3.1151 \(\int \frac {x^3 (a+b \tan ^{-1}(c x))}{d+e x^2} \, dx\)

Optimal. Leaf size=361 \[ \frac {d \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}-\frac {d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e^2}-\frac {d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e^2}+\frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 e}+\frac {b \tan ^{-1}(c x)}{2 c^2 e}-\frac {i b d \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 e^2}+\frac {i b d \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^2}+\frac {i b d \text {Li}_2\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^2}-\frac {b x}{2 c e} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4916, 4852, 321, 203, 4980, 4856, 2402, 2315, 2447} \[ -\frac {i b d \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e^2}+\frac {i b d \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{4 e^2}+\frac {i b d \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{4 e^2}+\frac {d \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}-\frac {d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e^2}-\frac {d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e^2}+\frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 e}+\frac {b \tan ^{-1}(c x)}{2 c^2 e}-\frac {b x}{2 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 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 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 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 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 4916

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 503, normalized size = 1.39 \[ -\frac {a d \log \left (d+e x^2\right )}{2 e^2}+\frac {a x^2}{2 e}+\frac {b \tan ^{-1}(c x)}{2 c^2 e}-\frac {i b d \text {Li}_2\left (-\frac {\sqrt {e} (1-i c x)}{i c \sqrt {-d}-\sqrt {e}}\right )}{4 e^2}-\frac {i b d \text {Li}_2\left (\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 e^2}+\frac {i b d \text {Li}_2\left (-\frac {\sqrt {e} (i c x+1)}{i c \sqrt {-d}-\sqrt {e}}\right )}{4 e^2}+\frac {i b d \text {Li}_2\left (\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 e^2}+\frac {i b d \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 e^2}-\frac {i b d \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 e^2}-\frac {i b d \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 e^2}+\frac {i b d \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 e^2}+\frac {b x^2 \tan ^{-1}(c x)}{2 e}-\frac {b x}{2 c e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \arctan \left (c x\right ) + a x^{3}}{e x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [C]  time = 0.46, size = 703, normalized size = 1.95 \[ \frac {a \,x^{2}}{2 e}-\frac {a d \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{2 e^{2}}+\frac {b \arctan \left (c x \right ) x^{2}}{2 e}-\frac {b \arctan \left (c x \right ) d \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{2 e^{2}}-\frac {b x}{2 c e}+\frac {b \arctan \left (c x \right )}{2 c^{2} e}-\frac {i b d \ln \left (c x +i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 e^{2}}+\frac {i b d \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 e^{2}}-\frac {i b d \ln \left (c x -i\right ) \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{4 e^{2}}+\frac {i b d \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 e^{2}}+\frac {i b d \ln \left (c x -i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 e^{2}}-\frac {i b d \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 e^{2}}+\frac {i b d \ln \left (c x -i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 e^{2}}-\frac {i b d \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 e^{2}}+\frac {i b d \ln \left (c x +i\right ) \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{4 e^{2}}-\frac {i b d \ln \left (c x +i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a/e*x^2-1/2*a*d/e^2*ln(c^2*e*x^2+c^2*d)+1/2*b*arctan(c*x)*x^2/e-1/2*b*arctan(c*x)/e^2*d*ln(c^2*e*x^2+c^2*d
)-1/2*b*x/c/e+1/2*b*arctan(c*x)/c^2/e+1/4*I*b/e^2*d*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*d*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*d*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*d*ln(I+c*x)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,in
dex=2)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2))+1/4*I*b/e^2*d*dilog((RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,ind
ex=2)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=2))-1/4*I*b/e^2*d*ln(c*x-I)*ln(c^2*e*x^2+c^2*d)-1/4*I*b/e^2*
d*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*d
*ln(I+c*x)*ln(c^2*e*x^2+c^2*d)-1/4*I*b/e^2*d*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*d*ln(I+c*x)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1)-c*x-I)/RootO
f(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a {\left (\frac {x^{2}}{e} - \frac {d \log \left (e x^{2} + d\right )}{e^{2}}\right )} + 2 \, b \int \frac {x^{3} \arctan \left (c x\right )}{2 \, {\left (e x^{2} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{e\,x^2+d} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*atan(c*x)))/(d + e*x^2),x)

[Out]

int((x^3*(a + b*atan(c*x)))/(d + e*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{d + e x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(a + b*atan(c*x))/(d + e*x**2), x)

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