3.109 \(\int \frac {\cot ^{-1}(a+b x)}{c+\frac {d}{x}} \, dx\)

Optimal. Leaf size=338 \[ -\frac {i d \text {Li}_2\left (-\frac {b (d+c x)}{(a+i) c-b d}\right )}{2 c^2}+\frac {i d \text {Li}_2\left (\frac {b (d+c x)}{-a c+i c+b d}\right )}{2 c^2}+\frac {i d \log (c x+d) \log \left (\frac {c (-a-b x+i)}{-a c+b d+i c}\right )}{2 c^2}-\frac {i d \log \left (-\frac {-a-b x+i}{a+b x}\right ) \log (c x+d)}{2 c^2}-\frac {i d \log (c x+d) \log \left (\frac {c (a+b x+i)}{-b d+(a+i) c}\right )}{2 c^2}+\frac {i d \log \left (\frac {a+b x+i}{a+b x}\right ) \log (c x+d)}{2 c^2}+\frac {\log (-a-b x+i)}{2 b c}+\frac {i (a+b x) \log \left (-\frac {-a-b x+i}{a+b x}\right )}{2 b c}+\frac {\log (a+b x+i)}{2 b c}-\frac {i (a+b x) \log \left (\frac {a+b x+i}{a+b x}\right )}{2 b c} \]

[Out]

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

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Rubi [A]  time = 0.50, antiderivative size = 422, normalized size of antiderivative = 1.25, number of steps used = 37, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5052, 2513, 2409, 2389, 2295, 2394, 2393, 2391, 193, 43} \[ -\frac {i d \text {PolyLog}\left (2,\frac {c (-a-b x+i)}{b d+(-a+i) c}\right )}{2 c^2}+\frac {i d \text {PolyLog}\left (2,\frac {c (a+b x+i)}{-b d+(a+i) c}\right )}{2 c^2}-\frac {i d \left (\log \left (-\frac {-a-b x+i}{a+b x}\right )+\log (a+b x)-\log (a+b x-i)\right ) \log (c x+d)}{2 c^2}+\frac {i d \left (\log (a+b x)-\log (a+b x+i)+\log \left (\frac {a+b x+i}{a+b x}\right )\right ) \log (c x+d)}{2 c^2}+\frac {i d \log (a+b x+i) \log \left (-\frac {b (c x+d)}{-b d+(a+i) c}\right )}{2 c^2}-\frac {i d \log (a+b x-i) \log \left (\frac {b (c x+d)}{b d+(-a+i) c}\right )}{2 c^2}+\frac {i x \left (\log \left (-\frac {-a-b x+i}{a+b x}\right )+\log (a+b x)-\log (a+b x-i)\right )}{2 c}-\frac {i (-a-b x+i) \log (a+b x-i)}{2 b c}-\frac {i (a+b x+i) \log (a+b x+i)}{2 b c}-\frac {i x \left (\log (a+b x)-\log (a+b x+i)+\log \left (\frac {a+b x+i}{a+b x}\right )\right )}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

Int[ArcCot[a + b*x]/(c + d/x),x]

[Out]

((I/2)*x*(Log[-((I - a - b*x)/(a + b*x))] + Log[a + b*x] - Log[-I + a + b*x]))/c - ((I/2)*(I - a - b*x)*Log[-I
+ a + b*x])/(b*c) - ((I/2)*(I + a + b*x)*Log[I + a + b*x])/(b*c) - ((I/2)*x*(Log[a + b*x] - Log[I + a + b*x]
+ Log[(I + a + b*x)/(a + b*x)]))/c - ((I/2)*d*(Log[-((I - a - b*x)/(a + b*x))] + Log[a + b*x] - Log[-I + a + b
*x])*Log[d + c*x])/c^2 + ((I/2)*d*(Log[a + b*x] - Log[I + a + b*x] + Log[(I + a + b*x)/(a + b*x)])*Log[d + c*x
])/c^2 + ((I/2)*d*Log[I + a + b*x]*Log[-((b*(d + c*x))/((I + a)*c - b*d))])/c^2 - ((I/2)*d*Log[-I + a + b*x]*L
og[(b*(d + c*x))/((I - a)*c + b*d)])/c^2 - ((I/2)*d*PolyLog[2, (c*(I - a - b*x))/((I - a)*c + b*d)])/c^2 + ((I
/2)*d*PolyLog[2, (c*(I + a + b*x))/((I + a)*c - b*d)])/c^2

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

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

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 2513

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*(RFx_.), x_Symbol] :> Dist[
p*r, Int[RFx*Log[a + b*x], x], x] + (Dist[q*r, Int[RFx*Log[c + d*x], x], x] - Dist[p*r*Log[a + b*x] + q*r*Log[
c + d*x] - Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r], Int[RFx, x], x]) /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] &&
RationalFunctionQ[RFx, x] && NeQ[b*c - a*d, 0] &&  !MatchQ[RFx, (u_.)*(a + b*x)^(m_.)*(c + d*x)^(n_.) /; Integ
ersQ[m, n]]

Rule 5052

Int[ArcCot[(a_) + (b_.)*(x_)]/((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Dist[I/2, Int[Log[(-I + a + b*x)/(a + b*
x)]/(c + d*x^n), x], x] - Dist[I/2, Int[Log[(I + a + b*x)/(a + b*x)]/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d}
, x] && RationalQ[n]

Rubi steps

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

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Mathematica [A]  time = 11.12, size = 602, normalized size = 1.78 \[ -\frac {\left ((a+b x)^2+1\right ) \left (a b c d \sqrt {\frac {\left (a^2+1\right ) c^2-2 a b c d+b^2 d^2}{(a c-b d)^2}} \cot ^{-1}(a+b x)^2 e^{-i \tan ^{-1}\left (\frac {c}{a c-b d}\right )}-b^2 d^2 \sqrt {\frac {\left (a^2+1\right ) c^2-2 a b c d+b^2 d^2}{(a c-b d)^2}} \cot ^{-1}(a+b x)^2 e^{-i \tan ^{-1}\left (\frac {c}{a c-b d}\right )}+b^2 d^2 \cot ^{-1}(a+b x)^2+2 c^2 \log \left (\frac {1}{(a+b x) \sqrt {\frac {1}{(a+b x)^2}+1}}\right )-2 c^2 (a+b x) \cot ^{-1}(a+b x)-i b c d \text {Li}_2\left (\exp \left (2 i \left (\cot ^{-1}(a+b x)-\tan ^{-1}\left (\frac {c}{a c-b d}\right )\right )\right )\right )+2 b c d \cot ^{-1}(a+b x) \log \left (1-\exp \left (2 i \left (\cot ^{-1}(a+b x)-\tan ^{-1}\left (\frac {c}{a c-b d}\right )\right )\right )\right )-2 b c d \tan ^{-1}\left (\frac {c}{a c-b d}\right ) \log \left (1-\exp \left (2 i \left (\cot ^{-1}(a+b x)-\tan ^{-1}\left (\frac {c}{a c-b d}\right )\right )\right )\right )+i b c d \text {Li}_2\left (e^{2 i \cot ^{-1}(a+b x)}\right )-\pi b c d \log \left (\frac {1}{\sqrt {\frac {1}{(a+b x)^2}+1}}\right )-a b c d \cot ^{-1}(a+b x)^2+i b c d \cot ^{-1}(a+b x)^2+i \pi b c d \cot ^{-1}(a+b x)+\pi b c d \log \left (1+e^{-2 i \cot ^{-1}(a+b x)}\right )-2 b c d \cot ^{-1}(a+b x) \log \left (1-e^{2 i \cot ^{-1}(a+b x)}\right )+2 i b c d \cot ^{-1}(a+b x) \tan ^{-1}\left (\frac {c}{a c-b d}\right )+2 b c d \tan ^{-1}\left (\frac {c}{a c-b d}\right ) \log \left (\sin \left (\cot ^{-1}(a+b x)-\tan ^{-1}\left (\frac {c}{a c-b d}\right )\right )\right )\right )}{2 b c^3 (a+b x)^2 \sqrt {\frac {a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}} \sqrt {\frac {1}{(a+b x)^2}+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCot[a + b*x]/(c + d/x),x]

[Out]

-1/2*((1 + (a + b*x)^2)*(I*b*c*d*Pi*ArcCot[a + b*x] - 2*c^2*(a + b*x)*ArcCot[a + b*x] + I*b*c*d*ArcCot[a + b*x
]^2 - a*b*c*d*ArcCot[a + b*x]^2 + b^2*d^2*ArcCot[a + b*x]^2 + (a*b*c*d*Sqrt[((1 + a^2)*c^2 - 2*a*b*c*d + b^2*d
^2)/(a*c - b*d)^2]*ArcCot[a + b*x]^2)/E^(I*ArcTan[c/(a*c - b*d)]) - (b^2*d^2*Sqrt[((1 + a^2)*c^2 - 2*a*b*c*d +
b^2*d^2)/(a*c - b*d)^2]*ArcCot[a + b*x]^2)/E^(I*ArcTan[c/(a*c - b*d)]) + (2*I)*b*c*d*ArcCot[a + b*x]*ArcTan[c
/(a*c - b*d)] + b*c*d*Pi*Log[1 + E^((-2*I)*ArcCot[a + b*x])] - 2*b*c*d*ArcCot[a + b*x]*Log[1 - E^((2*I)*ArcCot
[a + b*x])] + 2*b*c*d*ArcCot[a + b*x]*Log[1 - E^((2*I)*(ArcCot[a + b*x] - ArcTan[c/(a*c - b*d)]))] - 2*b*c*d*A
rcTan[c/(a*c - b*d)]*Log[1 - E^((2*I)*(ArcCot[a + b*x] - ArcTan[c/(a*c - b*d)]))] - b*c*d*Pi*Log[1/Sqrt[1 + (a
+ b*x)^(-2)]] + 2*c^2*Log[1/((a + b*x)*Sqrt[1 + (a + b*x)^(-2)])] + 2*b*c*d*ArcTan[c/(a*c - b*d)]*Log[Sin[Arc
Cot[a + b*x] - ArcTan[c/(a*c - b*d)]]] + I*b*c*d*PolyLog[2, E^((2*I)*ArcCot[a + b*x])] - I*b*c*d*PolyLog[2, E^
((2*I)*(ArcCot[a + b*x] - ArcTan[c/(a*c - b*d)]))]))/(b*c^3*(a + b*x)^2*Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a
+ b*x)^2]*Sqrt[1 + (a + b*x)^(-2)])

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fricas [F]  time = 1.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x \operatorname {arccot}\left (b x + a\right )}{c x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/(c+d/x),x, algorithm="fricas")

[Out]

integral(x*arccot(b*x + a)/(c*x + d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccot}\left (b x + a\right )}{c + \frac {d}{x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/(c+d/x),x, algorithm="giac")

[Out]

integrate(arccot(b*x + a)/(c + d/x), x)

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maple [A]  time = 0.07, size = 317, normalized size = 0.94 \[ \frac {\mathrm {arccot}\left (b x +a \right ) x}{c}+\frac {\mathrm {arccot}\left (b x +a \right ) a}{b c}-\frac {\mathrm {arccot}\left (b x +a \right ) d \ln \left (c \left (b x +a \right )-a c +b d \right )}{c^{2}}+\frac {\ln \left (a^{2} c^{2}-2 a b c d +b^{2} d^{2}+2 a c \left (c \left (b x +a \right )-a c +b d \right )-2 \left (c \left (b x +a \right )-a c +b d \right ) b d +\left (c \left (b x +a \right )-a c +b d \right )^{2}+c^{2}\right )}{2 b c}+\frac {i d \ln \left (c \left (b x +a \right )-a c +b d \right ) \ln \left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )}{2 c^{2}}-\frac {i d \ln \left (c \left (b x +a \right )-a c +b d \right ) \ln \left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )}{2 c^{2}}+\frac {i d \dilog \left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )}{2 c^{2}}-\frac {i d \dilog \left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )}{2 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(b*x+a)/(c+1/x*d),x)

[Out]

arccot(b*x+a)/c*x+1/b*arccot(b*x+a)/c*a-arccot(b*x+a)*d/c^2*ln(c*(b*x+a)-a*c+b*d)+1/2/b/c*ln(a^2*c^2-2*a*b*c*d
+b^2*d^2+2*a*c*(c*(b*x+a)-a*c+b*d)-2*(c*(b*x+a)-a*c+b*d)*b*d+(c*(b*x+a)-a*c+b*d)^2+c^2)+1/2*I/c^2*d*ln(c*(b*x+
a)-a*c+b*d)*ln((I*c-c*(b*x+a))/(I*c-a*c+b*d))-1/2*I/c^2*d*ln(c*(b*x+a)-a*c+b*d)*ln((I*c+c*(b*x+a))/(I*c+a*c-b*
d))+1/2*I/c^2*d*dilog((I*c-c*(b*x+a))/(I*c-a*c+b*d))-1/2*I/c^2*d*dilog((I*c+c*(b*x+a))/(I*c+a*c-b*d))

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maxima [A]  time = 0.55, size = 280, normalized size = 0.83 \[ \frac {2 \, b c x \arctan \left (1, b x + a\right ) - b d \arctan \left (1, b x + a\right ) \log \left (-\frac {b^{2} c^{2} x^{2} + 2 \, b^{2} c d x + b^{2} d^{2}}{2 \, a b c d - b^{2} d^{2} - {\left (a^{2} + 1\right )} c^{2}}\right ) - 2 \, a c \arctan \left (b x + a\right ) + i \, b d {\rm Li}_2\left (\frac {b c x + {\left (a + i\right )} c}{{\left (a + i\right )} c - b d}\right ) - i \, b d {\rm Li}_2\left (\frac {b c x + {\left (a - i\right )} c}{{\left (a - i\right )} c - b d}\right ) - {\left (b d \arctan \left (-\frac {b c^{2} x + b c d}{2 \, a b c d - b^{2} d^{2} - {\left (a^{2} + 1\right )} c^{2}}, \frac {a b c d - b^{2} d^{2} + {\left (a b c^{2} - b^{2} c d\right )} x}{2 \, a b c d - b^{2} d^{2} - {\left (a^{2} + 1\right )} c^{2}}\right ) - c\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/(c+d/x),x, algorithm="maxima")

[Out]

1/2*(2*b*c*x*arctan2(1, b*x + a) - b*d*arctan2(1, b*x + a)*log(-(b^2*c^2*x^2 + 2*b^2*c*d*x + b^2*d^2)/(2*a*b*c
*d - b^2*d^2 - (a^2 + 1)*c^2)) - 2*a*c*arctan(b*x + a) + I*b*d*dilog((b*c*x + (a + I)*c)/((a + I)*c - b*d)) -
I*b*d*dilog((b*c*x + (a - I)*c)/((a - I)*c - b*d)) - (b*d*arctan2(-(b*c^2*x + b*c*d)/(2*a*b*c*d - b^2*d^2 - (a
^2 + 1)*c^2), (a*b*c*d - b^2*d^2 + (a*b*c^2 - b^2*c*d)*x)/(2*a*b*c*d - b^2*d^2 - (a^2 + 1)*c^2)) - c)*log(b^2*
x^2 + 2*a*b*x + a^2 + 1))/(b*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a + b*x)/(c + d/x),x)

[Out]

int(acot(a + b*x)/(c + d/x), 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(acot(b*x+a)/(c+d/x),x)

[Out]

Timed out

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