Integrand size = 21, antiderivative size = 1518 \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^3} \, dx =\text {Too large to display} \]
b*c*ln(x)/d^3+(-a-b*arctan(c*x))/d^3/x+1/16*b*c*(5*c^2*d-3*e)*e*ln(c^2*x^2 +1)/d^3/(c^2*d-e)^2-1/16*b*c*(5*c^2*d-3*e)*e*ln(e*x^2+d)/d^3/(c^2*d-e)^2-7 /32*I*b*c*polylog(2,(-c^2)^(1/2)*(d^(1/2)-I*x*e^(1/2))/((-c^2)^(1/2)*d^(1/ 2)-I*e^(1/2)))*e^(1/2)/d^(7/2)/(-c^2)^(1/2)-7/32*I*b*c*polylog(2,(-c^2)^(1 /2)*(d^(1/2)+I*x*e^(1/2))/((-c^2)^(1/2)*d^(1/2)-I*e^(1/2)))*e^(1/2)/d^(7/2 )/(-c^2)^(1/2)-7/32*I*b*c*ln((1-x*(-c^2)^(1/2))*e^(1/2)/(I*(-c^2)^(1/2)*d^ (1/2)+e^(1/2)))*ln(1-I*x*e^(1/2)/d^(1/2))*e^(1/2)/d^(7/2)/(-c^2)^(1/2)-7/3 2*I*b*c*ln((1+x*(-c^2)^(1/2))*e^(1/2)/(I*(-c^2)^(1/2)*d^(1/2)+e^(1/2)))*ln (1+I*x*e^(1/2)/d^(1/2))*e^(1/2)/d^(7/2)/(-c^2)^(1/2)+7/32*I*b*c*ln(-(1+x*( -c^2)^(1/2))*e^(1/2)/(I*(-c^2)^(1/2)*d^(1/2)-e^(1/2)))*ln(1-I*x*e^(1/2)/d^ (1/2))*e^(1/2)/d^(7/2)/(-c^2)^(1/2)+7/32*I*b*c*ln(-(1-x*(-c^2)^(1/2))*e^(1 /2)/(I*(-c^2)^(1/2)*d^(1/2)-e^(1/2)))*ln(1+I*x*e^(1/2)/d^(1/2))*e^(1/2)/d^ (7/2)/(-c^2)^(1/2)-1/4*e*x*(a+b*arctan(c*x))/d^2/(e*x^2+d)^2-7/8*e*x*(a+b* arctan(c*x))/d^3/(e*x^2+d)-1/4*I*b*polylog(2,(1-I*c*x)*e^(1/2)/(I*c*(-d)^( 1/2)+e^(1/2)))*e^(1/2)/(-d)^(7/2)-1/4*I*b*polylog(2,(1+I*c*x)*e^(1/2)/(I*c *(-d)^(1/2)+e^(1/2)))*e^(1/2)/(-d)^(7/2)-1/2*b*c*ln(c^2*x^2+1)/d^3+1/4*I*b *ln(1-I*c*x)*ln(c*((-d)^(1/2)-x*e^(1/2))/(c*(-d)^(1/2)+I*e^(1/2)))*e^(1/2) /(-d)^(7/2)+1/4*I*b*ln(1+I*c*x)*ln(c*((-d)^(1/2)+x*e^(1/2))/(c*(-d)^(1/2)+ I*e^(1/2)))*e^(1/2)/(-d)^(7/2)+1/8*b*c*e/d^2/(c^2*d-e)/(e*x^2+d)+1/4*b*c*e *ln(c^2*x^2+1)/d^3/(c^2*d-e)-1/4*b*c*e*ln(e*x^2+d)/d^3/(c^2*d-e)-1/4*I*...
Time = 13.08 (sec) , antiderivative size = 2005, normalized size of antiderivative = 1.32 \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^3} \, dx=\text {Result too large to show} \]
-(a/(d^3*x)) - (a*e*x)/(4*d^2*(d + e*x^2)^2) - (7*a*e*x)/(8*d^3*(d + e*x^2 )) - (15*a*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(7/2)) + b*c^7*(-(Arc Tan[c*x]/(c^7*d^3*x)) + Log[(c*x)/Sqrt[1 + c^2*x^2]]/(c^6*d^3) - (9*e*Log[ 1 - ((-(c^2*d) + e)*Cos[2*ArcTan[c*x]])/(c^2*d + e)])/(16*c^4*d^2*(c^2*d - e)^2) + (7*e^2*Log[1 - ((-(c^2*d) + e)*Cos[2*ArcTan[c*x]])/(c^2*d + e)])/ (16*c^6*d^3*(c^2*d - e)^2) - (15*e*(4*ArcTan[c*x]*ArcTanh[(c*d)/(Sqrt[-(c^ 2*d*e)]*x)] + 2*ArcCos[(-(c^2*d) - e)/(c^2*d - e)]*ArcTanh[(c*e*x)/Sqrt[-( c^2*d*e)]] - (ArcCos[(-(c^2*d) - e)/(c^2*d - e)] - (2*I)*ArcTanh[(c*e*x)/S qrt[-(c^2*d*e)]])*Log[1 - ((c^2*d + e - (2*I)*Sqrt[-(c^2*d*e)])*(2*c^2*d - 2*c*Sqrt[-(c^2*d*e)]*x))/((c^2*d - e)*(2*c^2*d + 2*c*Sqrt[-(c^2*d*e)]*x)) ] + (-ArcCos[(-(c^2*d) - e)/(c^2*d - e)] - (2*I)*ArcTanh[(c*e*x)/Sqrt[-(c^ 2*d*e)]])*Log[1 - ((c^2*d + e + (2*I)*Sqrt[-(c^2*d*e)])*(2*c^2*d - 2*c*Sqr t[-(c^2*d*e)]*x))/((c^2*d - e)*(2*c^2*d + 2*c*Sqrt[-(c^2*d*e)]*x))] + (Arc Cos[(-(c^2*d) - e)/(c^2*d - e)] - (2*I)*(ArcTanh[(c*d)/(Sqrt[-(c^2*d*e)]*x )] + ArcTanh[(c*e*x)/Sqrt[-(c^2*d*e)]]))*Log[(Sqrt[2]*Sqrt[-(c^2*d*e)])/(S qrt[c^2*d - e]*E^(I*ArcTan[c*x])*Sqrt[c^2*d + e + (c^2*d - e)*Cos[2*ArcTan [c*x]]])] + (ArcCos[(-(c^2*d) - e)/(c^2*d - e)] + (2*I)*(ArcTanh[(c*d)/(Sq rt[-(c^2*d*e)]*x)] + ArcTanh[(c*e*x)/Sqrt[-(c^2*d*e)]]))*Log[(Sqrt[2]*Sqrt [-(c^2*d*e)]*E^(I*ArcTan[c*x]))/(Sqrt[c^2*d - e]*Sqrt[c^2*d + e + (c^2*d - e)*Cos[2*ArcTan[c*x]]])] + I*(PolyLog[2, ((c^2*d + e - (2*I)*Sqrt[-(c^...
Time = 2.71 (sec) , antiderivative size = 1518, 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.
| \(\displaystyle \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (-\frac {e (a+b \arctan (c x))}{d^3 \left (d+e x^2\right )}+\frac {a+b \arctan (c x)}{d^3 x^2}-\frac {e (a+b \arctan (c x))}{d^2 \left (d+e x^2\right )^2}-\frac {e (a+b \arctan (c x))}{d \left (d+e x^2\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {7 x (a+b \arctan (c x)) e}{8 d^3 \left (e x^2+d\right )}-\frac {x (a+b \arctan (c x)) e}{4 d^2 \left (e x^2+d\right )^2}+\frac {b c \log \left (c^2 x^2+1\right ) e}{4 d^3 \left (c^2 d-e\right )}+\frac {b c \left (5 c^2 d-3 e\right ) \log \left (c^2 x^2+1\right ) e}{16 d^3 \left (c^2 d-e\right )^2}-\frac {b c \log \left (e x^2+d\right ) e}{4 d^3 \left (c^2 d-e\right )}-\frac {b c \left (5 c^2 d-3 e\right ) \log \left (e x^2+d\right ) e}{16 d^3 \left (c^2 d-e\right )^2}+\frac {b c e}{8 d^2 \left (c^2 d-e\right ) \left (e x^2+d\right )}-\frac {7 (a+b \arctan (c x)) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \sqrt {e}}{8 d^{7/2}}-\frac {a \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \sqrt {e}}{d^{7/2}}-\frac {i b \log (i c x+1) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right ) \sqrt {e}}{4 (-d)^{7/2}}+\frac {i b \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{\sqrt {-d} c+i \sqrt {e}}\right ) \sqrt {e}}{4 (-d)^{7/2}}-\frac {i b \log (1-i c x) \log \left (\frac {c \left (\sqrt {e} x+\sqrt {-d}\right )}{c \sqrt {-d}-i \sqrt {e}}\right ) \sqrt {e}}{4 (-d)^{7/2}}+\frac {i b \log (i c x+1) \log \left (\frac {c \left (\sqrt {e} x+\sqrt {-d}\right )}{\sqrt {-d} c+i \sqrt {e}}\right ) \sqrt {e}}{4 (-d)^{7/2}}-\frac {7 i b c \log \left (\frac {\sqrt {e} \left (1-\sqrt {-c^2} x\right )}{i \sqrt {-c^2} \sqrt {d}+\sqrt {e}}\right ) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right ) \sqrt {e}}{32 \sqrt {-c^2} d^{7/2}}+\frac {7 i b c \log \left (-\frac {\sqrt {e} \left (\sqrt {-c^2} x+1\right )}{i \sqrt {-c^2} \sqrt {d}-\sqrt {e}}\right ) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right ) \sqrt {e}}{32 \sqrt {-c^2} d^{7/2}}+\frac {7 i b c \log \left (-\frac {\sqrt {e} \left (1-\sqrt {-c^2} x\right )}{i \sqrt {-c^2} \sqrt {d}-\sqrt {e}}\right ) \log \left (\frac {i \sqrt {e} x}{\sqrt {d}}+1\right ) \sqrt {e}}{32 \sqrt {-c^2} d^{7/2}}-\frac {7 i b c \log \left (\frac {\sqrt {e} \left (\sqrt {-c^2} x+1\right )}{i \sqrt {-c^2} \sqrt {d}+\sqrt {e}}\right ) \log \left (\frac {i \sqrt {e} x}{\sqrt {d}}+1\right ) \sqrt {e}}{32 \sqrt {-c^2} d^{7/2}}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i-c x)}{\sqrt {-d} c+i \sqrt {e}}\right ) \sqrt {e}}{4 (-d)^{7/2}}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right ) \sqrt {e}}{4 (-d)^{7/2}}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right ) \sqrt {e}}{4 (-d)^{7/2}}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (c x+i)}{\sqrt {-d} c+i \sqrt {e}}\right ) \sqrt {e}}{4 (-d)^{7/2}}-\frac {7 i b c \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} \left (\sqrt {d}-i \sqrt {e} x\right )}{\sqrt {-c^2} \sqrt {d}-i \sqrt {e}}\right ) \sqrt {e}}{32 \sqrt {-c^2} d^{7/2}}+\frac {7 i b c \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} \left (\sqrt {d}-i \sqrt {e} x\right )}{\sqrt {-c^2} \sqrt {d}+i \sqrt {e}}\right ) \sqrt {e}}{32 \sqrt {-c^2} d^{7/2}}-\frac {7 i b c \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} \left (i \sqrt {e} x+\sqrt {d}\right )}{\sqrt {-c^2} \sqrt {d}-i \sqrt {e}}\right ) \sqrt {e}}{32 \sqrt {-c^2} d^{7/2}}+\frac {7 i b c \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} \left (i \sqrt {e} x+\sqrt {d}\right )}{\sqrt {-c^2} \sqrt {d}+i \sqrt {e}}\right ) \sqrt {e}}{32 \sqrt {-c^2} d^{7/2}}-\frac {a+b \arctan (c x)}{d^3 x}+\frac {b c \log (x)}{d^3}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d^3}\) |
(b*c*e)/(8*d^2*(c^2*d - e)*(d + e*x^2)) - (a + b*ArcTan[c*x])/(d^3*x) - (e *x*(a + b*ArcTan[c*x]))/(4*d^2*(d + e*x^2)^2) - (7*e*x*(a + b*ArcTan[c*x]) )/(8*d^3*(d + e*x^2)) - (a*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/d^(7/2) - (7*Sqrt[e]*(a + b*ArcTan[c*x])*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(7/2)) + (b*c*Log[x])/d^3 - ((I/4)*b*Sqrt[e]*Log[1 + I*c*x]*Log[(c*(Sqrt[-d] - Sqrt [e]*x))/(c*Sqrt[-d] - I*Sqrt[e])])/(-d)^(7/2) + ((I/4)*b*Sqrt[e]*Log[1 - I *c*x]*Log[(c*(Sqrt[-d] - Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])])/(-d)^(7/2) - ((I/4)*b*Sqrt[e]*Log[1 - I*c*x]*Log[(c*(Sqrt[-d] + Sqrt[e]*x))/(c*Sqrt[ -d] - I*Sqrt[e])])/(-d)^(7/2) + ((I/4)*b*Sqrt[e]*Log[1 + I*c*x]*Log[(c*(Sq rt[-d] + Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])])/(-d)^(7/2) - (((7*I)/32)*b *c*Sqrt[e]*Log[(Sqrt[e]*(1 - Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] + Sqrt[e ])]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]])/(Sqrt[-c^2]*d^(7/2)) + (((7*I)/32)*b*c *Sqrt[e]*Log[-((Sqrt[e]*(1 + Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] - Sqrt[e ]))]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]])/(Sqrt[-c^2]*d^(7/2)) + (((7*I)/32)*b* c*Sqrt[e]*Log[-((Sqrt[e]*(1 - Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] - Sqrt[ e]))]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]])/(Sqrt[-c^2]*d^(7/2)) - (((7*I)/32)*b *c*Sqrt[e]*Log[(Sqrt[e]*(1 + Sqrt[-c^2]*x))/(I*Sqrt[-c^2]*Sqrt[d] + Sqrt[e ])]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]])/(Sqrt[-c^2]*d^(7/2)) - (b*c*Log[1 + c^ 2*x^2])/(2*d^3) + (b*c*(5*c^2*d - 3*e)*e*Log[1 + c^2*x^2])/(16*d^3*(c^2*d - e)^2) + (b*c*e*Log[1 + c^2*x^2])/(4*d^3*(c^2*d - e)) - (b*c*(5*c^2*d ...
3.12.72.3.1 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 = 2.39 (sec) , antiderivative size = 5730, normalized size of antiderivative = 3.77
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(5730\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(5776\) |
| default | \(\text {Expression too large to display}\) | \(5776\) |
| risch | \(\text {Expression too large to display}\) | \(7503\) |
\[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan (c x)}{x^2 \left (d+e x^2\right )^3} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^2\,{\left (e\,x^2+d\right )}^3} \,d x \]