Integrand size = 10, antiderivative size = 83 \[ \int (a+b \arctan (c x))^2 \, dx=\frac {i (a+b \arctan (c x))^2}{c}+x (a+b \arctan (c x))^2+\frac {2 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c}+\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c} \]
I*(a+b*arctan(c*x))^2/c+x*(a+b*arctan(c*x))^2+2*b*(a+b*arctan(c*x))*ln(2/( 1+I*c*x))/c+I*b^2*polylog(2,1-2/(1+I*c*x))/c
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.08 \[ \int (a+b \arctan (c x))^2 \, dx=\frac {b^2 (-i+c x) \arctan (c x)^2+2 b \arctan (c x) \left (a c x+b \log \left (1+e^{2 i \arctan (c x)}\right )\right )+a \left (a c x-b \log \left (1+c^2 x^2\right )\right )-i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )}{c} \]
(b^2*(-I + c*x)*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]*(a*c*x + b*Log[1 + E^((2*I )*ArcTan[c*x])]) + a*(a*c*x - b*Log[1 + c^2*x^2]) - I*b^2*PolyLog[2, -E^(( 2*I)*ArcTan[c*x])])/c
Time = 0.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {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.
| \(\displaystyle \int (a+b \arctan (c x))^2 \, dx\) |
| \(\Big \downarrow \) 5345 |
| \(\displaystyle x (a+b \arctan (c x))^2-2 b c \int \frac {x (a+b \arctan (c x))}{c^2 x^2+1}dx\) |
| \(\Big \downarrow \) 5455 |
| \(\displaystyle x (a+b \arctan (c x))^2-2 b c \left (-\frac {\int \frac {a+b \arctan (c x)}{i-c x}dx}{c}-\frac {i (a+b \arctan (c x))^2}{2 b c^2}\right )\) |
| \(\Big \downarrow \) 5379 |
| \(\displaystyle x (a+b \arctan (c x))^2-2 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}-b \int \frac {\log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx}{c}-\frac {i (a+b \arctan (c x))^2}{2 b c^2}\right )\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle x (a+b \arctan (c x))^2-2 b c \left (-\frac {\frac {i b \int \frac {\log \left (\frac {2}{i c x+1}\right )}{1-\frac {2}{i c x+1}}d\frac {1}{i c x+1}}{c}+\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}}{c}-\frac {i (a+b \arctan (c x))^2}{2 b c^2}\right )\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle x (a+b \arctan (c x))^2-2 b c \left (-\frac {i (a+b \arctan (c x))^2}{2 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{2 c}}{c}\right )\) |
x*(a + b*ArcTan[c*x])^2 - 2*b*c*(((-1/2*I)*(a + b*ArcTan[c*x])^2)/(b*c^2) - (((a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c + ((I/2)*b*PolyLog[2, 1 - 2/ (1 + I*c*x)])/c)/c)
3.1.18.3.1 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_)]*(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 = 2.19 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.48
| method | result | size |
| derivativedivides | \(\frac {c x \,a^{2}-i \arctan \left (c x \right )^{2} b^{2}+\arctan \left (c x \right )^{2} b^{2} c x +2 \arctan \left (c x \right ) \ln \left (1+\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right ) b^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right ) b^{2}+2 a b c x \arctan \left (c x \right )-a b \ln \left (c^{2} x^{2}+1\right )}{c}\) | \(123\) |
| default | \(\frac {c x \,a^{2}-i \arctan \left (c x \right )^{2} b^{2}+\arctan \left (c x \right )^{2} b^{2} c x +2 \arctan \left (c x \right ) \ln \left (1+\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right ) b^{2}-i \operatorname {polylog}\left (2, -\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right ) b^{2}+2 a b c x \arctan \left (c x \right )-a b \ln \left (c^{2} x^{2}+1\right )}{c}\) | \(123\) |
| parts | \(a^{2} x +b^{2} \arctan \left (c x \right )^{2} x -\frac {i b^{2} \arctan \left (c x \right )^{2}}{c}-\frac {i b^{2} \operatorname {polylog}\left (2, -\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right )}{c}+\frac {2 b^{2} \arctan \left (c x \right ) \ln \left (1+\frac {\left (i c x +1\right )^{2}}{c^{2} x^{2}+1}\right )}{c}+2 a b x \arctan \left (c x \right )-\frac {a b \ln \left (c^{2} x^{2}+1\right )}{c}\) | \(128\) |
| risch | \(\frac {i b^{2} \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{c}-\frac {b a \ln \left (i c x +1\right )}{c}+\frac {b^{2} \ln \left (i c x +1\right ) \ln \left (-i c x +1\right ) x}{2}+\frac {i b^{2} \ln \left (i c x +1\right ) \ln \left (-i c x +1\right )}{2 c}+\frac {2 a b}{c}-\frac {b^{2} \ln \left (i c x +1\right )^{2} x}{4}-\frac {i \ln \left (-i c x +1\right )^{2} b^{2}}{4 c}+\frac {i b^{2} \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{c}+\frac {i b^{2} \ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {i b^{2} \ln \left (i c x +1\right )}{c}-i b a \ln \left (i c x +1\right ) x +a^{2} x -\frac {\ln \left (-i c x +1\right ) a b}{c}-\frac {i b^{2} \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{c}+i \ln \left (-i c x +1\right ) a b x -\frac {\ln \left (-i c x +1\right )^{2} b^{2} x}{4}+\frac {i a^{2}}{c}+\frac {i b^{2} \ln \left (i c x +1\right )^{2}}{4 c}+\frac {i b^{2}}{c}-\frac {b^{2} \arctan \left (c x \right )}{c}\) | \(322\) |
1/c*(c*x*a^2-I*arctan(c*x)^2*b^2+arctan(c*x)^2*b^2*c*x+2*arctan(c*x)*ln(1+ (1+I*c*x)^2/(c^2*x^2+1))*b^2-I*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))*b^2+2*a *b*c*x*arctan(c*x)-a*b*ln(c^2*x^2+1))
\[ \int (a+b \arctan (c x))^2 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{2} \,d x } \]
\[ \int (a+b \arctan (c x))^2 \, dx=\int \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}\, dx \]
\[ \int (a+b \arctan (c x))^2 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{2} \,d x } \]
1/16*(4*x*arctan(c*x)^2 + 192*c^2*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^ 2 + 1), x) + 16*c^2*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x ) + 64*c^2*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - x*log(c ^2*x^2 + 1)^2 + 4*arctan(c*x)^3/c - 128*c*integrate(1/16*x*arctan(c*x)/(c^ 2*x^2 + 1), x) + 16*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x))*b ^2 + a^2*x + (2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b/c
\[ \int (a+b \arctan (c x))^2 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int (a+b \arctan (c x))^2 \, dx=\int {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2 \,d x \]