3.1157 \(\int \frac {a+b \tan ^{-1}(c x)}{x^2 (d+e x^2)} \, dx\)

Optimal. Leaf size=561 \[ -\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d}+\frac {i b \sqrt {e} \text {Li}_2\left (\frac {\sqrt {e} (i-c x)}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \text {Li}_2\left (\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \text {Li}_2\left (\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \text {Li}_2\left (\frac {\sqrt {e} (c x+i)}{\sqrt {-d} c+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {b c \log (x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.53, antiderivative size = 561, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 13, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {4918, 4852, 266, 36, 29, 31, 4910, 205, 4908, 2409, 2394, 2393, 2391} \[ \frac {i b \sqrt {e} \text {PolyLog}\left (2,\frac {\sqrt {e} (-c x+i)}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \text {PolyLog}\left (2,\frac {\sqrt {e} (1-i c x)}{\sqrt {e}+i c \sqrt {-d}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \text {PolyLog}\left (2,\frac {\sqrt {e} (1+i c x)}{\sqrt {e}+i c \sqrt {-d}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \text {PolyLog}\left (2,\frac {\sqrt {e} (c x+i)}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d}-\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {b c \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2409

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

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 4908

Int[ArcTan[(c_.)*(x_)]/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[I/2, Int[Log[1 - I*c*x]/(d + e*x^2), x], x] -
Dist[I/2, Int[Log[1 + I*c*x]/(d + e*x^2), x], x] /; FreeQ[{c, d, e}, x]

Rule 4910

Int[(ArcTan[(c_.)*(x_)]*(b_.) + (a_))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[a, Int[1/(d + e*x^2), x], x] +
Dist[b, Int[ArcTan[c*x]/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]

Rule 4918

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

Rubi steps

\begin {align*} \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (d+e x^2\right )} \, dx &=\frac {\int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx}{d}-\frac {e \int \frac {a+b \tan ^{-1}(c x)}{d+e x^2} \, dx}{d}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x}+\frac {(b c) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d}-\frac {(a e) \int \frac {1}{d+e x^2} \, dx}{d}-\frac {(b e) \int \frac {\tan ^{-1}(c x)}{d+e x^2} \, dx}{d}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d}-\frac {(i b e) \int \frac {\log (1-i c x)}{d+e x^2} \, dx}{2 d}+\frac {(i b e) \int \frac {\log (1+i c x)}{d+e x^2} \, dx}{2 d}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}-\frac {\left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d}-\frac {(i b e) \int \left (\frac {\sqrt {-d} \log (1-i c x)}{2 d \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\sqrt {-d} \log (1-i c x)}{2 d \left (\sqrt {-d}+\sqrt {e} x\right )}\right ) \, dx}{2 d}+\frac {(i b e) \int \left (\frac {\sqrt {-d} \log (1+i c x)}{2 d \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\sqrt {-d} \log (1+i c x)}{2 d \left (\sqrt {-d}+\sqrt {e} x\right )}\right ) \, dx}{2 d}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b c \log (x)}{d}-\frac {b c \log \left (1+c^2 x^2\right )}{2 d}-\frac {(i b e) \int \frac {\log (1-i c x)}{\sqrt {-d}-\sqrt {e} x} \, dx}{4 (-d)^{3/2}}-\frac {(i b e) \int \frac {\log (1-i c x)}{\sqrt {-d}+\sqrt {e} x} \, dx}{4 (-d)^{3/2}}+\frac {(i b e) \int \frac {\log (1+i c x)}{\sqrt {-d}-\sqrt {e} x} \, dx}{4 (-d)^{3/2}}+\frac {(i b e) \int \frac {\log (1+i c x)}{\sqrt {-d}+\sqrt {e} x} \, dx}{4 (-d)^{3/2}}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b c \log (x)}{d}-\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {b c \log \left (1+c^2 x^2\right )}{2 d}-\frac {\left (b c \sqrt {e}\right ) \int \frac {\log \left (-\frac {i c \left (\sqrt {-d}-\sqrt {e} x\right )}{-i c \sqrt {-d}+\sqrt {e}}\right )}{1-i c x} \, dx}{4 (-d)^{3/2}}-\frac {\left (b c \sqrt {e}\right ) \int \frac {\log \left (\frac {i c \left (\sqrt {-d}-\sqrt {e} x\right )}{i c \sqrt {-d}+\sqrt {e}}\right )}{1+i c x} \, dx}{4 (-d)^{3/2}}+\frac {\left (b c \sqrt {e}\right ) \int \frac {\log \left (-\frac {i c \left (\sqrt {-d}+\sqrt {e} x\right )}{-i c \sqrt {-d}-\sqrt {e}}\right )}{1-i c x} \, dx}{4 (-d)^{3/2}}+\frac {\left (b c \sqrt {e}\right ) \int \frac {\log \left (\frac {i c \left (\sqrt {-d}+\sqrt {e} x\right )}{i c \sqrt {-d}-\sqrt {e}}\right )}{1+i c x} \, dx}{4 (-d)^{3/2}}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b c \log (x)}{d}-\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {b c \log \left (1+c^2 x^2\right )}{2 d}+\frac {\left (i b \sqrt {e}\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt {e} x}{-i c \sqrt {-d}-\sqrt {e}}\right )}{x} \, dx,x,1-i c x\right )}{4 (-d)^{3/2}}-\frac {\left (i b \sqrt {e}\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt {e} x}{i c \sqrt {-d}-\sqrt {e}}\right )}{x} \, dx,x,1+i c x\right )}{4 (-d)^{3/2}}-\frac {\left (i b \sqrt {e}\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {\sqrt {e} x}{-i c \sqrt {-d}+\sqrt {e}}\right )}{x} \, dx,x,1-i c x\right )}{4 (-d)^{3/2}}+\frac {\left (i b \sqrt {e}\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {\sqrt {e} x}{i c \sqrt {-d}+\sqrt {e}}\right )}{x} \, dx,x,1+i c x\right )}{4 (-d)^{3/2}}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x}-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b c \log (x)}{d}-\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {b c \log \left (1+c^2 x^2\right )}{2 d}+\frac {i b \sqrt {e} \text {Li}_2\left (\frac {\sqrt {e} (i-c x)}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \text {Li}_2\left (\frac {\sqrt {e} (1-i c x)}{i c \sqrt {-d}+\sqrt {e}}\right )}{4 (-d)^{3/2}}-\frac {i b \sqrt {e} \text {Li}_2\left (\frac {\sqrt {e} (1+i c x)}{i c \sqrt {-d}+\sqrt {e}}\right )}{4 (-d)^{3/2}}+\frac {i b \sqrt {e} \text {Li}_2\left (\frac {\sqrt {e} (i+c x)}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 (-d)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.82, size = 468, normalized size = 0.83 \[ \frac {-\frac {\sqrt {e} \left (4 a \sqrt {-d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+i b \sqrt {d} \left (\text {Li}_2\left (\frac {\sqrt {e} (i-c x)}{\sqrt {-d} c+i \sqrt {e}}\right )+\log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )\right )-i b \sqrt {d} \left (\text {Li}_2\left (\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )+\log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )\right )-i b \sqrt {d} \left (\text {Li}_2\left (\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )+\log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )\right )+i b \sqrt {d} \left (\text {Li}_2\left (\frac {\sqrt {e} (c x+i)}{\sqrt {-d} c+i \sqrt {e}}\right )+\log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )\right )\right )}{4 \sqrt {-d^2}}-\frac {a+b \tan ^{-1}(c x)}{x}-\frac {1}{2} b c \log \left (c^2 x^2+1\right )+b c \log (x)}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (c x\right ) + a}{e x^{4} + d x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [C]  time = 0.84, size = 2439, normalized size = 4.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b/c^2*(d*e)^(1/2)/d^3*e^3*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e)^(1/2))/(c^
2*d-e)^2+3/8*b*c^6*(d*e)^(1/2)*d/e*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e)^(1/2)
)/(c^2*d-e)^2-1/2*b*c^2*(d*e)^(1/2)/d*e*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e)^
(1/2))/(c^2*d-e)^2+c*b/d*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+c*b/d*ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)-1/4*c*b/d*l
n((1+I*c*x)^4/(c^2*x^2+1)^2*c^2*d+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-(1+I*c*x)^4/(c^2*x^2+1)^2*e+c^2*d+2*(1+I*c*x
)^2/(c^2*x^2+1)*e-e)-2*b*c^3/(c^2*d-e)*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))-1/8*b*c^3/(c^2*d-e)*ln((1+I*c*x)^4/(c^2
*x^2+1)^2*c^2*d+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-(1+I*c*x)^4/(c^2*x^2+1)^2*e+c^2*d+2*(1+I*c*x)^2/(c^2*x^2+1)*e-
e)+3/4*b*(d*e)^(1/2)/d^2*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e)^(1/2))-b*arctan
(c*x)/x/d-a/d/x+5/4*b*(d*e)^(1/2)/d^2*e*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e)^
(1/2))/(c^2*d-e)+1/4*b*(d*e)^(1/2)/d^2*e^2*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*
e)^(1/2))/(c^2*d-e)^2-1/8*b/c^2*(d*e)^(1/2)/d^3*e*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e
)/c/(d*e)^(1/2))+1/4*c*b/d*e/(c^2*d-e)*ln((1+I*c*x)^4/(c^2*x^2+1)^2*c^2*d+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-(1+I
*c*x)^4/(c^2*x^2+1)^2*e+c^2*d+2*(1+I*c*x)^2/(c^2*x^2+1)*e-e)+1/8*c*b/d*e^2/(c^2*d-e)^2*ln((1+I*c*x)^4/(c^2*x^2
+1)^2*c^2*d+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-(1+I*c*x)^4/(c^2*x^2+1)^2*e+c^2*d+2*(1+I*c*x)^2/(c^2*x^2+1)*e-e)+2
*c*b/d*e/(c^2*d-e)*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))-3/4*b*c^4*(d*e)^(1/2)/e*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^
2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e)^(1/2))/(c^2*d-e)+3/8*b*c^2*(d*e)^(1/2)/d/e*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*
x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e)^(1/2))-1/2*b*c^2*(d*e)^(1/2)/d*arctanh(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2
*x^2+1)+2*c^2*d+2*e)/c/(d*e)^(1/2))/(c^2*d-e)+1/8*b/c/d^2*e^3/(c^2*d-e)^2*ln((1+I*c*x)^4/(c^2*x^2+1)^2*c^2*d+2
*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-(1+I*c*x)^4/(c^2*x^2+1)^2*e+c^2*d+2*(1+I*c*x)^2/(c^2*x^2+1)*e-e)-1/8*b/c/d^2*e^
2/(c^2*d-e)*ln((1+I*c*x)^4/(c^2*x^2+1)^2*c^2*d+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-(1+I*c*x)^4/(c^2*x^2+1)^2*e+c^2
*d+2*(1+I*c*x)^2/(c^2*x^2+1)*e-e)+1/4*b/c/d^2*e*sum((_R1^2*c^2*d-_R1^2*e+3*c^2*d+e)/(_R1^2*c^2*d-_R1^2*e+c^2*d
+e)*(I*arctan(c*x)*ln((_R1-(1+I*c*x)/(c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-(1+I*c*x)/(c^2*x^2+1)^(1/2))/_R1)),_R1
=RootOf((c^2*d-e)*_Z^4+(2*c^2*d+2*e)*_Z^2+c^2*d-e))-a*e/d/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))-1/4*b/c/d^2*e*ln
((1+I*c*x)^4/(c^2*x^2+1)^2*c^2*d+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-(1+I*c*x)^4/(c^2*x^2+1)^2*e+c^2*d+2*(1+I*c*x)
^2/(c^2*x^2+1)*e-e)-1/4*b/c/d^2*e*sum((_R1^2*c^2*d-_R1^2*e-c^2*d+e)/(_R1^2*c^2*d-_R1^2*e+c^2*d+e)*(I*arctan(c*
x)*ln((_R1-(1+I*c*x)/(c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-(1+I*c*x)/(c^2*x^2+1)^(1/2))/_R1)),_R1=RootOf((c^2*d-e
)*_Z^4+(2*c^2*d+2*e)*_Z^2+c^2*d-e))-5/8*b*c^3*e/(c^2*d-e)^2*ln((1+I*c*x)^4/(c^2*x^2+1)^2*c^2*d+2*c^2*d*(1+I*c*
x)^2/(c^2*x^2+1)-(1+I*c*x)^4/(c^2*x^2+1)^2*e+c^2*d+2*(1+I*c*x)^2/(c^2*x^2+1)*e-e)-1/4*b*c^4*(d*e)^(1/2)*arctan
h(1/4*(2*(c^2*d-e)*(1+I*c*x)^2/(c^2*x^2+1)+2*c^2*d+2*e)/c/(d*e)^(1/2))/(c^2*d-e)^2+3/8*b*c^5*d/(c^2*d-e)^2*ln(
(1+I*c*x)^4/(c^2*x^2+1)^2*c^2*d+2*c^2*d*(1+I*c*x)^2/(c^2*x^2+1)-(1+I*c*x)^4/(c^2*x^2+1)^2*e+c^2*d+2*(1+I*c*x)^
2/(c^2*x^2+1)*e-e)+I*c*b*arctan(c*x)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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