Integrand size = 21, antiderivative size = 629 \[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^3} \, dx=-\frac {b c}{2 d^3 x}-\frac {b c e^2 x}{8 d^3 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b c^2 \arctan (c x)}{2 d^3}+\frac {b c^4 e \arctan (c x)}{4 d^2 \left (c^2 d-e\right )^2}+\frac {b c^2 e \arctan (c x)}{d^3 \left (c^2 d-e\right )}-\frac {a+b \arctan (c x)}{2 d^3 x^2}-\frac {e (a+b \arctan (c x))}{4 d^2 \left (d+e x^2\right )^2}-\frac {e (a+b \arctan (c x))}{d^3 \left (d+e x^2\right )}-\frac {b c e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{7/2} \left (c^2 d-e\right )}-\frac {b c \left (3 c^2 d-e\right ) e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} \left (c^2 d-e\right )^2}-\frac {3 a e \log (x)}{d^4}-\frac {3 e (a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right )}{d^4}+\frac {3 e (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^4}+\frac {3 e (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^4}-\frac {3 i b e \operatorname {PolyLog}(2,-i c x)}{2 d^4}+\frac {3 i b e \operatorname {PolyLog}(2,i c x)}{2 d^4}+\frac {3 i b e \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 d^4}-\frac {3 i b e \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^4}-\frac {3 i b e \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^4} \]
-1/2*b*c/d^3/x-1/8*b*c*e^2*x/d^3/(c^2*d-e)/(e*x^2+d)-1/2*b*c^2*arctan(c*x) /d^3+1/4*b*c^4*e*arctan(c*x)/d^2/(c^2*d-e)^2+b*c^2*e*arctan(c*x)/d^3/(c^2* d-e)+1/2*(-a-b*arctan(c*x))/d^3/x^2-1/4*e*(a+b*arctan(c*x))/d^2/(e*x^2+d)^ 2-e*(a+b*arctan(c*x))/d^3/(e*x^2+d)-b*c*e^(3/2)*arctan(x*e^(1/2)/d^(1/2))/ d^(7/2)/(c^2*d-e)-1/8*b*c*(3*c^2*d-e)*e^(3/2)*arctan(x*e^(1/2)/d^(1/2))/d^ (7/2)/(c^2*d-e)^2-3*a*e*ln(x)/d^4-3*e*(a+b*arctan(c*x))*ln(2/(1-I*c*x))/d^ 4+3/2*e*(a+b*arctan(c*x))*ln(2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^ (1/2)-I*e^(1/2)))/d^4+3/2*e*(a+b*arctan(c*x))*ln(2*c*((-d)^(1/2)+x*e^(1/2) )/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/d^4-3/2*I*b*e*polylog(2,-I*c*x)/d^4+ 3/2*I*b*e*polylog(2,1-2/(1-I*c*x))/d^4+3/2*I*b*e*polylog(2,I*c*x)/d^4-3/4* I*b*e*polylog(2,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^( 1/2)))/d^4-3/4*I*b*e*polylog(2,1-2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*( -d)^(1/2)+I*e^(1/2)))/d^4
Time = 12.59 (sec) , antiderivative size = 723, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^3} \, dx=\frac {-a \left (\frac {d \left (2 d^2+9 d e x^2+6 e^2 x^4\right )}{x^2 \left (d+e x^2\right )^2}+12 e \log (x)-6 e \log \left (d+e x^2\right )\right )+b \left (-\frac {2 c d}{x}-\frac {c d e^2 x}{2 \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {c^2 d \left (-2 c^4 d^2+9 c^2 d e-6 e^2\right ) \arctan (c x)}{\left (-c^2 d+e\right )^2}-\frac {d \left (2 d^2+9 d e x^2+6 e^2 x^4\right ) \arctan (c x)}{x^2 \left (d+e x^2\right )^2}+\frac {c \sqrt {d} e^{3/2} \left (-11 c^2 d+9 e\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \left (-c^2 d+e\right )^2}-12 e \arctan (c x) \log (x)+6 e \arctan (c x) \left (\log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right )+\log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right )-\log \left (d+e x^2\right )\right )+6 e \arctan (c x) \log \left (d+e x^2\right )-6 i e (\log (x) \log (1+i c x)+\operatorname {PolyLog}(2,-i c x))+6 i e (\log (x) \log (1-i c x)+\operatorname {PolyLog}(2,i c x))-3 i e \left (\log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (-1+i c x)}{c \sqrt {d}-\sqrt {e}}\right )+\operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )\right )+3 i e \left (\log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (1+i c x)}{c \sqrt {d}+\sqrt {e}}\right )+\operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}+\sqrt {e}}\right )\right )+3 i e \left (\log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (-1-i c x)}{c \sqrt {d}-\sqrt {e}}\right )+\operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )\right )-3 i e \left (\log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (1-i c x)}{c \sqrt {d}+\sqrt {e}}\right )+\operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{c \sqrt {d}+\sqrt {e}}\right )\right )\right )}{4 d^4} \]
(-(a*((d*(2*d^2 + 9*d*e*x^2 + 6*e^2*x^4))/(x^2*(d + e*x^2)^2) + 12*e*Log[x ] - 6*e*Log[d + e*x^2])) + b*((-2*c*d)/x - (c*d*e^2*x)/(2*(c^2*d - e)*(d + e*x^2)) + (c^2*d*(-2*c^4*d^2 + 9*c^2*d*e - 6*e^2)*ArcTan[c*x])/(-(c^2*d) + e)^2 - (d*(2*d^2 + 9*d*e*x^2 + 6*e^2*x^4)*ArcTan[c*x])/(x^2*(d + e*x^2)^ 2) + (c*Sqrt[d]*e^(3/2)*(-11*c^2*d + 9*e)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2* (-(c^2*d) + e)^2) - 12*e*ArcTan[c*x]*Log[x] + 6*e*ArcTan[c*x]*(Log[((-I)*S qrt[d])/Sqrt[e] + x] + Log[(I*Sqrt[d])/Sqrt[e] + x] - Log[d + e*x^2]) + 6* e*ArcTan[c*x]*Log[d + e*x^2] - (6*I)*e*(Log[x]*Log[1 + I*c*x] + PolyLog[2, (-I)*c*x]) + (6*I)*e*(Log[x]*Log[1 - I*c*x] + PolyLog[2, I*c*x]) - (3*I)* e*(Log[(I*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(-1 + I*c*x))/(c*Sqrt[d] - Sq rt[e])] + PolyLog[2, (c*(Sqrt[d] - I*Sqrt[e]*x))/(c*Sqrt[d] - Sqrt[e])]) + (3*I)*e*(Log[(I*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(1 + I*c*x))/(c*Sqrt[d ] + Sqrt[e])] + PolyLog[2, (c*(Sqrt[d] - I*Sqrt[e]*x))/(c*Sqrt[d] + Sqrt[e ])]) + (3*I)*e*(Log[((-I)*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(-1 - I*c*x)) /(c*Sqrt[d] - Sqrt[e])] + PolyLog[2, (c*(Sqrt[d] + I*Sqrt[e]*x))/(c*Sqrt[d ] - Sqrt[e])]) - (3*I)*e*(Log[((-I)*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(1 - I*c*x))/(c*Sqrt[d] + Sqrt[e])] + PolyLog[2, (c*(Sqrt[d] + I*Sqrt[e]*x))/ (c*Sqrt[d] + Sqrt[e])])))/(4*d^4)
Time = 0.96 (sec) , antiderivative size = 629, 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.
| \(\displaystyle \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {3 e^2 x (a+b \arctan (c x))}{d^4 \left (d+e x^2\right )}-\frac {3 e (a+b \arctan (c x))}{d^4 x}+\frac {2 e^2 x (a+b \arctan (c x))}{d^3 \left (d+e x^2\right )^2}+\frac {a+b \arctan (c x)}{d^3 x^3}+\frac {e^2 x (a+b \arctan (c x))}{d^2 \left (d+e x^2\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 e \log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{d^4}+\frac {3 e (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d^4}+\frac {3 e (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d^4}-\frac {e (a+b \arctan (c x))}{d^3 \left (d+e x^2\right )}-\frac {a+b \arctan (c x)}{2 d^3 x^2}-\frac {e (a+b \arctan (c x))}{4 d^2 \left (d+e x^2\right )^2}-\frac {3 a e \log (x)}{d^4}-\frac {b c e^{3/2} \left (3 c^2 d-e\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} \left (c^2 d-e\right )^2}-\frac {b c e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{7/2} \left (c^2 d-e\right )}+\frac {b c^2 e \arctan (c x)}{d^3 \left (c^2 d-e\right )}-\frac {b c^2 \arctan (c x)}{2 d^3}+\frac {b c^4 e \arctan (c x)}{4 d^2 \left (c^2 d-e\right )^2}-\frac {b c e^2 x}{8 d^3 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {3 i b e \operatorname {PolyLog}(2,-i c x)}{2 d^4}+\frac {3 i b e \operatorname {PolyLog}(2,i c x)}{2 d^4}+\frac {3 i b e \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 d^4}-\frac {3 i b e \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^4}-\frac {3 i b e \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^4}-\frac {b c}{2 d^3 x}\) |
-1/2*(b*c)/(d^3*x) - (b*c*e^2*x)/(8*d^3*(c^2*d - e)*(d + e*x^2)) - (b*c^2* ArcTan[c*x])/(2*d^3) + (b*c^4*e*ArcTan[c*x])/(4*d^2*(c^2*d - e)^2) + (b*c^ 2*e*ArcTan[c*x])/(d^3*(c^2*d - e)) - (a + b*ArcTan[c*x])/(2*d^3*x^2) - (e* (a + b*ArcTan[c*x]))/(4*d^2*(d + e*x^2)^2) - (e*(a + b*ArcTan[c*x]))/(d^3* (d + e*x^2)) - (b*c*e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(d^(7/2)*(c^2*d - e)) - (b*c*(3*c^2*d - e)*e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(7/2)* (c^2*d - e)^2) - (3*a*e*Log[x])/d^4 - (3*e*(a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/d^4 + (3*e*(a + b*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/( (c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*d^4) + (3*e*(a + b*ArcTan[c*x]) *Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))]) /(2*d^4) - (((3*I)/2)*b*e*PolyLog[2, (-I)*c*x])/d^4 + (((3*I)/2)*b*e*PolyL og[2, I*c*x])/d^4 + (((3*I)/2)*b*e*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^4 - (( (3*I)/4)*b*e*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I* Sqrt[e])*(1 - I*c*x))])/d^4 - (((3*I)/4)*b*e*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/d^4
3.12.69.3.1 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])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.00 (sec) , antiderivative size = 951, normalized size of antiderivative = 1.51
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(951\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(987\) |
| default | \(\text {Expression too large to display}\) | \(987\) |
| risch | \(\text {Expression too large to display}\) | \(1858\) |
-1/2*a/d^3/x^2-3*a*e*ln(x)/d^4-a*e/d^3/(e*x^2+d)+3/2*a*e/d^4*ln(e*x^2+d)-1 /4*a*e/d^2/(e*x^2+d)^2+b*c^2*(-1/2*arctan(c*x)/d^3/c^2/x^2-3/c^2*arctan(c* x)/d^4*e*ln(c*x)+3/2/c^2*arctan(c*x)*e/d^4*ln(c^2*e*x^2+c^2*d)-1/4*c^2*arc tan(c*x)*e/d^2/(c^2*e*x^2+c^2*d)^2-arctan(c*x)*e/d^3/(c^2*e*x^2+c^2*d)-1/2 *c^6*(-6/d^4/c^8*e*(-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))+3/d^4/c^8*e*(-1/2*I*(ln(c*x-I)*l n(c^2*e*x^2+c^2*d)-2*e*(1/2*ln(c*x-I)*(ln((RootOf(e*_Z^2+2*I*e*_Z+c^2*d-e, index=1)-c*x+I)/RootOf(e*_Z^2+2*I*e*_Z+c^2*d-e,index=1))+ln((RootOf(e*_Z^2 +2*I*e*_Z+c^2*d-e,index=2)-c*x+I)/RootOf(e*_Z^2+2*I*e*_Z+c^2*d-e,index=2)) )/e+1/2*(dilog((RootOf(e*_Z^2+2*I*e*_Z+c^2*d-e,index=1)-c*x+I)/RootOf(e*_Z ^2+2*I*e*_Z+c^2*d-e,index=1))+dilog((RootOf(e*_Z^2+2*I*e*_Z+c^2*d-e,index= 2)-c*x+I)/RootOf(e*_Z^2+2*I*e*_Z+c^2*d-e,index=2)))/e))+1/2*I*(ln(I+c*x)*l n(c^2*e*x^2+c^2*d)-2*e*(1/2*ln(I+c*x)*(ln((RootOf(e*_Z^2-2*I*e*_Z+c^2*d-e, index=1)-c*x-I)/RootOf(e*_Z^2-2*I*e*_Z+c^2*d-e,index=1))+ln((RootOf(e*_Z^2 -2*I*e*_Z+c^2*d-e,index=2)-c*x-I)/RootOf(e*_Z^2-2*I*e*_Z+c^2*d-e,index=2)) )/e+1/2*(dilog((RootOf(e*_Z^2-2*I*e*_Z+c^2*d-e,index=1)-c*x-I)/RootOf(e*_Z ^2-2*I*e*_Z+c^2*d-e,index=1))+dilog((RootOf(e*_Z^2-2*I*e*_Z+c^2*d-e,index= 2)-c*x-I)/RootOf(e*_Z^2-2*I*e*_Z+c^2*d-e,index=2)))/e)))-1/2/d^3/c^6*(-e^2 /(c^2*d-e)^2*((1/2*c^2*d-1/2*e)*c*x/(c^2*e*x^2+c^2*d)+1/2*(11*c^2*d-9*e)/c /(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2)))+(-2*c^4*d^2+9*c^2*d*e-6*e^2)/(c^2...
\[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x^{3}} \,d x } \]
-1/4*a*((6*e^2*x^4 + 9*d*e*x^2 + 2*d^2)/(d^3*e^2*x^6 + 2*d^4*e*x^4 + d^5*x ^2) - 6*e*log(e*x^2 + d)/d^4 + 12*e*log(x)/d^4) + 2*b*integrate(1/2*arctan (c*x)/(e^3*x^9 + 3*d*e^2*x^7 + 3*d^2*e*x^5 + d^3*x^3), x)
\[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^3} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^3\,{\left (e\,x^2+d\right )}^3} \,d x \]