Integrand size = 23, antiderivative size = 192 \[ \int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx=\frac {(a+b \arctan (c x))^2}{c^2 d}-\frac {i x (a+b \arctan (c x))^2}{c d}-\frac {2 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^2 d}-\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^2 d}+\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^2 d}-\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^2 d}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^2 d} \]
(a+b*arctan(c*x))^2/c^2/d-I*x*(a+b*arctan(c*x))^2/c/d-2*I*b*(a+b*arctan(c* x))*ln(2/(1+I*c*x))/c^2/d-(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/c^2/d+b^2*po lylog(2,1-2/(1+I*c*x))/c^2/d-I*b*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x) )/c^2/d-1/2*b^2*polylog(3,1-2/(1+I*c*x))/c^2/d
Time = 0.73 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.24 \[ \int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx=-\frac {i \left (6 a^2 c x-6 a^2 \arctan (c x)+12 a b c x \arctan (c x)-12 a b \arctan (c x)^2-6 i b^2 \arctan (c x)^2+6 b^2 c x \arctan (c x)^2-4 b^2 \arctan (c x)^3-12 i a b \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+12 b^2 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-6 i b^2 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+3 i a^2 \log \left (1+c^2 x^2\right )-6 a b \log \left (1+c^2 x^2\right )-6 b (a+i b+b \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-3 i b^2 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )}{6 c^2 d} \]
((-1/6*I)*(6*a^2*c*x - 6*a^2*ArcTan[c*x] + 12*a*b*c*x*ArcTan[c*x] - 12*a*b *ArcTan[c*x]^2 - (6*I)*b^2*ArcTan[c*x]^2 + 6*b^2*c*x*ArcTan[c*x]^2 - 4*b^2 *ArcTan[c*x]^3 - (12*I)*a*b*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 1 2*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - (6*I)*b^2*ArcTan[c*x]^2 *Log[1 + E^((2*I)*ArcTan[c*x])] + (3*I)*a^2*Log[1 + c^2*x^2] - 6*a*b*Log[1 + c^2*x^2] - 6*b*(a + I*b + b*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c* x])] - (3*I)*b^2*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/(c^2*d)
Time = 1.27 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {5401, 27, 5345, 5379, 5455, 5379, 2849, 2752, 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.
\(\displaystyle \int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx\) |
\(\Big \downarrow \) 5401 |
\(\displaystyle \frac {i \int \frac {(a+b \arctan (c x))^2}{d (i c x+1)}dx}{c}-\frac {i \int (a+b \arctan (c x))^2dx}{c d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i \int \frac {(a+b \arctan (c x))^2}{i c x+1}dx}{c d}-\frac {i \int (a+b \arctan (c x))^2dx}{c d}\) |
\(\Big \downarrow \) 5345 |
\(\displaystyle \frac {i \int \frac {(a+b \arctan (c x))^2}{i c x+1}dx}{c d}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \int \frac {x (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c d}\) |
\(\Big \downarrow \) 5379 |
\(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 i b \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \int \frac {x (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c d}\) |
\(\Big \downarrow \) 5455 |
\(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 i b \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \left (-\frac {\int \frac {a+b \arctan (c x)}{i-c x}dx}{c}-\frac {i (a+b \arctan (c x))^2}{2 b c^2}\right )\right )}{c d}\) |
\(\Big \downarrow \) 5379 |
\(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 i b \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}-b \int \frac {\log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx}{c}-\frac {i (a+b \arctan (c x))^2}{2 b c^2}\right )\right )}{c d}\) |
\(\Big \downarrow \) 2849 |
\(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 i b \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \left (-\frac {\frac {i b \int \frac {\log \left (\frac {2}{i c x+1}\right )}{1-\frac {2}{i c x+1}}d\frac {1}{i c x+1}}{c}+\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}}{c}-\frac {i (a+b \arctan (c x))^2}{2 b c^2}\right )\right )}{c d}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 i b \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \left (-\frac {i (a+b \arctan (c x))^2}{2 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{2 c}}{c}\right )\right )}{c d}\) |
\(\Big \downarrow \) 5529 |
\(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 i b \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}\right )\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \left (-\frac {i (a+b \arctan (c x))^2}{2 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{2 c}}{c}\right )\right )}{c d}\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 i b \left (-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}-\frac {b \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{4 c}\right )\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \left (-\frac {i (a+b \arctan (c x))^2}{2 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{2 c}}{c}\right )\right )}{c d}\) |
((-I)*(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)))/(c*d) + (I*((I*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/c - (2*I)*b*(((-1/2*I)*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/c - (b*PolyLog[3, 1 - 2/(1 + I*c*x)])/(4*c))))/(c*d)
3.1.97.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*((f_.)*(x_))^(m_.))/((d_) + ( e_.)*(x_)), x_Symbol] :> Simp[f/e Int[(f*x)^(m - 1)*(a + b*ArcTan[c*x])^p , x], x] - Simp[d*(f/e) Int[(f*x)^(m - 1)*((a + b*ArcTan[c*x])^p/(d + e*x )), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e ^2, 0] && GtQ[m, 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 6.85 (sec) , antiderivative size = 2330, normalized size of antiderivative = 12.14
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2330\) |
default | \(\text {Expression too large to display}\) | \(2330\) |
parts | \(\text {Expression too large to display}\) | \(2363\) |
1/c^2*(-I*a^2/d*c*x+1/2*a^2/d*ln(c^2*x^2+1)+I*a^2/d*arctan(c*x)+b^2/d*(-1/ 2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-1/2*polylog(2,-(1+I*c*x)^2/(c^2*x^2+ 1))-arctan(c*x)^2+I*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-dilog( 1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-Pi *arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-Pi*arctan(c*x)*ln(1-I*(1+ I*c*x)/(c^2*x^2+1)^(1/2))+Pi*arctan(c*x)*ln(1+(1+I*c*x)^2/(c^2*x^2+1))-I*a rctan(c*x)^2*c*x-arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))+arctan(c*x) ^2*ln(c*x-I)-1/2*I*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((1+I*c*x)^2 /(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*(I *arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*arctan(c*x)*ln(1-I*(1+I *c*x)/(c^2*x^2+1)^(1/2))+dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+dilog(1-I* (1+I*c*x)/(c^2*x^2+1)^(1/2)))-1/2*I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+ I*c*x)^2/(c^2*x^2+1)))^3*(I*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2) )+I*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+dilog(1+I*(1+I*c*x)/(c ^2*x^2+1)^(1/2))+dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2)))+1/4*I*Pi*csgn((1+ I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*(2*I*arctan(c*x)*ln(1+ (1+I*c*x)^2/(c^2*x^2+1))+2*arctan(c*x)^2+polylog(2,-(1+I*c*x)^2/(c^2*x^2+1 )))-I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*(I*ar ctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*arctan(c*x)*ln(1-I*(1+I*c* x)/(c^2*x^2+1)^(1/2))+dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+dilog(1-I*...
\[ \int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x}{i \, c d x + d} \,d x } \]
integral(1/4*(I*b^2*x*log(-(c*x + I)/(c*x - I))^2 + 4*a*b*x*log(-(c*x + I) /(c*x - I)) - 4*I*a^2*x)/(c*d*x - I*d), x)
Timed out. \[ \int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx=\text {Timed out} \]
\[ \int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x}{i \, c d x + d} \,d x } \]
a^2*(-I*x/(c*d) + log(I*c*x + 1)/(c^2*d)) + 1/96*(-24*I*b^2*c*x*arctan(c*x )^2 + 24*I*b^2*arctan(c*x)^3 - 3*b^2*log(c^2*x^2 + 1)^3 - 16*I*(72*b^2*c^2 *integrate(1/16*x^2*arctan(c*x)^2/(c^3*d*x^2 + c*d), x) + 6*b^2*c^2*integr ate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^3*d*x^2 + c*d), x) + 192*a*b*c^2*integr ate(1/16*x^2*arctan(c*x)/(c^3*d*x^2 + c*d), x) + 24*b^2*c^2*integrate(1/16 *x^2*log(c^2*x^2 + 1)/(c^3*d*x^2 + c*d), x) + 24*b^2*c*integrate(1/16*x*ar ctan(c*x)*log(c^2*x^2 + 1)/(c^3*d*x^2 + c*d), x) - 48*b^2*c*integrate(1/16 *x*arctan(c*x)/(c^3*d*x^2 + c*d), x) + 12*b^2*integrate(1/16*log(c^2*x^2 + 1)^2/(c^3*d*x^2 + c*d), x) + b^2*arctan(c*x)^3/(c^2*d))*c^2*d + 96*c^2*d* integrate(1/16*(20*b^2*x*arctan(c*x)^2 + 3*b^2*x*log(c^2*x^2 + 1)^2 - 8*(b ^2*c*x^2 - 4*a*b*x)*arctan(c*x) - 4*(b^2*c*x^2*arctan(c*x) + b^2*x)*log(c^ 2*x^2 + 1))/(c^2*d*x^2 + d), x) + 6*(I*b^2*c*x + I*b^2*arctan(c*x))*log(c^ 2*x^2 + 1)^2 + 12*(2*b^2*c*x*arctan(c*x) - b^2*arctan(c*x)^2)*log(c^2*x^2 + 1))/(c^2*d)
\[ \int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x}{i \, c d x + d} \,d x } \]
Timed out. \[ \int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int \frac {x\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \]