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

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

[Out]

((-I/4)*Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] - (I - a)*Sqrt[d])])/(Sqrt[-c]*Sqrt[d]
) + ((I/4)*Log[1 - I*a - I*b*x]*Log[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])])/(Sqrt[-c]*Sqrt
[d]) + ((I/4)*Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[-c] + Sqrt[d]*x))/(b*Sqrt[-c] + (I - a)*Sqrt[d])])/(Sqrt[-c]*S
qrt[d]) - ((I/4)*Log[1 - I*a - I*b*x]*Log[(b*(Sqrt[-c] + Sqrt[d]*x))/(b*Sqrt[-c] - (I + a)*Sqrt[d])])/(Sqrt[-c
]*Sqrt[d]) - ((I/4)*PolyLog[2, -((Sqrt[d]*(I - a - b*x))/(b*Sqrt[-c] - (I - a)*Sqrt[d]))])/(Sqrt[-c]*Sqrt[d])
+ ((I/4)*PolyLog[2, (Sqrt[d]*(I - a - b*x))/(b*Sqrt[-c] + (I - a)*Sqrt[d])])/(Sqrt[-c]*Sqrt[d]) - ((I/4)*PolyL
og[2, -((Sqrt[d]*(I + a + b*x))/(b*Sqrt[-c] - (I + a)*Sqrt[d]))])/(Sqrt[-c]*Sqrt[d]) + ((I/4)*PolyLog[2, (Sqrt
[d]*(I + a + b*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])])/(Sqrt[-c]*Sqrt[d])

________________________________________________________________________________________

Rubi [A]  time = 0.607336, 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} \[ -\frac{i \text{PolyLog}\left (2,-\frac{\sqrt{d} (-a-b x+i)}{b \sqrt{-c}-(-a+i) \sqrt{d}}\right )}{4 \sqrt{-c} \sqrt{d}}+\frac{i \text{PolyLog}\left (2,\frac{\sqrt{d} (-a-b x+i)}{b \sqrt{-c}+(-a+i) \sqrt{d}}\right )}{4 \sqrt{-c} \sqrt{d}}-\frac{i \text{PolyLog}\left (2,-\frac{\sqrt{d} (a+b x+i)}{b \sqrt{-c}-(a+i) \sqrt{d}}\right )}{4 \sqrt{-c} \sqrt{d}}+\frac{i \text{PolyLog}\left (2,\frac{\sqrt{d} (a+b x+i)}{b \sqrt{-c}+(a+i) \sqrt{d}}\right )}{4 \sqrt{-c} \sqrt{d}}-\frac{i \log (i a+i b x+1) \log \left (\frac{b \left (\sqrt{-c}-\sqrt{d} x\right )}{b \sqrt{-c}-(-a+i) \sqrt{d}}\right )}{4 \sqrt{-c} \sqrt{d}}+\frac{i \log (-i a-i b x+1) \log \left (\frac{b \left (\sqrt{-c}-\sqrt{d} x\right )}{b \sqrt{-c}+(a+i) \sqrt{d}}\right )}{4 \sqrt{-c} \sqrt{d}}+\frac{i \log (i a+i b x+1) \log \left (\frac{b \left (\sqrt{-c}+\sqrt{d} x\right )}{b \sqrt{-c}+(-a+i) \sqrt{d}}\right )}{4 \sqrt{-c} \sqrt{d}}-\frac{i \log (-i a-i b x+1) \log \left (\frac{b \left (\sqrt{-c}+\sqrt{d} x\right )}{b \sqrt{-c}-(a+i) \sqrt{d}}\right )}{4 \sqrt{-c} \sqrt{d}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/(c + d*x^2),x]

[Out]

((-I/4)*Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] - (I - a)*Sqrt[d])])/(Sqrt[-c]*Sqrt[d]
) + ((I/4)*Log[1 - I*a - I*b*x]*Log[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])])/(Sqrt[-c]*Sqrt
[d]) + ((I/4)*Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[-c] + Sqrt[d]*x))/(b*Sqrt[-c] + (I - a)*Sqrt[d])])/(Sqrt[-c]*S
qrt[d]) - ((I/4)*Log[1 - I*a - I*b*x]*Log[(b*(Sqrt[-c] + Sqrt[d]*x))/(b*Sqrt[-c] - (I + a)*Sqrt[d])])/(Sqrt[-c
]*Sqrt[d]) - ((I/4)*PolyLog[2, -((Sqrt[d]*(I - a - b*x))/(b*Sqrt[-c] - (I - a)*Sqrt[d]))])/(Sqrt[-c]*Sqrt[d])
+ ((I/4)*PolyLog[2, (Sqrt[d]*(I - a - b*x))/(b*Sqrt[-c] + (I - a)*Sqrt[d])])/(Sqrt[-c]*Sqrt[d]) - ((I/4)*PolyL
og[2, -((Sqrt[d]*(I + a + b*x))/(b*Sqrt[-c] - (I + a)*Sqrt[d]))])/(Sqrt[-c]*Sqrt[d]) + ((I/4)*PolyLog[2, (Sqrt
[d]*(I + a + b*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])])/(Sqrt[-c]*Sqrt[d])

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

Mathematica [A]  time = 0.329175, size = 409, normalized size = 0.75 \[ -\frac{i \left (-\text{PolyLog}\left (2,\frac{\sqrt{d} (a+b x-i)}{-b \sqrt{-c}+(a-i) \sqrt{d}}\right )+\text{PolyLog}\left (2,\frac{\sqrt{d} (a+b x-i)}{b \sqrt{-c}+(a-i) \sqrt{d}}\right )+\text{PolyLog}\left (2,\frac{\sqrt{d} (a+b x+i)}{-b \sqrt{-c}+(a+i) \sqrt{d}}\right )-\text{PolyLog}\left (2,\frac{\sqrt{d} (a+b x+i)}{b \sqrt{-c}+(a+i) \sqrt{d}}\right )+\log (i a+i b x+1) \log \left (\frac{b \left (\sqrt{-c}-\sqrt{d} x\right )}{b \sqrt{-c}+(a-i) \sqrt{d}}\right )-\log (-i (a+b x+i)) \log \left (\frac{b \left (\sqrt{-c}-\sqrt{d} x\right )}{b \sqrt{-c}+(a+i) \sqrt{d}}\right )-\log (i a+i b x+1) \log \left (\frac{b \left (\sqrt{-c}+\sqrt{d} x\right )}{b \sqrt{-c}-(a-i) \sqrt{d}}\right )+\log (-i (a+b x+i)) \log \left (\frac{b \left (\sqrt{-c}+\sqrt{d} x\right )}{b \sqrt{-c}-(a+i) \sqrt{d}}\right )\right )}{4 \sqrt{-c} \sqrt{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/4)*(Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] + (-I + a)*Sqrt[d])] - Log[(-I)*(I +
a + b*x)]*Log[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])] - Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[-
c] + Sqrt[d]*x))/(b*Sqrt[-c] - (-I + a)*Sqrt[d])] + Log[(-I)*(I + a + b*x)]*Log[(b*(Sqrt[-c] + Sqrt[d]*x))/(b*
Sqrt[-c] - (I + a)*Sqrt[d])] - PolyLog[2, (Sqrt[d]*(-I + a + b*x))/(-(b*Sqrt[-c]) + (-I + a)*Sqrt[d])] + PolyL
og[2, (Sqrt[d]*(-I + a + b*x))/(b*Sqrt[-c] + (-I + a)*Sqrt[d])] + PolyLog[2, (Sqrt[d]*(I + a + b*x))/(-(b*Sqrt
[-c]) + (I + a)*Sqrt[d])] - PolyLog[2, (Sqrt[d]*(I + a + b*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])]))/(Sqrt[-c]*Sqr
t[d])

________________________________________________________________________________________

Maple [B]  time = 0.713, 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(b*x+a)/(d*x^2+c),x)

[Out]

1/2*I/b*(b^2*c*d)^(1/2)/c/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*ln(1-(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b
*x+a)^2)/(-a^2*d-c*b^2-2*(b^2*c*d)^(1/2)-d))*arctan(b*x+a)-1/2*I/b*(b^2*c*d)^(1/2)/c/(a^2*d+c*b^2-2*(b^2*c*d)^
(1/2)+d)*ln(1-(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2+2*(b^2*c*d)^(1/2)-d))*arctan
(b*x+a)+I*b/(a^2*d+c*b^2-2*(b^2*c*d)^(1/2)+d)*ln(1-(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2
*d-c*b^2+2*(b^2*c*d)^(1/2)-d))*arctan(b*x+a)+I*b/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*ln(1-(2*I*a*d+a^2*d+c*b^2-d
)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2-2*(b^2*c*d)^(1/2)-d))*arctan(b*x+a)-1/2*b/d*(b^2*c*d)^(1/2)/(a^2
*d+c*b^2-2*(b^2*c*d)^(1/2)+d)*arctan(b*x+a)^2+b/(a^2*d+c*b^2-2*(b^2*c*d)^(1/2)+d)*arctan(b*x+a)^2-1/2/b*(b^2*c
*d)^(1/2)/c/(a^2*d+c*b^2-2*(b^2*c*d)^(1/2)+d)*arctan(b*x+a)^2-1/2/b*(b^2*c*d)^(1/2)/c/(a^2*d+c*b^2-2*(b^2*c*d)
^(1/2)+d)*arctan(b*x+a)^2*a^2-1/4*b/d*(b^2*c*d)^(1/2)/(a^2*d+c*b^2-2*(b^2*c*d)^(1/2)+d)*polylog(2,(2*I*a*d+a^2
*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2+2*(b^2*c*d)^(1/2)-d))+1/2*b/(a^2*d+c*b^2-2*(b^2*c*d)^(
1/2)+d)*polylog(2,(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2+2*(b^2*c*d)^(1/2)-d))-1/
4/b*(b^2*c*d)^(1/2)/c/(a^2*d+c*b^2-2*(b^2*c*d)^(1/2)+d)*polylog(2,(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(
b*x+a)^2)/(-a^2*d-c*b^2+2*(b^2*c*d)^(1/2)-d))-1/4/b*(b^2*c*d)^(1/2)/c/(a^2*d+c*b^2-2*(b^2*c*d)^(1/2)+d)*polylo
g(2,(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2+2*(b^2*c*d)^(1/2)-d))*a^2+1/2*I/b*(b^2
*c*d)^(1/2)/c/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*ln(1-(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a
^2*d-c*b^2-2*(b^2*c*d)^(1/2)-d))*arctan(b*x+a)*a^2+1/2*I*b/d*(b^2*c*d)^(1/2)/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)
*ln(1-(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2-2*(b^2*c*d)^(1/2)-d))*arctan(b*x+a)-
1/2*I/b*(b^2*c*d)^(1/2)/c/(a^2*d+c*b^2-2*(b^2*c*d)^(1/2)+d)*ln(1-(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b
*x+a)^2)/(-a^2*d-c*b^2+2*(b^2*c*d)^(1/2)-d))*arctan(b*x+a)*a^2-1/2*I*b/d*(b^2*c*d)^(1/2)/(a^2*d+c*b^2-2*(b^2*c
*d)^(1/2)+d)*ln(1-(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2+2*(b^2*c*d)^(1/2)-d))*ar
ctan(b*x+a)+1/2*b/d*(b^2*c*d)^(1/2)/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*arctan(b*x+a)^2+b/(a^2*d+c*b^2+2*(b^2*c*
d)^(1/2)+d)*arctan(b*x+a)^2+1/2/b*(b^2*c*d)^(1/2)/c/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*arctan(b*x+a)^2+1/2/b*(b
^2*c*d)^(1/2)/c/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*arctan(b*x+a)^2*a^2+1/4*b/d*(b^2*c*d)^(1/2)/(a^2*d+c*b^2+2*(
b^2*c*d)^(1/2)+d)*polylog(2,(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2-2*(b^2*c*d)^(1
/2)-d))+1/2*b/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*polylog(2,(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2
)/(-a^2*d-c*b^2-2*(b^2*c*d)^(1/2)-d))+1/4/b*(b^2*c*d)^(1/2)/c/(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*polylog(2,(2*I
*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^2-2*(b^2*c*d)^(1/2)-d))+1/4/b*(b^2*c*d)^(1/2)/c/
(a^2*d+c*b^2+2*(b^2*c*d)^(1/2)+d)*polylog(2,(2*I*a*d+a^2*d+c*b^2-d)*(1+I*(b*x+a))^2/(1+(b*x+a)^2)/(-a^2*d-c*b^
2-2*(b^2*c*d)^(1/2)-d))*a^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(b*x+a)/(d*x^2+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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