Integrand size = 23, antiderivative size = 197 \[ \int \frac {a+b \arctan (c x)}{x^4 (d+i c d x)} \, dx=-\frac {b c}{6 d x^2}+\frac {i b c^2}{2 d x}+\frac {i b c^3 \arctan (c x)}{2 d}-\frac {a+b \arctan (c x)}{3 d x^3}+\frac {i c (a+b \arctan (c x))}{2 d x^2}+\frac {c^2 (a+b \arctan (c x))}{d x}-\frac {4 b c^3 \log (x)}{3 d}+\frac {2 b c^3 \log \left (1+c^2 x^2\right )}{3 d}+\frac {i c^3 (a+b \arctan (c x)) \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{2 d} \]
-1/6*b*c/d/x^2+1/2*I*b*c^2/d/x+1/2*I*b*c^3*arctan(c*x)/d+1/3*(-a-b*arctan( c*x))/d/x^3+1/2*I*c*(a+b*arctan(c*x))/d/x^2+c^2*(a+b*arctan(c*x))/d/x-4/3* b*c^3*ln(x)/d+2/3*b*c^3*ln(c^2*x^2+1)/d+I*c^3*(a+b*arctan(c*x))*ln(2-2/(1+ I*c*x))/d-1/2*b*c^3*polylog(2,-1+2/(1+I*c*x))/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.10 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.29 \[ \int \frac {a+b \arctan (c x)}{x^4 (d+i c d x)} \, dx=\frac {-2 a+3 i a c x-b c x+6 a c^2 x^2-2 b \arctan (c x)+3 i b c x \arctan (c x)+6 b c^2 x^2 \arctan (c x)+3 i b c^2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )+6 i a c^3 x^3 \log (x)-8 b c^3 x^3 \log (x)+6 i a c^3 x^3 \log \left (\frac {2 i}{i-c x}\right )+6 i b c^3 x^3 \arctan (c x) \log \left (\frac {2 i}{i-c x}\right )+4 b c^3 x^3 \log \left (1+c^2 x^2\right )-3 b c^3 x^3 \operatorname {PolyLog}(2,-i c x)+3 b c^3 x^3 \operatorname {PolyLog}(2,i c x)-3 b c^3 x^3 \operatorname {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )}{6 d x^3} \]
(-2*a + (3*I)*a*c*x - b*c*x + 6*a*c^2*x^2 - 2*b*ArcTan[c*x] + (3*I)*b*c*x* ArcTan[c*x] + 6*b*c^2*x^2*ArcTan[c*x] + (3*I)*b*c^2*x^2*Hypergeometric2F1[ -1/2, 1, 1/2, -(c^2*x^2)] + (6*I)*a*c^3*x^3*Log[x] - 8*b*c^3*x^3*Log[x] + (6*I)*a*c^3*x^3*Log[(2*I)/(I - c*x)] + (6*I)*b*c^3*x^3*ArcTan[c*x]*Log[(2* I)/(I - c*x)] + 4*b*c^3*x^3*Log[1 + c^2*x^2] - 3*b*c^3*x^3*PolyLog[2, (-I) *c*x] + 3*b*c^3*x^3*PolyLog[2, I*c*x] - 3*b*c^3*x^3*PolyLog[2, (I + c*x)/( -I + c*x)])/(6*d*x^3)
Time = 1.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.98, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {5405, 27, 5361, 243, 54, 2009, 5405, 5361, 264, 216, 5405, 5361, 243, 47, 14, 16, 5403, 2897}
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.
| \(\displaystyle \int \frac {a+b \arctan (c x)}{x^4 (d+i c d x)} \, dx\) |
| \(\Big \downarrow \) 5405 |
| \(\displaystyle \frac {\int \frac {a+b \arctan (c x)}{x^4}dx}{d}-i c \int \frac {a+b \arctan (c x)}{d x^3 (i c x+1)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a+b \arctan (c x)}{x^4}dx}{d}-\frac {i c \int \frac {a+b \arctan (c x)}{x^3 (i c x+1)}dx}{d}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {\frac {1}{3} b c \int \frac {1}{x^3 \left (c^2 x^2+1\right )}dx-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \int \frac {a+b \arctan (c x)}{x^3 (i c x+1)}dx}{d}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {1}{6} b c \int \frac {1}{x^4 \left (c^2 x^2+1\right )}dx^2-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \int \frac {a+b \arctan (c x)}{x^3 (i c x+1)}dx}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {\frac {1}{6} b c \int \left (\frac {c^4}{c^2 x^2+1}-\frac {c^2}{x^2}+\frac {1}{x^4}\right )dx^2-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \int \frac {a+b \arctan (c x)}{x^3 (i c x+1)}dx}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \int \frac {a+b \arctan (c x)}{x^3 (i c x+1)}dx}{d}\) |
| \(\Big \downarrow \) 5405 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (\int \frac {a+b \arctan (c x)}{x^3}dx-i c \int \frac {a+b \arctan (c x)}{x^2 (i c x+1)}dx\right )}{d}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \int \frac {a+b \arctan (c x)}{x^2 (i c x+1)}dx+\frac {1}{2} b c \int \frac {1}{x^2 \left (c^2 x^2+1\right )}dx-\frac {a+b \arctan (c x)}{2 x^2}\right )}{d}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \int \frac {a+b \arctan (c x)}{x^2 (i c x+1)}dx+\frac {1}{2} b c \left (c^2 \left (-\int \frac {1}{c^2 x^2+1}dx\right )-\frac {1}{x}\right )-\frac {a+b \arctan (c x)}{2 x^2}\right )}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \int \frac {a+b \arctan (c x)}{x^2 (i c x+1)}dx-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
| \(\Big \downarrow \) 5405 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \left (\int \frac {a+b \arctan (c x)}{x^2}dx-i c \int \frac {a+b \arctan (c x)}{x (i c x+1)}dx\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \left (-i c \int \frac {a+b \arctan (c x)}{x (i c x+1)}dx+b c \int \frac {1}{x \left (c^2 x^2+1\right )}dx-\frac {a+b \arctan (c x)}{x}\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \left (-i c \int \frac {a+b \arctan (c x)}{x (i c x+1)}dx+\frac {1}{2} b c \int \frac {1}{x^2 \left (c^2 x^2+1\right )}dx^2-\frac {a+b \arctan (c x)}{x}\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
| \(\Big \downarrow \) 47 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \left (-i c \int \frac {a+b \arctan (c x)}{x (i c x+1)}dx+\frac {1}{2} b c \left (\int \frac {1}{x^2}dx^2-c^2 \int \frac {1}{c^2 x^2+1}dx^2\right )-\frac {a+b \arctan (c x)}{x}\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
| \(\Big \downarrow \) 14 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \left (-i c \int \frac {a+b \arctan (c x)}{x (i c x+1)}dx+\frac {1}{2} b c \left (\log \left (x^2\right )-c^2 \int \frac {1}{c^2 x^2+1}dx^2\right )-\frac {a+b \arctan (c x)}{x}\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \left (-i c \int \frac {a+b \arctan (c x)}{x (i c x+1)}dx-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
| \(\Big \downarrow \) 5403 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \left (-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))-b c \int \frac {\log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )-\frac {a+b \arctan (c x)}{3 x^3}}{d}-\frac {i c \left (-i c \left (-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))+\frac {1}{2} i b \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right )\right )-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {a+b \arctan (c x)}{2 x^2}+\frac {1}{2} b c \left (-c \arctan (c x)-\frac {1}{x}\right )\right )}{d}\) |
(-1/3*(a + b*ArcTan[c*x])/x^3 + (b*c*(-x^(-2) - c^2*Log[x^2] + c^2*Log[1 + c^2*x^2]))/6)/d - (I*c*(-1/2*(a + b*ArcTan[c*x])/x^2 + (b*c*(-x^(-1) - c* ArcTan[c*x]))/2 - I*c*(-((a + b*ArcTan[c*x])/x) + (b*c*(Log[x^2] - Log[1 + c^2*x^2]))/2 - I*c*((a + b*ArcTan[c*x])*Log[2 - 2/(1 + I*c*x)] + (I/2)*b* PolyLog[2, -1 + 2/(1 + I*c*x)]))))/d
3.1.50.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
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_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_ Symbol] :> Simp[(a + b*ArcTan[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Si mp[b*c*(p/d) Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 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[1/d Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x ] - Simp[e/(d*f) 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] && LtQ[m, -1]
Time = 1.03 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.41
| method | result | size |
| derivativedivides | \(c^{3} \left (-\frac {a}{3 d \,c^{3} x^{3}}+\frac {i a}{2 d \,c^{2} x^{2}}+\frac {i a \ln \left (c x \right )}{d}+\frac {a}{d c x}-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {a \arctan \left (c x \right )}{d}+\frac {b \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}+\frac {i \arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {i}{2 c x}+\frac {\arctan \left (c x \right )}{c x}+i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {\ln \left (c x -i\right )^{2}}{4}-\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i \arctan \left (c x \right )}{2}-\frac {1}{6 c^{2} x^{2}}-\frac {4 \ln \left (c x \right )}{3}+\frac {2 \ln \left (c^{2} x^{2}+1\right )}{3}-i \arctan \left (c x \right ) \ln \left (c x -i\right )\right )}{d}\right )\) | \(278\) |
| default | \(c^{3} \left (-\frac {a}{3 d \,c^{3} x^{3}}+\frac {i a}{2 d \,c^{2} x^{2}}+\frac {i a \ln \left (c x \right )}{d}+\frac {a}{d c x}-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {a \arctan \left (c x \right )}{d}+\frac {b \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}+\frac {i \arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {i}{2 c x}+\frac {\arctan \left (c x \right )}{c x}+i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {\ln \left (c x -i\right )^{2}}{4}-\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i \arctan \left (c x \right )}{2}-\frac {1}{6 c^{2} x^{2}}-\frac {4 \ln \left (c x \right )}{3}+\frac {2 \ln \left (c^{2} x^{2}+1\right )}{3}-i \arctan \left (c x \right ) \ln \left (c x -i\right )\right )}{d}\right )\) | \(278\) |
| parts | \(-\frac {a}{3 d \,x^{3}}+\frac {i c a}{2 d \,x^{2}}+\frac {i a \,c^{3} \ln \left (x \right )}{d}+\frac {c^{2} a}{d x}-\frac {i c^{3} a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {c^{3} a \arctan \left (c x \right )}{d}+\frac {b \,c^{3} \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}+\frac {i \arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {i}{2 c x}+\frac {\arctan \left (c x \right )}{c x}+i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {\ln \left (c x -i\right )^{2}}{4}-\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i \arctan \left (c x \right )}{2}-\frac {1}{6 c^{2} x^{2}}-\frac {4 \ln \left (c x \right )}{3}+\frac {2 \ln \left (c^{2} x^{2}+1\right )}{3}-i \arctan \left (c x \right ) \ln \left (c x -i\right )\right )}{d}\) | \(279\) |
| risch | \(-\frac {b c}{6 d \,x^{2}}+\frac {11 b \,c^{3} \ln \left (c^{2} x^{2}+1\right )}{24 d}+\frac {i b \,c^{2}}{2 d x}+\frac {i b \ln \left (i c x +1\right )}{6 d \,x^{3}}+\frac {i c^{3} \ln \left (-i c x \right ) a}{d}-\frac {c^{3} \ln \left (\frac {1}{2}-\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) b}{2 d}+\frac {c^{3} b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{2 d}-\frac {c b \ln \left (-i c x +1\right )}{4 d \,x^{2}}-\frac {i b \ln \left (-i c x +1\right )}{6 d \,x^{3}}+\frac {11 i c^{3} b \arctan \left (c x \right )}{12 d}+\frac {b c \ln \left (i c x +1\right )}{4 d \,x^{2}}+\frac {i c a}{2 d \,x^{2}}-\frac {i c^{3} a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {i c^{2} b \ln \left (-i c x +1\right )}{2 d x}-\frac {i b \,c^{2} \ln \left (i c x +1\right )}{2 d x}+\frac {c^{3} \operatorname {dilog}\left (-i c x +1\right ) b}{2 d}-\frac {5 c^{3} b \ln \left (-i c x \right )}{12 d}+\frac {5 c^{3} b \ln \left (-i c x +1\right )}{12 d}-\frac {c^{3} b \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d}-\frac {b \,c^{3} \ln \left (i c x +1\right )^{2}}{4 d}-\frac {b \,c^{3} \operatorname {dilog}\left (i c x +1\right )}{2 d}-\frac {11 b \,c^{3} \ln \left (i c x \right )}{12 d}-\frac {a}{3 d \,x^{3}}+\frac {c^{2} a}{d x}+\frac {c^{3} a \arctan \left (c x \right )}{d}\) | \(418\) |
c^3*(-1/3/d*a/c^3/x^3+1/2*I/d*a/c^2/x^2+I/d*a*ln(c*x)+a/d/c/x-1/2*I/d*a*ln (c^2*x^2+1)+1/d*a*arctan(c*x)+1/d*b*(-1/3*arctan(c*x)/c^3/x^3+1/2*I*arctan (c*x)/c^2/x^2+1/2*I/c/x+1/c/x*arctan(c*x)+I*arctan(c*x)*ln(c*x)-1/2*ln(c*x -I)*ln(-1/2*I*(c*x+I))-1/2*dilog(-1/2*I*(c*x+I))+1/4*ln(c*x-I)^2-1/2*ln(c* x)*ln(1+I*c*x)+1/2*ln(c*x)*ln(1-I*c*x)-1/2*dilog(1+I*c*x)+1/2*dilog(1-I*c* x)+1/2*I*arctan(c*x)-1/6/c^2/x^2-4/3*ln(c*x)+2/3*ln(c^2*x^2+1)-I*arctan(c* x)*ln(c*x-I)))
Time = 0.26 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \arctan (c x)}{x^4 (d+i c d x)} \, dx=\frac {6 \, b c^{3} x^{3} {\rm Li}_2\left (\frac {c x + i}{c x - i} + 1\right ) - 4 \, {\left (-3 i \, a + 4 \, b\right )} c^{3} x^{3} \log \left (x\right ) + 5 \, b c^{3} x^{3} \log \left (\frac {c x + i}{c}\right ) + {\left (-12 i \, a + 11 \, b\right )} c^{3} x^{3} \log \left (\frac {c x - i}{c}\right ) + 6 \, {\left (2 \, a + i \, b\right )} c^{2} x^{2} - 2 \, {\left (-3 i \, a + b\right )} c x + {\left (6 i \, b c^{2} x^{2} - 3 \, b c x - 2 i \, b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) - 4 \, a}{12 \, d x^{3}} \]
1/12*(6*b*c^3*x^3*dilog((c*x + I)/(c*x - I) + 1) - 4*(-3*I*a + 4*b)*c^3*x^ 3*log(x) + 5*b*c^3*x^3*log((c*x + I)/c) + (-12*I*a + 11*b)*c^3*x^3*log((c* x - I)/c) + 6*(2*a + I*b)*c^2*x^2 - 2*(-3*I*a + b)*c*x + (6*I*b*c^2*x^2 - 3*b*c*x - 2*I*b)*log(-(c*x + I)/(c*x - I)) - 4*a)/(d*x^3)
\[ \int \frac {a+b \arctan (c x)}{x^4 (d+i c d x)} \, dx=- \frac {i \left (\int \frac {a}{c x^{5} - i x^{4}}\, dx + \int \frac {b \operatorname {atan}{\left (c x \right )}}{c x^{5} - i x^{4}}\, dx\right )}{d} \]
\[ \int \frac {a+b \arctan (c x)}{x^4 (d+i c d x)} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (i \, c d x + d\right )} x^{4}} \,d x } \]
-1/6*(6*I*c^3*log(I*c*x + 1)/d - 6*I*c^3*log(x)/d - (6*c^2*x^2 + 3*I*c*x - 2)/(d*x^3))*a + (-I*c*integrate(arctan(c*x)/(c^2*d*x^5 + d*x^3), x) + int egrate(arctan(c*x)/(c^2*d*x^6 + d*x^4), x))*b
\[ \int \frac {a+b \arctan (c x)}{x^4 (d+i c d x)} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (i \, c d x + d\right )} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x^4 (d+i c d x)} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^4\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \]