3.2.26 \(\int \frac {(a+b \text {ArcTan}(c x^3))^3}{x^7} \, dx\) [126]

Optimal. Leaf size=146 \[ -\frac {1}{2} i b c^2 \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2-\frac {b c \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2}{2 x^3}-\frac {1}{6} c^2 \left (a+b \text {ArcTan}\left (c x^3\right )\right )^3-\frac {\left (a+b \text {ArcTan}\left (c x^3\right )\right )^3}{6 x^6}+b^2 c^2 \left (a+b \text {ArcTan}\left (c x^3\right )\right ) \log \left (2-\frac {2}{1-i c x^3}\right )-\frac {1}{2} i b^3 c^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x^3}\right ) \]

[Out]

-1/2*I*b*c^2*(a+b*arctan(c*x^3))^2-1/2*b*c*(a+b*arctan(c*x^3))^2/x^3-1/6*c^2*(a+b*arctan(c*x^3))^3-1/6*(a+b*ar
ctan(c*x^3))^3/x^6+b^2*c^2*(a+b*arctan(c*x^3))*ln(2-2/(1-I*c*x^3))-1/2*I*b^3*c^2*polylog(2,-1+2/(1-I*c*x^3))

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Rubi [A]
time = 0.23, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4948, 4946, 5038, 5044, 4988, 2497, 5004} \begin {gather*} b^2 c^2 \log \left (2-\frac {2}{1-i c x^3}\right ) \left (a+b \text {ArcTan}\left (c x^3\right )\right )-\frac {1}{2} i b c^2 \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2-\frac {1}{6} c^2 \left (a+b \text {ArcTan}\left (c x^3\right )\right )^3-\frac {b c \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2}{2 x^3}-\frac {\left (a+b \text {ArcTan}\left (c x^3\right )\right )^3}{6 x^6}-\frac {1}{2} i b^3 c^2 \text {Li}_2\left (\frac {2}{1-i c x^3}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^3])^3/x^7,x]

[Out]

(-1/2*I)*b*c^2*(a + b*ArcTan[c*x^3])^2 - (b*c*(a + b*ArcTan[c*x^3])^2)/(2*x^3) - (c^2*(a + b*ArcTan[c*x^3])^3)
/6 - (a + b*ArcTan[c*x^3])^3/(6*x^6) + b^2*c^2*(a + b*ArcTan[c*x^3])*Log[2 - 2/(1 - I*c*x^3)] - (I/2)*b^3*c^2*
PolyLog[2, -1 + 2/(1 - I*c*x^3)]

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 4948

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTan[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Sim
plify[(m + 1)/n]]

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}\left (c x^3\right )\right )^3}{x^7} \, dx &=\int \left (\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{8 x^7}+\frac {3 i b \left (-2 i a+b \log \left (1-i c x^3\right )\right )^2 \log \left (1+i c x^3\right )}{8 x^7}-\frac {3 i b^2 \left (-2 i a+b \log \left (1-i c x^3\right )\right ) \log ^2\left (1+i c x^3\right )}{8 x^7}+\frac {i b^3 \log ^3\left (1+i c x^3\right )}{8 x^7}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{x^7} \, dx+\frac {1}{8} (3 i b) \int \frac {\left (-2 i a+b \log \left (1-i c x^3\right )\right )^2 \log \left (1+i c x^3\right )}{x^7} \, dx-\frac {1}{8} \left (3 i b^2\right ) \int \frac {\left (-2 i a+b \log \left (1-i c x^3\right )\right ) \log ^2\left (1+i c x^3\right )}{x^7} \, dx+\frac {1}{8} \left (i b^3\right ) \int \frac {\log ^3\left (1+i c x^3\right )}{x^7} \, dx\\ &=\frac {1}{24} \text {Subst}\left (\int \frac {(2 a+i b \log (1-i c x))^3}{x^3} \, dx,x,x^3\right )+\frac {1}{8} (i b) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{24} \left (i b^3\right ) \text {Subst}\left (\int \frac {\log ^3(1+i c x)}{x^3} \, dx,x,x^3\right )\\ &=-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} (i b) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} (b c) \text {Subst}\left (\int \frac {(2 a+i b \log (1-i c x))^2}{x^2 (1-i c x)} \, dx,x,x^3\right )-\frac {1}{16} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(1+i c x)}{x^2 (1+i c x)} \, dx,x,x^3\right )\\ &=-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{16} (i b) \text {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{x \left (-\frac {i}{c}+\frac {i x}{c}\right )^2} \, dx,x,1-i c x^3\right )+\frac {1}{8} (i b) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} \left (i b^3\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{x \left (\frac {i}{c}-\frac {i x}{c}\right )^2} \, dx,x,1+i c x^3\right )\\ &=-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{16} (i b) \text {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{\left (-\frac {i}{c}+\frac {i x}{c}\right )^2} \, dx,x,1-i c x^3\right )+\frac {1}{8} (i b) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} \left (i b^3\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{\left (\frac {i}{c}-\frac {i x}{c}\right )^2} \, dx,x,1+i c x^3\right )-\frac {1}{16} (b c) \text {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{x \left (-\frac {i}{c}+\frac {i x}{c}\right )} \, dx,x,1-i c x^3\right )+\frac {1}{16} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{x \left (\frac {i}{c}-\frac {i x}{c}\right )} \, dx,x,1+i c x^3\right )\\ &=-\frac {b c \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{16 x^3}-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}+\frac {b^3 c \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{16 x^3}-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} (i b) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{16} (b c) \text {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{-\frac {i}{c}+\frac {i x}{c}} \, dx,x,1-i c x^3\right )+\frac {1}{8} \left (i b^2 c\right ) \text {Subst}\left (\int \frac {2 a+i b \log (x)}{-\frac {i}{c}+\frac {i x}{c}} \, dx,x,1-i c x^3\right )+\frac {1}{16} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{\frac {i}{c}-\frac {i x}{c}} \, dx,x,1+i c x^3\right )-\frac {1}{8} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log (x)}{\frac {i}{c}-\frac {i x}{c}} \, dx,x,1+i c x^3\right )-\frac {1}{16} \left (i b c^2\right ) \text {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{x} \, dx,x,1-i c x^3\right )-\frac {1}{16} \left (i b^3 c^2\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{x} \, dx,x,1+i c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} i b c^2 \log \left (i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}+\frac {b^3 c \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{16 x^3}+\frac {1}{16} i b^3 c^2 \log \left (-i c x^3\right ) \log ^2\left (1+i c x^3\right )-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} i b^3 c^2 \text {Li}_2\left (-i c x^3\right )+\frac {1}{8} (i b) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log (x)}{-\frac {i}{c}+\frac {i x}{c}} \, dx,x,1-i c x^3\right )-\frac {1}{16} c^2 \text {Subst}\left (\int x^2 \, dx,x,2 a+i b \log \left (1-i c x^3\right )\right )+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (1-x) (2 a+i b \log (x))}{x} \, dx,x,1-i c x^3\right )-\frac {1}{16} \left (i b^3 c^2\right ) \text {Subst}\left (\int x^2 \, dx,x,\log \left (1+i c x^3\right )\right )-\frac {1}{8} \left (i b^3 c^2\right ) \text {Subst}\left (\int \frac {\log (1-x) \log (x)}{x} \, dx,x,1+i c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} i b c^2 \log \left (i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2-\frac {1}{48} c^2 \left (2 a+i b \log \left (1-i c x^3\right )\right )^3-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}+\frac {b^3 c \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{16 x^3}+\frac {1}{16} i b^3 c^2 \log \left (-i c x^3\right ) \log ^2\left (1+i c x^3\right )-\frac {1}{48} i b^3 c^2 \log ^3\left (1+i c x^3\right )-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} i b^3 c^2 \text {Li}_2\left (-i c x^3\right )-\frac {1}{8} i b^3 c^2 \text {Li}_2\left (i c x^3\right )-\frac {1}{8} b^2 c^2 \left (2 a+i b \log \left (1-i c x^3\right )\right ) \text {Li}_2\left (1-i c x^3\right )+\frac {1}{8} i b^3 c^2 \log \left (1+i c x^3\right ) \text {Li}_2\left (1+i c x^3\right )+\frac {1}{8} (i b) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{8} \left (i b^3 c^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-i c x^3\right )-\frac {1}{8} \left (i b^3 c^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1+i c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} i b c^2 \log \left (i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2-\frac {1}{48} c^2 \left (2 a+i b \log \left (1-i c x^3\right )\right )^3-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}+\frac {b^3 c \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{16 x^3}+\frac {1}{16} i b^3 c^2 \log \left (-i c x^3\right ) \log ^2\left (1+i c x^3\right )-\frac {1}{48} i b^3 c^2 \log ^3\left (1+i c x^3\right )-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} i b^3 c^2 \text {Li}_2\left (-i c x^3\right )-\frac {1}{8} i b^3 c^2 \text {Li}_2\left (i c x^3\right )-\frac {1}{8} b^2 c^2 \left (2 a+i b \log \left (1-i c x^3\right )\right ) \text {Li}_2\left (1-i c x^3\right )+\frac {1}{8} i b^3 c^2 \log \left (1+i c x^3\right ) \text {Li}_2\left (1+i c x^3\right )+\frac {1}{8} i b^3 c^2 \text {Li}_3\left (1-i c x^3\right )-\frac {1}{8} i b^3 c^2 \text {Li}_3\left (1+i c x^3\right )+\frac {1}{8} (i b) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \text {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 196, normalized size = 1.34 \begin {gather*} -\frac {3 b^2 \left (a+a c^2 x^6+b c x^3 \left (1+i c x^3\right )\right ) \text {ArcTan}\left (c x^3\right )^2+b^3 \left (1+c^2 x^6\right ) \text {ArcTan}\left (c x^3\right )^3+3 b \text {ArcTan}\left (c x^3\right ) \left (a \left (a+2 b c x^3+a c^2 x^6\right )-2 b^2 c^2 x^6 \log \left (1-e^{2 i \text {ArcTan}\left (c x^3\right )}\right )\right )+a \left (a \left (a+3 b c x^3\right )-6 b^2 c^2 x^6 \log \left (\frac {c x^3}{\sqrt {1+c^2 x^6}}\right )\right )+3 i b^3 c^2 x^6 \text {PolyLog}\left (2,e^{2 i \text {ArcTan}\left (c x^3\right )}\right )}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^3])^3/x^7,x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctan \left (c \,x^{3}\right )\right )^{3}}{x^{7}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^3))^3/x^7,x)

[Out]

int((a+b*arctan(c*x^3))^3/x^7,x)

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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(c*x^3))^3/x^7,x, algorithm="maxima")

[Out]

-1/2*((c*arctan(c*x^3) + 1/x^3)*c + arctan(c*x^3)/x^6)*a^2*b + 1/2*((arctan(c*x^3)^2 - log(c^2*x^6 + 1) + 6*lo
g(x))*c^2 - 2*(c*arctan(c*x^3) + 1/x^3)*c*arctan(c*x^3))*a*b^2 - 1/2*a*b^2*arctan(c*x^3)^2/x^6 + 1/192*(192*x^
6*integrate(-1/64*(12*c^2*x^6*arctan(c*x^3)*log(c^2*x^6 + 1) - 12*c*x^3*arctan(c*x^3)^2 - 56*(c^2*x^6 + 1)*arc
tan(c*x^3)^3 + 3*(c*x^3 - 2*(c^2*x^6 + 1)*arctan(c*x^3))*log(c^2*x^6 + 1)^2)/(c^2*x^13 + x^7), x) - 4*arctan(c
*x^3)^3 + 3*arctan(c*x^3)*log(c^2*x^6 + 1)^2)*b^3/x^6 - 1/6*a^3/x^6

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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(c*x^3))^3/x^7,x, algorithm="fricas")

[Out]

integral((b^3*arctan(c*x^3)^3 + 3*a*b^2*arctan(c*x^3)^2 + 3*a^2*b*arctan(c*x^3) + a^3)/x^7, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atan}{\left (c x^{3} \right )}\right )^{3}}{x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**3))**3/x**7,x)

[Out]

Integral((a + b*atan(c*x**3))**3/x**7, x)

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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((a+b*arctan(c*x^3))^3/x^7,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^3) + a)^3/x^7, x)

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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\,x^3\right )\right )}^3}{x^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))^3/x^7,x)

[Out]

int((a + b*atan(c*x^3))^3/x^7, x)

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