Integrand size = 14, antiderivative size = 144 \[ \int x \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\frac {i \left (a+b \arctan \left (c x^2\right )\right )^3}{2 c}+\frac {1}{2} x^2 \left (a+b \arctan \left (c x^2\right )\right )^3+\frac {3 b \left (a+b \arctan \left (c x^2\right )\right )^2 \log \left (\frac {2}{1+i c x^2}\right )}{2 c}+\frac {3 i b^2 \left (a+b \arctan \left (c x^2\right )\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x^2}\right )}{2 c}+\frac {3 b^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x^2}\right )}{4 c} \]
1/2*I*(a+b*arctan(c*x^2))^3/c+1/2*x^2*(a+b*arctan(c*x^2))^3+3/2*b*(a+b*arc tan(c*x^2))^2*ln(2/(1+I*c*x^2))/c+3/2*I*b^2*(a+b*arctan(c*x^2))*polylog(2, 1-2/(1+I*c*x^2))/c+3/4*b^3*polylog(3,1-2/(1+I*c*x^2))/c
Time = 0.13 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.56 \[ \int x \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\frac {2 a^3 c x^2+6 a^2 b c x^2 \arctan \left (c x^2\right )-6 i a b^2 \arctan \left (c x^2\right )^2+6 a b^2 c x^2 \arctan \left (c x^2\right )^2-2 i b^3 \arctan \left (c x^2\right )^3+2 b^3 c x^2 \arctan \left (c x^2\right )^3+12 a b^2 \arctan \left (c x^2\right ) \log \left (1+e^{2 i \arctan \left (c x^2\right )}\right )+6 b^3 \arctan \left (c x^2\right )^2 \log \left (1+e^{2 i \arctan \left (c x^2\right )}\right )-3 a^2 b \log \left (1+c^2 x^4\right )-6 i b^2 \left (a+b \arctan \left (c x^2\right )\right ) \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (c x^2\right )}\right )+3 b^3 \operatorname {PolyLog}\left (3,-e^{2 i \arctan \left (c x^2\right )}\right )}{4 c} \]
(2*a^3*c*x^2 + 6*a^2*b*c*x^2*ArcTan[c*x^2] - (6*I)*a*b^2*ArcTan[c*x^2]^2 + 6*a*b^2*c*x^2*ArcTan[c*x^2]^2 - (2*I)*b^3*ArcTan[c*x^2]^3 + 2*b^3*c*x^2*A rcTan[c*x^2]^3 + 12*a*b^2*ArcTan[c*x^2]*Log[1 + E^((2*I)*ArcTan[c*x^2])] + 6*b^3*ArcTan[c*x^2]^2*Log[1 + E^((2*I)*ArcTan[c*x^2])] - 3*a^2*b*Log[1 + c^2*x^4] - (6*I)*b^2*(a + b*ArcTan[c*x^2])*PolyLog[2, -E^((2*I)*ArcTan[c*x ^2])] + 3*b^3*PolyLog[3, -E^((2*I)*ArcTan[c*x^2])])/(4*c)
Time = 0.69 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5363, 5345, 5455, 5379, 5529, 7164}
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.
| \(\displaystyle \int x \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx\) |
| \(\Big \downarrow \) 5363 |
| \(\displaystyle \frac {1}{2} \int \left (a+b \arctan \left (c x^2\right )\right )^3dx^2\) |
| \(\Big \downarrow \) 5345 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^3-3 b c \int \frac {x^2 \left (a+b \arctan \left (c x^2\right )\right )^2}{c^2 x^4+1}dx^2\right )\) |
| \(\Big \downarrow \) 5455 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^3-3 b c \left (-\frac {\int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{i-c x^2}dx^2}{c}-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^3}{3 b c^2}\right )\right )\) |
| \(\Big \downarrow \) 5379 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^3-3 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )^2}{c}-2 b \int \frac {\left (a+b \arctan \left (c x^2\right )\right ) \log \left (\frac {2}{i c x^2+1}\right )}{c^2 x^4+1}dx^2}{c}-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^3}{3 b c^2}\right )\right )\) |
| \(\Big \downarrow \) 5529 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^3-3 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )^2}{c}-2 b \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{i c x^2+1}\right )}{c^2 x^4+1}dx^2-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x^2+1}\right ) \left (a+b \arctan \left (c x^2\right )\right )}{2 c}\right )}{c}-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^3}{3 b c^2}\right )\right )\) |
| \(\Big \downarrow \) 7164 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^3-3 b c \left (-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^3}{3 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )^2}{c}-2 b \left (-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x^2+1}\right ) \left (a+b \arctan \left (c x^2\right )\right )}{2 c}-\frac {b \operatorname {PolyLog}\left (3,1-\frac {2}{i c x^2+1}\right )}{4 c}\right )}{c}\right )\right )\) |
(x^2*(a + b*ArcTan[c*x^2])^3 - 3*b*c*(((-1/3*I)*(a + b*ArcTan[c*x^2])^3)/( b*c^2) - (((a + b*ArcTan[c*x^2])^2*Log[2/(1 + I*c*x^2)])/c - 2*b*(((-1/2*I )*(a + b*ArcTan[c*x^2])*PolyLog[2, 1 - 2/(1 + I*c*x^2)])/c - (b*PolyLog[3, 1 - 2/(1 + I*c*x^2)])/(4*c)))/c))/2
3.1.87.3.1 Defintions of rubi rules used
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[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_.)*(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]
Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 ), x_Symbol] :> Simp[(-I)*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)) , x] + Simp[b*p*(I/2) Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 - u]/ (d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c ^2*d] && EqQ[(1 - u)^2 - (1 - 2*(I/(I - c*x)))^2, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (129 ) = 258\).
Time = 9.39 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.91
| method | result | size |
| derivativedivides | \(\frac {a^{3} c \,x^{2}+b^{3} \left (\arctan \left (c \,x^{2}\right )^{3} \left (c \,x^{2}+i\right )-2 i \arctan \left (c \,x^{2}\right )^{3}+3 \arctan \left (c \,x^{2}\right )^{2} \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )-3 i \arctan \left (c \,x^{2}\right ) \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )+\frac {3 \operatorname {polylog}\left (3, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )}{2}\right )+3 a \,b^{2} \left (\arctan \left (c \,x^{2}\right )^{2} \left (c \,x^{2}+i\right )+2 \arctan \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )-2 i \arctan \left (c \,x^{2}\right )^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )\right )+3 a^{2} b \left (c \,x^{2} \arctan \left (c \,x^{2}\right )-\frac {\ln \left (c^{2} x^{4}+1\right )}{2}\right )}{2 c}\) | \(275\) |
| default | \(\frac {a^{3} c \,x^{2}+b^{3} \left (\arctan \left (c \,x^{2}\right )^{3} \left (c \,x^{2}+i\right )-2 i \arctan \left (c \,x^{2}\right )^{3}+3 \arctan \left (c \,x^{2}\right )^{2} \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )-3 i \arctan \left (c \,x^{2}\right ) \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )+\frac {3 \operatorname {polylog}\left (3, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )}{2}\right )+3 a \,b^{2} \left (\arctan \left (c \,x^{2}\right )^{2} \left (c \,x^{2}+i\right )+2 \arctan \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )-2 i \arctan \left (c \,x^{2}\right )^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )\right )+3 a^{2} b \left (c \,x^{2} \arctan \left (c \,x^{2}\right )-\frac {\ln \left (c^{2} x^{4}+1\right )}{2}\right )}{2 c}\) | \(275\) |
| parts | \(\frac {a^{3} x^{2}}{2}+\frac {b^{3} \left (\arctan \left (c \,x^{2}\right )^{3} \left (c \,x^{2}+i\right )-2 i \arctan \left (c \,x^{2}\right )^{3}+3 \arctan \left (c \,x^{2}\right )^{2} \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )-3 i \arctan \left (c \,x^{2}\right ) \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )+\frac {3 \operatorname {polylog}\left (3, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )}{2}\right )}{2 c}+\frac {3 a \,b^{2} \left (\arctan \left (c \,x^{2}\right )^{2} \left (c \,x^{2}+i\right )+2 \arctan \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )-2 i \arctan \left (c \,x^{2}\right )^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )\right )}{2 c}+\frac {3 a^{2} b \arctan \left (c \,x^{2}\right ) x^{2}}{2}-\frac {3 a^{2} b \ln \left (c^{2} x^{4}+1\right )}{4 c}\) | \(281\) |
1/2/c*(a^3*c*x^2+b^3*(arctan(c*x^2)^3*(c*x^2+I)-2*I*arctan(c*x^2)^3+3*arct an(c*x^2)^2*ln(1+(1+I*c*x^2)^2/(c^2*x^4+1))-3*I*arctan(c*x^2)*polylog(2,-( 1+I*c*x^2)^2/(c^2*x^4+1))+3/2*polylog(3,-(1+I*c*x^2)^2/(c^2*x^4+1)))+3*a*b ^2*(arctan(c*x^2)^2*(c*x^2+I)+2*arctan(c*x^2)*ln(1+(1+I*c*x^2)^2/(c^2*x^4+ 1))-2*I*arctan(c*x^2)^2-I*polylog(2,-(1+I*c*x^2)^2/(c^2*x^4+1)))+3*a^2*b*( c*x^2*arctan(c*x^2)-1/2*ln(c^2*x^4+1)))
\[ \int x \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3} x \,d x } \]
integral(b^3*x*arctan(c*x^2)^3 + 3*a*b^2*x*arctan(c*x^2)^2 + 3*a^2*b*x*arc tan(c*x^2) + a^3*x, x)
\[ \int x \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int x \left (a + b \operatorname {atan}{\left (c x^{2} \right )}\right )^{3}\, dx \]
\[ \int x \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3} x \,d x } \]
1/16*b^3*x^2*arctan(c*x^2)^3 - 3/64*b^3*x^2*arctan(c*x^2)*log(c^2*x^4 + 1) ^2 + 7/64*b^3*arctan(c*x^2)^4/c + 28*b^3*c^2*integrate(1/32*x^5*arctan(c*x ^2)^3/(c^2*x^4 + 1), x) + 3*b^3*c^2*integrate(1/32*x^5*arctan(c*x^2)*log(c ^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 96*a*b^2*c^2*integrate(1/32*x^5*arctan(c *x^2)^2/(c^2*x^4 + 1), x) + 12*b^3*c^2*integrate(1/32*x^5*arctan(c*x^2)*lo g(c^2*x^4 + 1)/(c^2*x^4 + 1), x) + 1/2*a^3*x^2 + 1/2*a*b^2*arctan(c*x^2)^3 /c - 12*b^3*c*integrate(1/32*x^3*arctan(c*x^2)^2/(c^2*x^4 + 1), x) + 3*b^3 *c*integrate(1/32*x^3*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 3*b^3*integra te(1/32*x*arctan(c*x^2)*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 3/4*(2*c*x^ 2*arctan(c*x^2) - log(c^2*x^4 + 1))*a^2*b/c
\[ \int x \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3} x \,d x } \]
Timed out. \[ \int x \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int x\,{\left (a+b\,\mathrm {atan}\left (c\,x^2\right )\right )}^3 \,d x \]