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

Optimal. Leaf size=543 \[ -\frac{i \text{PolyLog}\left (2,\frac{\sqrt{b} (-d-e x+i)}{-\sqrt{-a} e+\sqrt{b} (-d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{i \text{PolyLog}\left (2,\frac{\sqrt{b} (-d-e x+i)}{\sqrt{-a} e+\sqrt{b} (-d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{i \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x+i)}{-\sqrt{-a} e+\sqrt{b} (d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{i \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x+i)}{\sqrt{-a} e+\sqrt{b} (d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{i \log (-i d-i e x+1) \log \left (\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} (d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{i \log (-i d-i e x+1) \log \left (-\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{-\sqrt{-a} e+\sqrt{b} (d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{i \log (i d+i e x+1) \log \left (-\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{-\sqrt{-a} e+\sqrt{b} (-d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{i \log (i d+i e x+1) \log \left (\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} (-d+i)}\right )}{4 \sqrt{-a} \sqrt{b}} \]

[Out]

((I/4)*Log[(e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*(I + d) + Sqrt[-a]*e)]*Log[1 - I*d - I*e*x])/(Sqrt[-a]*Sqrt[b])
- ((I/4)*Log[-((e*(Sqrt[-a] + Sqrt[b]*x))/(Sqrt[b]*(I + d) - Sqrt[-a]*e))]*Log[1 - I*d - I*e*x])/(Sqrt[-a]*Sq
rt[b]) - ((I/4)*Log[-((e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*(I - d) - Sqrt[-a]*e))]*Log[1 + I*d + I*e*x])/(Sqrt[
-a]*Sqrt[b]) + ((I/4)*Log[(e*(Sqrt[-a] + Sqrt[b]*x))/(Sqrt[b]*(I - d) + Sqrt[-a]*e)]*Log[1 + I*d + I*e*x])/(Sq
rt[-a]*Sqrt[b]) - ((I/4)*PolyLog[2, (Sqrt[b]*(I - d - e*x))/(Sqrt[b]*(I - d) - Sqrt[-a]*e)])/(Sqrt[-a]*Sqrt[b]
) + ((I/4)*PolyLog[2, (Sqrt[b]*(I - d - e*x))/(Sqrt[b]*(I - d) + Sqrt[-a]*e)])/(Sqrt[-a]*Sqrt[b]) - ((I/4)*Pol
yLog[2, (Sqrt[b]*(I + d + e*x))/(Sqrt[b]*(I + d) - Sqrt[-a]*e)])/(Sqrt[-a]*Sqrt[b]) + ((I/4)*PolyLog[2, (Sqrt[
b]*(I + d + e*x))/(Sqrt[b]*(I + d) + Sqrt[-a]*e)])/(Sqrt[-a]*Sqrt[b])

________________________________________________________________________________________

Rubi [A]  time = 0.600126, antiderivative size = 543, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5051, 2409, 2394, 2393, 2391} \[ -\frac{i \text{PolyLog}\left (2,\frac{\sqrt{b} (-d-e x+i)}{-\sqrt{-a} e+\sqrt{b} (-d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{i \text{PolyLog}\left (2,\frac{\sqrt{b} (-d-e x+i)}{\sqrt{-a} e+\sqrt{b} (-d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{i \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x+i)}{-\sqrt{-a} e+\sqrt{b} (d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{i \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x+i)}{\sqrt{-a} e+\sqrt{b} (d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{i \log (-i d-i e x+1) \log \left (\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} (d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{i \log (-i d-i e x+1) \log \left (-\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{-\sqrt{-a} e+\sqrt{b} (d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{i \log (i d+i e x+1) \log \left (-\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{-\sqrt{-a} e+\sqrt{b} (-d+i)}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{i \log (i d+i e x+1) \log \left (\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} (-d+i)}\right )}{4 \sqrt{-a} \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[d + e*x]/(a + b*x^2),x]

[Out]

((I/4)*Log[(e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*(I + d) + Sqrt[-a]*e)]*Log[1 - I*d - I*e*x])/(Sqrt[-a]*Sqrt[b])
- ((I/4)*Log[-((e*(Sqrt[-a] + Sqrt[b]*x))/(Sqrt[b]*(I + d) - Sqrt[-a]*e))]*Log[1 - I*d - I*e*x])/(Sqrt[-a]*Sq
rt[b]) - ((I/4)*Log[-((e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*(I - d) - Sqrt[-a]*e))]*Log[1 + I*d + I*e*x])/(Sqrt[
-a]*Sqrt[b]) + ((I/4)*Log[(e*(Sqrt[-a] + Sqrt[b]*x))/(Sqrt[b]*(I - d) + Sqrt[-a]*e)]*Log[1 + I*d + I*e*x])/(Sq
rt[-a]*Sqrt[b]) - ((I/4)*PolyLog[2, (Sqrt[b]*(I - d - e*x))/(Sqrt[b]*(I - d) - Sqrt[-a]*e)])/(Sqrt[-a]*Sqrt[b]
) + ((I/4)*PolyLog[2, (Sqrt[b]*(I - d - e*x))/(Sqrt[b]*(I - d) + Sqrt[-a]*e)])/(Sqrt[-a]*Sqrt[b]) - ((I/4)*Pol
yLog[2, (Sqrt[b]*(I + d + e*x))/(Sqrt[b]*(I + d) - Sqrt[-a]*e)])/(Sqrt[-a]*Sqrt[b]) + ((I/4)*PolyLog[2, (Sqrt[
b]*(I + d + e*x))/(Sqrt[b]*(I + d) + Sqrt[-a]*e)])/(Sqrt[-a]*Sqrt[b])

Rule 5051

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

Rule 2409

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[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]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

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

Rule 2391

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

Rubi steps

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

Mathematica [A]  time = 0.354393, size = 409, normalized size = 0.75 \[ \frac{i \left (\text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x-i)}{-\sqrt{-a} e+\sqrt{b} (d-i)}\right )-\text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x-i)}{\sqrt{-a} e+\sqrt{b} (d-i)}\right )-\text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x+i)}{-\sqrt{-a} e+\sqrt{b} (d+i)}\right )+\text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x+i)}{\sqrt{-a} e+\sqrt{b} (d+i)}\right )+\log (i d+i e x+1) \left (-\log \left (\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} (d-i)}\right )\right )+\log (i d+i e x+1) \log \left (\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{\sqrt{-a} e-\sqrt{b} (d-i)}\right )+\log (-i (d+e x+i)) \log \left (\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} (d+i)}\right )-\log (-i (d+e x+i)) \log \left (\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{\sqrt{-a} e-\sqrt{b} (d+i)}\right )\right )}{4 \sqrt{-a} \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*(-(Log[(e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*(-I + d) + Sqrt[-a]*e)]*Log[1 + I*d + I*e*x]) + Log[(e*(Sqrt
[-a] + Sqrt[b]*x))/(-(Sqrt[b]*(-I + d)) + Sqrt[-a]*e)]*Log[1 + I*d + I*e*x] + Log[(e*(Sqrt[-a] - Sqrt[b]*x))/(
Sqrt[b]*(I + d) + Sqrt[-a]*e)]*Log[(-I)*(I + d + e*x)] - Log[(e*(Sqrt[-a] + Sqrt[b]*x))/(-(Sqrt[b]*(I + d)) +
Sqrt[-a]*e)]*Log[(-I)*(I + d + e*x)] + PolyLog[2, (Sqrt[b]*(-I + d + e*x))/(Sqrt[b]*(-I + d) - Sqrt[-a]*e)] -
PolyLog[2, (Sqrt[b]*(-I + d + e*x))/(Sqrt[b]*(-I + d) + Sqrt[-a]*e)] - PolyLog[2, (Sqrt[b]*(I + d + e*x))/(Sqr
t[b]*(I + d) - Sqrt[-a]*e)] + PolyLog[2, (Sqrt[b]*(I + d + e*x))/(Sqrt[b]*(I + d) + Sqrt[-a]*e)]))/(Sqrt[-a]*S
qrt[b])

________________________________________________________________________________________

Maple [B]  time = 0.702, size = 2192, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (e x + d\right )}{b x^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (e x + d\right )}{b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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