Integrand size = 21, antiderivative size = 555 \[ \int \frac {x^2 (a+b \arctan (c x))}{d+e x^2} \, dx=\frac {a x}{e}+\frac {b x \arctan (c x)}{e}-\frac {a \sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}-\frac {i b \sqrt {-d} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 e^{3/2}}+\frac {i b \sqrt {-d} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 e^{3/2}}-\frac {i b \sqrt {-d} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 e^{3/2}}+\frac {i b \sqrt {-d} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 e^{3/2}}-\frac {b \log \left (1+c^2 x^2\right )}{2 c e}+\frac {i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i-c x)}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 e^{3/2}}-\frac {i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1-i c x)}{i c \sqrt {-d}+\sqrt {e}}\right )}{4 e^{3/2}}-\frac {i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1+i c x)}{i c \sqrt {-d}+\sqrt {e}}\right )}{4 e^{3/2}}+\frac {i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i+c x)}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 e^{3/2}} \]
a*x/e+b*x*arctan(c*x)/e-1/2*b*ln(c^2*x^2+1)/c/e-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)))*(-d)^(1/2)/e^(3/2)+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)))*(-d)^(1 /2)/e^(3/2)-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)))*(-d)^(1/2)/e^(3/2)+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)))*(-d)^(1/2)/e^(3/2)+1/4*I*b*polylog(2,(I-c*x )*e^(1/2)/(c*(-d)^(1/2)+I*e^(1/2)))*(-d)^(1/2)/e^(3/2)+1/4*I*b*polylog(2,( c*x+I)*e^(1/2)/(c*(-d)^(1/2)+I*e^(1/2)))*(-d)^(1/2)/e^(3/2)-1/4*I*b*polylo g(2,(1-I*c*x)*e^(1/2)/(I*c*(-d)^(1/2)+e^(1/2)))*(-d)^(1/2)/e^(3/2)-1/4*I*b *polylog(2,(1+I*c*x)*e^(1/2)/(I*c*(-d)^(1/2)+e^(1/2)))*(-d)^(1/2)/e^(3/2)- a*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/e^(3/2)
Time = 2.55 (sec) , antiderivative size = 766, normalized size of antiderivative = 1.38 \[ \int \frac {x^2 (a+b \arctan (c x))}{d+e x^2} \, dx=\frac {a x}{e}-\frac {a \sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}+\frac {b \left (4 c x \arctan (c x)-2 \log \left (1+c^2 x^2\right )+\frac {c^2 d \left (-4 \arctan (c x) \text {arctanh}\left (\frac {c d}{\sqrt {-c^2 d e} x}\right )-2 \arccos \left (\frac {c^2 d+e}{-c^2 d+e}\right ) \text {arctanh}\left (\frac {c e x}{\sqrt {-c^2 d e}}\right )-\left (\arccos \left (\frac {c^2 d+e}{-c^2 d+e}\right )-2 i \text {arctanh}\left (\frac {c e x}{\sqrt {-c^2 d e}}\right )\right ) \log \left (\frac {2 c d \left (i e+\sqrt {-c^2 d e}\right ) (-i+c x)}{\left (c^2 d-e\right ) \left (-c d+\sqrt {-c^2 d e} x\right )}\right )-\left (\arccos \left (\frac {c^2 d+e}{-c^2 d+e}\right )+2 i \text {arctanh}\left (\frac {c e x}{\sqrt {-c^2 d e}}\right )\right ) \log \left (\frac {2 c d \left (-i e+\sqrt {-c^2 d e}\right ) (i+c x)}{\left (c^2 d-e\right ) \left (-c d+\sqrt {-c^2 d e} x\right )}\right )+\left (\arccos \left (\frac {c^2 d+e}{-c^2 d+e}\right )+2 i \text {arctanh}\left (\frac {c d}{\sqrt {-c^2 d e} x}\right )+2 i \text {arctanh}\left (\frac {c e x}{\sqrt {-c^2 d e}}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-c^2 d e} e^{-i \arctan (c x)}}{\sqrt {-c^2 d+e} \sqrt {-c^2 d-e+\left (-c^2 d+e\right ) \cos (2 \arctan (c x))}}\right )+\left (\arccos \left (\frac {c^2 d+e}{-c^2 d+e}\right )-2 i \text {arctanh}\left (\frac {c d}{\sqrt {-c^2 d e} x}\right )-2 i \text {arctanh}\left (\frac {c e x}{\sqrt {-c^2 d e}}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-c^2 d e} e^{i \arctan (c x)}}{\sqrt {-c^2 d+e} \sqrt {-c^2 d-e+\left (-c^2 d+e\right ) \cos (2 \arctan (c x))}}\right )+i \left (-\operatorname {PolyLog}\left (2,\frac {\left (c^2 d+e-2 i \sqrt {-c^2 d e}\right ) \left (c d+\sqrt {-c^2 d e} x\right )}{\left (c^2 d-e\right ) \left (c d-\sqrt {-c^2 d e} x\right )}\right )+\operatorname {PolyLog}\left (2,\frac {\left (c^2 d+e+2 i \sqrt {-c^2 d e}\right ) \left (c d+\sqrt {-c^2 d e} x\right )}{\left (c^2 d-e\right ) \left (c d-\sqrt {-c^2 d e} x\right )}\right )\right )\right )}{\sqrt {-c^2 d e}}\right )}{4 c e} \]
(a*x)/e - (a*Sqrt[d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/e^(3/2) + (b*(4*c*x*ArcT an[c*x] - 2*Log[1 + c^2*x^2] + (c^2*d*(-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)/Sqr t[-(c^2*d*e)]] - (ArcCos[(c^2*d + e)/(-(c^2*d) + e)] - (2*I)*ArcTanh[(c*e* x)/Sqrt[-(c^2*d*e)]])*Log[(2*c*d*(I*e + Sqrt[-(c^2*d*e)])*(-I + c*x))/((c^ 2*d - e)*(-(c*d) + 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[(2*c*d*((-I)*e + Sqrt[ -(c^2*d*e)])*(I + c*x))/((c^2*d - e)*(-(c*d) + Sqrt[-(c^2*d*e)]*x))] + (Ar cCos[(c^2*d + e)/(-(c^2*d) + e)] + (2*I)*ArcTanh[(c*d)/(Sqrt[-(c^2*d*e)]*x )] + (2*I)*ArcTanh[(c*e*x)/Sqrt[-(c^2*d*e)]])*Log[(Sqrt[2]*Sqrt[-(c^2*d*e) ])/(Sqrt[-(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)*ArcT anh[(c*d)/(Sqrt[-(c^2*d*e)]*x)] - (2*I)*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^2*d*e)])*(c*d + Sqrt[-(c^2*d*e)]*x))/((c^2*d - e) *(c*d - Sqrt[-(c^2*d*e)]*x))] + PolyLog[2, ((c^2*d + e + (2*I)*Sqrt[-(c^2* d*e)])*(c*d + Sqrt[-(c^2*d*e)]*x))/((c^2*d - e)*(c*d - Sqrt[-(c^2*d*e)]*x) )])))/Sqrt[-(c^2*d*e)]))/(4*c*e)
Time = 0.98 (sec) , antiderivative size = 550, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5451, 2009, 5445, 218, 5443, 2856, 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.
| \(\displaystyle \int \frac {x^2 (a+b \arctan (c x))}{d+e x^2} \, dx\) |
| \(\Big \downarrow \) 5451 |
| \(\displaystyle \frac {\int (a+b \arctan (c x))dx}{e}-\frac {d \int \frac {a+b \arctan (c x)}{e x^2+d}dx}{e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{e}-\frac {d \int \frac {a+b \arctan (c x)}{e x^2+d}dx}{e}\) |
| \(\Big \downarrow \) 5445 |
| \(\displaystyle \frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{e}-\frac {d \left (a \int \frac {1}{e x^2+d}dx+b \int \frac {\arctan (c x)}{e x^2+d}dx\right )}{e}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{e}-\frac {d \left (b \int \frac {\arctan (c x)}{e x^2+d}dx+\frac {a \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}\right )}{e}\) |
| \(\Big \downarrow \) 5443 |
| \(\displaystyle \frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{e}-\frac {d \left (\frac {a \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}+b \left (\frac {1}{2} i \int \frac {\log (1-i c x)}{e x^2+d}dx-\frac {1}{2} i \int \frac {\log (i c x+1)}{e x^2+d}dx\right )\right )}{e}\) |
| \(\Big \downarrow \) 2856 |
| \(\displaystyle \frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{e}-\frac {d \left (\frac {a \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}+b \left (\frac {1}{2} i \int \left (\frac {\sqrt {-d} \log (1-i c x)}{2 d \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\sqrt {-d} \log (1-i c x)}{2 d \left (\sqrt {e} x+\sqrt {-d}\right )}\right )dx-\frac {1}{2} i \int \left (\frac {\sqrt {-d} \log (i c x+1)}{2 d \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\sqrt {-d} \log (i c x+1)}{2 d \left (\sqrt {e} x+\sqrt {-d}\right )}\right )dx\right )\right )}{e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{e}-\frac {d \left (\frac {a \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}+b \left (\frac {1}{2} i \left (-\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {e} (c x+i)}{\sqrt {-d} c+i \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}\right )-\frac {1}{2} i \left (-\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i-c x)}{\sqrt {-d} c+i \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}\right )\right )\right )}{e}\) |
(a*x + b*x*ArcTan[c*x] - (b*Log[1 + c^2*x^2])/(2*c))/e - (d*((a*ArcTan[(Sq rt[e]*x)/Sqrt[d]])/(Sqrt[d]*Sqrt[e]) + b*((-1/2*I)*((Log[1 + I*c*x]*Log[(c *(Sqrt[-d] - Sqrt[e]*x))/(c*Sqrt[-d] - I*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e]) - (Log[1 + I*c*x]*Log[(c*(Sqrt[-d] + Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])]) /(2*Sqrt[-d]*Sqrt[e]) - PolyLog[2, (Sqrt[e]*(I - c*x))/(c*Sqrt[-d] + I*Sqr t[e])]/(2*Sqrt[-d]*Sqrt[e]) + PolyLog[2, (Sqrt[e]*(1 + I*c*x))/(I*c*Sqrt[- d] + Sqrt[e])]/(2*Sqrt[-d]*Sqrt[e])) + (I/2)*((Log[1 - I*c*x]*Log[(c*(Sqrt [-d] - Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e]) - (Log[ 1 - I*c*x]*Log[(c*(Sqrt[-d] + Sqrt[e]*x))/(c*Sqrt[-d] - I*Sqrt[e])])/(2*Sq rt[-d]*Sqrt[e]) - PolyLog[2, (Sqrt[e]*(1 - I*c*x))/(I*c*Sqrt[-d] + Sqrt[e] )]/(2*Sqrt[-d]*Sqrt[e]) + PolyLog[2, (Sqrt[e]*(I + c*x))/(c*Sqrt[-d] + I*S qrt[e])]/(2*Sqrt[-d]*Sqrt[e])))))/e
3.12.55.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) ^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
Int[ArcTan[(c_.)*(x_)]/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Simp[I/2 Int[ Log[1 - I*c*x]/(d + e*x^2), x], x] - Simp[I/2 Int[Log[1 + I*c*x]/(d + e*x ^2), x], x] /; FreeQ[{c, d, e}, x]
Int[(ArcTan[(c_.)*(x_)]*(b_.) + (a_))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Simp[a Int[1/(d + e*x^2), x], x] + Simp[b Int[ArcTan[c*x]/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x] )^p, x], x] - Simp[d*(f^2/e) Int[(f*x)^(m - 2)*((a + b*ArcTan[c*x])^p/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Time = 0.94 (sec) , antiderivative size = 526, normalized size of antiderivative = 0.95
| method | result | size |
| risch | \(-\frac {i a d \,\operatorname {arctanh}\left (\frac {2 \left (-i c x +1\right ) e -2 e}{2 c \sqrt {e d}}\right )}{e \sqrt {e d}}-\frac {i b \ln \left (i c x +1\right ) x}{2 e}-\frac {b \ln \left (-i c x +1\right )}{2 c e}+\frac {b}{c e}-\frac {b d \ln \left (-i c x +1\right ) \ln \left (\frac {c \sqrt {e d}-\left (-i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{4 e \sqrt {e d}}+\frac {b d \ln \left (-i c x +1\right ) \ln \left (\frac {c \sqrt {e d}+\left (-i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{4 e \sqrt {e d}}-\frac {b d \operatorname {dilog}\left (\frac {c \sqrt {e d}-\left (-i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{4 e \sqrt {e d}}+\frac {b d \operatorname {dilog}\left (\frac {c \sqrt {e d}+\left (-i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{4 e \sqrt {e d}}+\frac {a x}{e}+\frac {i a}{c e}+\frac {i b \ln \left (-i c x +1\right ) x}{2 e}-\frac {b \ln \left (i c x +1\right )}{2 c e}-\frac {b d \ln \left (i c x +1\right ) \ln \left (\frac {c \sqrt {e d}-\left (i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{4 e \sqrt {e d}}+\frac {b d \ln \left (i c x +1\right ) \ln \left (\frac {c \sqrt {e d}+\left (i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{4 e \sqrt {e d}}-\frac {b d \operatorname {dilog}\left (\frac {c \sqrt {e d}-\left (i c x +1\right ) e +e}{c \sqrt {e d}+e}\right )}{4 e \sqrt {e d}}+\frac {b d \operatorname {dilog}\left (\frac {c \sqrt {e d}+\left (i c x +1\right ) e -e}{c \sqrt {e d}-e}\right )}{4 e \sqrt {e d}}\) | \(526\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(2390\) |
| default | \(\text {Expression too large to display}\) | \(2390\) |
| parts | \(\text {Expression too large to display}\) | \(2398\) |
-I*a*d/e/(e*d)^(1/2)*arctanh(1/2*(2*(1-I*c*x)*e-2*e)/c/(e*d)^(1/2))-1/2*I* b/e*ln(1+I*c*x)*x-1/2/c*b/e*ln(1-I*c*x)+b/c/e-1/4*b*d/e*ln(1-I*c*x)/(e*d)^ (1/2)*ln((c*(e*d)^(1/2)-(1-I*c*x)*e+e)/(c*(e*d)^(1/2)+e))+1/4*b*d/e*ln(1-I *c*x)/(e*d)^(1/2)*ln((c*(e*d)^(1/2)+(1-I*c*x)*e-e)/(c*(e*d)^(1/2)-e))-1/4* b*d/e/(e*d)^(1/2)*dilog((c*(e*d)^(1/2)-(1-I*c*x)*e+e)/(c*(e*d)^(1/2)+e))+1 /4*b*d/e/(e*d)^(1/2)*dilog((c*(e*d)^(1/2)+(1-I*c*x)*e-e)/(c*(e*d)^(1/2)-e) )+a*x/e+I/c*a/e+1/2*I*b/e*ln(1-I*c*x)*x-1/2*b/c/e*ln(1+I*c*x)-1/4*b*d/e*ln (1+I*c*x)/(e*d)^(1/2)*ln((c*(e*d)^(1/2)-(1+I*c*x)*e+e)/(c*(e*d)^(1/2)+e))+ 1/4*b*d/e*ln(1+I*c*x)/(e*d)^(1/2)*ln((c*(e*d)^(1/2)+(1+I*c*x)*e-e)/(c*(e*d )^(1/2)-e))-1/4*b*d/e/(e*d)^(1/2)*dilog((c*(e*d)^(1/2)-(1+I*c*x)*e+e)/(c*( e*d)^(1/2)+e))+1/4*b*d/e/(e*d)^(1/2)*dilog((c*(e*d)^(1/2)+(1+I*c*x)*e-e)/( c*(e*d)^(1/2)-e))
\[ \int \frac {x^2 (a+b \arctan (c x))}{d+e x^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} x^{2}}{e x^{2} + d} \,d x } \]
\[ \int \frac {x^2 (a+b \arctan (c x))}{d+e x^2} \, dx=\int \frac {x^{2} \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{d + e x^{2}}\, dx \]
Exception generated. \[ \int \frac {x^2 (a+b \arctan (c x))}{d+e x^2} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^2 (a+b \arctan (c x))}{d+e x^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} x^{2}}{e x^{2} + d} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \arctan (c x))}{d+e x^2} \, dx=\int \frac {x^2\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{e\,x^2+d} \,d x \]