3.2.14 \(\int x^8 (a+b \arctan (c x^3))^2 \, dx\) [114]

3.2.14.1 Optimal result

Integrand size = 16, antiderivative size = 154 \[ \int x^8 \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\frac {b^2 x^3}{9 c^2}-\frac {b^2 \arctan \left (c x^3\right )}{9 c^3}-\frac {b x^6 \left (a+b \arctan \left (c x^3\right )\right )}{9 c}-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{9 c^3}+\frac {1}{9} x^9 \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {2 b \left (a+b \arctan \left (c x^3\right )\right ) \log \left (\frac {2}{1+i c x^3}\right )}{9 c^3}-\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x^3}\right )}{9 c^3} \]

output

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

3.2.14.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.92 \[ \int x^8 \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\frac {b^2 c x^3-a b c^2 x^6+a^2 c^3 x^9+b^2 \left (i+c^3 x^9\right ) \arctan \left (c x^3\right )^2-b \arctan \left (c x^3\right ) \left (b+b c^2 x^6-2 a c^3 x^9+2 b \log \left (1+e^{2 i \arctan \left (c x^3\right )}\right )\right )+a b \log \left (1+c^2 x^6\right )+i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (c x^3\right )}\right )}{9 c^3} \]

input

Integrate[x^8*(a + b*ArcTan[c*x^3])^2,x]
 

output

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

3.2.14.3 Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5363, 5361, 5451, 5361, 262, 216, 5455, 5379, 2849, 2752}

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.

\(\displaystyle \int x^8 \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx\)

\(\Big \downarrow \) 5363

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5451

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 5455

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

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

input

Int[x^8*(a + b*ArcTan[c*x^3])^2,x]
 

output

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

3.2.14.3.1 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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 2752

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

rule 2849

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp 
[-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 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 5379

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] + Simp[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 5451

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[f^2/e   Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x] 
)^p, x], x] - Simp[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 5455

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

3.2.14.4 Maple [C] (warning: unable to verify)

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

Time = 3.69 (sec) , antiderivative size = 11449, normalized size of antiderivative = 74.34

method	result	size
default	\(\text {Expression too large to display}\)	\(11449\)
parts	\(\text {Expression too large to display}\)	\(11449\)

input

int(x^8*(a+b*arctan(c*x^3))^2,x,method=_RETURNVERBOSE)
 

output

result too large to display
 

3.2.14.5 Fricas [F]

\[ \int x^8 \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2} x^{8} \,d x } \]

input

integrate(x^8*(a+b*arctan(c*x^3))^2,x, algorithm="fricas")
 

output

integral(b^2*x^8*arctan(c*x^3)^2 + 2*a*b*x^8*arctan(c*x^3) + a^2*x^8, x)
 

3.2.14.6 Sympy [F]

\[ \int x^8 \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\int x^{8} \left (a + b \operatorname {atan}{\left (c x^{3} \right )}\right )^{2}\, dx \]

input

integrate(x**8*(a+b*atan(c*x**3))**2,x)
 

output

Integral(x**8*(a + b*atan(c*x**3))**2, x)
 

3.2.14.7 Maxima [F]

\[ \int x^8 \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2} x^{8} \,d x } \]

input

integrate(x^8*(a+b*arctan(c*x^3))^2,x, algorithm="maxima")
 

output

1/9*a^2*x^9 + 1/9*(2*x^9*arctan(c*x^3) - (x^6/c^2 - log(c^2*x^6 + 1)/c^4)* 
c)*a*b + 1/144*(4*x^9*arctan(c*x^3)^2 - x^9*log(c^2*x^6 + 1)^2 + 144*integ 
rate(1/48*(4*c^2*x^14*log(c^2*x^6 + 1) - 8*c*x^11*arctan(c*x^3) + 36*(c^2* 
x^14 + x^8)*arctan(c*x^3)^2 + 3*(c^2*x^14 + x^8)*log(c^2*x^6 + 1)^2)/(c^2* 
x^6 + 1), x))*b^2
 

3.2.14.8 Giac [F]

\[ \int x^8 \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2} x^{8} \,d x } \]

input

integrate(x^8*(a+b*arctan(c*x^3))^2,x, algorithm="giac")
 

output

integrate((b*arctan(c*x^3) + a)^2*x^8, x)
 

3.2.14.9 Mupad [F(-1)]

Timed out. \[ \int x^8 \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\int x^8\,{\left (a+b\,\mathrm {atan}\left (c\,x^3\right )\right )}^2 \,d x \]

input

int(x^8*(a + b*atan(c*x^3))^2,x)
 

output

int(x^8*(a + b*atan(c*x^3))^2, x)