3.1168 \(\int \frac {a+b \tan ^{-1}(c x)}{x (d+e x^2)^3} \, dx\)

Optimal. Leaf size=574 \[ -\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d^3}+\frac {\log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {a+b \tan ^{-1}(c x)}{2 d^2 \left (d+e x^2\right )}+\frac {a+b \tan ^{-1}(c x)}{4 d \left (d+e x^2\right )^2}+\frac {a \log (x)}{d^3}+\frac {b c \sqrt {e} \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} \left (c^2 d-e\right )^2}+\frac {b c \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}+\frac {b c e x}{8 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}-\frac {b c^4 \tan ^{-1}(c x)}{4 d \left (c^2 d-e\right )^2}+\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^3}+\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}-\frac {i b \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.63, antiderivative size = 574, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 13, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {4980, 4848, 2391, 4974, 414, 522, 203, 205, 391, 4856, 2402, 2315, 2447} \[ \frac {i b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{4 d^3}+\frac {i b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{4 d^3}+\frac {i b \text {PolyLog}(2,-i c x)}{2 d^3}-\frac {i b \text {PolyLog}(2,i c x)}{2 d^3}-\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 d^3}+\frac {a+b \tan ^{-1}(c x)}{2 d^2 \left (d+e x^2\right )}-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d^3}+\frac {\log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {a+b \tan ^{-1}(c x)}{4 d \left (d+e x^2\right )^2}+\frac {a \log (x)}{d^3}+\frac {b c e x}{8 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}+\frac {b c \sqrt {e} \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} \left (c^2 d-e\right )^2}+\frac {b c \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}-\frac {b c^4 \tan ^{-1}(c x)}{4 d \left (c^2 d-e\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*e*x)/(8*d^2*(c^2*d - e)*(d + e*x^2)) - (b*c^4*ArcTan[c*x])/(4*d*(c^2*d - e)^2) - (b*c^2*ArcTan[c*x])/(2*d
^2*(c^2*d - e)) + (a + b*ArcTan[c*x])/(4*d*(d + e*x^2)^2) + (a + b*ArcTan[c*x])/(2*d^2*(d + e*x^2)) + (b*c*Sqr
t[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(5/2)*(c^2*d - e)) + (b*c*(3*c^2*d - e)*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt
[d]])/(8*d^(5/2)*(c^2*d - e)^2) + (a*Log[x])/d^3 + ((a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/d^3 - ((a + b*ArcT
an[c*x])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*d^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*d^3) + ((I/2)*b*PolyLog[2, (-
I)*c*x])/d^3 - ((I/2)*b*PolyLog[2, I*c*x])/d^3 - ((I/2)*b*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^3 + ((I/4)*b*PolyLo
g[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/d^3 + ((I/4)*b*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 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 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 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 \left (d+e x^2\right )^3} \, dx &=\int \left (\frac {a+b \tan ^{-1}(c x)}{d^3 x}-\frac {e x \left (a+b \tan ^{-1}(c x)\right )}{d \left (d+e x^2\right )^3}-\frac {e x \left (a+b \tan ^{-1}(c x)\right )}{d^2 \left (d+e x^2\right )^2}-\frac {e 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} \, dx}{d^3}-\frac {e \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{d+e x^2} \, dx}{d^3}-\frac {e \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx}{d^2}-\frac {e \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx}{d}\\ &=\frac {a+b \tan ^{-1}(c x)}{4 d \left (d+e x^2\right )^2}+\frac {a+b \tan ^{-1}(c x)}{2 d^2 \left (d+e x^2\right )}+\frac {a \log (x)}{d^3}+\frac {(i b) \int \frac {\log (1-i c x)}{x} \, dx}{2 d^3}-\frac {(i b) \int \frac {\log (1+i c x)}{x} \, dx}{2 d^3}-\frac {(b c) \int \frac {1}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )} \, dx}{2 d^2}-\frac {(b c) \int \frac {1}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )^2} \, dx}{4 d}-\frac {e \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 e x}{8 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {a+b \tan ^{-1}(c x)}{4 d \left (d+e x^2\right )^2}+\frac {a+b \tan ^{-1}(c x)}{2 d^2 \left (d+e x^2\right )}+\frac {a \log (x)}{d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}-\frac {(b c) \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 d^2 \left (c^2 d-e\right )}-\frac {\left (b c^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^2 \left (c^2 d-e\right )}+\frac {\sqrt {e} \int \frac {a+b \tan ^{-1}(c x)}{\sqrt {-d}-\sqrt {e} x} \, dx}{2 d^3}-\frac {\sqrt {e} \int \frac {a+b \tan ^{-1}(c x)}{\sqrt {-d}+\sqrt {e} x} \, dx}{2 d^3}+\frac {(b c e) \int \frac {1}{d+e x^2} \, dx}{2 d^2 \left (c^2 d-e\right )}\\ &=\frac {b c e x}{8 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}+\frac {a+b \tan ^{-1}(c x)}{4 d \left (d+e x^2\right )^2}+\frac {a+b \tan ^{-1}(c x)}{2 d^2 \left (d+e x^2\right )}+\frac {b c \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{d^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 d^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 d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}-2 \frac {(b c) \int \frac {\log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{2 d^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 d^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 d^3}-\frac {\left (b c^5\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 d \left (c^2 d-e\right )^2}+\frac {\left (b c \left (3 c^2 d-e\right ) e\right ) \int \frac {1}{d+e x^2} \, dx}{8 d^2 \left (c^2 d-e\right )^2}\\ &=\frac {b c e x}{8 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b c^4 \tan ^{-1}(c x)}{4 d \left (c^2 d-e\right )^2}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}+\frac {a+b \tan ^{-1}(c x)}{4 d \left (d+e x^2\right )^2}+\frac {a+b \tan ^{-1}(c x)}{2 d^2 \left (d+e x^2\right )}+\frac {b c \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}+\frac {b c \left (3 c^2 d-e\right ) \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} \left (c^2 d-e\right )^2}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{d^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 d^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 d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^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 d^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 d^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 d^3}\\ &=\frac {b c e x}{8 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b c^4 \tan ^{-1}(c x)}{4 d \left (c^2 d-e\right )^2}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^2 \left (c^2 d-e\right )}+\frac {a+b \tan ^{-1}(c x)}{4 d \left (d+e x^2\right )^2}+\frac {a+b \tan ^{-1}(c x)}{2 d^2 \left (d+e x^2\right )}+\frac {b c \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \left (c^2 d-e\right )}+\frac {b c \left (3 c^2 d-e\right ) \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} \left (c^2 d-e\right )^2}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{d^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 d^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 d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}-\frac {i b \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 d^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 d^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 d^3}\\ \end {align*}

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Mathematica [A]  time = 14.42, size = 645, normalized size = 1.12 \[ \frac {2 a \left (\frac {d \left (3 d+2 e x^2\right )}{\left (d+e x^2\right )^2}-2 \log \left (d+e x^2\right )+4 \log (x)\right )+b \left (\frac {c d e x}{\left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {2 c^2 d \left (2 e-3 c^2 d\right ) \tan ^{-1}(c x)}{\left (e-c^2 d\right )^2}+\frac {c \sqrt {d} \sqrt {e} \left (7 c^2 d-5 e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\left (e-c^2 d\right )^2}+2 i \text {Li}_2\left (\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{c \sqrt {d}-\sqrt {e}}\right )-2 i \text {Li}_2\left (\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{\sqrt {d} c+\sqrt {e}}\right )-2 i \text {Li}_2\left (\frac {c \left (i \sqrt {e} x+\sqrt {d}\right )}{c \sqrt {d}-\sqrt {e}}\right )+2 i \text {Li}_2\left (\frac {c \left (i \sqrt {e} x+\sqrt {d}\right )}{\sqrt {d} c+\sqrt {e}}\right )+\frac {2 d \tan ^{-1}(c x) \left (3 d+2 e x^2\right )}{\left (d+e x^2\right )^2}-2 i \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (-1-i c x)}{c \sqrt {d}-\sqrt {e}}\right )+2 i \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (1-i c x)}{c \sqrt {d}+\sqrt {e}}\right )+2 i \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (-1+i c x)}{c \sqrt {d}-\sqrt {e}}\right )-2 i \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right ) \log \left (\frac {\sqrt {e} (1+i c x)}{c \sqrt {d}+\sqrt {e}}\right )-4 \tan ^{-1}(c x) \log \left (x-\frac {i \sqrt {d}}{\sqrt {e}}\right )-4 \tan ^{-1}(c x) \log \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )-4 i (-\text {Li}_2(-i c x)+\text {Li}_2(i c x)+\log (x) (\log (1-i c x)-\log (1+i c x)))+8 \log (x) \tan ^{-1}(c x)\right )}{8 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*((d*(3*d + 2*e*x^2))/(d + e*x^2)^2 + 4*Log[x] - 2*Log[d + e*x^2]) + b*((c*d*e*x)/((c^2*d - e)*(d + e*x^2)
) + (2*c^2*d*(-3*c^2*d + 2*e)*ArcTan[c*x])/(-(c^2*d) + e)^2 + (2*d*(3*d + 2*e*x^2)*ArcTan[c*x])/(d + e*x^2)^2
+ (c*Sqrt[d]*(7*c^2*d - 5*e)*Sqrt[e]*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(-(c^2*d) + e)^2 + 8*ArcTan[c*x]*Log[x] - 4*
ArcTan[c*x]*Log[((-I)*Sqrt[d])/Sqrt[e] + x] - 4*ArcTan[c*x]*Log[(I*Sqrt[d])/Sqrt[e] + x] - (2*I)*Log[((-I)*Sqr
t[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(-1 - I*c*x))/(c*Sqrt[d] - Sqrt[e])] + (2*I)*Log[((-I)*Sqrt[d])/Sqrt[e] + x]*L
og[(Sqrt[e]*(1 - I*c*x))/(c*Sqrt[d] + Sqrt[e])] + (2*I)*Log[(I*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(-1 + I*c*x)
)/(c*Sqrt[d] - Sqrt[e])] - (2*I)*Log[(I*Sqrt[d])/Sqrt[e] + x]*Log[(Sqrt[e]*(1 + I*c*x))/(c*Sqrt[d] + Sqrt[e])]
- (4*I)*(Log[x]*(Log[1 - I*c*x] - Log[1 + I*c*x]) - PolyLog[2, (-I)*c*x] + PolyLog[2, I*c*x]) + (2*I)*PolyLog
[2, (c*(Sqrt[d] - I*Sqrt[e]*x))/(c*Sqrt[d] - Sqrt[e])] - (2*I)*PolyLog[2, (c*(Sqrt[d] - I*Sqrt[e]*x))/(c*Sqrt[
d] + Sqrt[e])] - (2*I)*PolyLog[2, (c*(Sqrt[d] + I*Sqrt[e]*x))/(c*Sqrt[d] - Sqrt[e])] + (2*I)*PolyLog[2, (c*(Sq
rt[d] + I*Sqrt[e]*x))/(c*Sqrt[d] + Sqrt[e])]))/(8*d^3)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (c x\right ) + a}{e^{3} x^{7} + 3 \, d e^{2} x^{5} + 3 \, d^{2} e x^{3} + d^{3} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [C]  time = 0.36, size = 1041, normalized size = 1.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*b/d^3*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))+1/4*I*b/d^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,i
ndex=2))+1/2*I*b/d^3*ln(c*x)*ln(1+I*c*x)-1/2*I*b/d^3*ln(c*x)*ln(1-I*c*x)+1/4*I*b/d^3*ln(I+c*x)*ln(c^2*e*x^2+c^
2*d)+1/2*b*c^2*arctan(c*x)/d^2/(c^2*e*x^2+c^2*d)+1/4*b*c^4*arctan(c*x)/d/(c^2*e*x^2+c^2*d)^2-1/4*I*b/d^3*ln(c*
x-I)*ln(c^2*e*x^2+c^2*d)-1/4*I*b/d^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/d^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))-5/8*b*c/d^2/(c^2*d-e)^2*e^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+1/8*b*c^5*e/(
c^2*d-e)^2*x/d/(c^2*e*x^2+c^2*d)-1/8*b*c^3/d^2*e^2/(c^2*d-e)^2*x/(c^2*e*x^2+c^2*d)+7/8*b*c^3*e/(c^2*d-e)^2/d/(
d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+1/4*I*b/d^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))+1/2*I*b/d^3*dilog(1+I*c*x)-1/2*b*arctan(c*x)/d^3*ln(c^2*e*x^2+c^2*d)+b*arctan(
c*x)/d^3*ln(c*x)+1/2*a*c^2/d^2/(c^2*e*x^2+c^2*d)+1/4*a*c^4/d/(c^2*e*x^2+c^2*d)^2+1/4*I*b/d^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/d^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/d^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))-1/2*I*b/d^3*dilog(1-I*c*x)+a/d^3*ln(c*
x)+1/2*b*c^2/d^2/(c^2*d-e)^2*arctan(c*x)*e-1/2*a/d^3*ln(c^2*e*x^2+c^2*d)-3/4*b*c^4*arctan(c*x)/d/(c^2*d-e)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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