(a x)^3}{2 a^4 c}+\frac {x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac {i \tan ^{-1}(a x)^4}{4 a^4 c}-\frac {3 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c}+\frac {3 i \tan ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}+\frac {3 \tan ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}-\frac {3 i \text {Li}_4\left (1-\frac {2}{1+i a x}\right )}{4 a^4 c}+\frac {3 \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c}\\ &=-\frac {3 i \tan ^{-1}(a x)^2}{2 a^4 c}-\frac {3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac {\tan ^{-1}(a x)^3}{2 a^4 c}+\frac {x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac {i \tan ^{-1}(a x)^4}{4 a^4 c}-\frac {3 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c}+\frac {3 i \tan ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}+\frac {3 \tan ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}-\frac {3 i \text {Li}_4\left (1-\frac {2}{1+i a x}\right )}{4 a^4 c}-\frac {(3 i) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )}{a^4 c}\\ &=-\frac {3 i \tan ^{-1}(a x)^2}{2 a^4 c}-\frac {3 x \tan ^{-1}(a x)^2}{2 a^3 c}+\frac {\tan ^{-1}(a x)^3}{2 a^4 c}+\frac {x^2 \tan ^{-1}(a x)^3}{2 a^2 c}+\frac {i \tan ^{-1}(a x)^4}{4 a^4 c}-\frac {3 \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c}-\frac {3 i \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}+\frac {3 i \tan ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}+\frac {3 \tan ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1+i a x}\right )}{2 a^4 c}-\frac {3 i \text {Li}_4\left (1-\frac {2}{1+i a x}\right )}{4 a^4 c}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 162, normalized size = 0.62 \begin {gather*} \frac {6 i \text {ArcTan}(a x)^2-6 a x \text {ArcTan}(a x)^2+2 \left (1+a^2 x^2\right ) \text {ArcTan}(a x)^3-i \text {ArcTan}(a x)^4-12 \text {ArcTan}(a x) \log \left (1+e^{2 i \text {ArcTan}(a x)}\right )+4 \text {ArcTan}(a x)^3 \log \left (1+e^{2 i \text {ArcTan}(a x)}\right )-6 i \left (-1+\text {ArcTan}(a x)^2\right ) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(a x)}\right )+6 \text {ArcTan}(a x) \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(a x)}\right )+3 i \text {PolyLog}\left (4,-e^{2 i \text {ArcTan}(a x)}\right )}{4 a^4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcTan[a*x]^3)/(c + a^2*c*x^2),x]

[Out]

((6*I)*ArcTan[a*x]^2 - 6*a*x*ArcTan[a*x]^2 + 2*(1 + a^2*x^2)*ArcTan[a*x]^3 - I*ArcTan[a*x]^4 - 12*ArcTan[a*x]*
Log[1 + E^((2*I)*ArcTan[a*x])] + 4*ArcTan[a*x]^3*Log[1 + E^((2*I)*ArcTan[a*x])] - (6*I)*(-1 + ArcTan[a*x]^2)*P
olyLog[2, -E^((2*I)*ArcTan[a*x])] + 6*ArcTan[a*x]*PolyLog[3, -E^((2*I)*ArcTan[a*x])] + (3*I)*PolyLog[4, -E^((2
*I)*ArcTan[a*x])])/(4*a^4*c)

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Maple [A]
time = 73.76, size = 259, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {-\frac {i \arctan \left (a x \right )^{4}}{4 c}+\frac {\arctan \left (a x \right )^{2} \left (-i \arctan \left (a x \right )+\arctan \left (a x \right ) a x -3\right ) \left (a x +i\right )}{2 c}+\frac {\arctan \left (a x \right )^{3} \ln \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )}{c}-\frac {3 i \arctan \left (a x \right )^{2} \polylog \left (2, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{2 c}+\frac {3 \arctan \left (a x \right ) \polylog \left (3, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{2 c}+\frac {3 i \polylog \left (4, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{4 c}+\frac {3 i \arctan \left (a x \right )^{2}}{c}-\frac {3 \arctan \left (a x \right ) \ln \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )}{c}+\frac {3 i \polylog \left (2, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{2 c}}{a^{4}}\)	\(259\)
default	\(\frac {-\frac {i \arctan \left (a x \right )^{4}}{4 c}+\frac {\arctan \left (a x \right )^{2} \left (-i \arctan \left (a x \right )+\arctan \left (a x \right ) a x -3\right ) \left (a x +i\right )}{2 c}+\frac {\arctan \left (a x \right )^{3} \ln \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )}{c}-\frac {3 i \arctan \left (a x \right )^{2} \polylog \left (2, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{2 c}+\frac {3 \arctan \left (a x \right ) \polylog \left (3, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{2 c}+\frac {3 i \polylog \left (4, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{4 c}+\frac {3 i \arctan \left (a x \right )^{2}}{c}-\frac {3 \arctan \left (a x \right ) \ln \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )}{c}+\frac {3 i \polylog \left (2, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{2 c}}{a^{4}}\)	\(259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(a*x)^3/(a^2*c*x^2+c),x,method=_RETURNVERBOSE)

[Out]

1/a^4*(-1/4*I*arctan(a*x)^4/c+1/2/c*arctan(a*x)^2*(-I*arctan(a*x)+arctan(a*x)*a*x-3)*(I+a*x)+1/c*arctan(a*x)^3
*ln((1+I*a*x)^2/(a^2*x^2+1)+1)-3/2*I/c*arctan(a*x)^2*polylog(2,-(1+I*a*x)^2/(a^2*x^2+1))+3/2/c*arctan(a*x)*pol
ylog(3,-(1+I*a*x)^2/(a^2*x^2+1))+3/4*I/c*polylog(4,-(1+I*a*x)^2/(a^2*x^2+1))+3*I/c*arctan(a*x)^2-3/c*arctan(a*
x)*ln((1+I*a*x)^2/(a^2*x^2+1)+1)+3/2*I/c*polylog(2,-(1+I*a*x)^2/(a^2*x^2+1)))

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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(x^3*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="maxima")

[Out]

integrate(x^3*arctan(a*x)^3/(a^2*c*x^2 + c), x)

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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(x^3*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

integral(x^3*arctan(a*x)^3/(a^2*c*x^2 + c), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {x^{3} \operatorname {atan}^{3}{\left (a x \right )}}{a^{2} x^{2} + 1}\, dx}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(a*x)**3/(a**2*c*x**2+c),x)

[Out]

Integral(x**3*atan(a*x)**3/(a**2*x**2 + 1), x)/c

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Giac [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(x^3*arctan(a*x)^3/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,{\mathrm {atan}\left (a\,x\right )}^3}{c\,a^2\,x^2+c} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*atan(a*x)^3)/(c + a^2*c*x^2),x)

[Out]

int((x^3*atan(a*x)^3)/(c + a^2*c*x^2), x)

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