3.1152 \(\int \frac{x (a+b \tan ^{-1}(c x))}{d+e x^2} \, dx\)

Optimal. Leaf size=311 \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 e}-\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 e}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.243335, antiderivative size = 311, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {4980, 4856, 2402, 2315, 2447} \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 e}-\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 e}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/e) + ((a + b*ArcTan[c*x])*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])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] +
I*Sqrt[e])*(1 - I*c*x))])/(2*e) + ((I/2)*b*PolyLog[2, 1 - 2/(1 - I*c*x)])/e - ((I/4)*b*PolyLog[2, 1 - (2*c*(Sq
rt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/e - ((I/4)*b*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqr
t[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/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 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{x \left (a+b \tan ^{-1}(c x)\right )}{d+e x^2} \, dx &=\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\\ &=-\frac{\int \frac{a+b \tan ^{-1}(c x)}{\sqrt{-d}-\sqrt{e} x} \, dx}{2 \sqrt{e}}+\frac{\int \frac{a+b \tan ^{-1}(c x)}{\sqrt{-d}+\sqrt{e} x} \, dx}{2 \sqrt{e}}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e}+\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 e}+\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 e}+2 \frac{(b c) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{2 e}-\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 e}-\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 e}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e}+\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 e}+\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 e}-\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 e}-\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 e}+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 e}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e}+\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 e}+\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 e}+\frac{i b \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{2 e}-\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 e}-\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 e}\\ \end{align*}

Mathematica [A]  time = 0.113027, size = 441, normalized size = 1.42 \[ \frac{i b \text{PolyLog}\left (2,-\frac{\sqrt{e} (1-i c x)}{-\sqrt{e}+i c \sqrt{-d}}\right )}{4 e}+\frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (1-i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )}{4 e}-\frac{i b \text{PolyLog}\left (2,-\frac{\sqrt{e} (1+i c x)}{-\sqrt{e}+i c \sqrt{-d}}\right )}{4 e}-\frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (1+i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )}{4 e}+\frac{a \log \left (d+e x^2\right )}{2 e}-\frac{i b \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )}{4 e}+\frac{i b \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 e}+\frac{i b \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )}{4 e}-\frac{i b \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.193, size = 646, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a/e*ln(c^2*e*x^2+c^2*d)+1/2*b/e*ln(c^2*e*x^2+c^2*d)*arctan(c*x)-1/4*I*b/e*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/e*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/e*ln(c*x-I)*ln(c^2*e*x^2+c^2*d
)-1/4*I*b/e*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))-1/4
*I*b/e*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/
e*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/e*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/e*ln(c*x+I)*ln(c^2*e*x^2+c^2*d)+1/4*I*b/e*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))+1/4*I*b/e*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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

[Out]

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