Integrand size = 14, antiderivative size = 101 \[ \int x \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 c}+\frac {1}{2} x^2 \left (a+b \arctan \left (c x^2\right )\right )^2+\frac {b \left (a+b \arctan \left (c x^2\right )\right ) \log \left (\frac {2}{1+i c x^2}\right )}{c}+\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x^2}\right )}{2 c} \]
1/2*I*(a+b*arctan(c*x^2))^2/c+1/2*x^2*(a+b*arctan(c*x^2))^2+b*(a+b*arctan( c*x^2))*ln(2/(1+I*c*x^2))/c+1/2*I*b^2*polylog(2,1-2/(1+I*c*x^2))/c
Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int x \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\frac {b^2 \left (-i+c x^2\right ) \arctan \left (c x^2\right )^2+2 b \arctan \left (c x^2\right ) \left (a c x^2+b \log \left (1+e^{2 i \arctan \left (c x^2\right )}\right )\right )+a \left (a c x^2-b \log \left (1+c^2 x^4\right )\right )-i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (c x^2\right )}\right )}{2 c} \]
(b^2*(-I + c*x^2)*ArcTan[c*x^2]^2 + 2*b*ArcTan[c*x^2]*(a*c*x^2 + b*Log[1 + E^((2*I)*ArcTan[c*x^2])]) + a*(a*c*x^2 - b*Log[1 + c^2*x^4]) - I*b^2*Poly Log[2, -E^((2*I)*ArcTan[c*x^2])])/(2*c)
Time = 0.48 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.12, 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, 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.
| \(\displaystyle \int x \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 5363 |
| \(\displaystyle \frac {1}{2} \int \left (a+b \arctan \left (c x^2\right )\right )^2dx^2\) |
| \(\Big \downarrow \) 5345 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^2-2 b c \int \frac {x^2 \left (a+b \arctan \left (c x^2\right )\right )}{c^2 x^4+1}dx^2\right )\) |
| \(\Big \downarrow \) 5455 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^2-2 b c \left (-\frac {\int \frac {a+b \arctan \left (c x^2\right )}{i-c x^2}dx^2}{c}-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b c^2}\right )\right )\) |
| \(\Big \downarrow \) 5379 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^2-2 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )}{c}-b \int \frac {\log \left (\frac {2}{i c x^2+1}\right )}{c^2 x^4+1}dx^2}{c}-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b c^2}\right )\right )\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^2-2 b c \left (-\frac {\frac {i b \int \frac {\log \left (\frac {2}{i c x^2+1}\right )}{1-\frac {2}{i c x^2+1}}d\frac {1}{i c x^2+1}}{c}+\frac {\log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )}{c}}{c}-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b c^2}\right )\right )\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \arctan \left (c x^2\right )\right )^2-2 b c \left (-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )}{c}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x^2+1}\right )}{2 c}}{c}\right )\right )\) |
(x^2*(a + b*ArcTan[c*x^2])^2 - 2*b*c*(((-1/2*I)*(a + b*ArcTan[c*x^2])^2)/( b*c^2) - (((a + b*ArcTan[c*x^2])*Log[2/(1 + I*c*x^2)])/c + ((I/2)*b*PolyLo g[2, 1 - 2/(1 + I*c*x^2)])/c)/c))/2
3.1.77.3.1 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[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]
Time = 4.44 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.39
| method | result | size |
| parts | \(\frac {a^{2} x^{2}}{2}+\frac {b^{2} \left (\arctan \left (c \,x^{2}\right )^{2} \left (c \,x^{2}+i\right )+2 \arctan \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )-2 i \arctan \left (c \,x^{2}\right )^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )\right )}{2 c}+\frac {a b \left (c \,x^{2} \arctan \left (c \,x^{2}\right )-\frac {\ln \left (c^{2} x^{4}+1\right )}{2}\right )}{c}\) | \(140\) |
| derivativedivides | \(\frac {a^{2} c \,x^{2}-i \arctan \left (c \,x^{2}\right )^{2} b^{2}+\arctan \left (c \,x^{2}\right )^{2} b^{2} c \,x^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) b^{2}+2 \arctan \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) b^{2}+2 a b c \,x^{2} \arctan \left (c \,x^{2}\right )-a b \ln \left (c^{2} x^{4}+1\right )}{2 c}\) | \(142\) |
| default | \(\frac {a^{2} c \,x^{2}-i \arctan \left (c \,x^{2}\right )^{2} b^{2}+\arctan \left (c \,x^{2}\right )^{2} b^{2} c \,x^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) b^{2}+2 \arctan \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) b^{2}+2 a b c \,x^{2} \arctan \left (c \,x^{2}\right )-a b \ln \left (c^{2} x^{4}+1\right )}{2 c}\) | \(142\) |
| risch | \(\frac {i a^{2}}{2 c}-\frac {b^{2} \arctan \left (c \,x^{2}\right )}{2 c}+\frac {i \ln \left (-i c \,x^{2}+1\right ) a b \,x^{2}}{2}-\frac {\ln \left (-i c \,x^{2}+1\right ) a b}{2 c}-\frac {i \ln \left (-i c \,x^{2}+1\right )^{2} b^{2}}{8 c}+\frac {i b^{2} \ln \left (i c \,x^{2}+1\right )^{2}}{8 c}-\frac {b a \ln \left (i c \,x^{2}+1\right )}{2 c}+\frac {b^{2} \ln \left (i c \,x^{2}+1\right ) \ln \left (-i c \,x^{2}+1\right ) x^{2}}{4}-\frac {i b^{2} \ln \left (i c \,x^{2}+1\right )}{2 c}+\frac {i b^{2} \ln \left (i c \,x^{2}+1\right ) \ln \left (-i c \,x^{2}+1\right )}{4 c}+\frac {i b^{2} \operatorname {dilog}\left (\frac {1}{2}-\frac {i c \,x^{2}}{2}\right )}{2 c}+\frac {a^{2} x^{2}}{2}+\frac {a b}{c}-\frac {i b^{2} \ln \left (\frac {1}{2}+\frac {i c \,x^{2}}{2}\right ) \ln \left (-i c \,x^{2}+1\right )}{2 c}+\frac {i b^{2} \ln \left (c^{2} x^{4}+1\right )}{4 c}+\frac {i b^{2} \ln \left (\frac {1}{2}+\frac {i c \,x^{2}}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c \,x^{2}}{2}\right )}{2 c}+\frac {i b^{2}}{2 c}-\frac {b^{2} \ln \left (i c \,x^{2}+1\right )^{2} x^{2}}{8}-\frac {\ln \left (-i c \,x^{2}+1\right )^{2} b^{2} x^{2}}{8}-\frac {i b a \ln \left (i c \,x^{2}+1\right ) x^{2}}{2}\) | \(372\) |
1/2*a^2*x^2+1/2*b^2/c*(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)))+a*b/c*(c*x^2*arctan(c*x^2)-1/2*ln(c^2*x^4+1))
\[ \int x \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2} x \,d x } \]
\[ \int x \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\int x \left (a + b \operatorname {atan}{\left (c x^{2} \right )}\right )^{2}\, dx \]
\[ \int x \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2} x \,d x } \]
1/2*a^2*x^2 + 1/32*(4*x^2*arctan(c*x^2)^2 - x^2*log(c^2*x^4 + 1)^2 + 384*c ^2*integrate(1/16*x^5*arctan(c*x^2)^2/(c^2*x^4 + 1), x) + 32*c^2*integrate (1/16*x^5*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 128*c^2*integrate(1/16*x^ 5*log(c^2*x^4 + 1)/(c^2*x^4 + 1), x) + 4*arctan(c*x^2)^3/c - 256*c*integra te(1/16*x^3*arctan(c*x^2)/(c^2*x^4 + 1), x) + 32*integrate(1/16*x*log(c^2* x^4 + 1)^2/(c^2*x^4 + 1), x))*b^2 + 1/2*(2*c*x^2*arctan(c*x^2) - log(c^2*x ^4 + 1))*a*b/c
\[ \int x \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2} x \,d x } \]
Timed out. \[ \int x \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\int x\,{\left (a+b\,\mathrm {atan}\left (c\,x^2\right )\right )}^2 \,d x \]