3.1156 \(\int \frac{a+b \tan ^{-1}(c x)}{d+e x^2} \, dx\)

Optimal. Leaf size=517 \[ \frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (-c x+i)}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}-\frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (1-i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )}{4 \sqrt{-d} \sqrt{e}}-\frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (1+i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )}{4 \sqrt{-d} \sqrt{e}}+\frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (c x+i)}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}+\frac{a \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \sqrt{e}}-\frac{i b \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}+\frac{i b \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}-\frac{i b \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}+\frac{i b \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}} \]

[Out]

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

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Rubi [A]  time = 0.405913, antiderivative size = 517, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {4910, 205, 4908, 2409, 2394, 2393, 2391} \[ \frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (-c x+i)}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}-\frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (1-i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )}{4 \sqrt{-d} \sqrt{e}}-\frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (1+i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )}{4 \sqrt{-d} \sqrt{e}}+\frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (c x+i)}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}+\frac{a \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \sqrt{e}}-\frac{i b \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}+\frac{i b \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}-\frac{i b \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}}+\frac{i b \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d} \sqrt{e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4910

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 4908

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

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

Mathematica [A]  time = 0.232943, size = 461, normalized size = 0.89 \[ \frac{i b \sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{e} (-c x+i)}{c \sqrt{-d}+i \sqrt{e}}\right )-i b \sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{e} (1-i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )-i b \sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{e} (1+i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )+i b \sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{e} (c x+i)}{c \sqrt{-d}+i \sqrt{e}}\right )+4 a \sqrt{-d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )-i b \sqrt{d} \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )+i b \sqrt{d} \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )-i b \sqrt{d} \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )+i b \sqrt{d} \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 \sqrt{-d^2} \sqrt{e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.219, size = 886, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))-I*c*b*ln(1-(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*d-2*(c^2*e*d)^(1/2)-e
))*arctan(c*x)/(c^4*d^2-2*c^2*d*e+e^2)*(c^2*e*d)^(1/2)+1/2*I/c*b*ln(1-(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*
d-2*(c^2*e*d)^(1/2)-e))*arctan(c*x)/d/(c^4*d^2-2*c^2*d*e+e^2)*(c^2*e*d)^(1/2)*e-1/2/c*b*(c^2*e*d)^(1/2)/e/d*ar
ctan(c*x)^2-1/4/c*b*(c^2*e*d)^(1/2)/e/d*polylog(2,(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*d+2*(c^2*e*d)^(1/2)-
e))+1/2*c^3*b/e/(c^4*d^2-2*c^2*d*e+e^2)*arctan(c*x)^2*(c^2*e*d)^(1/2)*d-c*b/(c^4*d^2-2*c^2*d*e+e^2)*arctan(c*x
)^2*(c^2*e*d)^(1/2)+1/2*I*c^3*b*ln(1-(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*d-2*(c^2*e*d)^(1/2)-e))*arctan(c*
x)/e/(c^4*d^2-2*c^2*d*e+e^2)*(c^2*e*d)^(1/2)*d-1/2*I/c*b*(c^2*e*d)^(1/2)/e/d*arctan(c*x)*ln(1-(c^2*d-e)*(1+I*c
*x)^2/(c^2*x^2+1)/(-c^2*d+2*(c^2*e*d)^(1/2)-e))+1/4*c^3*b/e/(c^4*d^2-2*c^2*d*e+e^2)*polylog(2,(c^2*d-e)*(1+I*c
*x)^2/(c^2*x^2+1)/(-c^2*d-2*(c^2*e*d)^(1/2)-e))*(c^2*e*d)^(1/2)*d-1/2*c*b/(c^4*d^2-2*c^2*d*e+e^2)*polylog(2,(c
^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)/(-c^2*d-2*(c^2*e*d)^(1/2)-e))*(c^2*e*d)^(1/2)+1/2/c*b/d/(c^4*d^2-2*c^2*d*e+e^2
)*arctan(c*x)^2*(c^2*e*d)^(1/2)*e+1/4/c*b/d/(c^4*d^2-2*c^2*d*e+e^2)*polylog(2,(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1
)/(-c^2*d-2*(c^2*e*d)^(1/2)-e))*(c^2*e*d)^(1/2)*e

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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