3.1161 \(\int \frac {a+b \tan ^{-1}(c x)}{x^3 (d+e x^2)^2} \, dx\)

Optimal. Leaf size=489 \[ -\frac {2 e \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{d^3}+\frac {e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{d^3}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 \left (d+e x^2\right )}-\frac {a+b \tan ^{-1}(c x)}{2 d^2 x^2}-\frac {2 a e \log (x)}{d^3}-\frac {b c e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}+\frac {b c^2 e \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2}-\frac {i b e \text {Li}_2(-i c x)}{d^3}+\frac {i b e \text {Li}_2(i c x)}{d^3}+\frac {i b e \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b e \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^3}-\frac {i b e \text {Li}_2\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{2 d^3}-\frac {b c}{2 d^2 x} \]

[Out]

-1/2*b*c/d^2/x-1/2*b*c^2*arctan(c*x)/d^2+1/2*b*c^2*e*arctan(c*x)/d^2/(c^2*d-e)+1/2*(-a-b*arctan(c*x))/d^2/x^2-
1/2*e*(a+b*arctan(c*x))/d^2/(e*x^2+d)-1/2*b*c*e^(3/2)*arctan(x*e^(1/2)/d^(1/2))/d^(5/2)/(c^2*d-e)-2*a*e*ln(x)/
d^3-2*e*(a+b*arctan(c*x))*ln(2/(1-I*c*x))/d^3+e*(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)))/d^3+e*(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))
)/d^3-I*b*e*polylog(2,-I*c*x)/d^3+I*b*e*polylog(2,I*c*x)/d^3+I*b*e*polylog(2,1-2/(1-I*c*x))/d^3-1/2*I*b*e*poly
log(2,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/d^3-1/2*I*b*e*polylog(2,1-2*c*((-d)^(1/
2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/d^3

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Rubi [A]  time = 0.51, antiderivative size = 489, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 13, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {4980, 4852, 325, 203, 4848, 2391, 4974, 391, 205, 4856, 2402, 2315, 2447} \[ -\frac {i b e \text {PolyLog}(2,-i c x)}{d^3}+\frac {i b e \text {PolyLog}(2,i c x)}{d^3}+\frac {i b e \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b e \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d^3}-\frac {i b e \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d^3}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 \left (d+e x^2\right )}-\frac {2 e \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{d^3}+\frac {e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{d^3}-\frac {a+b \tan ^{-1}(c x)}{2 d^2 x^2}-\frac {2 a e \log (x)}{d^3}-\frac {b c e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}+\frac {b c^2 e \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2}-\frac {b c}{2 d^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(2*d^2*x) - (b*c^2*ArcTan[c*x])/(2*d^2) + (b*c^2*e*ArcTan[c*x])/(2*d^2*(c^2*d - e)) - (a + b*ArcTan[c*x
])/(2*d^2*x^2) - (e*(a + b*ArcTan[c*x]))/(2*d^2*(d + e*x^2)) - (b*c*e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^
(5/2)*(c^2*d - e)) - (2*a*e*Log[x])/d^3 - (2*e*(a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/d^3 + (e*(a + b*ArcTan[
c*x])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/d^3 + (e*(a + b*ArcTan[c*x])*L
og[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/d^3 - (I*b*e*PolyLog[2, (-I)*c*x])/d^
3 + (I*b*e*PolyLog[2, I*c*x])/d^3 + (I*b*e*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^3 - ((I/2)*b*e*PolyLog[2, 1 - (2*c
*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/d^3 - ((I/2)*b*e*PolyLog[2, 1 - (2*c*(Sqrt[-
d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/d^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 325

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

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 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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, 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 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 {a+b \tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^2} \, dx &=\int \left (\frac {a+b \tan ^{-1}(c x)}{d^2 x^3}-\frac {2 e \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac {e^2 x \left (a+b \tan ^{-1}(c x)\right )}{d^2 \left (d+e x^2\right )^2}+\frac {2 e^2 x \left (a+b \tan ^{-1}(c x)\right )}{d^3 \left (d+e x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx}{d^2}-\frac {(2 e) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx}{d^3}+\frac {\left (2 e^2\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{d+e x^2} \, dx}{d^3}+\frac {e^2 \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx}{d^2}\\ &=-\frac {a+b \tan ^{-1}(c x)}{2 d^2 x^2}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 \left (d+e x^2\right )}-\frac {2 a e \log (x)}{d^3}+\frac {(b c) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{2 d^2}-\frac {(i b e) \int \frac {\log (1-i c x)}{x} \, dx}{d^3}+\frac {(i b e) \int \frac {\log (1+i c x)}{x} \, dx}{d^3}+\frac {(b c e) \int \frac {1}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )} \, dx}{2 d^2}+\frac {\left (2 e^2\right ) \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}{d^3}\\ &=-\frac {b c}{2 d^2 x}-\frac {a+b \tan ^{-1}(c x)}{2 d^2 x^2}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 \left (d+e x^2\right )}-\frac {2 a e \log (x)}{d^3}-\frac {i b e \text {Li}_2(-i c x)}{d^3}+\frac {i b e \text {Li}_2(i c x)}{d^3}-\frac {\left (b c^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^2}+\frac {\left (b c^3 e\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^2 \left (c^2 d-e\right )}-\frac {e^{3/2} \int \frac {a+b \tan ^{-1}(c x)}{\sqrt {-d}-\sqrt {e} x} \, dx}{d^3}+\frac {e^{3/2} \int \frac {a+b \tan ^{-1}(c x)}{\sqrt {-d}+\sqrt {e} x} \, dx}{d^3}-\frac {\left (b c e^2\right ) \int \frac {1}{d+e x^2} \, dx}{2 d^2 \left (c^2 d-e\right )}\\ &=-\frac {b c}{2 d^2 x}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2}+\frac {b c^2 e \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {a+b \tan ^{-1}(c x)}{2 d^2 x^2}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 \left (d+e x^2\right )}-\frac {b c e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}-\frac {2 a e \log (x)}{d^3}-\frac {2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{d^3}+\frac {e \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 )}{d^3}+\frac {e \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 )}{d^3}-\frac {i b e \text {Li}_2(-i c x)}{d^3}+\frac {i b e \text {Li}_2(i c x)}{d^3}+2 \frac {(b c e) \int \frac {\log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d^3}-\frac {(b c e) \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}{d^3}-\frac {(b c e) \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}{d^3}\\ &=-\frac {b c}{2 d^2 x}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2}+\frac {b c^2 e \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {a+b \tan ^{-1}(c x)}{2 d^2 x^2}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 \left (d+e x^2\right )}-\frac {b c e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}-\frac {2 a e \log (x)}{d^3}-\frac {2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{d^3}+\frac {e \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 )}{d^3}+\frac {e \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 )}{d^3}-\frac {i b e \text {Li}_2(-i c x)}{d^3}+\frac {i b e \text {Li}_2(i c x)}{d^3}-\frac {i b e \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 )}{2 d^3}-\frac {i b e \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 )}{2 d^3}+2 \frac {(i b e) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i c x}\right )}{d^3}\\ &=-\frac {b c}{2 d^2 x}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2}+\frac {b c^2 e \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {a+b \tan ^{-1}(c x)}{2 d^2 x^2}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 \left (d+e x^2\right )}-\frac {b c e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}-\frac {2 a e \log (x)}{d^3}-\frac {2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{d^3}+\frac {e \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 )}{d^3}+\frac {e \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 )}{d^3}-\frac {i b e \text {Li}_2(-i c x)}{d^3}+\frac {i b e \text {Li}_2(i c x)}{d^3}+\frac {i b e \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b e \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 )}{2 d^3}-\frac {i b e \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 )}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 13.70, size = 643, normalized size = 1.31 \[ -\frac {a \left (d \left (\frac {e}{d+e x^2}+\frac {1}{x^2}\right )-2 e \log \left (d+e x^2\right )+4 e \log (x)\right )+b \left (\frac {c \sqrt {d} e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{c^2 d-e}+\frac {c^2 d \left (c^2 d-2 e\right ) \tan ^{-1}(c x)}{c^2 d-e}-e \left (-i \text {Li}_2\left (\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )+i \text {Li}_2\left (\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{\sqrt {d} c+\sqrt {e}}\right )+i \text {Li}_2\left (\frac {c \left (i \sqrt {e} x+\sqrt {d}\right )}{c \sqrt {d}-\sqrt {e}}\right )-i \text {Li}_2\left (\frac {c \left (i \sqrt {e} x+\sqrt {d}\right )}{\sqrt {d} c+\sqrt {e}}\right )-2 \tan ^{-1}(c x) \log \left (d+e x^2\right )+i \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (-1-i c x)}{c \sqrt {d}-\sqrt {e}}\right )-i \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (1-i c x)}{c \sqrt {d}+\sqrt {e}}\right )-i \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (-1+i c x)}{c \sqrt {d}-\sqrt {e}}\right )+i \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (1+i c x)}{c \sqrt {d}+\sqrt {e}}\right )+2 \tan ^{-1}(c x) \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right )+2 \tan ^{-1}(c x) \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )\right )+d \tan ^{-1}(c x) \left (\frac {e}{d+e x^2}+\frac {1}{x^2}\right )-2 e \tan ^{-1}(c x) \log \left (d+e x^2\right )+\frac {c d}{x}-2 i e (-\text {Li}_2(-i c x)+\text {Li}_2(i c x)+\log (x) (\log (1-i c x)-\log (1+i c x)))+4 e \log (x) \tan ^{-1}(c x)\right )}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (c x\right ) + a}{e^{2} x^{7} + 2 \, d e x^{5} + d^{2} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctan(c*x) + a)/(e^2*x^7 + 2*d*e*x^5 + d^2*x^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((a+b*arctan(c*x))/x^3/(e*x^2+d)^2,x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 0.37, size = 925, normalized size = 1.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^4*b/d/(c^2*d-e)*arctan(c*x)+b*arctan(c*x)*e/d^3*ln(c^2*e*x^2+c^2*d)-2*b*arctan(c*x)/d^3*e*ln(c*x)+I*b/d
^3*e*dilog(1-I*c*x)+1/2*I*b/d^3*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))+1/2*I*b/d^3*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))-1/2*I*b/d^3*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))-I*b/d^3*e*dilog(1+I*c*x)-1/2*I*b/d^3*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))-1/2*a/d^2/x^2-I*b/d^3*e*ln(c*x)*ln(1+I*c*x)-1/2*I*b/d^3*e*ln(c*x-I)*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))-1/2*I*b/d^3*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))-1/2*I*b/d^3*e
*ln(I+c*x)*ln(c^2*e*x^2+c^2*d)-1/2*c^2*b*arctan(c*x)*e/d^2/(c^2*e*x^2+c^2*d)+1/2*I*b/d^3*e*ln(c*x-I)*ln(c^2*e*
x^2+c^2*d)+1/2*I*b/d^3*e*ln(I+c*x)*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))+I*b/d^3*e*ln(c*x)*ln(1-I*c*x)+1/2*I*b/d^3*e*ln(I+c*x)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,inde
x=2)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2))-1/2*c*b/d^2*e^2/(c^2*d-e)/(d*e)^(1/2)*arctan(e*x/(d*e)^(1
/2))+b*c^2*e*arctan(c*x)/d^2/(c^2*d-e)-1/2*c^2*a*e/d^2/(c^2*e*x^2+c^2*d)+a*e/d^3*ln(c^2*e*x^2+c^2*d)-2*a/d^3*e
*ln(c*x)-1/2*b*c/d^2/x-1/2*b*arctan(c*x)/d^2/x^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {2 \, e x^{2} + d}{d^{2} e x^{4} + d^{3} x^{2}} - \frac {2 \, e \log \left (e x^{2} + d\right )}{d^{3}} + \frac {4 \, e \log \relax (x)}{d^{3}}\right )} + 2 \, b \int \frac {\arctan \left (c x\right )}{2 \, {\left (e^{2} x^{7} + 2 \, d e x^{5} + d^{2} x^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*((2*e*x^2 + d)/(d^2*e*x^4 + d^3*x^2) - 2*e*log(e*x^2 + d)/d^3 + 4*e*log(x)/d^3) + 2*b*integrate(1/2*arc
tan(c*x)/(e^2*x^7 + 2*d*e*x^5 + d^2*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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