3.1160 \(\int \frac{a+b \tan ^{-1}(c x)}{x (d+e x^2)^2} \, dx\)

Optimal. Leaf size=443 \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 d^2}+\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 d^2}+\frac{i b \text{PolyLog}(2,-i c x)}{2 d^2}-\frac{i b \text{PolyLog}(2,i c x)}{2 d^2}-\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 d^2}+\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}+\frac{a+b \tan ^{-1}(c x)}{2 d \left (d+e x^2\right )}+\frac{a \log (x)}{d^2}+\frac{b c \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} \left (c^2 d-e\right )}-\frac{b c^2 \tan ^{-1}(c x)}{2 d \left (c^2 d-e\right )} \]

[Out]

-(b*c^2*ArcTan[c*x])/(2*d*(c^2*d - e)) + (a + b*ArcTan[c*x])/(2*d*(d + e*x^2)) + (b*c*Sqrt[e]*ArcTan[(Sqrt[e]*
x)/Sqrt[d]])/(2*d^(3/2)*(c^2*d - e)) + (a*Log[x])/d^2 + ((a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/d^2 - ((a + b
*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*d^2) - ((a + b*ArcT
an[c*x])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(2*d^2) + ((I/2)*b*PolyLog[
2, (-I)*c*x])/d^2 - ((I/2)*b*PolyLog[2, I*c*x])/d^2 - ((I/2)*b*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^2 + ((I/4)*b*P
olyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/d^2 + ((I/4)*b*PolyLog[2,
1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/d^2

________________________________________________________________________________________

Rubi [A]  time = 0.48941, antiderivative size = 443, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 11, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.524, Rules used = {4980, 4848, 2391, 4974, 391, 203, 205, 4856, 2402, 2315, 2447} \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 d^2}+\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 d^2}+\frac{i b \text{PolyLog}(2,-i c x)}{2 d^2}-\frac{i b \text{PolyLog}(2,i c x)}{2 d^2}-\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 d^2}+\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}+\frac{a+b \tan ^{-1}(c x)}{2 d \left (d+e x^2\right )}+\frac{a \log (x)}{d^2}+\frac{b c \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} \left (c^2 d-e\right )}-\frac{b c^2 \tan ^{-1}(c x)}{2 d \left (c^2 d-e\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c^2*ArcTan[c*x])/(2*d*(c^2*d - e)) + (a + b*ArcTan[c*x])/(2*d*(d + e*x^2)) + (b*c*Sqrt[e]*ArcTan[(Sqrt[e]*
x)/Sqrt[d]])/(2*d^(3/2)*(c^2*d - e)) + (a*Log[x])/d^2 + ((a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/d^2 - ((a + b
*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*d^2) - ((a + b*ArcT
an[c*x])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(2*d^2) + ((I/2)*b*PolyLog[
2, (-I)*c*x])/d^2 - ((I/2)*b*PolyLog[2, I*c*x])/d^2 - ((I/2)*b*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^2 + ((I/4)*b*P
olyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/d^2 + ((I/4)*b*PolyLog[2,
1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/d^2

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 4848

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

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 4974

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

Rule 391

Int[1/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^n),
x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 4856

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

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{a+b \tan ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx &=\int \left (\frac{a+b \tan ^{-1}(c x)}{d^2 x}-\frac{e x \left (a+b \tan ^{-1}(c x)\right )}{d \left (d+e x^2\right )^2}-\frac{e x \left (a+b \tan ^{-1}(c x)\right )}{d^2 \left (d+e x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{a+b \tan ^{-1}(c x)}{x} \, dx}{d^2}-\frac{e \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{d+e x^2} \, dx}{d^2}-\frac{e \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx}{d}\\ &=\frac{a+b \tan ^{-1}(c x)}{2 d \left (d+e x^2\right )}+\frac{a \log (x)}{d^2}+\frac{(i b) \int \frac{\log (1-i c x)}{x} \, dx}{2 d^2}-\frac{(i b) \int \frac{\log (1+i c x)}{x} \, dx}{2 d^2}-\frac{(b c) \int \frac{1}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )} \, dx}{2 d}-\frac{e \int \left (-\frac{a+b \tan ^{-1}(c x)}{2 \sqrt{e} \left (\sqrt{-d}-\sqrt{e} x\right )}+\frac{a+b \tan ^{-1}(c x)}{2 \sqrt{e} \left (\sqrt{-d}+\sqrt{e} x\right )}\right ) \, dx}{d^2}\\ &=\frac{a+b \tan ^{-1}(c x)}{2 d \left (d+e x^2\right )}+\frac{a \log (x)}{d^2}+\frac{i b \text{Li}_2(-i c x)}{2 d^2}-\frac{i b \text{Li}_2(i c x)}{2 d^2}-\frac{\left (b c^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{2 d \left (c^2 d-e\right )}+\frac{\sqrt{e} \int \frac{a+b \tan ^{-1}(c x)}{\sqrt{-d}-\sqrt{e} x} \, dx}{2 d^2}-\frac{\sqrt{e} \int \frac{a+b \tan ^{-1}(c x)}{\sqrt{-d}+\sqrt{e} x} \, dx}{2 d^2}+\frac{(b c e) \int \frac{1}{d+e x^2} \, dx}{2 d \left (c^2 d-e\right )}\\ &=-\frac{b c^2 \tan ^{-1}(c x)}{2 d \left (c^2 d-e\right )}+\frac{a+b \tan ^{-1}(c x)}{2 d \left (d+e x^2\right )}+\frac{b c \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} \left (c^2 d-e\right )}+\frac{a \log (x)}{d^2}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \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^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \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^2}+\frac{i b \text{Li}_2(-i c x)}{2 d^2}-\frac{i b \text{Li}_2(i c x)}{2 d^2}-2 \frac{(b c) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{2 d^2}+\frac{(b c) \int \frac{\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 )}{1+c^2 x^2} \, dx}{2 d^2}+\frac{(b c) \int \frac{\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 )}{1+c^2 x^2} \, dx}{2 d^2}\\ &=-\frac{b c^2 \tan ^{-1}(c x)}{2 d \left (c^2 d-e\right )}+\frac{a+b \tan ^{-1}(c x)}{2 d \left (d+e x^2\right )}+\frac{b c \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} \left (c^2 d-e\right )}+\frac{a \log (x)}{d^2}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \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^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \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^2}+\frac{i b \text{Li}_2(-i c x)}{2 d^2}-\frac{i b \text{Li}_2(i c x)}{2 d^2}+\frac{i b \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 )}{4 d^2}+\frac{i b \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 )}{4 d^2}-2 \frac{(i b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )}{2 d^2}\\ &=-\frac{b c^2 \tan ^{-1}(c x)}{2 d \left (c^2 d-e\right )}+\frac{a+b \tan ^{-1}(c x)}{2 d \left (d+e x^2\right )}+\frac{b c \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} \left (c^2 d-e\right )}+\frac{a \log (x)}{d^2}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \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^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \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^2}+\frac{i b \text{Li}_2(-i c x)}{2 d^2}-\frac{i b \text{Li}_2(i c x)}{2 d^2}-\frac{i b \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{2 d^2}+\frac{i b \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 )}{4 d^2}+\frac{i b \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 )}{4 d^2}\\ \end{align*}

Mathematica [A]  time = 5.87583, size = 590, normalized size = 1.33 \[ \frac{2 a \left (\frac{d}{d+e x^2}-\log \left (d+e x^2\right )+2 \log (x)\right )+b \left (i \text{PolyLog}\left (2,\frac{c \left (\sqrt{d}-i \sqrt{e} x\right )}{c \sqrt{d}-\sqrt{e}}\right )-i \text{PolyLog}\left (2,\frac{c \left (\sqrt{d}-i \sqrt{e} x\right )}{c \sqrt{d}+\sqrt{e}}\right )-i \text{PolyLog}\left (2,\frac{c \left (\sqrt{d}+i \sqrt{e} x\right )}{c \sqrt{d}-\sqrt{e}}\right )+i \text{PolyLog}\left (2,\frac{c \left (\sqrt{d}+i \sqrt{e} x\right )}{c \sqrt{d}+\sqrt{e}}\right )+2 i \text{PolyLog}(2,-i c x)-2 i \text{PolyLog}(2,i c x)-\frac{2 c^2 d \tan ^{-1}(c x)}{c^2 d-e}+\frac{2 c \sqrt{d} \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{c^2 d-e}+\frac{2 d \tan ^{-1}(c x)}{d+e x^2}-i \log \left (x-\frac{i \sqrt{d}}{\sqrt{e}}\right ) \log \left (\frac{\sqrt{e} (-1-i c x)}{c \sqrt{d}-\sqrt{e}}\right )+i \log \left (x-\frac{i \sqrt{d}}{\sqrt{e}}\right ) \log \left (\frac{\sqrt{e} (1-i c x)}{c \sqrt{d}+\sqrt{e}}\right )+i \log \left (x+\frac{i \sqrt{d}}{\sqrt{e}}\right ) \log \left (\frac{\sqrt{e} (-1+i c x)}{c \sqrt{d}-\sqrt{e}}\right )-i \log \left (x+\frac{i \sqrt{d}}{\sqrt{e}}\right ) \log \left (\frac{\sqrt{e} (1+i c x)}{c \sqrt{d}+\sqrt{e}}\right )-2 \tan ^{-1}(c x) \log \left (x-\frac{i \sqrt{d}}{\sqrt{e}}\right )-2 \tan ^{-1}(c x) \log \left (x+\frac{i \sqrt{d}}{\sqrt{e}}\right )-2 i \log (x) \log (1-i c x)+2 i \log (x) \log (1+i c x)+4 \log (x) \tan ^{-1}(c x)\right )}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.212, size = 847, normalized size = 1.9 \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))/x/(e*x^2+d)^2,x)

[Out]

-1/2*a/d^2*ln(c^2*e*x^2+c^2*d)+1/2*a*c^2/d/(c^2*e*x^2+c^2*d)+a/d^2*ln(c*x)-1/2*b*arctan(c*x)/d^2*ln(c^2*e*x^2+
c^2*d)+1/2*b*c^2*arctan(c*x)/d/(c^2*e*x^2+c^2*d)+b*arctan(c*x)/d^2*ln(c*x)+1/2*b*c/d*e/(c^2*d-e)/(d*e)^(1/2)*a
rctan(e*x/(d*e)^(1/2))-1/2*b*c^2*arctan(c*x)/d/(c^2*d-e)+1/4*I*b/d^2*ln(c*x-I)*ln((RootOf(e*_Z^2+2*I*_Z*e+c^2*
d-e,index=1)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1))-1/4*I*b/d^2*dilog((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e
,index=2)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2))-1/4*I*b/d^2*ln(c*x+I)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2
*d-e,index=1)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1))+1/4*I*b/d^2*dilog((RootOf(e*_Z^2+2*I*_Z*e+c^2*d-
e,index=2)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=2))+1/2*I*b/d^2*ln(c*x)*ln(1+I*c*x)-1/2*I*b/d^2*dilog(1
-I*c*x)+1/4*I*b/d^2*ln(c*x-I)*ln((RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=2)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-
e,index=2))-1/4*I*b/d^2*ln(c*x+I)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^
2*d-e,index=2))-1/4*I*b/d^2*ln(c*x-I)*ln(c^2*e*x^2+c^2*d)-1/2*I*b/d^2*ln(c*x)*ln(1-I*c*x)-1/4*I*b/d^2*dilog((R
ootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1))+1/4*I*b/d^2*ln(c*x+I)*l
n(c^2*e*x^2+c^2*d)+1/2*I*b/d^2*dilog(1+I*c*x)+1/4*I*b/d^2*dilog((RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1)-c*x+I
)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a{\left (\frac{1}{d e x^{2} + d^{2}} - \frac{\log \left (e x^{2} + d\right )}{d^{2}} + \frac{2 \, \log \left (x\right )}{d^{2}}\right )} + 2 \, b \int \frac{\arctan \left (c x\right )}{2 \,{\left (e^{2} x^{5} + 2 \, d e x^{3} + d^{2} x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(1/(d*e*x^2 + d^2) - log(e*x^2 + d)/d^2 + 2*log(x)/d^2) + 2*b*integrate(1/2*arctan(c*x)/(e^2*x^5 + 2*d*e
*x^3 + d^2*x), x)

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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^{2} x^{5} + 2 \, d e x^{3} + d^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctan(c*x) + a)/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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