3.1265 \(\int \frac {(a+b \tan ^{-1}(c x))^2}{x (d+e x^2)} \, dx\)

Optimal. Leaf size=637 \[ \frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d}-\frac {i b \text {Li}_2\left (1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {i b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {i b \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {\log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac {b^2 \text {Li}_3\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 d}+\frac {b^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )}{2 d} \]

[Out]

-2*(a+b*arctan(c*x))^2*arctanh(-1+2/(1+I*c*x))/d+(a+b*arctan(c*x))^2*ln(2/(1-I*c*x))/d-1/2*(a+b*arctan(c*x))^2
*ln(2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/d-1/2*(a+b*arctan(c*x))^2*ln(2*c*((-d)^(1/2
)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/d-I*b*(a+b*arctan(c*x))*polylog(2,1-2/(1-I*c*x))/d-I*b*(a+b*a
rctan(c*x))*polylog(2,1-2/(1+I*c*x))/d+I*b*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))/d+1/2*I*b*(a+b*arctan(c
*x))*polylog(2,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/d+1/2*I*b*(a+b*arctan(c*x))*po
lylog(2,1-2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/d+1/2*b^2*polylog(3,1-2/(1-I*c*x))/d-
1/2*b^2*polylog(3,1-2/(1+I*c*x))/d+1/2*b^2*polylog(3,-1+2/(1+I*c*x))/d-1/4*b^2*polylog(3,1-2*c*((-d)^(1/2)-x*e
^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/d-1/4*b^2*polylog(3,1-2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)
^(1/2)+I*e^(1/2)))/d

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Rubi [A]  time = 0.67, antiderivative size = 637, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4980, 4850, 4988, 4884, 4994, 6610, 4858} \[ \frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d}-\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {i b \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{4 d}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{4 d}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 d}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 d}+\frac {b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d}+\frac {\log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)])/d + ((a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/d - ((a +
b*ArcTan[c*x])^2*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*d) - ((a + b*Arc
Tan[c*x])^2*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(2*d) - (I*b*(a + b*ArcT
an[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/d - (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/d + (I*b*(
a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d + ((I/2)*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(Sqrt[
-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/d + ((I/2)*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*
c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/d + (b^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*
d) - (b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*d) + (b^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/(2*d) - (b^2*PolyLog[3,
1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(4*d) - (b^2*PolyLog[3, 1 - (2*c*(Sq
rt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(4*d)

Rule 4850

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 +
I*c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTan[c*x])^(p - 1)*ArcTanh[1 - 2/(1 + I*c*x)])/(1 + c^2*x^2), x], x
] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 4858

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^2*Log[2/
(1 - I*c*x)])/e, x] + (Simp[((a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] + Sim
p[(I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e, x] - Simp[(I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 -
(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] - Simp[(b^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*e), x] + Simp
[(b^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && Ne
Q[c^2*d^2 + e^2, 0]

Rule 4884

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

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

Rule 4988

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[(
Log[1 + u]*(a + b*ArcTan[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[1 - u]*(a + b*ArcTan[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 - (2*I)/(I - c*x))^
2, 0]

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u
])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x \left (d+e x^2\right )} \, dx &=\int \left (\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac {e x \left (a+b \tan ^{-1}(c x)\right )^2}{d \left (d+e x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx}{d}-\frac {e \int \frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{d+e x^2} \, dx}{d}\\ &=\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d}-\frac {(4 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}-\frac {e \int \left (-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )}\right ) \, dx}{d}\\ &=\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d}+\frac {(2 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}-\frac {(2 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}+\frac {\sqrt {e} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{\sqrt {-d}-\sqrt {e} x} \, dx}{2 d}-\frac {\sqrt {e} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{\sqrt {-d}+\sqrt {e} x} \, dx}{2 d}\\ &=\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}-\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d}-\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d}+\frac {\left (i b^2 c\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}-\frac {\left (i b^2 c\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}-\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d}-\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 d}+\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d}\\ \end {align*}

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Mathematica [B]  time = 7.59, size = 1412, normalized size = 2.22 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(24*a^2*Log[x] - 12*a^2*Log[d + e*x^2] - 24*a*b*((-I)*ArcTan[c*x]^2 + (2*I)*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*
ArcTan[(c*e*x)/Sqrt[c^2*d*e]] - 2*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + (-ArcSin[Sqrt[(c^2*d)/(c^2*d -
e)]] + ArcTan[c*x])*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] + (ArcSin[Sqrt[
(c^2*d)/(c^2*d - e)]] + ArcTan[c*x])*Log[(-2*Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e*(-1 + E^((2*I)*ArcTan[c*x
])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] + I*(ArcTan[c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*x])])
- (I/2)*(PolyLog[2, ((-(c^2*d) - e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] + PolyLog[2, -(((c^2
*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e))])) + b^2*((-I)*Pi^3 + (16*I)*ArcTan[c*x]^3 + 24*
ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - 12*ArcTan[c*x]^2*Log[1 + ((c*Sqrt[d] - Sqrt[e])*E^((2*I)*ArcTa
n[c*x]))/(c*Sqrt[d] + Sqrt[e])] - 12*ArcTan[c*x]^2*Log[1 + ((c*Sqrt[d] + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sq
rt[d] - Sqrt[e])] + 12*ArcTan[c*x]^2*Log[1 + ((c^2*d + e - 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)
] + 24*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*
x]))/(c^2*d - e)] - 12*ArcTan[c*x]^2*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)
] - 24*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[(-2*Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e*(-1 + E^(
(2*I)*ArcTan[c*x])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] - 24*ArcTan[c*x]^2*Log[(-2*Sqrt[c^2*d*e]
*E^((2*I)*ArcTan[c*x]) + e*(-1 + E^((2*I)*ArcTan[c*x])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] + 24
*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[((2*I)*c^2*d - (2*I)*Sqrt[c^2*d*e] + 2*c*(-e + Sqrt[c^2*d*e
])*x)/((c^2*d - e)*(I + c*x))] + 12*ArcTan[c*x]^2*Log[((2*I)*c^2*d - (2*I)*Sqrt[c^2*d*e] + 2*c*(-e + Sqrt[c^2*
d*e])*x)/((c^2*d - e)*(I + c*x))] - 24*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[1 + ((c^2*d + e + 2*S
qrt[c^2*d*e])*(Cos[2*ArcTan[c*x]] + I*Sin[2*ArcTan[c*x]]))/(c^2*d - e)] + 12*ArcTan[c*x]^2*Log[1 + ((c^2*d + e
+ 2*Sqrt[c^2*d*e])*(Cos[2*ArcTan[c*x]] + I*Sin[2*ArcTan[c*x]]))/(c^2*d - e)] + (24*I)*ArcTan[c*x]*PolyLog[2,
E^((-2*I)*ArcTan[c*x])] + (12*I)*ArcTan[c*x]*PolyLog[2, ((-(c*Sqrt[d]) + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sq
rt[d] + Sqrt[e])] + (12*I)*ArcTan[c*x]*PolyLog[2, -(((c*Sqrt[d] + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] -
Sqrt[e]))] + 12*PolyLog[3, E^((-2*I)*ArcTan[c*x])] - 6*PolyLog[3, ((-(c*Sqrt[d]) + Sqrt[e])*E^((2*I)*ArcTan[c
*x]))/(c*Sqrt[d] + Sqrt[e])] - 6*PolyLog[3, -(((c*Sqrt[d] + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] - Sqrt[
e]))]))/(24*d)

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + a^{2}}{e x^{3} + d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctan(c*x)^2 + 2*a*b*arctan(c*x) + a^2)/(e*x^3 + d*x), x)

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

[Out]

Timed out

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maple [F]  time = 24.51, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctan \left (c x \right )\right )^{2}}{x \left (e \,x^{2}+d \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a^{2} {\left (\frac {\log \left (e x^{2} + d\right )}{d} - \frac {2 \, \log \relax (x)}{d}\right )} + \int \frac {b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right )}{e x^{3} + d x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2*(log(e*x^2 + d)/d - 2*log(x)/d) + integrate((b^2*arctan(c*x)^2 + 2*a*b*arctan(c*x))/(e*x^3 + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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