Integrand size = 23, antiderivative size = 203 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x} \, dx=4 i a c d^4 x+\frac {13}{4} b c d^4 x+\frac {2}{3} i b c^2 d^4 x^2-\frac {1}{12} b c^3 d^4 x^3-\frac {13}{4} b d^4 \arctan (c x)+4 i b c d^4 x \arctan (c x)-3 c^2 d^4 x^2 (a+b \arctan (c x))-\frac {4}{3} i c^3 d^4 x^3 (a+b \arctan (c x))+\frac {1}{4} c^4 d^4 x^4 (a+b \arctan (c x))+a d^4 \log (x)-\frac {8}{3} i b d^4 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^4 \operatorname {PolyLog}(2,i c x) \]
4*I*a*c*d^4*x+13/4*b*c*d^4*x+2/3*I*b*c^2*d^4*x^2-1/12*b*c^3*d^4*x^3-13/4*b *d^4*arctan(c*x)+4*I*b*c*d^4*x*arctan(c*x)-3*c^2*d^4*x^2*(a+b*arctan(c*x)) -4/3*I*c^3*d^4*x^3*(a+b*arctan(c*x))+1/4*c^4*d^4*x^4*(a+b*arctan(c*x))+a*d ^4*ln(x)-8/3*I*b*d^4*ln(c^2*x^2+1)+1/2*I*b*d^4*polylog(2,-I*c*x)-1/2*I*b*d ^4*polylog(2,I*c*x)
Time = 0.12 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.86 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x} \, dx=\frac {1}{12} d^4 \left (48 i a c x+39 b c x-36 a c^2 x^2+8 i b c^2 x^2-16 i a c^3 x^3-b c^3 x^3+3 a c^4 x^4-39 b \arctan (c x)+48 i b c x \arctan (c x)-36 b c^2 x^2 \arctan (c x)-16 i b c^3 x^3 \arctan (c x)+3 b c^4 x^4 \arctan (c x)+12 a \log (x)-32 i b \log \left (1+c^2 x^2\right )+6 i b \operatorname {PolyLog}(2,-i c x)-6 i b \operatorname {PolyLog}(2,i c x)\right ) \]
(d^4*((48*I)*a*c*x + 39*b*c*x - 36*a*c^2*x^2 + (8*I)*b*c^2*x^2 - (16*I)*a* c^3*x^3 - b*c^3*x^3 + 3*a*c^4*x^4 - 39*b*ArcTan[c*x] + (48*I)*b*c*x*ArcTan [c*x] - 36*b*c^2*x^2*ArcTan[c*x] - (16*I)*b*c^3*x^3*ArcTan[c*x] + 3*b*c^4* x^4*ArcTan[c*x] + 12*a*Log[x] - (32*I)*b*Log[1 + c^2*x^2] + (6*I)*b*PolyLo g[2, (-I)*c*x] - (6*I)*b*PolyLog[2, I*c*x]))/12
Time = 0.41 (sec) , antiderivative size = 203, 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.
| \(\displaystyle \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (c^4 d^4 x^3 (a+b \arctan (c x))-4 i c^3 d^4 x^2 (a+b \arctan (c x))-6 c^2 d^4 x (a+b \arctan (c x))+4 i c d^4 (a+b \arctan (c x))+\frac {d^4 (a+b \arctan (c x))}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} c^4 d^4 x^4 (a+b \arctan (c x))-\frac {4}{3} i c^3 d^4 x^3 (a+b \arctan (c x))-3 c^2 d^4 x^2 (a+b \arctan (c x))+4 i a c d^4 x+a d^4 \log (x)-\frac {13}{4} b d^4 \arctan (c x)+4 i b c d^4 x \arctan (c x)-\frac {1}{12} b c^3 d^4 x^3+\frac {2}{3} i b c^2 d^4 x^2-\frac {8}{3} i b d^4 \log \left (c^2 x^2+1\right )+\frac {1}{2} i b d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^4 \operatorname {PolyLog}(2,i c x)+\frac {13}{4} b c d^4 x\) |
(4*I)*a*c*d^4*x + (13*b*c*d^4*x)/4 + ((2*I)/3)*b*c^2*d^4*x^2 - (b*c^3*d^4* x^3)/12 - (13*b*d^4*ArcTan[c*x])/4 + (4*I)*b*c*d^4*x*ArcTan[c*x] - 3*c^2*d ^4*x^2*(a + b*ArcTan[c*x]) - ((4*I)/3)*c^3*d^4*x^3*(a + b*ArcTan[c*x]) + ( c^4*d^4*x^4*(a + b*ArcTan[c*x]))/4 + a*d^4*Log[x] - ((8*I)/3)*b*d^4*Log[1 + c^2*x^2] + (I/2)*b*d^4*PolyLog[2, (-I)*c*x] - (I/2)*b*d^4*PolyLog[2, I*c *x]
3.1.35.3.1 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 = 1.42 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.95
| method | result | size |
| parts | \(d^{4} a \left (\frac {c^{4} x^{4}}{4}-\frac {4 i c^{3} x^{3}}{3}-3 c^{2} x^{2}+4 i c x +\ln \left (x \right )\right )+d^{4} b \left (4 i \arctan \left (c x \right ) c x +\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}-\frac {4 i \arctan \left (c x \right ) c^{3} x^{3}}{3}-3 c^{2} x^{2} \arctan \left (c x \right )+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {13 c x}{4}-\frac {c^{3} x^{3}}{12}+\frac {2 i c^{2} x^{2}}{3}-\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {13 \arctan \left (c x \right )}{4}\right )\) | \(193\) |
| derivativedivides | \(d^{4} a \left (4 i c x +\frac {c^{4} x^{4}}{4}-\frac {4 i c^{3} x^{3}}{3}-3 c^{2} x^{2}+\ln \left (c x \right )\right )+d^{4} b \left (4 i \arctan \left (c x \right ) c x +\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}-\frac {4 i \arctan \left (c x \right ) c^{3} x^{3}}{3}-3 c^{2} x^{2} \arctan \left (c x \right )+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {13 c x}{4}-\frac {c^{3} x^{3}}{12}+\frac {2 i c^{2} x^{2}}{3}-\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {13 \arctan \left (c x \right )}{4}\right )\) | \(195\) |
| default | \(d^{4} a \left (4 i c x +\frac {c^{4} x^{4}}{4}-\frac {4 i c^{3} x^{3}}{3}-3 c^{2} x^{2}+\ln \left (c x \right )\right )+d^{4} b \left (4 i \arctan \left (c x \right ) c x +\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}-\frac {4 i \arctan \left (c x \right ) c^{3} x^{3}}{3}-3 c^{2} x^{2} \arctan \left (c x \right )+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {13 c x}{4}-\frac {c^{3} x^{3}}{12}+\frac {2 i c^{2} x^{2}}{3}-\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {13 \arctan \left (c x \right )}{4}\right )\) | \(195\) |
| risch | \(-3 x^{2} d^{4} c^{2} a +\frac {13 b c \,d^{4} x}{4}-\frac {103 d^{4} a}{12}-\frac {b \,c^{3} d^{4} x^{3}}{12}-\frac {i b \,d^{4} \ln \left (i c x +1\right ) c^{4} x^{4}}{8}+\frac {i b \,d^{4} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {a \,c^{4} d^{4} x^{4}}{4}-\frac {103 i d^{4} b \ln \left (-i c x +1\right )}{24}+\frac {58 i b \,d^{4}}{9}-\frac {2 b \,d^{4} \ln \left (i c x +1\right ) c^{3} x^{3}}{3}+\frac {3 i b \,d^{4} \ln \left (i c x +1\right ) c^{2} x^{2}}{2}+4 i a c \,d^{4} x -\frac {3 i d^{4} b \ln \left (-i c x +1\right ) c^{2} x^{2}}{2}+d^{4} a \ln \left (-i c x \right )+\frac {2 d^{4} b \,c^{3} x^{3} \ln \left (-i c x +1\right )}{3}-2 d^{4} b c x \ln \left (-i c x +1\right )-\frac {4 i x^{3} a \,c^{3} d^{4}}{3}+2 b \,d^{4} \ln \left (i c x +1\right ) c x -\frac {25 i b \,d^{4} \ln \left (i c x +1\right )}{24}+\frac {2 i x^{2} b \,c^{2} d^{4}}{3}+\frac {i d^{4} b \ln \left (-i c x +1\right ) c^{4} x^{4}}{8}-\frac {i d^{4} b \operatorname {dilog}\left (-i c x +1\right )}{2}\) | \(321\) |
d^4*a*(1/4*c^4*x^4-4/3*I*c^3*x^3-3*c^2*x^2+4*I*c*x+ln(x))+d^4*b*(4*I*arcta n(c*x)*c*x+1/4*c^4*x^4*arctan(c*x)-4/3*I*arctan(c*x)*c^3*x^3-3*c^2*x^2*arc tan(c*x)+arctan(c*x)*ln(c*x)+1/2*I*ln(c*x)*ln(1+I*c*x)-1/2*I*ln(c*x)*ln(1- I*c*x)+1/2*I*dilog(1+I*c*x)-1/2*I*dilog(1-I*c*x)+13/4*c*x-1/12*c^3*x^3+2/3 *I*c^2*x^2-8/3*I*ln(c^2*x^2+1)-13/4*arctan(c*x))
\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]
integral(1/2*(2*a*c^4*d^4*x^4 - 8*I*a*c^3*d^4*x^3 - 12*a*c^2*d^4*x^2 + 8*I *a*c*d^4*x + 2*a*d^4 + (I*b*c^4*d^4*x^4 + 4*b*c^3*d^4*x^3 - 6*I*b*c^2*d^4* x^2 - 4*b*c*d^4*x + I*b*d^4)*log(-(c*x + I)/(c*x - I)))/x, x)
Timed out. \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x} \, dx=\text {Timed out} \]
Time = 0.45 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.08 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x} \, dx=\frac {1}{4} \, a c^{4} d^{4} x^{4} - \frac {4}{3} i \, a c^{3} d^{4} x^{3} - \frac {1}{12} \, b c^{3} d^{4} x^{3} - 3 \, a c^{2} d^{4} x^{2} + \frac {2}{3} i \, b c^{2} d^{4} x^{2} + 4 i \, a c d^{4} x + \frac {13}{4} \, b c d^{4} x - \frac {1}{12} \, {\left (3 \, \pi + 8 i\right )} b d^{4} \log \left (c^{2} x^{2} + 1\right ) + b d^{4} \arctan \left (c x\right ) \log \left (c x\right ) + 2 i \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{4} - \frac {1}{2} i \, b d^{4} {\rm Li}_2\left (i \, c x + 1\right ) + \frac {1}{2} i \, b d^{4} {\rm Li}_2\left (-i \, c x + 1\right ) + a d^{4} \log \left (x\right ) + \frac {1}{12} \, {\left (3 \, b c^{4} d^{4} x^{4} - 16 i \, b c^{3} d^{4} x^{3} - 36 \, b c^{2} d^{4} x^{2} - 39 \, b d^{4}\right )} \arctan \left (c x\right ) \]
1/4*a*c^4*d^4*x^4 - 4/3*I*a*c^3*d^4*x^3 - 1/12*b*c^3*d^4*x^3 - 3*a*c^2*d^4 *x^2 + 2/3*I*b*c^2*d^4*x^2 + 4*I*a*c*d^4*x + 13/4*b*c*d^4*x - 1/12*(3*pi + 8*I)*b*d^4*log(c^2*x^2 + 1) + b*d^4*arctan(c*x)*log(c*x) + 2*I*(2*c*x*arc tan(c*x) - log(c^2*x^2 + 1))*b*d^4 - 1/2*I*b*d^4*dilog(I*c*x + 1) + 1/2*I* b*d^4*dilog(-I*c*x + 1) + a*d^4*log(x) + 1/12*(3*b*c^4*d^4*x^4 - 16*I*b*c^ 3*d^4*x^3 - 36*b*c^2*d^4*x^2 - 39*b*d^4)*arctan(c*x)
\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]
Time = 1.05 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.22 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x} \, dx=\left \{\begin {array}{cl} a\,d^4\,\ln \left (x\right ) & \text {\ if\ \ }c=0\\ a\,d^4\,\ln \left (x\right )-b\,d^4\,\ln \left (c^2\,x^2+1\right )\,2{}\mathrm {i}-\frac {b\,d^4\,\left (3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3\right )}{12}-\frac {b\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-3\,a\,c^2\,d^4\,x^2-\frac {a\,c^3\,d^4\,x^3\,4{}\mathrm {i}}{3}+\frac {a\,c^4\,d^4\,x^4}{4}+a\,c\,d^4\,x\,4{}\mathrm {i}+3\,b\,c\,d^4\,x+\frac {b\,c^2\,d^4\,\left (\frac {x^2}{2}-\frac {\ln \left (c^2\,x^2+1\right )}{2\,c^2}\right )\,4{}\mathrm {i}}{3}-6\,b\,c^2\,d^4\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )-\frac {b\,c^3\,d^4\,x^3\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{3}+\frac {b\,c^4\,d^4\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}+b\,c\,d^4\,x\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
piecewise(c == 0, a*d^4*log(x), c ~= 0, - (b*d^4*(3*atan(c*x) - 3*c*x + c^ 3*x^3))/12 - b*d^4*log(c^2*x^2 + 1)*2i + a*d^4*log(x) - (b*d^4*dilog(- c*x *1i + 1)*1i)/2 + (b*d^4*dilog(c*x*1i + 1)*1i)/2 - 3*a*c^2*d^4*x^2 - (a*c^3 *d^4*x^3*4i)/3 + (a*c^4*d^4*x^4)/4 + a*c*d^4*x*4i + 3*b*c*d^4*x + (b*c^2*d ^4*(x^2/2 - log(c^2*x^2 + 1)/(2*c^2))*4i)/3 - 6*b*c^2*d^4*atan(c*x)*(1/(2* c^2) + x^2/2) - (b*c^3*d^4*x^3*atan(c*x)*4i)/3 + (b*c^4*d^4*x^4*atan(c*x)) /4 + b*c*d^4*x*atan(c*x)*4i)