Integrand size = 21, antiderivative size = 200 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^5} \, dx=-\frac {b c d^3}{12 x^3}+\frac {b c^3 d^3}{4 x}-\frac {3 b c d^2 e}{2 x}-\frac {b e^3 x}{2 c}+\frac {1}{4} b c^4 d^3 \arctan (c x)-\frac {3}{2} b c^2 d^2 e \arctan (c x)+\frac {b e^3 \arctan (c x)}{2 c^2}-\frac {d^3 (a+b \arctan (c x))}{4 x^4}-\frac {3 d^2 e (a+b \arctan (c x))}{2 x^2}+\frac {1}{2} e^3 x^2 (a+b \arctan (c x))+3 a d e^2 \log (x)+\frac {3}{2} i b d e^2 \operatorname {PolyLog}(2,-i c x)-\frac {3}{2} i b d e^2 \operatorname {PolyLog}(2,i c x) \]
-1/12*b*c*d^3/x^3+1/4*b*c^3*d^3/x-3/2*b*c*d^2*e/x-1/2*b*e^3*x/c+1/4*b*c^4* d^3*arctan(c*x)-3/2*b*c^2*d^2*e*arctan(c*x)+1/2*b*e^3*arctan(c*x)/c^2-1/4* d^3*(a+b*arctan(c*x))/x^4-3/2*d^2*e*(a+b*arctan(c*x))/x^2+1/2*e^3*x^2*(a+b *arctan(c*x))+3*a*d*e^2*ln(x)+3/2*I*b*d*e^2*polylog(2,-I*c*x)-3/2*I*b*d*e^ 2*polylog(2,I*c*x)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.18 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.84 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^5} \, dx=\frac {1}{12} \left (-\frac {6 b e^3 (c x-\arctan (c x))}{c^2}-\frac {3 d^3 (a+b \arctan (c x))}{x^4}-\frac {18 d^2 e (a+b \arctan (c x))}{x^2}+6 e^3 x^2 (a+b \arctan (c x))-\frac {b c d^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-c^2 x^2\right )}{x^3}-\frac {18 b c d^2 e \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )}{x}+36 a d e^2 \log (x)+18 i b d e^2 \operatorname {PolyLog}(2,-i c x)-18 i b d e^2 \operatorname {PolyLog}(2,i c x)\right ) \]
((-6*b*e^3*(c*x - ArcTan[c*x]))/c^2 - (3*d^3*(a + b*ArcTan[c*x]))/x^4 - (1 8*d^2*e*(a + b*ArcTan[c*x]))/x^2 + 6*e^3*x^2*(a + b*ArcTan[c*x]) - (b*c*d^ 3*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)])/x^3 - (18*b*c*d^2*e*Hyperg eometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/x + 36*a*d*e^2*Log[x] + (18*I)*b*d* e^2*PolyLog[2, (-I)*c*x] - (18*I)*b*d*e^2*PolyLog[2, I*c*x])/12
Time = 0.40 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5515, 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^5} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {d^3 (a+b \arctan (c x))}{x^5}+\frac {3 d^2 e (a+b \arctan (c x))}{x^3}+\frac {3 d e^2 (a+b \arctan (c x))}{x}+e^3 x (a+b \arctan (c x))\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^3 (a+b \arctan (c x))}{4 x^4}-\frac {3 d^2 e (a+b \arctan (c x))}{2 x^2}+\frac {1}{2} e^3 x^2 (a+b \arctan (c x))+3 a d e^2 \log (x)+\frac {1}{4} b c^4 d^3 \arctan (c x)-\frac {3}{2} b c^2 d^2 e \arctan (c x)+\frac {b e^3 \arctan (c x)}{2 c^2}+\frac {b c^3 d^3}{4 x}-\frac {b c d^3}{12 x^3}-\frac {3 b c d^2 e}{2 x}+\frac {3}{2} i b d e^2 \operatorname {PolyLog}(2,-i c x)-\frac {3}{2} i b d e^2 \operatorname {PolyLog}(2,i c x)-\frac {b e^3 x}{2 c}\) |
-1/12*(b*c*d^3)/x^3 + (b*c^3*d^3)/(4*x) - (3*b*c*d^2*e)/(2*x) - (b*e^3*x)/ (2*c) + (b*c^4*d^3*ArcTan[c*x])/4 - (3*b*c^2*d^2*e*ArcTan[c*x])/2 + (b*e^3 *ArcTan[c*x])/(2*c^2) - (d^3*(a + b*ArcTan[c*x]))/(4*x^4) - (3*d^2*e*(a + b*ArcTan[c*x]))/(2*x^2) + (e^3*x^2*(a + b*ArcTan[c*x]))/2 + 3*a*d*e^2*Log[ x] + ((3*I)/2)*b*d*e^2*PolyLog[2, (-I)*c*x] - ((3*I)/2)*b*d*e^2*PolyLog[2, I*c*x]
3.12.45.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*ArcTan[c*x] )^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d , e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])
Time = 0.53 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.20
| method | result | size |
| parts | \(a \left (\frac {e^{3} x^{2}}{2}+3 e^{2} d \ln \left (x \right )-\frac {d^{3}}{4 x^{4}}-\frac {3 e \,d^{2}}{2 x^{2}}\right )+b \,c^{4} \left (\frac {\arctan \left (c x \right ) x^{2} e^{3}}{2 c^{4}}+\frac {3 \arctan \left (c x \right ) e^{2} d \ln \left (c x \right )}{c^{4}}-\frac {\arctan \left (c x \right ) d^{3}}{4 c^{4} x^{4}}-\frac {3 \arctan \left (c x \right ) d^{2} e}{2 c^{4} x^{2}}-\frac {2 c x \,e^{3}-\frac {c^{3} d^{2} \left (c^{2} d -6 e \right )}{x}+\frac {c^{3} d^{3}}{3 x^{3}}+\left (-c^{6} d^{3}+6 c^{4} d^{2} e -2 e^{3}\right ) \arctan \left (c x \right )+12 c^{2} d \,e^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )}{4 c^{6}}\right )\) | \(241\) |
| derivativedivides | \(c^{4} \left (\frac {a \left (\frac {e^{3} c^{2} x^{2}}{2}+3 c^{2} d \,e^{2} \ln \left (c x \right )-\frac {c^{2} d^{3}}{4 x^{4}}-\frac {3 c^{2} d^{2} e}{2 x^{2}}\right )}{c^{6}}+\frac {b \left (\frac {\arctan \left (c x \right ) e^{3} c^{2} x^{2}}{2}+3 \arctan \left (c x \right ) c^{2} d \,e^{2} \ln \left (c x \right )-\frac {\arctan \left (c x \right ) c^{2} d^{3}}{4 x^{4}}-\frac {3 \arctan \left (c x \right ) c^{2} d^{2} e}{2 x^{2}}-\frac {c x \,e^{3}}{2}+\frac {\left (c^{6} d^{3}-6 c^{4} d^{2} e +2 e^{3}\right ) \arctan \left (c x \right )}{4}-\frac {c^{3} d^{3}}{12 x^{3}}+\frac {c^{3} d^{2} \left (c^{2} d -6 e \right )}{4 x}-3 c^{2} d \,e^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{6}}\right )\) | \(256\) |
| default | \(c^{4} \left (\frac {a \left (\frac {e^{3} c^{2} x^{2}}{2}+3 c^{2} d \,e^{2} \ln \left (c x \right )-\frac {c^{2} d^{3}}{4 x^{4}}-\frac {3 c^{2} d^{2} e}{2 x^{2}}\right )}{c^{6}}+\frac {b \left (\frac {\arctan \left (c x \right ) e^{3} c^{2} x^{2}}{2}+3 \arctan \left (c x \right ) c^{2} d \,e^{2} \ln \left (c x \right )-\frac {\arctan \left (c x \right ) c^{2} d^{3}}{4 x^{4}}-\frac {3 \arctan \left (c x \right ) c^{2} d^{2} e}{2 x^{2}}-\frac {c x \,e^{3}}{2}+\frac {\left (c^{6} d^{3}-6 c^{4} d^{2} e +2 e^{3}\right ) \arctan \left (c x \right )}{4}-\frac {c^{3} d^{3}}{12 x^{3}}+\frac {c^{3} d^{2} \left (c^{2} d -6 e \right )}{4 x}-3 c^{2} d \,e^{2} \left (-\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )}{c^{6}}\right )\) | \(256\) |
| risch | \(-\frac {b \,e^{3} x}{2 c}+\frac {b \,c^{4} d^{3} \arctan \left (c x \right )}{8}+\frac {b \,e^{3} \arctan \left (c x \right )}{4 c^{2}}-\frac {3 b c \,d^{2} e}{2 x}-\frac {3 b \,c^{2} d^{2} e \arctan \left (c x \right )}{4}-\frac {b c \,d^{3}}{12 x^{3}}+\frac {b \,c^{3} d^{3}}{4 x}-\frac {3 a \,d^{2} e}{2 x^{2}}+\frac {a \,e^{3}}{2 c^{2}}-\frac {a \,d^{3}}{4 x^{4}}+\frac {a \,x^{2} e^{3}}{2}-\frac {3 i b \,c^{2} e \,d^{2} \ln \left (i c x \right )}{4}+\frac {3 i b \,c^{2} e \,d^{2} \ln \left (i c x +1\right )}{4}-\frac {3 i c^{2} b \,d^{2} e \ln \left (c^{2} x^{2}+1\right )}{8}+\frac {3 i b e \,d^{2} \ln \left (i c x +1\right )}{4 x^{2}}+\frac {3 i c^{2} b \,d^{2} e \ln \left (-i c x \right )}{4}-\frac {3 i b \,d^{2} e \ln \left (-i c x +1\right )}{4 x^{2}}-\frac {i b \,e^{3} \ln \left (i c x +1\right ) x^{2}}{4}-\frac {i b \,e^{3} \ln \left (i c x +1\right )}{4 c^{2}}+\frac {i b \,c^{4} d^{3} \ln \left (i c x \right )}{8}+\frac {i c^{4} b \,d^{3} \ln \left (c^{2} x^{2}+1\right )}{16}-\frac {i b \,c^{4} d^{3} \ln \left (i c x +1\right )}{8}+\frac {i b \,d^{3} \ln \left (i c x +1\right )}{8 x^{4}}+\frac {3 i b \,e^{2} d \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {i b \,e^{3} \ln \left (c^{2} x^{2}+1\right )}{8 c^{2}}-\frac {3 i b \,e^{2} d \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i b \,e^{3} \ln \left (-i c x +1\right ) x^{2}}{4}-\frac {i c^{4} b \,d^{3} \ln \left (-i c x \right )}{8}-\frac {i b \,d^{3} \ln \left (-i c x +1\right )}{8 x^{4}}+3 a \,e^{2} d \ln \left (-i c x \right )\) | \(459\) |
a*(1/2*e^3*x^2+3*e^2*d*ln(x)-1/4*d^3/x^4-3/2*e*d^2/x^2)+b*c^4*(1/2*arctan( c*x)/c^4*x^2*e^3+3*arctan(c*x)/c^4*e^2*d*ln(c*x)-1/4*arctan(c*x)*d^3/c^4/x ^4-3/2*arctan(c*x)/c^4*d^2*e/x^2-1/4/c^6*(2*c*x*e^3-c^3*d^2*(c^2*d-6*e)/x+ 1/3*c^3*d^3/x^3+(-c^6*d^3+6*c^4*d^2*e-2*e^3)*arctan(c*x)+12*c^2*d*e^2*(-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))))
\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]
integral((a*e^3*x^6 + 3*a*d*e^2*x^4 + 3*a*d^2*e*x^2 + a*d^3 + (b*e^3*x^6 + 3*b*d*e^2*x^4 + 3*b*d^2*e*x^2 + b*d^3)*arctan(c*x))/x^5, x)
\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x^{5}}\, dx \]
Time = 0.46 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.09 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^5} \, dx=\frac {1}{2} \, a e^{3} x^{2} + \frac {1}{12} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b d^{3} - \frac {3}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b d^{2} e + 3 \, a d e^{2} \log \left (x\right ) - \frac {3 \, a d^{2} e}{2 \, x^{2}} - \frac {a d^{3}}{4 \, x^{4}} - \frac {3 \, \pi b c^{2} d e^{2} \log \left (c^{2} x^{2} + 1\right ) - 12 \, b c^{2} d e^{2} \arctan \left (c x\right ) \log \left (c x\right ) + 6 i \, b c^{2} d e^{2} {\rm Li}_2\left (i \, c x + 1\right ) - 6 i \, b c^{2} d e^{2} {\rm Li}_2\left (-i \, c x + 1\right ) + 2 \, b c e^{3} x - 2 \, {\left (b c^{2} e^{3} x^{2} + b e^{3}\right )} \arctan \left (c x\right )}{4 \, c^{2}} \]
1/2*a*e^3*x^2 + 1/12*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arct an(c*x)/x^4)*b*d^3 - 3/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*d^2 *e + 3*a*d*e^2*log(x) - 3/2*a*d^2*e/x^2 - 1/4*a*d^3/x^4 - 1/4*(3*pi*b*c^2* d*e^2*log(c^2*x^2 + 1) - 12*b*c^2*d*e^2*arctan(c*x)*log(c*x) + 6*I*b*c^2*d *e^2*dilog(I*c*x + 1) - 6*I*b*c^2*d*e^2*dilog(-I*c*x + 1) + 2*b*c*e^3*x - 2*(b*c^2*e^3*x^2 + b*e^3)*arctan(c*x))/c^2
\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]
Time = 0.83 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.17 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^5} \, dx=\left \{\begin {array}{cl} -\frac {a\,\left (d^3-2\,e^3\,x^6+6\,d^2\,e\,x^2-12\,d\,e^2\,x^4\,\ln \left (x\right )\right )}{4\,x^4} & \text {\ if\ \ }c=0\\ -b\,e^3\,\left (\frac {x}{2\,c}-\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\right )-\frac {a\,\left (d^3-2\,e^3\,x^6+6\,d^2\,e\,x^2-12\,d\,e^2\,x^4\,\ln \left (x\right )\right )}{4\,x^4}-\frac {b\,d^3\,\left (\frac {\frac {c^2}{3}-c^4\,x^2}{x^3}-c^5\,\mathrm {atan}\left (c\,x\right )\right )}{4\,c}-3\,b\,d^2\,e\,\left (\frac {c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}}{2\,c}+\frac {\mathrm {atan}\left (c\,x\right )}{2\,x^2}\right )-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{4\,x^4}-\frac {b\,d\,e^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {b\,d\,e^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
piecewise(c == 0, -(a*(d^3 - 2*e^3*x^6 + 6*d^2*e*x^2 - 12*d*e^2*x^4*log(x) ))/(4*x^4), c ~= 0, - b*e^3*(x/(2*c) - atan(c*x)*(1/(2*c^2) + x^2/2)) - (a *(d^3 - 2*e^3*x^6 + 6*d^2*e*x^2 - 12*d*e^2*x^4*log(x)))/(4*x^4) - (b*d^3*( (c^2/3 - c^4*x^2)/x^3 - c^5*atan(c*x)))/(4*c) - 3*b*d^2*e*((c^3*atan(c*x) + c^2/x)/(2*c) + atan(c*x)/(2*x^2)) - (b*d*e^2*dilog(- c*x*1i + 1)*3i)/2 + (b*d*e^2*dilog(c*x*1i + 1)*3i)/2 - (b*d^3*atan(c*x))/(4*x^4))