\(\int \frac {(a+b \arctan (c x^3))^2}{x^{10}} \, dx\) [120]

Optimal result

Integrand size = 16, antiderivative size = 154 \[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{10}} \, dx=-\frac {b^2 c^2}{9 x^3}-\frac {1}{9} b^2 c^3 \arctan \left (c x^3\right )-\frac {b c \left (a+b \arctan \left (c x^3\right )\right )}{9 x^6}+\frac {1}{9} i c^3 \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{9 x^9}-\frac {2}{9} b c^3 \left (a+b \arctan \left (c x^3\right )\right ) \log \left (2-\frac {2}{1-i c x^3}\right )+\frac {1}{9} i b^2 c^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x^3}\right ) \]

[Out]

-1/9*b^2*c^2/x^3-1/9*b^2*c^3*arctan(c*x^3)-1/9*b*c*(a+b*arctan(c*x^3))/x^6+1/9*I*c^3*(a+b*arctan(c*x^3))^2-1/9
*(a+b*arctan(c*x^3))^2/x^9-2/9*b*c^3*(a+b*arctan(c*x^3))*ln(2-2/(1-I*c*x^3))+1/9*I*b^2*c^3*polylog(2,-1+2/(1-I
*c*x^3))

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4948, 4946, 5038, 331, 209, 5044, 4988, 2497} \[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{10}} \, dx=\frac {1}{9} i c^3 \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {2}{9} b c^3 \log \left (2-\frac {2}{1-i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{9 x^9}-\frac {b c \left (a+b \arctan \left (c x^3\right )\right )}{9 x^6}-\frac {1}{9} b^2 c^3 \arctan \left (c x^3\right )+\frac {1}{9} i b^2 c^3 \operatorname {PolyLog}\left (2,\frac {2}{1-i c x^3}-1\right )-\frac {b^2 c^2}{9 x^3} \]

[In]

Int[(a + b*ArcTan[c*x^3])^2/x^10,x]

[Out]

-1/9*(b^2*c^2)/x^3 - (b^2*c^3*ArcTan[c*x^3])/9 - (b*c*(a + b*ArcTan[c*x^3]))/(9*x^6) + (I/9)*c^3*(a + b*ArcTan
[c*x^3])^2 - (a + b*ArcTan[c*x^3])^2/(9*x^9) - (2*b*c^3*(a + b*ArcTan[c*x^3])*Log[2 - 2/(1 - I*c*x^3)])/9 + (I
/9)*b^2*c^3*PolyLog[2, -1 + 2/(1 - I*c*x^3)]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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 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 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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {(a+b \arctan (c x))^2}{x^4} \, dx,x,x^3\right ) \\ & = -\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{9 x^9}+\frac {1}{9} (2 b c) \text {Subst}\left (\int \frac {a+b \arctan (c x)}{x^3 \left (1+c^2 x^2\right )} \, dx,x,x^3\right ) \\ & = -\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{9 x^9}+\frac {1}{9} (2 b c) \text {Subst}\left (\int \frac {a+b \arctan (c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{9} \left (2 b c^3\right ) \text {Subst}\left (\int \frac {a+b \arctan (c x)}{x \left (1+c^2 x^2\right )} \, dx,x,x^3\right ) \\ & = -\frac {b c \left (a+b \arctan \left (c x^3\right )\right )}{9 x^6}+\frac {1}{9} i c^3 \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{9 x^9}+\frac {1}{9} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx,x,x^3\right )-\frac {1}{9} \left (2 i b c^3\right ) \text {Subst}\left (\int \frac {a+b \arctan (c x)}{x (i+c x)} \, dx,x,x^3\right ) \\ & = -\frac {b^2 c^2}{9 x^3}-\frac {b c \left (a+b \arctan \left (c x^3\right )\right )}{9 x^6}+\frac {1}{9} i c^3 \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{9 x^9}-\frac {2}{9} b c^3 \left (a+b \arctan \left (c x^3\right )\right ) \log \left (2-\frac {2}{1-i c x^3}\right )-\frac {1}{9} \left (b^2 c^4\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x^2} \, dx,x,x^3\right )+\frac {1}{9} \left (2 b^2 c^4\right ) \text {Subst}\left (\int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx,x,x^3\right ) \\ & = -\frac {b^2 c^2}{9 x^3}-\frac {1}{9} b^2 c^3 \arctan \left (c x^3\right )-\frac {b c \left (a+b \arctan \left (c x^3\right )\right )}{9 x^6}+\frac {1}{9} i c^3 \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{9 x^9}-\frac {2}{9} b c^3 \left (a+b \arctan \left (c x^3\right )\right ) \log \left (2-\frac {2}{1-i c x^3}\right )+\frac {1}{9} i b^2 c^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{10}} \, dx=-\frac {a^2+a b c x^3+b^2 c^2 x^6+b^2 \left (1-i c^3 x^9\right ) \arctan \left (c x^3\right )^2+b \arctan \left (c x^3\right ) \left (2 a+b c x^3+b c^3 x^9+2 b c^3 x^9 \log \left (1-e^{2 i \arctan \left (c x^3\right )}\right )\right )+2 a b c^3 x^9 \log \left (c x^3\right )-a b c^3 x^9 \log \left (1+c^2 x^6\right )-i b^2 c^3 x^9 \operatorname {PolyLog}\left (2,e^{2 i \arctan \left (c x^3\right )}\right )}{9 x^9} \]

[In]

Integrate[(a + b*ArcTan[c*x^3])^2/x^10,x]

[Out]

-1/9*(a^2 + a*b*c*x^3 + b^2*c^2*x^6 + b^2*(1 - I*c^3*x^9)*ArcTan[c*x^3]^2 + b*ArcTan[c*x^3]*(2*a + b*c*x^3 + b
*c^3*x^9 + 2*b*c^3*x^9*Log[1 - E^((2*I)*ArcTan[c*x^3])]) + 2*a*b*c^3*x^9*Log[c*x^3] - a*b*c^3*x^9*Log[1 + c^2*
x^6] - I*b^2*c^3*x^9*PolyLog[2, E^((2*I)*ArcTan[c*x^3])])/x^9

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 5.34 (sec) , antiderivative size = 11496, normalized size of antiderivative = 74.65

method	result	size
default	\(\text {Expression too large to display}\)	\(11496\)
parts	\(\text {Expression too large to display}\)	\(11496\)

[In]

int((a+b*arctan(c*x^3))^2/x^10,x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{10}} \, dx=\int { \frac {{\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2}}{x^{10}} \,d x } \]

[In]

integrate((a+b*arctan(c*x^3))^2/x^10,x, algorithm="fricas")

[Out]

integral((b^2*arctan(c*x^3)^2 + 2*a*b*arctan(c*x^3) + a^2)/x^10, x)

Sympy [F]

\[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{10}} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x^{3} \right )}\right )^{2}}{x^{10}}\, dx \]

[In]

integrate((a+b*atan(c*x**3))**2/x**10,x)

[Out]

Integral((a + b*atan(c*x**3))**2/x**10, x)

Maxima [F]

\[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{10}} \, dx=\int { \frac {{\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2}}{x^{10}} \,d x } \]

[In]

integrate((a+b*arctan(c*x^3))^2/x^10,x, algorithm="maxima")

[Out]

1/9*((c^2*log(c^2*x^6 + 1) - c^2*log(x^6) - 1/x^6)*c - 2*arctan(c*x^3)/x^9)*a*b + 1/144*(144*x^9*integrate(-1/
48*(4*c^2*x^6*log(c^2*x^6 + 1) - 8*c*x^3*arctan(c*x^3) - 36*(c^2*x^6 + 1)*arctan(c*x^3)^2 - 3*(c^2*x^6 + 1)*lo
g(c^2*x^6 + 1)^2)/(c^2*x^16 + x^10), x) - 4*arctan(c*x^3)^2 + log(c^2*x^6 + 1)^2)*b^2/x^9 - 1/9*a^2/x^9

Giac [F]

\[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{10}} \, dx=\int { \frac {{\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2}}{x^{10}} \,d x } \]

[In]

integrate((a+b*arctan(c*x^3))^2/x^10,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^3) + a)^2/x^10, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{10}} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x^3\right )\right )}^2}{x^{10}} \,d x \]

[In]

int((a + b*atan(c*x^3))^2/x^10,x)

[Out]

int((a + b*atan(c*x^3))^2/x^10, x)