Integrand size = 23, antiderivative size = 189 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^4} \, dx=-\frac {b c d^3}{6 x^2}-\frac {3 i b c^2 d^3}{2 x}-\frac {3}{2} i b c^3 d^3 \arctan (c x)-\frac {d^3 (a+b \arctan (c x))}{3 x^3}-\frac {3 i c d^3 (a+b \arctan (c x))}{2 x^2}+\frac {3 c^2 d^3 (a+b \arctan (c x))}{x}-i a c^3 d^3 \log (x)-\frac {10}{3} b c^3 d^3 \log (x)+\frac {5}{3} b c^3 d^3 \log \left (1+c^2 x^2\right )+\frac {1}{2} b c^3 d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} b c^3 d^3 \operatorname {PolyLog}(2,i c x) \]
-1/6*b*c*d^3/x^2-3/2*I*b*c^2*d^3/x-3/2*I*b*c^3*d^3*arctan(c*x)-1/3*d^3*(a+ b*arctan(c*x))/x^3-3/2*I*c*d^3*(a+b*arctan(c*x))/x^2+3*c^2*d^3*(a+b*arctan (c*x))/x-I*a*c^3*d^3*ln(x)-10/3*b*c^3*d^3*ln(x)+5/3*b*c^3*d^3*ln(c^2*x^2+1 )+1/2*b*c^3*d^3*polylog(2,-I*c*x)-1/2*b*c^3*d^3*polylog(2,I*c*x)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.08 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.90 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^4} \, dx=\frac {d^3 \left (-2 a-9 i a c x-b c x+18 a c^2 x^2-2 b \arctan (c x)-9 i b c x \arctan (c x)+18 b c^2 x^2 \arctan (c x)-9 i b c^2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )-6 i a c^3 x^3 \log (x)-20 b c^3 x^3 \log (x)+10 b c^3 x^3 \log \left (1+c^2 x^2\right )+3 b c^3 x^3 \operatorname {PolyLog}(2,-i c x)-3 b c^3 x^3 \operatorname {PolyLog}(2,i c x)\right )}{6 x^3} \]
(d^3*(-2*a - (9*I)*a*c*x - b*c*x + 18*a*c^2*x^2 - 2*b*ArcTan[c*x] - (9*I)* b*c*x*ArcTan[c*x] + 18*b*c^2*x^2*ArcTan[c*x] - (9*I)*b*c^2*x^2*Hypergeomet ric2F1[-1/2, 1, 1/2, -(c^2*x^2)] - (6*I)*a*c^3*x^3*Log[x] - 20*b*c^3*x^3*L og[x] + 10*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]))/(6*x^3)
Time = 0.39 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {5411, 2009}
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.
| \(\displaystyle \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^4} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (-\frac {i c^3 d^3 (a+b \arctan (c x))}{x}-\frac {3 c^2 d^3 (a+b \arctan (c x))}{x^2}+\frac {d^3 (a+b \arctan (c x))}{x^4}+\frac {3 i c d^3 (a+b \arctan (c x))}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 c^2 d^3 (a+b \arctan (c x))}{x}-\frac {d^3 (a+b \arctan (c x))}{3 x^3}-\frac {3 i c d^3 (a+b \arctan (c x))}{2 x^2}-i a c^3 d^3 \log (x)-\frac {3}{2} i b c^3 d^3 \arctan (c x)+\frac {1}{2} b c^3 d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} b c^3 d^3 \operatorname {PolyLog}(2,i c x)-\frac {10}{3} b c^3 d^3 \log (x)-\frac {3 i b c^2 d^3}{2 x}+\frac {5}{3} b c^3 d^3 \log \left (c^2 x^2+1\right )-\frac {b c d^3}{6 x^2}\) |
-1/6*(b*c*d^3)/x^2 - (((3*I)/2)*b*c^2*d^3)/x - ((3*I)/2)*b*c^3*d^3*ArcTan[ c*x] - (d^3*(a + b*ArcTan[c*x]))/(3*x^3) - (((3*I)/2)*c*d^3*(a + b*ArcTan[ c*x]))/x^2 + (3*c^2*d^3*(a + b*ArcTan[c*x]))/x - I*a*c^3*d^3*Log[x] - (10* b*c^3*d^3*Log[x])/3 + (5*b*c^3*d^3*Log[1 + c^2*x^2])/3 + (b*c^3*d^3*PolyLo g[2, (-I)*c*x])/2 - (b*c^3*d^3*PolyLog[2, I*c*x])/2
3.1.27.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f* x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] & & IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
Time = 1.28 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.97
| method | result | size |
| parts | \(a \,d^{3} \left (-\frac {3 i c}{2 x^{2}}-i c^{3} \ln \left (x \right )+\frac {3 c^{2}}{x}-\frac {1}{3 x^{3}}\right )+b \,d^{3} c^{3} \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-i \arctan \left (c x \right ) \ln \left (c x \right )+\frac {3 \arctan \left (c x \right )}{c x}-\frac {3 i \arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {3 i}{2 c x}-\frac {1}{6 c^{2} x^{2}}-\frac {10 \ln \left (c x \right )}{3}+\frac {5 \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 i \arctan \left (c x \right )}{2}\right )\) | \(183\) |
| derivativedivides | \(c^{3} \left (a \,d^{3} \left (-\frac {1}{3 c^{3} x^{3}}-i \ln \left (c x \right )+\frac {3}{c x}-\frac {3 i}{2 c^{2} x^{2}}\right )+b \,d^{3} \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-i \arctan \left (c x \right ) \ln \left (c x \right )+\frac {3 \arctan \left (c x \right )}{c x}-\frac {3 i \arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {3 i}{2 c x}-\frac {1}{6 c^{2} x^{2}}-\frac {10 \ln \left (c x \right )}{3}+\frac {5 \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 i \arctan \left (c x \right )}{2}\right )\right )\) | \(188\) |
| default | \(c^{3} \left (a \,d^{3} \left (-\frac {1}{3 c^{3} x^{3}}-i \ln \left (c x \right )+\frac {3}{c x}-\frac {3 i}{2 c^{2} x^{2}}\right )+b \,d^{3} \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-i \arctan \left (c x \right ) \ln \left (c x \right )+\frac {3 \arctan \left (c x \right )}{c x}-\frac {3 i \arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {3 i}{2 c x}-\frac {1}{6 c^{2} x^{2}}-\frac {10 \ln \left (c x \right )}{3}+\frac {5 \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 i \arctan \left (c x \right )}{2}\right )\right )\) | \(188\) |
| risch | \(\frac {b \,d^{3} c^{3} \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {11 b \,d^{3} c^{3} \ln \left (i c x \right )}{12}+\frac {5 \ln \left (c^{2} x^{2}+1\right ) b \,c^{3} d^{3}}{3}-i c^{3} d^{3} a \ln \left (-i c x \right )+\frac {i b \,d^{3} \ln \left (i c x +1\right )}{6 x^{3}}-\frac {3 i b \,c^{3} d^{3} \arctan \left (c x \right )}{2}-\frac {3 b \,d^{3} c \ln \left (i c x +1\right )}{4 x^{2}}-\frac {b c \,d^{3}}{6 x^{2}}-\frac {3 i b \,d^{3} c^{2} \ln \left (i c x +1\right )}{2 x}-\frac {i d^{3} b \ln \left (-i c x +1\right )}{6 x^{3}}+\frac {3 c^{2} d^{3} a}{x}-\frac {3 i c \,d^{3} a}{2 x^{2}}-\frac {d^{3} a}{3 x^{3}}-\frac {c^{3} d^{3} b \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {29 c^{3} d^{3} b \ln \left (-i c x \right )}{12}+\frac {3 i c^{2} d^{3} b \ln \left (-i c x +1\right )}{2 x}-\frac {3 i b \,c^{2} d^{3}}{2 x}+\frac {3 c \,d^{3} b \ln \left (-i c x +1\right )}{4 x^{2}}\) | \(284\) |
a*d^3*(-3/2*I*c/x^2-I*c^3*ln(x)+3*c^2/x-1/3/x^3)+b*d^3*c^3*(-1/3*arctan(c* x)/c^3/x^3-I*arctan(c*x)*ln(c*x)+3/c/x*arctan(c*x)-3/2*I*arctan(c*x)/c^2/x ^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)-3/2*I/c/x-1/6/c^2/x^2-10/3*ln(c*x)+5/3*ln(c^2*x^2+1)-3/2*I* arctan(c*x))
\[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^4} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{4}} \,d x } \]
integral(1/2*(-2*I*a*c^3*d^3*x^3 - 6*a*c^2*d^3*x^2 + 6*I*a*c*d^3*x + 2*a*d ^3 + (b*c^3*d^3*x^3 - 3*I*b*c^2*d^3*x^2 - 3*b*c*d^3*x + I*b*d^3)*log(-(c*x + I)/(c*x - I)))/x^4, x)
Timed out. \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^4} \, dx=\text {Timed out} \]
\[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^4} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{4}} \,d x } \]
-I*b*c^3*d^3*integrate(arctan(c*x)/x, x) - I*a*c^3*d^3*log(x) + 3/2*(c*(lo g(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*c^2*d^3 - 3/2*I*((c*arctan (c*x) + 1/x)*c + arctan(c*x)/x^2)*b*c*d^3 + 1/6*((c^2*log(c^2*x^2 + 1) - c ^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*d^3 + 3*a*c^2*d^3/x - 3/2*I* a*c*d^3/x^2 - 1/3*a*d^3/x^3
\[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^4} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{4}} \,d x } \]
Time = 1.08 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.17 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^4} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^3}{3\,x^3} & \text {\ if\ \ }c=0\\ \frac {b\,c^3\,d^3\,\ln \left (-\frac {3\,c^6\,x^2}{2}-\frac {3\,c^4}{2}\right )}{6}-\frac {b\,c^3\,d^3\,\ln \left (x\right )}{3}-\frac {b\,c^3\,d^3\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )}{2}-3\,b\,c\,d^3\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )-\frac {b\,c\,d^3}{6\,x^2}-\frac {a\,d^3\,\left (2-18\,c^2\,x^2+c\,x\,9{}\mathrm {i}+c^3\,x^3\,\ln \left (x\right )\,6{}\mathrm {i}\right )}{6\,x^3}-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}+\frac {3\,b\,c^2\,d^3\,\mathrm {atan}\left (c\,x\right )}{x}-\frac {b\,d^3\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )\,3{}\mathrm {i}}{2}-\frac {b\,c\,d^3\,\mathrm {atan}\left (c\,x\right )\,3{}\mathrm {i}}{2\,x^2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
piecewise(c == 0, -(a*d^3)/(3*x^3), c ~= 0, - (b*d^3*(c^3*atan(c*x) + c^2/ x)*3i)/2 - (b*c^3*d^3*(dilog(- c*x*1i + 1) - dilog(c*x*1i + 1)))/2 - (b*c^ 3*d^3*log(x))/3 + (b*c^3*d^3*log(- (3*c^4)/2 - (3*c^6*x^2)/2))/6 - 3*b*c*d ^3*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2) - (b*c*d^3)/(6*x^2) - (a*d^3*(c *x*9i - 18*c^2*x^2 + c^3*x^3*log(x)*6i + 2))/(6*x^3) - (b*d^3*atan(c*x))/( 3*x^3) - (b*c*d^3*atan(c*x)*3i)/(2*x^2) + (3*b*c^2*d^3*atan(c*x))/x)