Integrand size = 23, antiderivative size = 227 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=-\frac {b c d^4}{12 x^3}-\frac {2 i b c^2 d^4}{3 x^2}+\frac {13 b c^3 d^4}{4 x}+\frac {13}{4} b c^4 d^4 \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{4 x^4}-\frac {4 i c d^4 (a+b \arctan (c x))}{3 x^3}+\frac {3 c^2 d^4 (a+b \arctan (c x))}{x^2}+\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}+a c^4 d^4 \log (x)-\frac {16}{3} i b c^4 d^4 \log (x)+\frac {8}{3} i b c^4 d^4 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,i c x) \]
-1/12*b*c*d^4/x^3-2/3*I*b*c^2*d^4/x^2+13/4*b*c^3*d^4/x+13/4*b*c^4*d^4*arct an(c*x)-1/4*d^4*(a+b*arctan(c*x))/x^4-4/3*I*c*d^4*(a+b*arctan(c*x))/x^3+3* c^2*d^4*(a+b*arctan(c*x))/x^2+4*I*c^3*d^4*(a+b*arctan(c*x))/x+a*c^4*d^4*ln (x)-16/3*I*b*c^4*d^4*ln(x)+8/3*I*b*c^4*d^4*ln(c^2*x^2+1)+1/2*I*b*c^4*d^4*p olylog(2,-I*c*x)-1/2*I*b*c^4*d^4*polylog(2,I*c*x)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.09 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\frac {d^4 \left (-3 a-16 i a c x+36 a c^2 x^2-8 i b c^2 x^2+48 i a c^3 x^3-3 b \arctan (c x)-16 i b c x \arctan (c x)+36 b c^2 x^2 \arctan (c x)+48 i b c^3 x^3 \arctan (c x)-b c x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-c^2 x^2\right )+36 b c^3 x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )+12 a c^4 x^4 \log (x)-64 i b c^4 x^4 \log (x)+32 i b c^4 x^4 \log \left (1+c^2 x^2\right )+6 i b c^4 x^4 \operatorname {PolyLog}(2,-i c x)-6 i b c^4 x^4 \operatorname {PolyLog}(2,i c x)\right )}{12 x^4} \]
(d^4*(-3*a - (16*I)*a*c*x + 36*a*c^2*x^2 - (8*I)*b*c^2*x^2 + (48*I)*a*c^3* x^3 - 3*b*ArcTan[c*x] - (16*I)*b*c*x*ArcTan[c*x] + 36*b*c^2*x^2*ArcTan[c*x ] + (48*I)*b*c^3*x^3*ArcTan[c*x] - b*c*x*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 36*b*c^3*x^3*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] + 1 2*a*c^4*x^4*Log[x] - (64*I)*b*c^4*x^4*Log[x] + (32*I)*b*c^4*x^4*Log[1 + c^ 2*x^2] + (6*I)*b*c^4*x^4*PolyLog[2, (-I)*c*x] - (6*I)*b*c^4*x^4*PolyLog[2, I*c*x]))/(12*x^4)
Time = 0.43 (sec) , antiderivative size = 227, 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.
| \(\displaystyle \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (\frac {c^4 d^4 (a+b \arctan (c x))}{x}-\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x^2}-\frac {6 c^2 d^4 (a+b \arctan (c x))}{x^3}+\frac {d^4 (a+b \arctan (c x))}{x^5}+\frac {4 i c d^4 (a+b \arctan (c x))}{x^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}+\frac {3 c^2 d^4 (a+b \arctan (c x))}{x^2}-\frac {d^4 (a+b \arctan (c x))}{4 x^4}-\frac {4 i c d^4 (a+b \arctan (c x))}{3 x^3}+a c^4 d^4 \log (x)+\frac {13}{4} b c^4 d^4 \arctan (c x)+\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,i c x)-\frac {16}{3} i b c^4 d^4 \log (x)+\frac {13 b c^3 d^4}{4 x}-\frac {2 i b c^2 d^4}{3 x^2}+\frac {8}{3} i b c^4 d^4 \log \left (c^2 x^2+1\right )-\frac {b c d^4}{12 x^3}\) |
-1/12*(b*c*d^4)/x^3 - (((2*I)/3)*b*c^2*d^4)/x^2 + (13*b*c^3*d^4)/(4*x) + ( 13*b*c^4*d^4*ArcTan[c*x])/4 - (d^4*(a + b*ArcTan[c*x]))/(4*x^4) - (((4*I)/ 3)*c*d^4*(a + b*ArcTan[c*x]))/x^3 + (3*c^2*d^4*(a + b*ArcTan[c*x]))/x^2 + ((4*I)*c^3*d^4*(a + b*ArcTan[c*x]))/x + a*c^4*d^4*Log[x] - ((16*I)/3)*b*c^ 4*d^4*Log[x] + ((8*I)/3)*b*c^4*d^4*Log[1 + c^2*x^2] + (I/2)*b*c^4*d^4*Poly Log[2, (-I)*c*x] - (I/2)*b*c^4*d^4*PolyLog[2, I*c*x]
3.1.39.3.1 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 = 1.90 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.94
| method | result | size |
| parts | \(d^{4} a \left (\frac {3 c^{2}}{x^{2}}+c^{4} \ln \left (x \right )-\frac {1}{4 x^{4}}+\frac {4 i c^{3}}{x}-\frac {4 i c}{3 x^{3}}\right )+d^{4} b \,c^{4} \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}-\frac {4 i \arctan \left (c x \right )}{3 c^{3} x^{3}}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {4 i \arctan \left (c x \right )}{c x}+\frac {3 \arctan \left (c x \right )}{c^{2} x^{2}}-\frac {2 i}{3 c^{2} x^{2}}-\frac {16 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {13}{4 c x}+\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {13 \arctan \left (c x \right )}{4}+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\) | \(214\) |
| derivativedivides | \(c^{4} \left (d^{4} a \left (-\frac {1}{4 c^{4} x^{4}}-\frac {4 i}{3 c^{3} x^{3}}+\ln \left (c x \right )+\frac {4 i}{c x}+\frac {3}{c^{2} x^{2}}\right )+d^{4} b \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}-\frac {4 i \arctan \left (c x \right )}{3 c^{3} x^{3}}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {4 i \arctan \left (c x \right )}{c x}+\frac {3 \arctan \left (c x \right )}{c^{2} x^{2}}-\frac {2 i}{3 c^{2} x^{2}}-\frac {16 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {13}{4 c x}+\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {13 \arctan \left (c x \right )}{4}+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )\) | \(218\) |
| default | \(c^{4} \left (d^{4} a \left (-\frac {1}{4 c^{4} x^{4}}-\frac {4 i}{3 c^{3} x^{3}}+\ln \left (c x \right )+\frac {4 i}{c x}+\frac {3}{c^{2} x^{2}}\right )+d^{4} b \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}-\frac {4 i \arctan \left (c x \right )}{3 c^{3} x^{3}}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {4 i \arctan \left (c x \right )}{c x}+\frac {3 \arctan \left (c x \right )}{c^{2} x^{2}}-\frac {2 i}{3 c^{2} x^{2}}-\frac {16 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {13}{4 c x}+\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {13 \arctan \left (c x \right )}{4}+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )\) | \(218\) |
| risch | \(-\frac {b c \,d^{4}}{12 x^{3}}+\frac {13 b \,c^{3} d^{4}}{4 x}-\frac {25 i b \,d^{4} c^{4} \ln \left (i c x \right )}{24}-\frac {3 i b \,d^{4} c^{2} \ln \left (i c x +1\right )}{2 x^{2}}+\frac {8 i b \,c^{4} d^{4} \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {3 i c^{2} d^{4} b \ln \left (-i c x +1\right )}{2 x^{2}}+\frac {i b \,d^{4} c^{4} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {13 b \,c^{4} d^{4} \arctan \left (c x \right )}{4}+\frac {i b \,d^{4} \ln \left (i c x +1\right )}{8 x^{4}}-\frac {103 i c^{4} d^{4} b \ln \left (-i c x \right )}{24}+\frac {4 i c^{3} d^{4} a}{x}-\frac {i d^{4} b \ln \left (-i c x +1\right )}{8 x^{4}}-\frac {i c^{4} d^{4} b \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {4 i c \,d^{4} a}{3 x^{3}}+\frac {2 b \,d^{4} c^{3} \ln \left (i c x +1\right )}{x}-\frac {2 b \,d^{4} c \ln \left (i c x +1\right )}{3 x^{3}}+\frac {2 c \,d^{4} b \ln \left (-i c x +1\right )}{3 x^{3}}-\frac {2 c^{3} d^{4} b \ln \left (-i c x +1\right )}{x}+c^{4} d^{4} a \ln \left (-i c x \right )+\frac {3 c^{2} d^{4} a}{x^{2}}-\frac {2 i b \,c^{2} d^{4}}{3 x^{2}}-\frac {d^{4} a}{4 x^{4}}\) | \(351\) |
d^4*a*(3*c^2/x^2+c^4*ln(x)-1/4/x^4+4*I*c^3/x-4/3*I*c/x^3)+d^4*b*c^4*(-1/4* arctan(c*x)/c^4/x^4-4/3*I*arctan(c*x)/c^3/x^3+arctan(c*x)*ln(c*x)+4*I*arct an(c*x)/c/x+3/c^2/x^2*arctan(c*x)-2/3*I/c^2/x^2-16/3*I*ln(c*x)-1/12/c^3/x^ 3+13/4/c/x+8/3*I*ln(c^2*x^2+1)+13/4*arctan(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))
\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]
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^5, x)
Timed out. \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\text {Timed out} \]
\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]
b*c^4*d^4*integrate(arctan(c*x)/x, x) + a*c^4*d^4*log(x) + 2*I*(c*(log(c^2 *x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*c^3*d^4 + 3*((c*arctan(c*x) + 1 /x)*c + arctan(c*x)/x^2)*b*c^2*d^4 + 2/3*I*((c^2*log(c^2*x^2 + 1) - c^2*lo g(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*c*d^4 + 4*I*a*c^3*d^4/x + 1/12*(( 3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d^4 + 3* a*c^2*d^4/x^2 - 4/3*I*a*c*d^4/x^3 - 1/4*a*d^4/x^4
\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]
Time = 1.01 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.31 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^4}{4\,x^4} & \text {\ if\ \ }c=0\\ 3\,b\,c\,d^4\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )-\frac {b\,d^4\,\left (\frac {\frac {c^2}{3}-c^4\,x^2}{x^3}-c^5\,\mathrm {atan}\left (c\,x\right )\right )}{4\,c}-\frac {b\,c^4\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,c^4\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-b\,c^2\,d^4\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )\,4{}\mathrm {i}+\frac {a\,d^4\,\left (36\,c^2\,x^2+12\,c^4\,x^4\,\ln \left (x\right )-3-c\,x\,16{}\mathrm {i}+c^3\,x^3\,48{}\mathrm {i}\right )}{12\,x^4}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{4\,x^4}-\frac {b\,d^4\,\left (c^4\,\ln \left (x\right )-\frac {c^4\,\ln \left (-\frac {c^4\,\left (3\,c^2\,x^2+1\right )}{2}-c^4\right )}{2}+\frac {c^2}{2\,x^2}\right )\,4{}\mathrm {i}}{3}-\frac {b\,c\,d^4\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{3\,x^3}+\frac {3\,b\,c^2\,d^4\,\mathrm {atan}\left (c\,x\right )}{x^2}+\frac {b\,c^3\,d^4\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{x} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
piecewise(c == 0, -(a*d^4)/(4*x^4), c ~= 0, - (b*d^4*(c^4*log(x) - (c^4*lo g(- (c^4*(3*c^2*x^2 + 1))/2 - c^4))/2 + c^2/(2*x^2))*4i)/3 - (b*d^4*((c^2/ 3 - c^4*x^2)/x^3 - c^5*atan(c*x)))/(4*c) - (b*c^4*d^4*dilog(- c*x*1i + 1)* 1i)/2 + (b*c^4*d^4*dilog(c*x*1i + 1)*1i)/2 - b*c^2*d^4*(c^2*log(x) - (c^2* log(c^2*x^2 + 1))/2)*4i + (a*d^4*(- c*x*16i + 36*c^2*x^2 + c^3*x^3*48i + 1 2*c^4*x^4*log(x) - 3))/(12*x^4) - (b*d^4*atan(c*x))/(4*x^4) + 3*b*c*d^4*(c ^3*atan(c*x) + c^2/x) - (b*c*d^4*atan(c*x)*4i)/(3*x^3) + (3*b*c^2*d^4*atan (c*x))/x^2 + (b*c^3*d^4*atan(c*x)*4i)/x)