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

3.2.20.1 Optimal result

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

output

-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))
 

3.2.20.2 Mathematica [A] (verified)

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

input

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

output

-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
 

3.2.20.3 Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5363, 5361, 5453, 5361, 264, 216, 5459, 5403, 2897}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5363

\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{x^{12}}dx^3\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \int \frac {a+b \arctan \left (c x^3\right )}{x^9 \left (c^2 x^6+1\right )}dx^3-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 5453

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (\int \frac {a+b \arctan \left (c x^3\right )}{x^9}dx^3-c^2 \int \frac {a+b \arctan \left (c x^3\right )}{x^3 \left (c^2 x^6+1\right )}dx^3\right )-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \left (-\int \frac {a+b \arctan \left (c x^3\right )}{x^3 \left (c^2 x^6+1\right )}dx^3\right )+\frac {1}{2} b c \int \frac {1}{x^6 \left (c^2 x^6+1\right )}dx^3-\frac {a+b \arctan \left (c x^3\right )}{2 x^6}\right )-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \left (-\int \frac {a+b \arctan \left (c x^3\right )}{x^3 \left (c^2 x^6+1\right )}dx^3\right )+\frac {1}{2} b c \left (c^2 \left (-\int \frac {1}{c^2 x^6+1}dx^3\right )-\frac {1}{x^3}\right )-\frac {a+b \arctan \left (c x^3\right )}{2 x^6}\right )-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \left (-\int \frac {a+b \arctan \left (c x^3\right )}{x^3 \left (c^2 x^6+1\right )}dx^3\right )-\frac {a+b \arctan \left (c x^3\right )}{2 x^6}+\frac {1}{2} b c \left (-c \arctan \left (c x^3\right )-\frac {1}{x^3}\right )\right )-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 5459

\(\displaystyle \frac {1}{3} \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^9}+\frac {2}{3} b c \left (-\left (c^2 \left (i \int \frac {a+b \arctan \left (c x^3\right )}{x^3 \left (c x^3+i\right )}dx^3-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b}\right )\right )-\frac {a+b \arctan \left (c x^3\right )}{2 x^6}+\frac {1}{2} b c \left (-c \arctan \left (c x^3\right )-\frac {1}{x^3}\right )\right )\right )\)

\(\Big \downarrow \) 5403

\(\displaystyle \frac {1}{3} \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^9}+\frac {2}{3} b c \left (-\left (c^2 \left (i \left (i b c \int \frac {\log \left (2-\frac {2}{1-i c x^3}\right )}{c^2 x^6+1}dx^3-i \log \left (2-\frac {2}{1-i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )\right )-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b}\right )\right )-\frac {a+b \arctan \left (c x^3\right )}{2 x^6}+\frac {1}{2} b c \left (-c \arctan \left (c x^3\right )-\frac {1}{x^3}\right )\right )\right )\)

\(\Big \downarrow \) 2897

\(\displaystyle \frac {1}{3} \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{3 x^9}+\frac {2}{3} b c \left (-\left (c^2 \left (i \left (-i \log \left (2-\frac {2}{1-i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{1-i c x^3}-1\right )\right )-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b}\right )\right )-\frac {a+b \arctan \left (c x^3\right )}{2 x^6}+\frac {1}{2} b c \left (-c \arctan \left (c x^3\right )-\frac {1}{x^3}\right )\right )\right )\)

input

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

output

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

3.2.20.3.1 Defintions of rubi rules used

rule 216

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

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 2897

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[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

rule 5361

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] - Simp[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 5363

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> 
Simp[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[Simplif 
y[(m + 1)/n]]
 

rule 5403

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] - Si 
mp[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 5453

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], 
 x] - Simp[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 5459

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

3.2.20.4 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	\(\text {Expression too large to display}\)	\(11496\)

input

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

output

result too large to display
 

3.2.20.5 Fricas [F]

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

input

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

output

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

3.2.20.6 Sympy [F]

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

input

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

output

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

3.2.20.7 Maxima [F]

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

input

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

output

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)*log( 
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
 

3.2.20.8 Giac [F]

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

input

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

output

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

3.2.20.9 Mupad [F(-1)]

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

input

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

output

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