\(\int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx\) [62]

Optimal result

Integrand size = 19, antiderivative size = 367 \[ \int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx=\frac {\arctan (d+e x) \log \left (\frac {2 e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c (i-d)+\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1-i (d+e x))}\right )}{\sqrt {b^2-4 a c}}-\frac {\arctan (d+e x) \log \left (\frac {2 e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c (i-d)+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1-i (d+e x))}\right )}{\sqrt {b^2-4 a c}}-\frac {i \operatorname {PolyLog}\left (2,1+\frac {2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e-2 c (d+e x)\right )}{\left (2 i c-2 c d+b e-\sqrt {b^2-4 a c} e\right ) (1-i (d+e x))}\right )}{2 \sqrt {b^2-4 a c}}+\frac {i \operatorname {PolyLog}\left (2,1+\frac {2 \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e-2 c (d+e x)\right )}{\left (2 c (i-d)+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1-i (d+e x))}\right )}{2 \sqrt {b^2-4 a c}} \]

[Out]

arctan(e*x+d)*ln(2*e*(b+2*c*x-(-4*a*c+b^2)^(1/2))/(1-I*(e*x+d))/(2*c*(I-d)+e*(b-(-4*a*c+b^2)^(1/2))))/(-4*a*c+
b^2)^(1/2)-arctan(e*x+d)*ln(2*e*(b+2*c*x+(-4*a*c+b^2)^(1/2))/(1-I*(e*x+d))/(2*c*(I-d)+e*(b+(-4*a*c+b^2)^(1/2))
))/(-4*a*c+b^2)^(1/2)-1/2*I*polylog(2,1+2*(2*c*d-2*c*(e*x+d)-e*(b-(-4*a*c+b^2)^(1/2)))/(1-I*(e*x+d))/(2*I*c-2*
c*d+b*e-e*(-4*a*c+b^2)^(1/2)))/(-4*a*c+b^2)^(1/2)+1/2*I*polylog(2,1+2*(2*c*d-2*c*(e*x+d)-e*(b+(-4*a*c+b^2)^(1/
2)))/(1-I*(e*x+d))/(2*c*(I-d)+e*(b+(-4*a*c+b^2)^(1/2))))/(-4*a*c+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {632, 212, 6860, 5155, 4966, 2449, 2352, 2497} \[ \int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx=\frac {\arctan (d+e x) \log \left (\frac {2 e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{(1-i (d+e x)) \left (e \left (b-\sqrt {b^2-4 a c}\right )+2 c (-d+i)\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {\arctan (d+e x) \log \left (\frac {2 e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{(1-i (d+e x)) \left (e \left (\sqrt {b^2-4 a c}+b\right )+2 c (-d+i)\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {i \operatorname {PolyLog}\left (2,\frac {2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e-2 c (d+e x)\right )}{\left (-2 d c+2 i c+b e-\sqrt {b^2-4 a c} e\right ) (1-i (d+e x))}+1\right )}{2 \sqrt {b^2-4 a c}}+\frac {i \operatorname {PolyLog}\left (2,\frac {2 \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e-2 c (d+e x)\right )}{\left (2 c (i-d)+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1-i (d+e x))}+1\right )}{2 \sqrt {b^2-4 a c}} \]

[In]

Int[ArcTan[d + e*x]/(a + b*x + c*x^2),x]

[Out]

(ArcTan[d + e*x]*Log[(2*e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*c*(I - d) + (b - Sqrt[b^2 - 4*a*c])*e)*(1 - I*(
d + e*x)))])/Sqrt[b^2 - 4*a*c] - (ArcTan[d + e*x]*Log[(2*e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*c*(I - d) + (b
+ Sqrt[b^2 - 4*a*c])*e)*(1 - I*(d + e*x)))])/Sqrt[b^2 - 4*a*c] - ((I/2)*PolyLog[2, 1 + (2*(2*c*d - (b - Sqrt[
b^2 - 4*a*c])*e - 2*c*(d + e*x)))/(((2*I)*c - 2*c*d + b*e - Sqrt[b^2 - 4*a*c]*e)*(1 - I*(d + e*x)))])/Sqrt[b^2
- 4*a*c] + ((I/2)*PolyLog[2, 1 + (2*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e - 2*c*(d + e*x)))/((2*c*(I - d) + (b +
Sqrt[b^2 - 4*a*c])*e)*(1 - I*(d + e*x)))])/Sqrt[b^2 - 4*a*c]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4966

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x]))*(Log[2/(1
- I*c*x)]/e), x] + (Dist[b*(c/e), Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[b*(c/e), Int[Log[2*c*((
d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(1 + c^2*x^2), x], x] + Simp[(a + b*ArcTan[c*x])*(Log[2*c*((d + e*x)/((c*
d + I*e)*(1 - I*c*x)))]/e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 5155

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x]
&& IGtQ[p, 0]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 c \arctan (d+e x)}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}-\frac {2 c \arctan (d+e x)}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}\right ) \, dx \\ & = \frac {(2 c) \int \frac {\arctan (d+e x)}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c) \int \frac {\arctan (d+e x)}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c}} \\ & = \frac {(2 c) \text {Subst}\left (\int \frac {\arctan (x)}{\frac {-2 c d+\left (b-\sqrt {b^2-4 a c}\right ) e}{e}+\frac {2 c x}{e}} \, dx,x,d+e x\right )}{\sqrt {b^2-4 a c} e}-\frac {(2 c) \text {Subst}\left (\int \frac {\arctan (x)}{\frac {-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}{e}+\frac {2 c x}{e}} \, dx,x,d+e x\right )}{\sqrt {b^2-4 a c} e} \\ & = \frac {\arctan (d+e x) \log \left (\frac {2 e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 i c-2 c d+b e-\sqrt {b^2-4 a c} e\right ) (1-i (d+e x))}\right )}{\sqrt {b^2-4 a c}}-\frac {\arctan (d+e x) \log \left (\frac {2 e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c (i-d)+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1-i (d+e x))}\right )}{\sqrt {b^2-4 a c}}-\frac {\text {Subst}\left (\int \frac {\log \left (\frac {2 \left (\frac {-2 c d+\left (b-\sqrt {b^2-4 a c}\right ) e}{e}+\frac {2 c x}{e}\right )}{\left (\frac {2 i c}{e}+\frac {-2 c d+\left (b-\sqrt {b^2-4 a c}\right ) e}{e}\right ) (1-i x)}\right )}{1+x^2} \, dx,x,d+e x\right )}{\sqrt {b^2-4 a c}}+\frac {\text {Subst}\left (\int \frac {\log \left (\frac {2 \left (\frac {-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}{e}+\frac {2 c x}{e}\right )}{\left (\frac {2 i c}{e}+\frac {-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}{e}\right ) (1-i x)}\right )}{1+x^2} \, dx,x,d+e x\right )}{\sqrt {b^2-4 a c}} \\ & = \frac {\arctan (d+e x) \log \left (\frac {2 e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 i c-2 c d+b e-\sqrt {b^2-4 a c} e\right ) (1-i (d+e x))}\right )}{\sqrt {b^2-4 a c}}-\frac {\arctan (d+e x) \log \left (\frac {2 e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c (i-d)+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1-i (d+e x))}\right )}{\sqrt {b^2-4 a c}}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 i c-2 c d+b e-\sqrt {b^2-4 a c} e\right ) (1-i (d+e x))}\right )}{2 \sqrt {b^2-4 a c}}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c (i-d)+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1-i (d+e x))}\right )}{2 \sqrt {b^2-4 a c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.21 \[ \int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx=\frac {i \left (\log \left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c (i+d)+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right ) \log (1-i (d+e x))-\log \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c (i+d)+\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \log (1-i (d+e x))-\log \left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c (-i+d)+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right ) \log (1+i (d+e x))+\log \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c (-i+d)+\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \log (1+i (d+e x))-\operatorname {PolyLog}\left (2,\frac {2 c (-i+d+e x)}{2 c (-i+d)+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )+\operatorname {PolyLog}\left (2,\frac {2 c (-i+d+e x)}{2 c (-i+d)-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )+\operatorname {PolyLog}\left (2,\frac {2 c (i+d+e x)}{2 c (i+d)+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )-\operatorname {PolyLog}\left (2,\frac {2 c (i+d+e x)}{2 c (i+d)-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )\right )}{2 \sqrt {b^2-4 a c}} \]

[In]

Integrate[ArcTan[d + e*x]/(a + b*x + c*x^2),x]

[Out]

((I/2)*(Log[(e*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))/(2*c*(I + d) + (-b + Sqrt[b^2 - 4*a*c])*e)]*Log[1 - I*(d + e*
x)] - Log[(e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*(I + d) + (b + Sqrt[b^2 - 4*a*c])*e)]*Log[1 - I*(d + e*x)]
- Log[(e*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))/(2*c*(-I + d) + (-b + Sqrt[b^2 - 4*a*c])*e)]*Log[1 + I*(d + e*x)]
+ Log[(e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*(-I + d) + (b + Sqrt[b^2 - 4*a*c])*e)]*Log[1 + I*(d + e*x)] -
PolyLog[2, (2*c*(-I + d + e*x))/(2*c*(-I + d) + (-b + Sqrt[b^2 - 4*a*c])*e)] + PolyLog[2, (2*c*(-I + d + e*x))
/(2*c*(-I + d) - (b + Sqrt[b^2 - 4*a*c])*e)] + PolyLog[2, (2*c*(I + d + e*x))/(2*c*(I + d) + (-b + Sqrt[b^2 -
4*a*c])*e)] - PolyLog[2, (2*c*(I + d + e*x))/(2*c*(I + d) - (b + Sqrt[b^2 - 4*a*c])*e)]))/Sqrt[b^2 - 4*a*c]

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 889 vs. \(2 (329 ) = 658\).

Time = 1.36 (sec) , antiderivative size = 890, normalized size of antiderivative = 2.43

method	result	size
risch	\(\frac {e \ln \left (-i e x -i d +1\right ) \ln \left (\frac {i b e -2 i c d -2 \left (-i e x -i d +1\right ) c +\sqrt {4 a c \,e^{2}-b^{2} e^{2}}+2 c}{i b e -2 i c d +2 c +\sqrt {4 a c \,e^{2}-b^{2} e^{2}}}\right )}{2 \sqrt {4 a c \,e^{2}-b^{2} e^{2}}}-\frac {e \ln \left (-i e x -i d +1\right ) \ln \left (\frac {i b e -2 i c d -2 \left (-i e x -i d +1\right ) c -\sqrt {4 a c \,e^{2}-b^{2} e^{2}}+2 c}{i b e -2 i c d +2 c -\sqrt {4 a c \,e^{2}-b^{2} e^{2}}}\right )}{2 \sqrt {4 a c \,e^{2}-b^{2} e^{2}}}+\frac {e \operatorname {dilog}\left (\frac {i b e -2 i c d -2 \left (-i e x -i d +1\right ) c +\sqrt {4 a c \,e^{2}-b^{2} e^{2}}+2 c}{i b e -2 i c d +2 c +\sqrt {4 a c \,e^{2}-b^{2} e^{2}}}\right )}{2 \sqrt {4 a c \,e^{2}-b^{2} e^{2}}}-\frac {e \operatorname {dilog}\left (\frac {i b e -2 i c d -2 \left (-i e x -i d +1\right ) c -\sqrt {4 a c \,e^{2}-b^{2} e^{2}}+2 c}{i b e -2 i c d +2 c -\sqrt {4 a c \,e^{2}-b^{2} e^{2}}}\right )}{2 \sqrt {4 a c \,e^{2}-b^{2} e^{2}}}+\frac {e \ln \left (i e x +i d +1\right ) \ln \left (\frac {i b e -2 i c d +2 \left (i e x +i d +1\right ) c -\sqrt {4 a c \,e^{2}-b^{2} e^{2}}-2 c}{i b e -2 i c d -\sqrt {4 a c \,e^{2}-b^{2} e^{2}}-2 c}\right )}{2 \sqrt {4 a c \,e^{2}-b^{2} e^{2}}}-\frac {e \ln \left (i e x +i d +1\right ) \ln \left (\frac {i b e -2 i c d +2 \left (i e x +i d +1\right ) c +\sqrt {4 a c \,e^{2}-b^{2} e^{2}}-2 c}{i b e -2 i c d +\sqrt {4 a c \,e^{2}-b^{2} e^{2}}-2 c}\right )}{2 \sqrt {4 a c \,e^{2}-b^{2} e^{2}}}+\frac {e \operatorname {dilog}\left (\frac {i b e -2 i c d +2 \left (i e x +i d +1\right ) c -\sqrt {4 a c \,e^{2}-b^{2} e^{2}}-2 c}{i b e -2 i c d -\sqrt {4 a c \,e^{2}-b^{2} e^{2}}-2 c}\right )}{2 \sqrt {4 a c \,e^{2}-b^{2} e^{2}}}-\frac {e \operatorname {dilog}\left (\frac {i b e -2 i c d +2 \left (i e x +i d +1\right ) c +\sqrt {4 a c \,e^{2}-b^{2} e^{2}}-2 c}{i b e -2 i c d +\sqrt {4 a c \,e^{2}-b^{2} e^{2}}-2 c}\right )}{2 \sqrt {4 a c \,e^{2}-b^{2} e^{2}}}\)	\(890\)
derivativedivides	\(\text {Expression too large to display}\)	\(4753\)
default	\(\text {Expression too large to display}\)	\(4753\)

[In]

int(arctan(e*x+d)/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/2*e*ln(1-I*d-I*e*x)/(4*a*c*e^2-b^2*e^2)^(1/2)*ln((I*b*e-2*I*c*d-2*(1-I*d-I*e*x)*c+(4*a*c*e^2-b^2*e^2)^(1/2)+
2*c)/(I*b*e-2*I*c*d+2*c+(4*a*c*e^2-b^2*e^2)^(1/2)))-1/2*e*ln(1-I*d-I*e*x)/(4*a*c*e^2-b^2*e^2)^(1/2)*ln((I*b*e-
2*I*c*d-2*(1-I*d-I*e*x)*c-(4*a*c*e^2-b^2*e^2)^(1/2)+2*c)/(I*b*e-2*I*c*d+2*c-(4*a*c*e^2-b^2*e^2)^(1/2)))+1/2*e/
(4*a*c*e^2-b^2*e^2)^(1/2)*dilog((I*b*e-2*I*c*d-2*(1-I*d-I*e*x)*c+(4*a*c*e^2-b^2*e^2)^(1/2)+2*c)/(I*b*e-2*I*c*d
+2*c+(4*a*c*e^2-b^2*e^2)^(1/2)))-1/2*e/(4*a*c*e^2-b^2*e^2)^(1/2)*dilog((I*b*e-2*I*c*d-2*(1-I*d-I*e*x)*c-(4*a*c
*e^2-b^2*e^2)^(1/2)+2*c)/(I*b*e-2*I*c*d+2*c-(4*a*c*e^2-b^2*e^2)^(1/2)))+1/2*e*ln(1+I*d+I*e*x)/(4*a*c*e^2-b^2*e
^2)^(1/2)*ln((I*b*e-2*I*c*d+2*(1+I*d+I*e*x)*c-(4*a*c*e^2-b^2*e^2)^(1/2)-2*c)/(I*b*e-2*I*c*d-(4*a*c*e^2-b^2*e^2
)^(1/2)-2*c))-1/2*e*ln(1+I*d+I*e*x)/(4*a*c*e^2-b^2*e^2)^(1/2)*ln((I*b*e-2*I*c*d+2*(1+I*d+I*e*x)*c+(4*a*c*e^2-b
^2*e^2)^(1/2)-2*c)/(I*b*e-2*I*c*d+(4*a*c*e^2-b^2*e^2)^(1/2)-2*c))+1/2*e/(4*a*c*e^2-b^2*e^2)^(1/2)*dilog((I*b*e
-2*I*c*d+2*(1+I*d+I*e*x)*c-(4*a*c*e^2-b^2*e^2)^(1/2)-2*c)/(I*b*e-2*I*c*d-(4*a*c*e^2-b^2*e^2)^(1/2)-2*c))-1/2*e
/(4*a*c*e^2-b^2*e^2)^(1/2)*dilog((I*b*e-2*I*c*d+2*(1+I*d+I*e*x)*c+(4*a*c*e^2-b^2*e^2)^(1/2)-2*c)/(I*b*e-2*I*c*
d+(4*a*c*e^2-b^2*e^2)^(1/2)-2*c))

Fricas [F]

\[ \int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx=\int { \frac {\arctan \left (e x + d\right )}{c x^{2} + b x + a} \,d x } \]

[In]

integrate(arctan(e*x+d)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

integral(arctan(e*x + d)/(c*x^2 + b*x + a), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx=\text {Timed out} \]

[In]

integrate(atan(e*x+d)/(c*x**2+b*x+a),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(arctan(e*x+d)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [F]

\[ \int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx=\int { \frac {\arctan \left (e x + d\right )}{c x^{2} + b x + a} \,d x } \]

[In]

integrate(arctan(e*x+d)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (d+e x)}{a+b x+c x^2} \, dx=\int \frac {\mathrm {atan}\left (d+e\,x\right )}{c\,x^2+b\,x+a} \,d x \]

[In]

int(atan(d + e*x)/(a + b*x + c*x^2),x)

[Out]

int(atan(d + e*x)/(a + b*x + c*x^2), x)