Integrand size = 21, antiderivative size = 574 \[ \int \frac {a+b \arctan (c x)}{x \left (d+e x^2\right )^3} \, dx=\frac {b c e x}{8 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b c^4 \arctan (c x)}{4 d \left (c^2 d-e\right )^2}-\frac {b c^2 \arctan (c x)}{2 d^2 \left (c^2 d-e\right )}+\frac {a+b \arctan (c x)}{4 d \left (d+e x^2\right )^2}+\frac {a+b \arctan (c x)}{2 d^2 \left (d+e x^2\right )}+\frac {b c \sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}+\frac {b c \left (3 c^2 d-e\right ) \sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} \left (c^2 d-e\right )^2}+\frac {a \log (x)}{d^3}+\frac {(a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {(a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^3}-\frac {(a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}(2,-i c x)}{2 d^3}-\frac {i b \operatorname {PolyLog}(2,i c x)}{2 d^3}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^3}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^3} \]
1/8*b*c*e*x/d^2/(c^2*d-e)/(e*x^2+d)-1/4*b*c^4*arctan(c*x)/d/(c^2*d-e)^2-1/ 2*b*c^2*arctan(c*x)/d^2/(c^2*d-e)+1/4*(a+b*arctan(c*x))/d/(e*x^2+d)^2+1/2* (a+b*arctan(c*x))/d^2/(e*x^2+d)+a*ln(x)/d^3+(a+b*arctan(c*x))*ln(2/(1-I*c* x))/d^3-1/2*(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^3-1/2*(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^3-1/2*I*b*polylog(2,I*c*x)/d^3+1 /4*I*b*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+1/4*I*b*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*polylog(2,1-2/(1-I*c*x))/d^3+1/2*I*b*poly log(2,-I*c*x)/d^3+1/2*b*c*arctan(x*e^(1/2)/d^(1/2))*e^(1/2)/d^(5/2)/(c^2*d -e)+1/8*b*c*(3*c^2*d-e)*arctan(x*e^(1/2)/d^(1/2))*e^(1/2)/d^(5/2)/(c^2*d-e )^2
Time = 10.06 (sec) , antiderivative size = 645, normalized size of antiderivative = 1.12 \[ \int \frac {a+b \arctan (c x)}{x \left (d+e x^2\right )^3} \, dx=\frac {2 a \left (\frac {d \left (3 d+2 e x^2\right )}{\left (d+e x^2\right )^2}+4 \log (x)-2 \log \left (d+e x^2\right )\right )+b \left (\frac {c d e x}{\left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {2 c^2 d \left (-3 c^2 d+2 e\right ) \arctan (c x)}{\left (-c^2 d+e\right )^2}+\frac {2 d \left (3 d+2 e x^2\right ) \arctan (c x)}{\left (d+e x^2\right )^2}+\frac {c \sqrt {d} \left (7 c^2 d-5 e\right ) \sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\left (-c^2 d+e\right )^2}+8 \arctan (c x) \log (x)-4 \arctan (c x) \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right )-4 \arctan (c x) \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right )-2 i \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (-1-i c x)}{c \sqrt {d}-\sqrt {e}}\right )+2 i \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (1-i c x)}{c \sqrt {d}+\sqrt {e}}\right )+2 i \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (-1+i c x)}{c \sqrt {d}-\sqrt {e}}\right )-2 i \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (1+i c x)}{c \sqrt {d}+\sqrt {e}}\right )-4 i (\log (x) (\log (1-i c x)-\log (1+i c x))-\operatorname {PolyLog}(2,-i c x)+\operatorname {PolyLog}(2,i c x))+2 i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )-2 i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}+\sqrt {e}}\right )-2 i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )+2 i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{c \sqrt {d}+\sqrt {e}}\right )\right )}{8 d^3} \]
(2*a*((d*(3*d + 2*e*x^2))/(d + e*x^2)^2 + 4*Log[x] - 2*Log[d + e*x^2]) + b *((c*d*e*x)/((c^2*d - e)*(d + e*x^2)) + (2*c^2*d*(-3*c^2*d + 2*e)*ArcTan[c *x])/(-(c^2*d) + e)^2 + (2*d*(3*d + 2*e*x^2)*ArcTan[c*x])/(d + e*x^2)^2 + (c*Sqrt[d]*(7*c^2*d - 5*e)*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(-(c^2*d) + e)^2 + 8*ArcTan[c*x]*Log[x] - 4*ArcTan[c*x]*Log[((-I)*Sqrt[d])/Sqrt[e] + x] - 4*ArcTan[c*x]*Log[(I*Sqrt[d])/Sqrt[e] + x] - (2*I)*Log[((-I)*Sqrt[d] )/Sqrt[e] + x]*Log[(Sqrt[e]*(-1 - I*c*x))/(c*Sqrt[d] - Sqrt[e])] + (2*I)*L og[((-I)*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(1 - I*c*x))/(c*Sqrt[d] + Sqrt [e])] + (2*I)*Log[(I*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(-1 + I*c*x))/(c*S qrt[d] - Sqrt[e])] - (2*I)*Log[(I*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(1 + I*c*x))/(c*Sqrt[d] + Sqrt[e])] - (4*I)*(Log[x]*(Log[1 - I*c*x] - Log[1 + I *c*x]) - PolyLog[2, (-I)*c*x] + PolyLog[2, I*c*x]) + (2*I)*PolyLog[2, (c*( Sqrt[d] - I*Sqrt[e]*x))/(c*Sqrt[d] - Sqrt[e])] - (2*I)*PolyLog[2, (c*(Sqrt [d] - I*Sqrt[e]*x))/(c*Sqrt[d] + Sqrt[e])] - (2*I)*PolyLog[2, (c*(Sqrt[d] + I*Sqrt[e]*x))/(c*Sqrt[d] - Sqrt[e])] + (2*I)*PolyLog[2, (c*(Sqrt[d] + I* Sqrt[e]*x))/(c*Sqrt[d] + Sqrt[e])]))/(8*d^3)
Time = 0.87 (sec) , antiderivative size = 574, 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.
| \(\displaystyle \int \frac {a+b \arctan (c x)}{x \left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (-\frac {e x (a+b \arctan (c x))}{d^3 \left (d+e x^2\right )}+\frac {a+b \arctan (c x)}{d^3 x}-\frac {e x (a+b \arctan (c x))}{d^2 \left (d+e x^2\right )^2}-\frac {e x (a+b \arctan (c x))}{d \left (d+e x^2\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {(a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d^3}-\frac {(a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d^3}+\frac {\log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{d^3}+\frac {a+b \arctan (c x)}{2 d^2 \left (d+e x^2\right )}+\frac {a+b \arctan (c x)}{4 d \left (d+e x^2\right )^2}+\frac {a \log (x)}{d^3}+\frac {b c \sqrt {e} \left (3 c^2 d-e\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} \left (c^2 d-e\right )^2}+\frac {b c \sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}-\frac {b c^2 \arctan (c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {b c^4 \arctan (c x)}{4 d \left (c^2 d-e\right )^2}+\frac {b c e x}{8 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^3}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^3}+\frac {i b \operatorname {PolyLog}(2,-i c x)}{2 d^3}-\frac {i b \operatorname {PolyLog}(2,i c x)}{2 d^3}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 d^3}\) |
(b*c*e*x)/(8*d^2*(c^2*d - e)*(d + e*x^2)) - (b*c^4*ArcTan[c*x])/(4*d*(c^2* d - e)^2) - (b*c^2*ArcTan[c*x])/(2*d^2*(c^2*d - e)) + (a + b*ArcTan[c*x])/ (4*d*(d + e*x^2)^2) + (a + b*ArcTan[c*x])/(2*d^2*(d + e*x^2)) + (b*c*Sqrt[ e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(5/2)*(c^2*d - e)) + (b*c*(3*c^2*d - e)*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(5/2)*(c^2*d - e)^2) + (a*Log [x])/d^3 + ((a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/d^3 - ((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^3) - ((a + b*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*S qrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(2*d^3) + ((I/2)*b*PolyLog[2, (-I)*c*x ])/d^3 - ((I/2)*b*PolyLog[2, I*c*x])/d^3 - ((I/2)*b*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^3 + ((I/4)*b*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sq rt[-d] - I*Sqrt[e])*(1 - I*c*x))])/d^3 + ((I/4)*b*PolyLog[2, 1 - (2*c*(Sqr t[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/d^3
3.12.68.3.1 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 = 0.60 (sec) , antiderivative size = 903, normalized size of antiderivative = 1.57
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(903\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(920\) |
| default | \(\text {Expression too large to display}\) | \(920\) |
| risch | \(\text {Expression too large to display}\) | \(1677\) |
a*ln(x)/d^3+1/2*a/d^2/(e*x^2+d)+1/4*a/d/(e*x^2+d)^2-1/2*a/d^3*ln(e*x^2+d)+ b*(arctan(c*x)/d^3*ln(c*x)-1/2*arctan(c*x)/d^3*ln(c^2*e*x^2+c^2*d)+1/4*c^4 *arctan(c*x)/d/(c^2*e*x^2+c^2*d)^2+1/2*c^2*arctan(c*x)/d^2/(c^2*e*x^2+c^2* d)-1/2*c^6*(-I/c^6/d^3*ln(c*x)*ln(1+I*c*x)+I/c^6/d^3*ln(c*x)*ln(1-I*c*x)-I /c^6/d^3*dilog(1+I*c*x)+I/c^6/d^3*dilog(1-I*c*x)-1/d^3/c^6*(-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,inde x=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,i ndex=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,inde x=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,i ndex=2)-c*x-I)/RootOf(e*_Z^2-2*I*e*_Z+c^2*d-e,index=2)))/e)))+1/2/d^2/c^4* (-1/(c^2*d-e)^2*e*((1/2*c^2*d-1/2*e)*c*x/(c^2*e*x^2+c^2*d)+1/2*(7*c^2*d-5* e)/c/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2)))+(3*c^2*d-2*e)/(c^2*d-e)^2*arctan (c*x))))
\[ \int \frac {a+b \arctan (c x)}{x \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x \left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \arctan (c x)}{x \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x} \,d x } \]
1/4*a*((2*e*x^2 + 3*d)/(d^2*e^2*x^4 + 2*d^3*e*x^2 + d^4) - 2*log(e*x^2 + d )/d^3 + 4*log(x)/d^3) + 2*b*integrate(1/2*arctan(c*x)/(e^3*x^7 + 3*d*e^2*x ^5 + 3*d^2*e*x^3 + d^3*x), x)
\[ \int \frac {a+b \arctan (c x)}{x \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x \left (d+e x^2\right )^3} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x\,{\left (e\,x^2+d\right )}^3} \,d x \]