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

Optimal. Leaf size=517 \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}+\frac {i b \text {Li}_2\left (\frac {\sqrt {e} (i-c x)}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 \sqrt {-d} \sqrt {e}}-\frac {i b \text {Li}_2\left (\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 \sqrt {-d} \sqrt {e}}-\frac {i b \text {Li}_2\left (\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 \sqrt {-d} \sqrt {e}}+\frac {i b \text {Li}_2\left (\frac {\sqrt {e} (c x+i)}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 \sqrt {-d} \sqrt {e}}-\frac {i b \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 \sqrt {-d} \sqrt {e}}+\frac {i b \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 \sqrt {-d} \sqrt {e}}-\frac {i b \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 \sqrt {-d} \sqrt {e}}+\frac {i b \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 \sqrt {-d} \sqrt {e}} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 517, normalized size of antiderivative = 1.00, 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 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.25, size = 461, normalized size = 0.89 \[ \frac {4 a \sqrt {-d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+i b \sqrt {d} \text {Li}_2\left (\frac {\sqrt {e} (i-c x)}{\sqrt {-d} c+i \sqrt {e}}\right )-i b \sqrt {d} \text {Li}_2\left (\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )-i b \sqrt {d} \text {Li}_2\left (\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )+i b \sqrt {d} \text {Li}_2\left (\frac {\sqrt {e} (c x+i)}{\sqrt {-d} c+i \sqrt {e}}\right )-i b \sqrt {d} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )+i b \sqrt {d} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )-i b \sqrt {d} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )+i b \sqrt {d} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 \sqrt {-d^2} \sqrt {e}} \]

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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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (c x\right ) + a}{e x^{2} + d}, x\right ) \]

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

[Out]

sage0*x

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maple [B]  time = 0.66, size = 886, normalized size = 1.71 \[ \frac {a \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}}+\frac {i c^{3} b \ln \left (1-\frac {\left (c^{2} d -e \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c^{2} d -2 \sqrt {c^{2} e d}-e \right )}\right ) \arctan \left (c x \right ) \sqrt {c^{2} e d}\, d}{2 e \left (d^{2} c^{4}-2 c^{2} e d +e^{2}\right )}-\frac {i c b \ln \left (1-\frac {\left (c^{2} d -e \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c^{2} d -2 \sqrt {c^{2} e d}-e \right )}\right ) \arctan \left (c x \right ) \sqrt {c^{2} e d}}{d^{2} c^{4}-2 c^{2} e d +e^{2}}-\frac {b \sqrt {c^{2} e d}\, \arctan \left (c x \right )^{2}}{2 c e d}-\frac {b \sqrt {c^{2} e d}\, \polylog \left (2, \frac {\left (c^{2} d -e \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c^{2} d +2 \sqrt {c^{2} e d}-e \right )}\right )}{4 c e d}-\frac {i b \sqrt {c^{2} e d}\, \arctan \left (c x \right ) \ln \left (1-\frac {\left (c^{2} d -e \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c^{2} d +2 \sqrt {c^{2} e d}-e \right )}\right )}{2 c e d}+\frac {c^{3} b \arctan \left (c x \right )^{2} \sqrt {c^{2} e d}\, d}{2 e \left (d^{2} c^{4}-2 c^{2} e d +e^{2}\right )}-\frac {c b \arctan \left (c x \right )^{2} \sqrt {c^{2} e d}}{d^{2} c^{4}-2 c^{2} e d +e^{2}}+\frac {i b \ln \left (1-\frac {\left (c^{2} d -e \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c^{2} d -2 \sqrt {c^{2} e d}-e \right )}\right ) \arctan \left (c x \right ) \sqrt {c^{2} e d}\, e}{2 c d \left (d^{2} c^{4}-2 c^{2} e d +e^{2}\right )}+\frac {c^{3} b \polylog \left (2, \frac {\left (c^{2} d -e \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c^{2} d -2 \sqrt {c^{2} e d}-e \right )}\right ) \sqrt {c^{2} e d}\, d}{4 e \left (d^{2} c^{4}-2 c^{2} e d +e^{2}\right )}-\frac {c b \polylog \left (2, \frac {\left (c^{2} d -e \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c^{2} d -2 \sqrt {c^{2} e d}-e \right )}\right ) \sqrt {c^{2} e d}}{2 \left (d^{2} c^{4}-2 c^{2} e d +e^{2}\right )}+\frac {b \arctan \left (c x \right )^{2} \sqrt {c^{2} e d}\, e}{2 c d \left (d^{2} c^{4}-2 c^{2} e d +e^{2}\right )}+\frac {b \polylog \left (2, \frac {\left (c^{2} d -e \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c^{2} d -2 \sqrt {c^{2} e d}-e \right )}\right ) \sqrt {c^{2} e d}\, e}{4 c d \left (d^{2} c^{4}-2 c^{2} e d +e^{2}\right )} \]

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))+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-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/c*b*(c^2*e*d)^(1/2)/e/d*
arctan(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*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))+1/2*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*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+1/4*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]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 \, b \int \frac {\arctan \left (c x\right )}{2 \, {\left (e x^{2} + d\right )}}\,{d x} + \frac {a \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}} \]

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]

2*b*integrate(1/2*arctan(c*x)/(e*x^2 + d), x) + a*arctan(e*x/sqrt(d*e))/sqrt(d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/(d + e*x^2),x)

[Out]

int((a + b*atan(c*x))/(d + e*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {atan}{\left (c x \right )}}{d + e x^{2}}\, dx \]

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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