Integrand size = 21, antiderivative size = 532 \[ \int \frac {x^5 (a+b \arctan (c x))}{\left (d+e x^2\right )^3} \, dx=-\frac {b c d x}{8 \left (c^2 d-e\right ) e^2 \left (d+e x^2\right )}+\frac {b c^4 d^2 \arctan (c x)}{4 \left (c^2 d-e\right )^2 e^3}-\frac {b c^2 d \arctan (c x)}{\left (c^2 d-e\right ) e^3}-\frac {d^2 (a+b \arctan (c x))}{4 e^3 \left (d+e x^2\right )^2}+\frac {d (a+b \arctan (c x))}{e^3 \left (d+e x^2\right )}+\frac {b c \sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\left (c^2 d-e\right ) e^{5/2}}-\frac {b c \sqrt {d} \left (3 c^2 d-e\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 \left (c^2 d-e\right )^2 e^{5/2}}-\frac {(a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right )}{e^3}+\frac {(a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 e^3}+\frac {(a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 e^3}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^3}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^3} \]
-1/8*b*c*d*x/(c^2*d-e)/e^2/(e*x^2+d)+1/4*b*c^4*d^2*arctan(c*x)/(c^2*d-e)^2 /e^3-b*c^2*d*arctan(c*x)/(c^2*d-e)/e^3-1/4*d^2*(a+b*arctan(c*x))/e^3/(e*x^ 2+d)^2+d*(a+b*arctan(c*x))/e^3/(e*x^2+d)-(a+b*arctan(c*x))*ln(2/(1-I*c*x)) /e^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)))/e^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)))/e^3+1/2*I*b*polylog(2,1-2/(1-I*c*x))/ e^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)))/e^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)))/e^3+b*c*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/(c^2*d -e)/e^(5/2)-1/8*b*c*(3*c^2*d-e)*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/(c^2*d-e )^2/e^(5/2)
Time = 9.84 (sec) , antiderivative size = 589, normalized size of antiderivative = 1.11 \[ \int \frac {x^5 (a+b \arctan (c x))}{\left (d+e x^2\right )^3} \, dx=\frac {a \left (\frac {d \left (3 d+4 e x^2\right )}{\left (d+e x^2\right )^2}+2 \log \left (d+e x^2\right )\right )+b \left (-\frac {c d e x}{2 \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {c^2 d \left (-3 c^2 d+4 e\right ) \arctan (c x)}{\left (-c^2 d+e\right )^2}+\frac {d \left (3 d+4 e x^2\right ) \arctan (c x)}{\left (d+e x^2\right )^2}+\frac {c \sqrt {d} \left (5 c^2 d-7 e\right ) \sqrt {e} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \left (-c^2 d+e\right )^2}+2 \arctan (c x) \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right )+2 \arctan (c x) \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right )+i \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (-1-i c x)}{c \sqrt {d}-\sqrt {e}}\right )-i \log \left (-\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (1-i c x)}{c \sqrt {d}+\sqrt {e}}\right )-i \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (-1+i c x)}{c \sqrt {d}-\sqrt {e}}\right )+i \log \left (\frac {i \sqrt {d}}{\sqrt {e}}+x\right ) \log \left (\frac {\sqrt {e} (1+i c x)}{c \sqrt {d}+\sqrt {e}}\right )-i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )+i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}+\sqrt {e}}\right )+i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )-i \operatorname {PolyLog}\left (2,\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{c \sqrt {d}+\sqrt {e}}\right )\right )}{4 e^3} \]
(a*((d*(3*d + 4*e*x^2))/(d + e*x^2)^2 + 2*Log[d + e*x^2]) + b*(-1/2*(c*d*e *x)/((c^2*d - e)*(d + e*x^2)) + (c^2*d*(-3*c^2*d + 4*e)*ArcTan[c*x])/(-(c^ 2*d) + e)^2 + (d*(3*d + 4*e*x^2)*ArcTan[c*x])/(d + e*x^2)^2 + (c*Sqrt[d]*( 5*c^2*d - 7*e)*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*(-(c^2*d) + e)^2) + 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[(Sqr t[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*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*S qrt[e]*x))/(c*Sqrt[d] - Sqrt[e])] - I*PolyLog[2, (c*(Sqrt[d] + I*Sqrt[e]*x ))/(c*Sqrt[d] + Sqrt[e])]))/(4*e^3)
Time = 0.88 (sec) , antiderivative size = 532, 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.
| \(\displaystyle \int \frac {x^5 (a+b \arctan (c x))}{\left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {d^2 x (a+b \arctan (c x))}{e^2 \left (d+e x^2\right )^3}-\frac {2 d x (a+b \arctan (c x))}{e^2 \left (d+e x^2\right )^2}+\frac {x (a+b \arctan (c x))}{e^2 \left (d+e x^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 (a+b \arctan (c x))}{4 e^3 \left (d+e x^2\right )^2}+\frac {d (a+b \arctan (c x))}{e^3 \left (d+e x^2\right )}+\frac {(a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e^3}+\frac {(a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e^3}-\frac {\log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{e^3}-\frac {b c \sqrt {d} \left (3 c^2 d-e\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 e^{5/2} \left (c^2 d-e\right )^2}+\frac {b c \sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{5/2} \left (c^2 d-e\right )}-\frac {b c^2 d \arctan (c x)}{e^3 \left (c^2 d-e\right )}+\frac {b c^4 d^2 \arctan (c x)}{4 e^3 \left (c^2 d-e\right )^2}-\frac {b c d x}{8 e^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^3}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^3}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e^3}\) |
-1/8*(b*c*d*x)/((c^2*d - e)*e^2*(d + e*x^2)) + (b*c^4*d^2*ArcTan[c*x])/(4* (c^2*d - e)^2*e^3) - (b*c^2*d*ArcTan[c*x])/((c^2*d - e)*e^3) - (d^2*(a + b *ArcTan[c*x]))/(4*e^3*(d + e*x^2)^2) + (d*(a + b*ArcTan[c*x]))/(e^3*(d + e *x^2)) + (b*c*Sqrt[d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/((c^2*d - e)*e^(5/2)) - (b*c*Sqrt[d]*(3*c^2*d - e)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*(c^2*d - e)^2* e^(5/2)) - ((a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/e^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*e^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*e^3) + ((I/2)*b*PolyLog[2, 1 - 2/(1 - I*c*x)])/e^3 - ((I/4)*b*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c *Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/e^3 - ((I/4)*b*PolyLog[2, 1 - (2*c*( Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/e^3
3.12.65.3.1 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 = 0.66 (sec) , antiderivative size = 817, normalized size of antiderivative = 1.54
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(817\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(843\) |
| default | \(\text {Expression too large to display}\) | \(843\) |
| risch | \(\text {Expression too large to display}\) | \(1666\) |
a*(1/e^3*d/(e*x^2+d)+1/2/e^3*ln(e*x^2+d)-1/4*d^2/e^3/(e*x^2+d)^2)+b/c^6*(1 /2*arctan(c*x)*c^6/e^3*ln(c^2*e*x^2+c^2*d)-1/4*arctan(c*x)*c^10*d^2/e^3/(c ^2*e*x^2+c^2*d)^2+arctan(c*x)*c^8*d/e^3/(c^2*e*x^2+c^2*d)-1/4*c^6*(d*c^2/e ^3*(-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*(5*c^2* d-7*e)/c/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2)))+1/(c^2*d-e)^2*(3*c^2*d-4*e)* arctan(c*x))+2/e^3*(-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)/R ootOf(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)*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)/R ootOf(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)))))
\[ \int \frac {x^5 (a+b \arctan (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^5 (a+b \arctan (c x))}{\left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {x^5 (a+b \arctan (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
1/4*a*((4*d*e*x^2 + 3*d^2)/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^3) + 2*log(e*x^2 + d)/e^3) + 2*b*integrate(1/2*x^5*arctan(c*x)/(e^3*x^6 + 3*d*e^2*x^4 + 3* d^2*e*x^2 + d^3), x)
\[ \int \frac {x^5 (a+b \arctan (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^5 (a+b \arctan (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^5\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]