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

Optimal. Leaf size=162 \[ \frac {\left (a+b \cot ^{-1}(c+d x)\right ) \log \left (\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}-\frac {\log \left (\frac {2}{1-i (c+d x)}\right ) \left (a+b \cot ^{-1}(c+d x)\right )}{f}+\frac {i b \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+i f) (1-i (c+d x))}\right )}{2 f}-\frac {i b \text {Li}_2\left (1-\frac {2}{1-i (c+d x)}\right )}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5048, 4857, 2402, 2315, 2447} \[ \frac {i b \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{2 f}-\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1-i (c+d x)}\right )}{2 f}+\frac {\left (a+b \cot ^{-1}(c+d x)\right ) \log \left (\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}-\frac {\log \left (\frac {2}{1-i (c+d x)}\right ) \left (a+b \cot ^{-1}(c+d x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCot[c + d*x])/(e + f*x),x]

[Out]

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

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

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Mathematica [A]  time = 0.11, size = 304, normalized size = 1.88 \[ \frac {a \log (f (c+d x)-c f+d e)}{f}-\frac {i b \text {Li}_2\left (\frac {d e-c f+f (c+d x)}{d e+(i-c) f}\right )}{2 f}+\frac {i b \text {Li}_2\left (\frac {d e-c f+f (c+d x)}{d e-(c+i) f}\right )}{2 f}-\frac {i b \log \left (\frac {f (-c-d x+i)}{d e+(-c+i) f}\right ) \log (f (c+d x)-c f+d e)}{2 f}+\frac {i b \log \left (-\frac {-c-d x+i}{c+d x}\right ) \log (f (c+d x)-c f+d e)}{2 f}+\frac {i b \log \left (-\frac {f (c+d x+i)}{d e-(c+i) f}\right ) \log (f (c+d x)-c f+d e)}{2 f}-\frac {i b \log \left (\frac {c+d x+i}{c+d x}\right ) \log (f (c+d x)-c f+d e)}{2 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*Log[d*e - c*f + f*(c + d*x)])/f - ((I/2)*b*Log[(f*(I - c - d*x))/(d*e + (I - c)*f)]*Log[d*e - c*f + f*(c +
d*x)])/f + ((I/2)*b*Log[-((I - c - d*x)/(c + d*x))]*Log[d*e - c*f + f*(c + d*x)])/f + ((I/2)*b*Log[-((f*(I + c
 + d*x))/(d*e - (I + c)*f))]*Log[d*e - c*f + f*(c + d*x)])/f - ((I/2)*b*Log[(I + c + d*x)/(c + d*x)]*Log[d*e -
 c*f + f*(c + d*x)])/f - ((I/2)*b*PolyLog[2, (d*e - c*f + f*(c + d*x))/(d*e + (I - c)*f)])/f + ((I/2)*b*PolyLo
g[2, (d*e - c*f + f*(c + d*x))/(d*e - (I + c)*f)])/f

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arccot}\left (d x + c\right ) + a}{f x + e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 224, normalized size = 1.38 \[ \frac {a \ln \left (f \left (d x +c \right )-c f +d e \right )}{f}+\frac {b \ln \left (f \left (d x +c \right )-c f +d e \right ) \mathrm {arccot}\left (d x +c \right )}{f}-\frac {i b \ln \left (f \left (d x +c \right )-c f +d e \right ) \ln \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )}{2 f}+\frac {i b \ln \left (f \left (d x +c \right )-c f +d e \right ) \ln \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )}{2 f}-\frac {i b \dilog \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )}{2 f}+\frac {i b \dilog \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )}{2 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*integrate(1/2*arctan2(1, d*x + c)/(f*x + e), x) + a*log(f*x + e)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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