3.1165 \(\int \frac {x^5 (a+b \tan ^{-1}(c x))}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=532 \[ -\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {d \left (a+b \tan ^{-1}(c x)\right )}{e^3 \left (d+e x^2\right )}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e^3}-\frac {\log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^3}-\frac {b c \sqrt {d} \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 e^{5/2} \left (c^2 d-e\right )^2}+\frac {b c \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{5/2} \left (c^2 d-e\right )}-\frac {b c^2 d \tan ^{-1}(c x)}{e^3 \left (c^2 d-e\right )}-\frac {b c d x}{8 e^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {b c^4 d^2 \tan ^{-1}(c x)}{4 e^3 \left (c^2 d-e\right )^2}-\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^3}-\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^3}+\frac {i b \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 e^3} \]

[Out]

-1/8*b*c*d*x/(c^2*d-e)/e^2/(e*x^2+d)+1/4*b*c^4*d^2*arctan(c*x)/(c^2*d-e)^2/e^3-b*c^2*d*arctan(c*x)/(c^2*d-e)/e
^3-1/4*d^2*(a+b*arctan(c*x))/e^3/(e*x^2+d)^2+d*(a+b*arctan(c*x))/e^3/(e*x^2+d)-(a+b*arctan(c*x))*ln(2/(1-I*c*x
))/e^3+1/2*(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^3+1/2*(a+b*ar
ctan(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^3+1/2*I*b*polylog(2,1-2/(1-I*c*
x))/e^3-1/4*I*b*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^3-1/4*I*b*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^3+b*c*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/
(c^2*d-e)/e^(5/2)-1/8*b*c*(3*c^2*d-e)*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/(c^2*d-e)^2/e^(5/2)

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Rubi [A]  time = 0.65, antiderivative size = 532, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 11, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {4980, 4974, 414, 522, 203, 205, 391, 4856, 2402, 2315, 2447} \[ -\frac {i b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{4 e^3}-\frac {i b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{4 e^3}+\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e^3}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {d \left (a+b \tan ^{-1}(c x)\right )}{e^3 \left (d+e x^2\right )}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e^3}-\frac {\log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^3}+\frac {b c^4 d^2 \tan ^{-1}(c x)}{4 e^3 \left (c^2 d-e\right )^2}-\frac {b c d x}{8 e^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b c^2 d \tan ^{-1}(c x)}{e^3 \left (c^2 d-e\right )}-\frac {b c \sqrt {d} \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 e^{5/2} \left (c^2 d-e\right )^2}+\frac {b c \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{5/2} \left (c^2 d-e\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcTan[c*x]))/(d + e*x^2)^3,x]

[Out]

-(b*c*d*x)/(8*(c^2*d - e)*e^2*(d + e*x^2)) + (b*c^4*d^2*ArcTan[c*x])/(4*(c^2*d - e)^2*e^3) - (b*c^2*d*ArcTan[c
*x])/((c^2*d - e)*e^3) - (d^2*(a + b*ArcTan[c*x]))/(4*e^3*(d + e*x^2)^2) + (d*(a + b*ArcTan[c*x]))/(e^3*(d + e
*x^2)) + (b*c*Sqrt[d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/((c^2*d - e)*e^(5/2)) - (b*c*Sqrt[d]*(3*c^2*d - e)*ArcTan[(
Sqrt[e]*x)/Sqrt[d]])/(8*(c^2*d - e)^2*e^(5/2)) - ((a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/e^3 + ((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^3) + ((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^3) + ((I/2)*b*PolyLog[2, 1 -
2/(1 - I*c*x)])/e^3 - ((I/4)*b*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*
x))])/e^3 - ((I/4)*b*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/e^3

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 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 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, 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 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 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 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^5 \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx &=\int \left (\frac {d^2 x \left (a+b \tan ^{-1}(c x)\right )}{e^2 \left (d+e x^2\right )^3}-\frac {2 d x \left (a+b \tan ^{-1}(c x)\right )}{e^2 \left (d+e x^2\right )^2}+\frac {x \left (a+b \tan ^{-1}(c x)\right )}{e^2 \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^2}-\frac {(2 d) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx}{e^2}+\frac {d^2 \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx}{e^2}\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {d \left (a+b \tan ^{-1}(c x)\right )}{e^3 \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}{e^3}+\frac {\left (b c d^2\right ) \int \frac {1}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )^2} \, dx}{4 e^3}+\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^2}\\ &=-\frac {b c d x}{8 \left (c^2 d-e\right ) e^2 \left (d+e x^2\right )}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {d \left (a+b \tan ^{-1}(c x)\right )}{e^3 \left (d+e x^2\right )}+\frac {(b c d) \int \frac {2 c^2 d-e-c^2 e x^2}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )} \, dx}{8 \left (c^2 d-e\right ) e^3}-\frac {\left (b c^3 d\right ) \int \frac {1}{1+c^2 x^2} \, dx}{\left (c^2 d-e\right ) e^3}-\frac {\int \frac {a+b \tan ^{-1}(c x)}{\sqrt {-d}-\sqrt {e} x} \, dx}{2 e^{5/2}}+\frac {\int \frac {a+b \tan ^{-1}(c x)}{\sqrt {-d}+\sqrt {e} x} \, dx}{2 e^{5/2}}+\frac {(b c d) \int \frac {1}{d+e x^2} \, dx}{\left (c^2 d-e\right ) e^2}\\ &=-\frac {b c d x}{8 \left (c^2 d-e\right ) e^2 \left (d+e x^2\right )}-\frac {b c^2 d \tan ^{-1}(c x)}{\left (c^2 d-e\right ) e^3}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {d \left (a+b \tan ^{-1}(c x)\right )}{e^3 \left (d+e x^2\right )}+\frac {b c \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\left (c^2 d-e\right ) e^{5/2}}-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{e^3}+\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^3}+\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^3}+2 \frac {(b c) \int \frac {\log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{2 e^3}-\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^3}-\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^3}+\frac {\left (b c^5 d^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 \left (c^2 d-e\right )^2 e^3}-\frac {\left (b c d \left (3 c^2 d-e\right )\right ) \int \frac {1}{d+e x^2} \, dx}{8 \left (c^2 d-e\right )^2 e^2}\\ &=-\frac {b c d x}{8 \left (c^2 d-e\right ) e^2 \left (d+e x^2\right )}+\frac {b c^4 d^2 \tan ^{-1}(c x)}{4 \left (c^2 d-e\right )^2 e^3}-\frac {b c^2 d \tan ^{-1}(c x)}{\left (c^2 d-e\right ) e^3}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {d \left (a+b \tan ^{-1}(c x)\right )}{e^3 \left (d+e x^2\right )}+\frac {b c \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\left (c^2 d-e\right ) e^{5/2}}-\frac {b c \sqrt {d} \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 \left (c^2 d-e\right )^2 e^{5/2}}-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{e^3}+\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^3}+\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^3}-\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^3}-\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^3}+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^3}\\ &=-\frac {b c d x}{8 \left (c^2 d-e\right ) e^2 \left (d+e x^2\right )}+\frac {b c^4 d^2 \tan ^{-1}(c x)}{4 \left (c^2 d-e\right )^2 e^3}-\frac {b c^2 d \tan ^{-1}(c x)}{\left (c^2 d-e\right ) e^3}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {d \left (a+b \tan ^{-1}(c x)\right )}{e^3 \left (d+e x^2\right )}+\frac {b c \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\left (c^2 d-e\right ) e^{5/2}}-\frac {b c \sqrt {d} \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 \left (c^2 d-e\right )^2 e^{5/2}}-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{e^3}+\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^3}+\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^3}+\frac {i b \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 e^3}-\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^3}-\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^3}\\ \end {align*}

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Mathematica [A]  time = 13.47, size = 589, normalized size = 1.11 \[ \frac {a \left (\frac {d \left (3 d+4 e x^2\right )}{\left (d+e x^2\right )^2}+2 \log \left (d+e x^2\right )\right )+b \left (-\frac {c d e x}{2 \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {c^2 d \left (4 e-3 c^2 d\right ) \tan ^{-1}(c x)}{\left (e-c^2 d\right )^2}+\frac {c \sqrt {d} \sqrt {e} \left (5 c^2 d-7 e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \left (e-c^2 d\right )^2}-i \text {Li}_2\left (\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )+i \text {Li}_2\left (\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{\sqrt {d} c+\sqrt {e}}\right )+i \text {Li}_2\left (\frac {c \left (i \sqrt {e} x+\sqrt {d}\right )}{c \sqrt {d}-\sqrt {e}}\right )-i \text {Li}_2\left (\frac {c \left (i \sqrt {e} x+\sqrt {d}\right )}{\sqrt {d} c+\sqrt {e}}\right )+\frac {d \tan ^{-1}(c x) \left (3 d+4 e x^2\right )}{\left (d+e x^2\right )^2}+i \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (-1-i c x)}{c \sqrt {d}-\sqrt {e}}\right )-i \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (1-i c x)}{c \sqrt {d}+\sqrt {e}}\right )-i \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (-1+i c x)}{c \sqrt {d}-\sqrt {e}}\right )+i \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (1+i c x)}{c \sqrt {d}+\sqrt {e}}\right )+2 \tan ^{-1}(c x) \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right )+2 \tan ^{-1}(c x) \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )\right )}{4 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*ArcTan[c*x]))/(d + e*x^2)^3,x]

[Out]

(a*((d*(3*d + 4*e*x^2))/(d + e*x^2)^2 + 2*Log[d + e*x^2]) + b*(-1/2*(c*d*e*x)/((c^2*d - e)*(d + e*x^2)) + (c^2
*d*(-3*c^2*d + 4*e)*ArcTan[c*x])/(-(c^2*d) + e)^2 + (d*(3*d + 4*e*x^2)*ArcTan[c*x])/(d + e*x^2)^2 + (c*Sqrt[d]
*(5*c^2*d - 7*e)*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*(-(c^2*d) + e)^2) + 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] + S
qrt[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] - 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])] - I*PolyLog[2, (c*(Sqrt[d] + I*Sqrt[e]*x))/(c*Sqrt[d] + Sqrt[e])]))/
(4*e^3)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{5} \arctan \left (c x\right ) + a x^{5}}{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctan(c*x))/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*x^5*arctan(c*x) + a*x^5)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), 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^5*(a+b*arctan(c*x))/(e*x^2+d)^3,x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 0.41, size = 959, normalized size = 1.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arctan(c*x))/(e*x^2+d)^3,x)

[Out]

-1/4*c^4*a*d^2/e^3/(c^2*e*x^2+c^2*d)^2+c^2*a/e^3*d/(c^2*e*x^2+c^2*d)+1/2*a/e^3*ln(c^2*e*x^2+c^2*d)-1/4*c^4*b*a
rctan(c*x)*d^2/e^3/(c^2*e*x^2+c^2*d)^2+c^2*b*arctan(c*x)/e^3*d/(c^2*e*x^2+c^2*d)+1/2*b*arctan(c*x)/e^3*ln(c^2*
e*x^2+c^2*d)-1/8*c^5*b/e^2*d^2/(c^2*d-e)^2*x/(c^2*e*x^2+c^2*d)+1/8*c^3*b/e*d/(c^2*d-e)^2*x/(c^2*e*x^2+c^2*d)+5
/8*c^3*b/e^2*d^2/(c^2*d-e)^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))-7/8*c*b/e*d/(c^2*d-e)^2/(d*e)^(1/2)*arctan(e*
x/(d*e)^(1/2))-3/4*b*c^4*d^2*arctan(c*x)/(c^2*d-e)^2/e^3+c^2*b/e^2*d/(c^2*d-e)^2*arctan(c*x)+1/4*I*b/e^3*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^3*dilog((R
ootOf(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^3*ln(I+c*x)*l
n((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^3*ln(c^2*
e*x^2+c^2*d)*ln(c*x-I)+1/4*I*b/e^3*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^3*ln(c^2*e*x^2+c^2*d)*ln(I+c*x)+1/4*I*b/e^3*ln(I+c*x)*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^3*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^3*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^3*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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, a {\left (\frac {4 \, d e x^{2} + 3 \, d^{2}}{e^{5} x^{4} + 2 \, d e^{4} x^{2} + d^{2} e^{3}} + \frac {2 \, \log \left (e x^{2} + d\right )}{e^{3}}\right )} + 2 \, b \int \frac {x^{5} \arctan \left (c x\right )}{2 \, {\left (e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctan(c*x))/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

1/4*a*((4*d*e*x^2 + 3*d^2)/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^3) + 2*log(e*x^2 + d)/e^3) + 2*b*integrate(1/2*x^5*a
rctan(c*x)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*atan(c*x)))/(d + e*x^2)^3,x)

[Out]

int((x^5*(a + b*atan(c*x)))/(d + e*x^2)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*atan(c*x))/(e*x**2+d)**3,x)

[Out]

Timed out

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