3.1158 \(\int \frac{x^3 (a+b \tan ^{-1}(c x))}{(d+e x^2)^2} \, dx\)

Optimal. Leaf size=403 \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 e^2}-\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 e^2}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e^2}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{2 e^2 \left (d+e x^2\right )}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e^2}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e^2}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}-\frac{b c^2 d \tan ^{-1}(c x)}{2 e^2 \left (c^2 d-e\right )}+\frac{b c \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 e^{3/2} \left (c^2 d-e\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.448857, antiderivative size = 403, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4980, 4974, 391, 203, 205, 4856, 2402, 2315, 2447} \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 e^2}-\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 e^2}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e^2}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{2 e^2 \left (d+e x^2\right )}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e^2}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e^2}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}-\frac{b c^2 d \tan ^{-1}(c x)}{2 e^2 \left (c^2 d-e\right )}+\frac{b c \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 e^{3/2} \left (c^2 d-e\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcTan[c*x]))/(d + e*x^2)^2,x]

[Out]

-(b*c^2*d*ArcTan[c*x])/(2*(c^2*d - e)*e^2) + (d*(a + b*ArcTan[c*x]))/(2*e^2*(d + e*x^2)) + (b*c*Sqrt[d]*ArcTan
[(Sqrt[e]*x)/Sqrt[d]])/(2*(c^2*d - e)*e^(3/2)) - ((a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/e^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*e^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*e^2) + ((I/2)*b*PolyLog[2, 1 -
2/(1 - I*c*x)])/e^2 - ((I/4)*b*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*
x))])/e^2 - ((I/4)*b*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/e^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 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{x^3 \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx &=\int \left (-\frac{d x \left (a+b \tan ^{-1}(c x)\right )}{e \left (d+e x^2\right )^2}+\frac{x \left (a+b \tan ^{-1}(c x)\right )}{e \left (d+e x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{d+e x^2} \, dx}{e}-\frac{d \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx}{e}\\ &=\frac{d \left (a+b \tan ^{-1}(c x)\right )}{2 e^2 \left (d+e x^2\right )}-\frac{(b c d) \int \frac{1}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )} \, dx}{2 e^2}+\frac{\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}{e}\\ &=\frac{d \left (a+b \tan ^{-1}(c x)\right )}{2 e^2 \left (d+e x^2\right )}-\frac{\left (b c^3 d\right ) \int \frac{1}{1+c^2 x^2} \, dx}{2 \left (c^2 d-e\right ) e^2}-\frac{\int \frac{a+b \tan ^{-1}(c x)}{\sqrt{-d}-\sqrt{e} x} \, dx}{2 e^{3/2}}+\frac{\int \frac{a+b \tan ^{-1}(c x)}{\sqrt{-d}+\sqrt{e} x} \, dx}{2 e^{3/2}}+\frac{(b c d) \int \frac{1}{d+e x^2} \, dx}{2 \left (c^2 d-e\right ) e}\\ &=-\frac{b c^2 d \tan ^{-1}(c x)}{2 \left (c^2 d-e\right ) e^2}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{2 e^2 \left (d+e x^2\right )}+\frac{b c \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 \left (c^2 d-e\right ) e^{3/2}}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e^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 e^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 e^2}+2 \frac{(b c) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{2 e^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 e^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 e^2}\\ &=-\frac{b c^2 d \tan ^{-1}(c x)}{2 \left (c^2 d-e\right ) e^2}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{2 e^2 \left (d+e x^2\right )}+\frac{b c \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 \left (c^2 d-e\right ) e^{3/2}}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e^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 e^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 e^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 e^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 e^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 e^2}\\ &=-\frac{b c^2 d \tan ^{-1}(c x)}{2 \left (c^2 d-e\right ) e^2}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{2 e^2 \left (d+e x^2\right )}+\frac{b c \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 \left (c^2 d-e\right ) e^{3/2}}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e^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 e^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 e^2}+\frac{i b \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{2 e^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 e^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 e^2}\\ \end{align*}

Mathematica [A]  time = 7.91023, size = 522, normalized size = 1.3 \[ \frac{2 a \left (\frac{d}{d+e x^2}+\log \left (d+e x^2\right )\right )+b \left (-i \text{PolyLog}\left (2,\frac{c \left (\sqrt{d}-i \sqrt{e} x\right )}{c \sqrt{d}-\sqrt{e}}\right )+i \text{PolyLog}\left (2,\frac{c \left (\sqrt{d}-i \sqrt{e} x\right )}{c \sqrt{d}+\sqrt{e}}\right )+i \text{PolyLog}\left (2,\frac{c \left (\sqrt{d}+i \sqrt{e} x\right )}{c \sqrt{d}-\sqrt{e}}\right )-i \text{PolyLog}\left (2,\frac{c \left (\sqrt{d}+i \sqrt{e} x\right )}{c \sqrt{d}+\sqrt{e}}\right )-\frac{2 c^2 d \tan ^{-1}(c x)}{c^2 d-e}+\frac{2 c \sqrt{d} \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{c^2 d-e}+\frac{2 d \tan ^{-1}(c x)}{d+e x^2}+i \log \left (x-\frac{i \sqrt{d}}{\sqrt{e}}\right ) \log \left (\frac{\sqrt{e} (-1-i c x)}{c \sqrt{d}-\sqrt{e}}\right )-i \log \left (x-\frac{i \sqrt{d}}{\sqrt{e}}\right ) \log \left (\frac{\sqrt{e} (1-i c x)}{c \sqrt{d}+\sqrt{e}}\right )-i \log \left (x+\frac{i \sqrt{d}}{\sqrt{e}}\right ) \log \left (\frac{\sqrt{e} (-1+i c x)}{c \sqrt{d}-\sqrt{e}}\right )+i \log \left (x+\frac{i \sqrt{d}}{\sqrt{e}}\right ) \log \left (\frac{\sqrt{e} (1+i c x)}{c \sqrt{d}+\sqrt{e}}\right )+2 \tan ^{-1}(c x) \log \left (x-\frac{i \sqrt{d}}{\sqrt{e}}\right )+2 \tan ^{-1}(c x) \log \left (x+\frac{i \sqrt{d}}{\sqrt{e}}\right )\right )}{4 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcTan[c*x]))/(d + e*x^2)^2,x]

[Out]

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

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Maple [C]  time = 0.203, size = 760, 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(x^3*(a+b*arctan(c*x))/(e*x^2+d)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{3} \arctan \left (c x\right ) + a x^{3}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*x^3*arctan(c*x) + a*x^3)/(e^2*x^4 + 2*d*e*x^2 + d^2), 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(x**3*(a+b*atan(c*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{{\left (b \arctan \left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))/(e*x^2+d)^2,x, algorithm="giac")

[Out]

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