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

Optimal. Leaf size=152 \[ \frac {i \text {Li}_2\left (1-\frac {2 b (c+d x)}{(b c-a d+i d) (1-i (a+b x))}\right )}{2 d}+\frac {\cot ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac {i \text {Li}_2\left (1-\frac {2}{1-i (a+b x)}\right )}{2 d}-\frac {\log \left (\frac {2}{1-i (a+b x)}\right ) \cot ^{-1}(a+b x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5048, 4857, 2402, 2315, 2447} \[ \frac {i \text {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{2 d}-\frac {i \text {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )}{2 d}+\frac {\cot ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac {\log \left (\frac {2}{1-i (a+b x)}\right ) \cot ^{-1}(a+b x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[a + b*x]/(c + d*x),x]

[Out]

-((ArcCot[a + b*x]*Log[2/(1 - I*(a + b*x))])/d) + (ArcCot[a + b*x]*Log[(2*b*(c + d*x))/((b*c + I*d - a*d)*(1 -
I*(a + b*x)))])/d - ((I/2)*PolyLog[2, 1 - 2/(1 - I*(a + b*x))])/d + ((I/2)*PolyLog[2, 1 - (2*b*(c + d*x))/((b
*c + I*d - a*d)*(1 - I*(a + b*x)))])/d

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 4857

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCot[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*ArcCot[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 5048

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCot[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x]
&& IGtQ[p, 0]

Rubi steps

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

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Mathematica [B]  time = 0.05, size = 345, normalized size = 2.27 \[ \frac {i \text {Li}_2\left (\frac {b \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{b c-a d-i d}\right )}{2 d}-\frac {i \text {Li}_2\left (\frac {b \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{b c-a d+i d}\right )}{2 d}-\frac {i \log \left (\frac {d (a+b x-i)}{b \left (-\frac {b c-a d}{b}-\frac {i d}{b}\right )}\right ) \log \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{2 d}+\frac {i \log \left (\frac {a+b x-i}{a+b x}\right ) \log \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{2 d}+\frac {i \log \left (\frac {d (a+b x+i)}{b \left (-\frac {b c-a d}{b}+\frac {i d}{b}\right )}\right ) \log \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{2 d}-\frac {i \log \left (\frac {a+b x+i}{a+b x}\right ) \log \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/2*I)*Log[(d*(-I + a + b*x))/(b*(((-I)*d)/b - (b*c - a*d)/b))]*Log[(b*c - a*d)/b + (d*(a + b*x))/b])/d + (
(I/2)*Log[(-I + a + b*x)/(a + b*x)]*Log[(b*c - a*d)/b + (d*(a + b*x))/b])/d + ((I/2)*Log[(d*(I + a + b*x))/(b*
((I*d)/b - (b*c - a*d)/b))]*Log[(b*c - a*d)/b + (d*(a + b*x))/b])/d - ((I/2)*Log[(I + a + b*x)/(a + b*x)]*Log[
(b*c - a*d)/b + (d*(a + b*x))/b])/d + ((I/2)*PolyLog[2, (b*((b*c - a*d)/b + (d*(a + b*x))/b))/(b*c - I*d - a*d
)])/d - ((I/2)*PolyLog[2, (b*((b*c - a*d)/b + (d*(a + b*x))/b))/(b*c + I*d - a*d)])/d

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fricas [F]  time = 2.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arccot}\left (b x + a\right )}{d x + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 198, normalized size = 1.30 \[ \frac {\ln \left (d \left (b x +a \right )-a d +b c \right ) \mathrm {arccot}\left (b x +a \right )}{d}-\frac {i \ln \left (d \left (b x +a \right )-a d +b c \right ) \ln \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )}{2 d}+\frac {i \ln \left (d \left (b x +a \right )-a d +b c \right ) \ln \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )}{2 d}-\frac {i \dilog \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )}{2 d}+\frac {i \dilog \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.55, size = 283, normalized size = 1.86 \[ \frac {\operatorname {arccot}\left (b x + a\right ) \log \left (d x + c\right )}{d} + \frac {\arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (d x + c\right )}{d} + \frac {\arctan \left (\frac {b d^{2} x + b c d}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}, \frac {b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - \arctan \left (b x + a\right ) \log \left (\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) + i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a + 1\right )} d}{-i \, b c + {\left (i \, a + 1\right )} d}\right ) - i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a - 1\right )} d}{-i \, b c + {\left (i \, a - 1\right )} d}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acot}\left (a+b\,x\right )}{c+d\,x} \,d x \]

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)/(d*x+c),x)

[Out]

Timed out

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