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

Optimal. Leaf size=323 \[ -\frac{i b d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^2}+\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^2}+\frac{i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c e}+\frac{d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^2}+\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{e}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c e}+\frac{2 b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c e} \]

[Out]

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

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Rubi [A]  time = 0.265561, antiderivative size = 323, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4858} \[ -\frac{i b d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^2}+\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^2}+\frac{i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c e}+\frac{d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^2}+\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{e}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c e}+\frac{2 b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4876

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

Rule 4846

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

Rule 4920

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[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 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 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} \, dx &=\int \left (\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{e}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{e (d+e x)}\right ) \, dx\\ &=\frac{\int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{e}-\frac{d \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d+e x} \, dx}{e}\\ &=\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{e}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}-\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^2}-\frac{(2 b c) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{e}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c e}+\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{e}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}-\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^2}+\frac{(2 b) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{e}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c e}+\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{e}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{e^2}+\frac{2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c e}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}-\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^2}-\frac{\left (2 b^2\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{e}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c e}+\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{e}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{e^2}+\frac{2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c e}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}-\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^2}+\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{c e}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c e}+\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{e}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{e^2}+\frac{2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c e}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}-\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{e^2}+\frac{i b^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c e}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^2}\\ \end{align*}

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

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [C]  time = 5.046, size = 16024, normalized size = 49.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2}{\left (\frac{x}{e} - \frac{d \log \left (e x + d\right )}{e^{2}}\right )} + \frac{4 \, b^{2} x \arctan \left (c x\right )^{2} - b^{2} x \log \left (c^{2} x^{2} + 1\right )^{2} + e \int \frac{12 \,{\left (b^{2} c^{2} e x^{3} + b^{2} e x\right )} \arctan \left (c x\right )^{2} +{\left (b^{2} c^{2} e x^{3} + b^{2} e x\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + 8 \,{\left (4 \, a b c^{2} e x^{3} - b^{2} c e x^{2} -{\left (b^{2} c d - 4 \, a b e\right )} x\right )} \arctan \left (c x\right ) + 4 \,{\left (b^{2} c^{2} e x^{3} + b^{2} c^{2} d x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{c^{2} e^{2} x^{3} + c^{2} d e x^{2} + e^{2} x + d e}\,{d x}}{16 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(x/e - d*log(e*x + d)/e^2) + 1/16*(4*b^2*x*arctan(c*x)^2 - b^2*x*log(c^2*x^2 + 1)^2 + 16*e*integrate(1/16*
(12*(b^2*c^2*e*x^3 + b^2*e*x)*arctan(c*x)^2 + (b^2*c^2*e*x^3 + b^2*e*x)*log(c^2*x^2 + 1)^2 + 8*(4*a*b*c^2*e*x^
3 - b^2*c*e*x^2 - (b^2*c*d - 4*a*b*e)*x)*arctan(c*x) + 4*(b^2*c^2*e*x^3 + b^2*c^2*d*x^2)*log(c^2*x^2 + 1))/(c^
2*e^2*x^3 + c^2*d*e*x^2 + e^2*x + d*e), x))/e

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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 + 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+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 + 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}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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