Integrand size = 16, antiderivative size = 147 \[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=-\frac {i b \left (a+b \arctan \left (c x^3\right )\right )^2}{2 c^2}-\frac {b x^3 \left (a+b \arctan \left (c x^3\right )\right )^2}{2 c}+\frac {\left (a+b \arctan \left (c x^3\right )\right )^3}{6 c^2}+\frac {1}{6} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {b^2 \left (a+b \arctan \left (c x^3\right )\right ) \log \left (\frac {2}{1+i c x^3}\right )}{c^2}-\frac {i b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x^3}\right )}{2 c^2} \]
-1/2*I*b*(a+b*arctan(c*x^3))^2/c^2-1/2*b*x^3*(a+b*arctan(c*x^3))^2/c+1/6*( a+b*arctan(c*x^3))^3/c^2+1/6*x^6*(a+b*arctan(c*x^3))^3-b^2*(a+b*arctan(c*x ^3))*ln(2/(1+I*c*x^3))/c^2-1/2*I*b^3*polylog(2,1-2/(1+I*c*x^3))/c^2
Time = 0.14 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.16 \[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\frac {3 b^2 \left (a+a c^2 x^6+b \left (i-c x^3\right )\right ) \arctan \left (c x^3\right )^2+b^3 \left (1+c^2 x^6\right ) \arctan \left (c x^3\right )^3+3 b \arctan \left (c x^3\right ) \left (a \left (a-2 b c x^3+a c^2 x^6\right )-2 b^2 \log \left (1+e^{2 i \arctan \left (c x^3\right )}\right )\right )+a \left (a c x^3 \left (-3 b+a c x^3\right )+3 b^2 \log \left (1+c^2 x^6\right )\right )+3 i b^3 \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (c x^3\right )}\right )}{6 c^2} \]
(3*b^2*(a + a*c^2*x^6 + b*(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*Log[1 + E^((2*I)*ArcTan[c*x^3])]) + a*(a*c*x^3*(-3*b + a*c*x^3) + 3*b^2* Log[1 + c^2*x^6]) + (3*I)*b^3*PolyLog[2, -E^((2*I)*ArcTan[c*x^3])])/(6*c^2 )
Time = 0.88 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5363, 5361, 5451, 5345, 5419, 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.
| \(\displaystyle \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx\) |
| \(\Big \downarrow \) 5363 |
| \(\displaystyle \frac {1}{3} \int x^3 \left (a+b \arctan \left (c x^3\right )\right )^3dx^3\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {3}{2} b c \int \frac {x^6 \left (a+b \arctan \left (c x^3\right )\right )^2}{c^2 x^6+1}dx^3\right )\) |
| \(\Big \downarrow \) 5451 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {\int \left (a+b \arctan \left (c x^3\right )\right )^2dx^3}{c^2}-\frac {\int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{c^2 x^6+1}dx^3}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 5345 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {x^3 \left (a+b \arctan \left (c x^3\right )\right )^2-2 b c \int \frac {x^3 \left (a+b \arctan \left (c x^3\right )\right )}{c^2 x^6+1}dx^3}{c^2}-\frac {\int \frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{c^2 x^6+1}dx^3}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 5419 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {x^3 \left (a+b \arctan \left (c x^3\right )\right )^2-2 b c \int \frac {x^3 \left (a+b \arctan \left (c x^3\right )\right )}{c^2 x^6+1}dx^3}{c^2}-\frac {\left (a+b \arctan \left (c x^3\right )\right )^3}{3 b c^3}\right )\right )\) |
| \(\Big \downarrow \) 5455 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {3}{2} b c \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^3}{3 b c^3}+\frac {x^3 \left (a+b \arctan \left (c x^3\right )\right )^2-2 b c \left (-\frac {\int \frac {a+b \arctan \left (c x^3\right )}{i-c x^3}dx^3}{c}-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 5379 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {3}{2} b c \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^3}{3 b c^3}+\frac {x^3 \left (a+b \arctan \left (c x^3\right )\right )^2-2 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )}{c}-b \int \frac {\log \left (\frac {2}{i c x^3+1}\right )}{c^2 x^6+1}dx^3}{c}-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {3}{2} b c \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^3}{3 b c^3}+\frac {x^3 \left (a+b \arctan \left (c x^3\right )\right )^2-2 b c \left (-\frac {\frac {i b \int \frac {\log \left (\frac {2}{i c x^3+1}\right )}{1-\frac {2}{i c x^3+1}}d\frac {1}{i c x^3+1}}{c}+\frac {\log \left (\frac {2}{1+i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )}{c}}{c}-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {3}{2} b c \left (-\frac {\left (a+b \arctan \left (c x^3\right )\right )^3}{3 b c^3}+\frac {x^3 \left (a+b \arctan \left (c x^3\right )\right )^2-2 b c \left (-\frac {i \left (a+b \arctan \left (c x^3\right )\right )^2}{2 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )}{c}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x^3+1}\right )}{2 c}}{c}\right )}{c^2}\right )\right )\) |
((x^6*(a + b*ArcTan[c*x^3])^3)/2 - (3*b*c*(-1/3*(a + b*ArcTan[c*x^3])^3/(b *c^3) + (x^3*(a + b*ArcTan[c*x^3])^2 - 2*b*c*(((-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))/2)/3
3.2.22.3.1 Defintions of rubi rules used
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]
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]
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Simp[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])
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]
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]]
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 ]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> 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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 7.57 (sec) , antiderivative size = 935, normalized size of antiderivative = 6.36
| method | result | size |
| risch | \(\text {Expression too large to display}\) | \(935\) |
| default | \(\text {Expression too large to display}\) | \(11515\) |
| parts | \(\text {Expression too large to display}\) | \(11515\) |
-1/16*b^2*(I*b*c^2*x^6*ln(1-I*c*x^3)+2*a*c^2*x^6-2*b*c*x^3+I*b*ln(1-I*c*x^ 3)+2*I*b+2*a)/c^2*ln(1+I*c*x^3)^2+1/2*a*b^2/c^2*ln(c^2*x^6+1)+1/2*a^2*b/c^ 2*arctan(c*x^3)-1/48*I*b^3*x^6*ln(1-I*c*x^3)^3+1/8*I/c^2*b^3*ln(c^2*x^6+1) +1/6*a^3*x^6+1/8*I/c^2*b^3*ln(1-I*c*x^3)^2-1/2/c*a^2*b*x^3+1/8*b^3/c*x^3*l n(1-I*c*x^3)^2+3/4*I/c*b^2*Sum(2/3*(ln(x-_alpha)*ln(1-I*c*x^3)+3*c*(-1/3*l n(x-_alpha)*(ln((RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index= 1)-x+_alpha)/RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index=1))+ ln((RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index=2)-x+_alpha)/ RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index=2))+ln(1/2*(2*(I/ c)^(1/3)+x-_alpha)/(I/c)^(1/3)))/c-1/3*(dilog((RootOf(_Z^2+_Z*RootOf(c*_Z^ 3-I)+RootOf(c*_Z^3-I)^2,index=1)-x+_alpha)/RootOf(_Z^2+_Z*RootOf(c*_Z^3-I) +RootOf(c*_Z^3-I)^2,index=1))+dilog((RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootO f(c*_Z^3-I)^2,index=2)-x+_alpha)/RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c* _Z^3-I)^2,index=2))+dilog(1/2*(2*(I/c)^(1/3)+x-_alpha)/(I/c)^(1/3)))/c))*b /c,_alpha=RootOf(c*_Z^3-RootOf(_Z^2+1,index=1)))+1/4*I*b*a^2*x^6*ln(1-I*c* x^3)-1/8*a*b^2*x^6*ln(1-I*c*x^3)^2-1/8/c^2*a*b^2*ln(1-I*c*x^3)^2+1/48*I*b^ 3*(c^2*x^6+1)/c^2*ln(1+I*c*x^3)^3-1/2*I/c*a*b^2*x^3*ln(1-I*c*x^3)-1/4/c^2* b^3*arctan(c*x^3)-1/48*I/c^2*b^3*ln(1-I*c*x^3)^3+(1/16*I*b^3*(c^2*x^6+1)/c ^2*ln(1-I*c*x^3)^2+1/16*b^2*(2*a*c*x^3-b)^2/c^2/a*ln(1-I*c*x^3)-1/16*b*(4* I*a^3*c^2*x^6-8*I*a^2*b*c*x^3+4*I*ln(1-I*c*x^3)*a*b^2+4*I*a*b^2-4*ln(1-...
\[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \]
integral(b^3*x^5*arctan(c*x^3)^3 + 3*a*b^2*x^5*arctan(c*x^3)^2 + 3*a^2*b*x ^5*arctan(c*x^3) + a^3*x^5, x)
\[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int x^{5} \left (a + b \operatorname {atan}{\left (c x^{3} \right )}\right )^{3}\, dx \]
\[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \]
1/2*a*b^2*x^6*arctan(c*x^3)^2 + 1/6*a^3*x^6 + 1/2*(x^6*arctan(c*x^3) - c*( x^3/c^2 - arctan(c*x^3)/c^3))*a^2*b - 1/2*(2*c*(x^3/c^2 - arctan(c*x^3)/c^ 3)*arctan(c*x^3) + (arctan(c*x^3)^2 - log(6*c^5*x^6 + 6*c^3))/c^2)*a*b^2 + 1/192*(4*x^6*arctan(c*x^3)^3 - 3*x^6*arctan(c*x^3)*log(c^2*x^6 + 1)^2 + 1 92*integrate(1/64*(12*c^2*x^11*arctan(c*x^3)*log(c^2*x^6 + 1) - 12*c*x^8*a rctan(c*x^3)^2 + 56*(c^2*x^11 + x^5)*arctan(c*x^3)^3 + 3*(c*x^8 + 2*(c^2*x ^11 + x^5)*arctan(c*x^3))*log(c^2*x^6 + 1)^2)/(c^2*x^6 + 1), x))*b^3
\[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \]
Timed out. \[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int x^5\,{\left (a+b\,\mathrm {atan}\left (c\,x^3\right )\right )}^3 \,d x \]