3.62 \(\int \frac {\tan ^{-1}(d+e x)}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=367 \[ -\frac {i \text {Li}_2\left (\frac {2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e-2 c (d+e x)\right )}{\left (-2 d c+2 i c+b e-\sqrt {b^2-4 a c} e\right ) (1-i (d+e x))}+1\right )}{2 \sqrt {b^2-4 a c}}+\frac {i \text {Li}_2\left (\frac {2 \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e-2 c (d+e x)\right )}{\left (2 c (i-d)+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1-i (d+e x))}+1\right )}{2 \sqrt {b^2-4 a c}}+\frac {\tan ^{-1}(d+e x) \log \left (\frac {2 e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{(1-i (d+e x)) \left (e \left (b-\sqrt {b^2-4 a c}\right )+2 c (-d+i)\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {\tan ^{-1}(d+e x) \log \left (\frac {2 e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{(1-i (d+e x)) \left (e \left (\sqrt {b^2-4 a c}+b\right )+2 c (-d+i)\right )}\right )}{\sqrt {b^2-4 a c}} \]

[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)

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Rubi [A]  time = 0.67, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {618, 206, 6728, 5047, 4856, 2402, 2315, 2447} \[ -\frac {i \text {PolyLog}\left (2,1+\frac {2 \left (-e \left (b-\sqrt {b^2-4 a c}\right )-2 c (d+e x)+2 c d\right )}{(1-i (d+e x)) \left (-e \sqrt {b^2-4 a c}+b e-2 c d+2 i c\right )}\right )}{2 \sqrt {b^2-4 a c}}+\frac {i \text {PolyLog}\left (2,1+\frac {2 \left (-e \left (\sqrt {b^2-4 a c}+b\right )-2 c (d+e x)+2 c d\right )}{(1-i (d+e x)) \left (e \left (\sqrt {b^2-4 a c}+b\right )+2 c (-d+i)\right )}\right )}{2 \sqrt {b^2-4 a c}}+\frac {\tan ^{-1}(d+e x) \log \left (\frac {2 e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{(1-i (d+e x)) \left (e \left (b-\sqrt {b^2-4 a c}\right )+2 c (-d+i)\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {\tan ^{-1}(d+e x) \log \left (\frac {2 e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{(1-i (d+e x)) \left (e \left (\sqrt {b^2-4 a c}+b\right )+2 c (-d+i)\right )}\right )}{\sqrt {b^2-4 a c}} \]

Antiderivative was successfully verified.

[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 206

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

Rule 618

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 2315

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

Rule 2402

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 2447

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 4856

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 5047

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 6728

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*} \int \frac {\tan ^{-1}(d+e x)}{a+b x+c x^2} \, dx &=\int \left (\frac {2 c \tan ^{-1}(d+e x)}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}-\frac {2 c \tan ^{-1}(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 {\tan ^{-1}(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 {\tan ^{-1}(d+e x)}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c}}\\ &=\frac {(2 c) \operatorname {Subst}\left (\int \frac {\tan ^{-1}(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) \operatorname {Subst}\left (\int \frac {\tan ^{-1}(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 {\tan ^{-1}(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 {\tan ^{-1}(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 {\operatorname {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 {\operatorname {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 {\tan ^{-1}(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 {\tan ^{-1}(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 \text {Li}_2\left (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 \text {Li}_2\left (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*}

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Mathematica [A]  time = 0.46, size = 443, normalized size = 1.21 \[ \frac {i \left (-\text {Li}_2\left (\frac {2 c (d+e x-i)}{2 c (d-i)+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )+\text {Li}_2\left (\frac {2 c (d+e x-i)}{2 c (d-i)-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )+\text {Li}_2\left (\frac {2 c (d+e x+i)}{2 c (d+i)+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )-\text {Li}_2\left (\frac {2 c (d+e x+i)}{2 c (d+i)-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )+\log (1-i (d+e x)) \log \left (\frac {e \left (\sqrt {b^2-4 a c}-b-2 c x\right )}{e \left (\sqrt {b^2-4 a c}-b\right )+2 c (d+i)}\right )-\log (1-i (d+e x)) \log \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{e \left (\sqrt {b^2-4 a c}+b\right )-2 c (d+i)}\right )-\log (1+i (d+e x)) \log \left (\frac {e \left (\sqrt {b^2-4 a c}-b-2 c x\right )}{e \left (\sqrt {b^2-4 a c}-b\right )+2 c (d-i)}\right )+\log (1+i (d+e x)) \log \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{e \left (\sqrt {b^2-4 a c}+b\right )-2 c (d-i)}\right )\right )}{2 \sqrt {b^2-4 a c}} \]

Warning: Unable to verify antiderivative.

[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]

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (e x + d\right )}{c x^{2} + b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 1.62, size = 4743, normalized size = 12.92 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*e*(e^2*(4*a*c-b^2))^(1/2)/c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*e+2*
I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))*b^2+1
/2/e*(e^2*(4*a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*e+2*I
*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))*d^2-1/
4*I*e*(e^2*(4*a*c-b^2))^(1/2)/c/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d
+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)+1/4*I*e*
(e^2*(4*a*c-b^2))^(1/2)/c/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2
-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)-1/2/e*(e^2*(4*
a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*e+2*I*d*c+a*e^2-b*
e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2))^(1/2)-c))*d^2-1/e*(e^2*(4*a*c
-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d)^2*d^2+1/e*(e^2*(4*a*c-b
^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d)^2*d^2-1/8*e*(e^2*(4*a*c-b
^2))^(1/2)/c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*e+2*I*d*c+a*e^2-b*e*d+c
*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2))^(1/2)-c))*b^2+1/4*e*(e^2*(4*a*c-b^
2))^(1/2)/c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d)^2*b^2-1/4*e*(e^2*(4*a*c-b^
2))^(1/2)/c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d)^2*b^2+1/2*e/(a*e^2-b*e*d+c
*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(
-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2))^(1/2)-c))+1/2*e/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(
-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-
c))+e/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d)^2+e/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/
2)+c)*arctan(e*x+d)^2-I*(e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1
-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2
)-c))*arctan(e*x+d)*b*d-I/e*(e^2*(4*a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c
)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2)
)^(1/2)-c))*arctan(e*x+d)-1/8*e*(e^2*(4*a*c-b^2))^(1/2)/c/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*polylo
g(2,(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2))^(
1/2)-c))+I*e/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x
+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)+1/8*e*(e^2*(4*a*c-b^2))^(1/
2)/c/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d)
)^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))-1/4*e*(e^2*(4*a*c-b^2))^(1/2)/c/(a*e^2-b*e*d
+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d)^2-I/e*(e^2*(4*a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2
-(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*
e*d-c*d^2+(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)*d^2+I/e*(e^2*(4*a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d
+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*
e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)*d^2-1/4*I*e*(e^2*(4*a*c-b^2))^(1/2)/c/(4*a*c-b^2)/(a
*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)
^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)*b^2+1/4*I*e*(e^2*(4*a*c-b^2))^(1/2)/c/(4*a
*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^
2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)*b^2+I*e/(a*e^2-b*e*d+c*d^2-(e^2*
(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*
d^2+(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)+1/e*(e^2*(4*a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^
2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d)^2+1/2/e*(e^2*(4*a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4
*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d
-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))-1/e*(e^2*(4*a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^
2))^(1/2)+c)*arctan(e*x+d)^2-1/2/e*(e^2*(4*a*c-b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^
(1/2)+c)*polylog(2,(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2
*(4*a*c-b^2))^(1/2)-c))+(e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*arct
an(e*x+d)^2*b*d-(e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d
)^2*b*d+1/2*(e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*
e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2))^(1/2)-c))*b
*d-1/2*(e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*polylog(2,(-I*b*e+2*I
*d*c+a*e^2-b*e*d+c*d^2-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))*b*d+1/
4*e*(e^2*(4*a*c-b^2))^(1/2)/c/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*arctan(e*x+d)^2+I*(e^2*(4*a*c-b^2)
)^(1/2)/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2-(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2-c)*(1
+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2+(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)*b*d+I/e*(e^2*(4*a*c-
b^2))^(1/2)*c/(4*a*c-b^2)/(a*e^2-b*e*d+c*d^2+(e^2*(4*a*c-b^2))^(1/2)+c)*ln(1-(-I*b*e+2*I*d*c+a*e^2-b*e*d+c*d^2
-c)*(1+I*(e*x+d))^2/((e*x+d)^2+1)/(-a*e^2+b*e*d-c*d^2-(e^2*(4*a*c-b^2))^(1/2)-c))*arctan(e*x+d)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is 4*a*c-b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {atan}\left (d+e\,x\right )}{c\,x^2+b\,x+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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