Integrand size = 21, antiderivative size = 489 \[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^2} \, dx=-\frac {b c}{2 d^2 x}-\frac {b c^2 \arctan (c x)}{2 d^2}+\frac {b c^2 e \arctan (c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {a+b \arctan (c x)}{2 d^2 x^2}-\frac {e (a+b \arctan (c x))}{2 d^2 \left (d+e x^2\right )}-\frac {b c e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}-\frac {2 a e \log (x)}{d^3}-\frac {2 e (a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right )}{d^3}+\frac {e (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{d^3}+\frac {e (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{d^3}-\frac {i b e \operatorname {PolyLog}(2,-i c x)}{d^3}+\frac {i b e \operatorname {PolyLog}(2,i c x)}{d^3}+\frac {i b e \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b e \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^3}-\frac {i b e \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^3} \]
-1/2*b*c/d^2/x-1/2*b*c^2*arctan(c*x)/d^2+1/2*b*c^2*e*arctan(c*x)/d^2/(c^2* d-e)+1/2*(-a-b*arctan(c*x))/d^2/x^2-1/2*e*(a+b*arctan(c*x))/d^2/(e*x^2+d)- 1/2*b*c*e^(3/2)*arctan(x*e^(1/2)/d^(1/2))/d^(5/2)/(c^2*d-e)-2*a*e*ln(x)/d^ 3-2*e*(a+b*arctan(c*x))*ln(2/(1-I*c*x))/d^3+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^3+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^ 3-I*b*e*polylog(2,-I*c*x)/d^3+I*b*e*polylog(2,I*c*x)/d^3+I*b*e*polylog(2,1 -2/(1-I*c*x))/d^3-1/2*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^3-1/2*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^3
Time = 9.89 (sec) , antiderivative size = 643, normalized size of antiderivative = 1.31 \[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^2} \, dx=-\frac {a \left (d \left (\frac {1}{x^2}+\frac {e}{d+e x^2}\right )+4 e \log (x)-2 e \log \left (d+e x^2\right )\right )+b \left (\frac {c d}{x}+\frac {c^2 d \left (c^2 d-2 e\right ) \arctan (c x)}{c^2 d-e}+d \left (\frac {1}{x^2}+\frac {e}{d+e x^2}\right ) \arctan (c x)+\frac {c \sqrt {d} e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{c^2 d-e}+4 e \arctan (c x) \log (x)-2 e \arctan (c x) \log \left (d+e x^2\right )-2 i e (\log (x) (\log (1-i c x)-\log (1+i c x))-\operatorname {PolyLog}(2,-i c x)+\operatorname {PolyLog}(2,i c x))-e \left (2 \arctan (c x) \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right )+2 \arctan (c x) \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right )+i \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (-1-i c x)}{c \sqrt {d}-\sqrt {e}}\right )-i \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (1-i c x)}{c \sqrt {d}+\sqrt {e}}\right )-i \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (-1+i c x)}{c \sqrt {d}-\sqrt {e}}\right )+i \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (1+i c x)}{c \sqrt {d}+\sqrt {e}}\right )-2 \arctan (c x) \log \left (d+e x^2\right )-i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )+i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}+\sqrt {e}}\right )+i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )-i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{c \sqrt {d}+\sqrt {e}}\right )\right )\right )}{2 d^3} \]
-1/2*(a*(d*(x^(-2) + e/(d + e*x^2)) + 4*e*Log[x] - 2*e*Log[d + e*x^2]) + b *((c*d)/x + (c^2*d*(c^2*d - 2*e)*ArcTan[c*x])/(c^2*d - e) + d*(x^(-2) + e/ (d + e*x^2))*ArcTan[c*x] + (c*Sqrt[d]*e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]]) /(c^2*d - e) + 4*e*ArcTan[c*x]*Log[x] - 2*e*ArcTan[c*x]*Log[d + e*x^2] - ( 2*I)*e*(Log[x]*(Log[1 - I*c*x] - Log[1 + I*c*x]) - PolyLog[2, (-I)*c*x] + PolyLog[2, I*c*x]) - e*(2*ArcTan[c*x]*Log[((-I)*Sqrt[d])/Sqrt[e] + x] + 2* ArcTan[c*x]*Log[(I*Sqrt[d])/Sqrt[e] + x] + I*Log[((-I)*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(-1 - I*c*x))/(c*Sqrt[d] - Sqrt[e])] - I*Log[((-I)*Sqrt[d] )/Sqrt[e] + x]*Log[(Sqrt[e]*(1 - I*c*x))/(c*Sqrt[d] + Sqrt[e])] - I*Log[(I *Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(-1 + I*c*x))/(c*Sqrt[d] - Sqrt[e])] + I*Log[(I*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(1 + I*c*x))/(c*Sqrt[d] + Sqr t[e])] - 2*ArcTan[c*x]*Log[d + e*x^2] - I*PolyLog[2, (c*(Sqrt[d] - I*Sqrt[ e]*x))/(c*Sqrt[d] - Sqrt[e])] + I*PolyLog[2, (c*(Sqrt[d] - I*Sqrt[e]*x))/( c*Sqrt[d] + Sqrt[e])] + I*PolyLog[2, (c*(Sqrt[d] + I*Sqrt[e]*x))/(c*Sqrt[d ] - Sqrt[e])] - I*PolyLog[2, (c*(Sqrt[d] + I*Sqrt[e]*x))/(c*Sqrt[d] + Sqrt [e])])))/d^3
Time = 0.80 (sec) , antiderivative size = 489, 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.
| \(\displaystyle \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {2 e^2 x (a+b \arctan (c x))}{d^3 \left (d+e x^2\right )}-\frac {2 e (a+b \arctan (c x))}{d^3 x}+\frac {e^2 x (a+b \arctan (c x))}{d^2 \left (d+e x^2\right )^2}+\frac {a+b \arctan (c x)}{d^2 x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 e \log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{d^3}+\frac {e (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{d^3}+\frac {e (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{d^3}-\frac {e (a+b \arctan (c x))}{2 d^2 \left (d+e x^2\right )}-\frac {a+b \arctan (c x)}{2 d^2 x^2}-\frac {2 a e \log (x)}{d^3}-\frac {b c e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}+\frac {b c^2 e \arctan (c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {b c^2 \arctan (c x)}{2 d^2}-\frac {i b e \operatorname {PolyLog}(2,-i c x)}{d^3}+\frac {i b e \operatorname {PolyLog}(2,i c x)}{d^3}+\frac {i b e \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b e \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^3}-\frac {i b e \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^3}-\frac {b c}{2 d^2 x}\) |
-1/2*(b*c)/(d^2*x) - (b*c^2*ArcTan[c*x])/(2*d^2) + (b*c^2*e*ArcTan[c*x])/( 2*d^2*(c^2*d - e)) - (a + b*ArcTan[c*x])/(2*d^2*x^2) - (e*(a + b*ArcTan[c* x]))/(2*d^2*(d + e*x^2)) - (b*c*e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^ (5/2)*(c^2*d - e)) - (2*a*e*Log[x])/d^3 - (2*e*(a + b*ArcTan[c*x])*Log[2/( 1 - I*c*x)])/d^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))])/d^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))])/d^3 - (I*b*e*PolyLog[2, (-I)*c*x])/d^3 + (I*b*e*PolyLog[2, I*c*x])/d^3 + (I*b *e*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^3 - ((I/2)*b*e*PolyLog[2, 1 - (2*c*(Sq rt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/d^3 - ((I/2) *b*e*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e]) *(1 - I*c*x))])/d^3
3.12.61.3.1 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 = 0.61 (sec) , antiderivative size = 851, normalized size of antiderivative = 1.74
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(851\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(877\) |
| default | \(\text {Expression too large to display}\) | \(877\) |
| risch | \(-\frac {b c}{2 d^{2} x}-\frac {i c^{4} b \,e^{2} \ln \left (-i c x +1\right ) x^{2}}{4 d^{2} \left (c^{2} d -e \right ) \left (-e \,c^{2} x^{2}-c^{2} d \right )}+\frac {i b \,c^{4} e^{2} \ln \left (i c x +1\right ) x^{2}}{4 d^{2} \left (c^{2} d -e \right ) \left (-e \,c^{2} x^{2}-c^{2} d \right )}-\frac {a}{2 d^{2} x^{2}}+\frac {c^{2} a e}{2 d^{2} \left (-e \,c^{2} x^{2}-c^{2} d \right )}-\frac {i c^{2} b \,e^{2} \ln \left (-i c x +1\right )}{4 d^{2} \left (c^{2} d -e \right ) \left (-e \,c^{2} x^{2}-c^{2} d \right )}-\frac {i c b \,e^{2} \operatorname {arctanh}\left (\frac {2 \left (-i c x +1\right ) e -2 e}{2 c \sqrt {e d}}\right )}{4 d^{2} \left (c^{2} d -e \right ) \sqrt {e d}}+\frac {i b \,c^{2} e^{2} \ln \left (i c x +1\right )}{4 d^{2} \left (c^{2} d -e \right ) \left (-e \,c^{2} x^{2}-c^{2} d \right )}+\frac {i b c \,e^{2} \operatorname {arctanh}\left (\frac {2 \left (i c x +1\right ) e -2 e}{2 c \sqrt {e d}}\right )}{4 d^{2} \left (c^{2} d -e \right ) \sqrt {e d}}+\frac {i c^{2} b \ln \left (-i c x \right )}{4 d^{2}}+\frac {i b e \operatorname {dilog}\left (\frac {c \sqrt {e d}+\left (-i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{2 d^{3}}-\frac {i c^{2} b \ln \left (-i c x +1\right )}{4 d^{2}}+\frac {i b e \operatorname {dilog}\left (-i c x +1\right )}{d^{3}}-\frac {i b \ln \left (-i c x +1\right )}{4 d^{2} x^{2}}+\frac {i b e \operatorname {dilog}\left (\frac {c \sqrt {e d}-\left (-i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{2 d^{3}}-\frac {i b \,c^{2} \ln \left (i c x \right )}{4 d^{2}}+\frac {i b \,c^{2} \ln \left (i c x +1\right )}{4 d^{2}}+\frac {i b \ln \left (i c x +1\right )}{4 d^{2} x^{2}}-\frac {i b e \operatorname {dilog}\left (i c x +1\right )}{d^{3}}-\frac {i b e \operatorname {dilog}\left (\frac {c \sqrt {e d}-\left (i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{2 d^{3}}-\frac {i b e \operatorname {dilog}\left (\frac {c \sqrt {e d}+\left (i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{2 d^{3}}-\frac {i b e \ln \left (i c x +1\right ) \ln \left (\frac {c \sqrt {e d}-\left (i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{2 d^{3}}-\frac {i b e \ln \left (i c x +1\right ) \ln \left (\frac {c \sqrt {e d}+\left (i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{2 d^{3}}+\frac {i b e \ln \left (-i c x +1\right ) \ln \left (\frac {c \sqrt {e d}+\left (-i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{2 d^{3}}+\frac {i b e \ln \left (-i c x +1\right ) \ln \left (\frac {c \sqrt {e d}-\left (-i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{2 d^{3}}+\frac {a e \ln \left (\left (-i c x +1\right )^{2} e -c^{2} d -2 \left (-i c x +1\right ) e +e \right )}{d^{3}}-\frac {2 a e \ln \left (-i c x \right )}{d^{3}}+\frac {i b \,c^{2} e \ln \left (\left (i c x +1\right )^{2} e -c^{2} d -2 \left (i c x +1\right ) e +e \right )}{8 d^{2} \left (c^{2} d -e \right )}-\frac {i c^{2} b e \ln \left (\left (-i c x +1\right )^{2} e -c^{2} d -2 \left (-i c x +1\right ) e +e \right )}{8 d^{2} \left (c^{2} d -e \right )}\) | \(1012\) |
-1/2*a/d^2/x^2-2*a*e*ln(x)/d^3-1/2*a*e/d^2/(e*x^2+d)+a*e/d^3*ln(e*x^2+d)+b *c^2*(-1/2*arctan(c*x)/d^2/c^2/x^2-2/c^2*arctan(c*x)/d^3*e*ln(c*x)-1/2*arc tan(c*x)*e/d^2/(c^2*e*x^2+c^2*d)+1/c^2*arctan(c*x)*e/d^3*ln(c^2*e*x^2+c^2* d)-1/2*c^4*(-4/d^3/c^6*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))+2/d^3/c^6*e*(-1/2*I*(ln(c* x-I)*ln(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,ind ex=2)))/e+1/2*(dilog((RootOf(e*_Z^2+2*I*e*_Z+c^2*d-e,index=1)-c*x+I)/RootO f(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)*ln(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,ind ex=2)))/e+1/2*(dilog((RootOf(e*_Z^2-2*I*e*_Z+c^2*d-e,index=1)-c*x-I)/RootO f(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/d^2/c^4*( -e^2/(c^2*d-e)/c/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2))-1/c/x+(-c^2*d+2*e)/(c ^2*d-e)*arctan(c*x))))
\[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^2} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{2} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^2} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{2} x^{3}} \,d x } \]
-1/2*a*((2*e*x^2 + d)/(d^2*e*x^4 + d^3*x^2) - 2*e*log(e*x^2 + d)/d^3 + 4*e *log(x)/d^3) + 2*b*integrate(1/2*arctan(c*x)/(e^2*x^7 + 2*d*e*x^5 + d^2*x^ 3), x)
\[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^2} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{2} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x^3 \left (d+e x^2\right )^2} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^3\,{\left (e\,x^2+d\right )}^2} \,d x \]