3.1266 \(\int \frac{(a+b \tan ^{-1}(c x))^2}{x^2 (d+e x^2)} \, dx\)

Optimal. Leaf size=553 \[ -\frac{i b \sqrt{e} \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 (-d)^{3/2}}+\frac{i b \sqrt{e} \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 (-d)^{3/2}}+\frac{b^2 \sqrt{e} \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 (-d)^{3/2}}-\frac{b^2 \sqrt{e} \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 (-d)^{3/2}}-\frac{i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d}+\frac{\sqrt{e} \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 (-d)^{3/2}}-\frac{\sqrt{e} \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 (-d)^{3/2}}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.516037, antiderivative size = 553, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4918, 4852, 4924, 4868, 2447, 4914, 4858} \[ -\frac{i b \sqrt{e} \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 (-d)^{3/2}}+\frac{i b \sqrt{e} \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 (-d)^{3/2}}+\frac{b^2 \sqrt{e} \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 (-d)^{3/2}}-\frac{b^2 \sqrt{e} \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 (-d)^{3/2}}-\frac{i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d}+\frac{\sqrt{e} \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 (-d)^{3/2}}-\frac{\sqrt{e} \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 (-d)^{3/2}}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTan[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4924

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

Rule 4868

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

Rule 4914

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcTan[c*x])^p, (d + e*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[q] && IGtQ[p, 0]

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]

Rubi steps

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

Mathematica [F]  time = 180.002, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

________________________________________________________________________________________

Maple [F]  time = 3.365, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arctan \left ( cx \right ) \right ) ^{2}}{{x}^{2} \left ( e{x}^{2}+d \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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))^2/x^2/(e*x^2+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + a^{2}}{e x^{4} + d x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/x^2/(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^4 + d*x^2), 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((a+b*atan(c*x))**2/x**2/(e*x**2+d),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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