3.1263 \(\int \frac{x (a+b \tan ^{-1}(c x))^2}{d+e x^2} \, dx\)

Optimal. Leaf size=492 \[ -\frac{i b \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e}-\frac{i b \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 e}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 e}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e} \]

[Out]

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

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Rubi [A]  time = 0.249505, antiderivative size = 492, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4980, 4858} \[ -\frac{i b \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e}-\frac{i b \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 e}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 e}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 8.21787, size = 1527, normalized size = 3.1 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((8*I)*a*b*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[(c*e*x)/Sqrt[c^2*d*e]] - 8*a*b*ArcTan[c*x]*Log[1 + E^((2*I
)*ArcTan[c*x])] - 4*b^2*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + 2*b^2*ArcTan[c*x]^2*Log[1 + ((c*Sqrt[d]
- Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] + Sqrt[e])] + 2*b^2*ArcTan[c*x]^2*Log[1 + ((c*Sqrt[d] + Sqrt[e])
*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] - Sqrt[e])] - 2*b^2*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)] - 4*a*b*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d
*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] + 4*a*b*ArcTan[c*x]*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*A
rcTan[c*x]))/(c^2*d - e)] - 4*b^2*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)] + 2*b^2*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)] + 4*a*b*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*Log[(-2*Sqrt[c^2*d*e]*E^((2*I)*ArcTa
n[c*x]) + e*(-1 + E^((2*I)*ArcTan[c*x])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] + 4*a*b*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)] + 4*b^2*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)] + 4*b^2*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)] - 4*b^2*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))] - 2*b^2*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))] + 2*a^2*Log[d + e*x^2] + 4*b^2*ArcSin[Sqrt[(c^2*
d)/(c^2*d - e)]]*ArcTan[c*x]*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)] - 2*b^2*ArcTan[c*x]^2*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*(Cos[2*ArcTan[c*x]] + I*Sin[2*Arc
Tan[c*x]]))/(c^2*d - e)] + (4*I)*b*(a + b*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - (2*I)*b^2*ArcTan[c
*x]*PolyLog[2, ((-(c*Sqrt[d]) + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] + Sqrt[e])] - (2*I)*b^2*ArcTan[c*x]
*PolyLog[2, -(((c*Sqrt[d] + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] - Sqrt[e]))] - (2*I)*a*b*PolyLog[2, -((
(c^2*d + e - 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e))] - (2*I)*a*b*PolyLog[2, -(((c^2*d + e + 2*Sq
rt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e))] - 2*b^2*PolyLog[3, -E^((2*I)*ArcTan[c*x])] + b^2*PolyLog[3,
((-(c*Sqrt[d]) + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] + Sqrt[e])] + b^2*PolyLog[3, -(((c*Sqrt[d] + Sqrt[
e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] - Sqrt[e]))])/(4*e)

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Maple [F]  time = 6.373, size = 0, normalized size = 0. \begin{align*} \int{\frac{x \left ( a+b\arctan \left ( cx \right ) \right ) ^{2}}{e{x}^{2}+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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