3.1.55 \(\int \frac {\text {ArcTan}(a+b x)}{c+\frac {d}{x}} \, dx\) [55]

Optimal. Leaf size=244 \[ -\frac {(1+i a+i b x) \log (1+i a+i b x)}{2 b c}-\frac {(1-i a-i b x) \log (-i (i+a+b x))}{2 b c}-\frac {i d \log (1-i a-i b x) \log \left (-\frac {b (d+c x)}{(i+a) c-b d}\right )}{2 c^2}+\frac {i d \log (1+i a+i b x) \log \left (\frac {b (d+c x)}{(i-a) c+b d}\right )}{2 c^2}+\frac {i d \text {PolyLog}\left (2,\frac {c (i-a-b x)}{i c-a c+b d}\right )}{2 c^2}-\frac {i d \text {PolyLog}\left (2,\frac {c (i+a+b x)}{(i+a) c-b d}\right )}{2 c^2} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5159, 2456, 2436, 2332, 2441, 2440, 2438} \begin {gather*} \frac {i d \text {Li}_2\left (\frac {c (-a-b x+i)}{-a c+i c+b d}\right )}{2 c^2}-\frac {i d \text {Li}_2\left (\frac {c (a+b x+i)}{(a+i) c-b d}\right )}{2 c^2}+\frac {i d \log (i a+i b x+1) \log \left (\frac {b (c x+d)}{b d+(-a+i) c}\right )}{2 c^2}-\frac {i d \log (-i a-i b x+1) \log \left (-\frac {b (c x+d)}{-b d+(a+i) c}\right )}{2 c^2}-\frac {(i a+i b x+1) \log (i a+i b x+1)}{2 b c}-\frac {(-i a-i b x+1) \log (-i (a+b x+i))}{2 b c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/(c + d/x),x]

[Out]

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

Rule 2332

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

Rule 2436

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 2438

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 2440

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 2441

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 2456

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 5159

Int[ArcTan[(a_) + (b_.)*(x_)]/((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Dist[I/2, Int[Log[1 - I*a - I*b*x]/(c +
d*x^n), x], x] - Dist[I/2, Int[Log[1 + I*a + I*b*x]/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ
[n]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a+b x)}{c+\frac {d}{x}} \, dx &=\frac {1}{2} i \int \frac {\log (1-i a-i b x)}{c+\frac {d}{x}} \, dx-\frac {1}{2} i \int \frac {\log (1+i a+i b x)}{c+\frac {d}{x}} \, dx\\ &=\frac {1}{2} i \int \left (\frac {\log (1-i a-i b x)}{c}-\frac {d \log (1-i a-i b x)}{c (d+c x)}\right ) \, dx-\frac {1}{2} i \int \left (\frac {\log (1+i a+i b x)}{c}-\frac {d \log (1+i a+i b x)}{c (d+c x)}\right ) \, dx\\ &=\frac {i \int \log (1-i a-i b x) \, dx}{2 c}-\frac {i \int \log (1+i a+i b x) \, dx}{2 c}-\frac {(i d) \int \frac {\log (1-i a-i b x)}{d+c x} \, dx}{2 c}+\frac {(i d) \int \frac {\log (1+i a+i b x)}{d+c x} \, dx}{2 c}\\ &=-\frac {i d \log (1-i a-i b x) \log \left (-\frac {b (d+c x)}{(i+a) c-b d}\right )}{2 c^2}+\frac {i d \log (1+i a+i b x) \log \left (\frac {b (d+c x)}{(i-a) c+b d}\right )}{2 c^2}-\frac {\text {Subst}(\int \log (x) \, dx,x,1-i a-i b x)}{2 b c}-\frac {\text {Subst}(\int \log (x) \, dx,x,1+i a+i b x)}{2 b c}+\frac {(b d) \int \frac {\log \left (-\frac {i b (d+c x)}{-(1-i a) c-i b d}\right )}{1-i a-i b x} \, dx}{2 c^2}+\frac {(b d) \int \frac {\log \left (\frac {i b (d+c x)}{-(1+i a) c+i b d}\right )}{1+i a+i b x} \, dx}{2 c^2}\\ &=-\frac {(1+i a+i b x) \log (1+i a+i b x)}{2 b c}-\frac {(1-i a-i b x) \log (-i (i+a+b x))}{2 b c}-\frac {i d \log (1-i a-i b x) \log \left (-\frac {b (d+c x)}{(i+a) c-b d}\right )}{2 c^2}+\frac {i d \log (1+i a+i b x) \log \left (\frac {b (d+c x)}{(i-a) c+b d}\right )}{2 c^2}+\frac {(i d) \text {Subst}\left (\int \frac {\log \left (1+\frac {c x}{-(1-i a) c-i b d}\right )}{x} \, dx,x,1-i a-i b x\right )}{2 c^2}-\frac {(i d) \text {Subst}\left (\int \frac {\log \left (1+\frac {c x}{-(1+i a) c+i b d}\right )}{x} \, dx,x,1+i a+i b x\right )}{2 c^2}\\ &=-\frac {(1+i a+i b x) \log (1+i a+i b x)}{2 b c}-\frac {(1-i a-i b x) \log (-i (i+a+b x))}{2 b c}-\frac {i d \log (1-i a-i b x) \log \left (-\frac {b (d+c x)}{(i+a) c-b d}\right )}{2 c^2}+\frac {i d \log (1+i a+i 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 = 7.67, size = 405, normalized size = 1.66 \begin {gather*} \frac {2 c (a+b x) \text {ArcTan}(a+b x)+\frac {b c d \text {ArcTan}(a+b x)^2}{a c-b d}+2 c \log \left (\frac {1}{\sqrt {1+(a+b x)^2}}\right )-i b d \left (\text {ArcTan}(a+b x) \left (\text {ArcTan}(a+b x)+2 i \log \left (1+e^{2 i \text {ArcTan}(a+b x)}\right )\right )+\text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(a+b x)}\right )\right )+\frac {b d \left (c \sqrt {1+a^2-\frac {2 a b d}{c}+\frac {b^2 d^2}{c^2}} e^{-i \text {ArcTan}\left (a-\frac {b d}{c}\right )} \text {ArcTan}(a+b x)^2+(a c-b d) \text {ArcTan}(a+b x) \left (i \pi +2 i \text {ArcTan}\left (a-\frac {b d}{c}\right )+2 \log \left (1-e^{2 i \left (-\text {ArcTan}\left (a-\frac {b d}{c}\right )+\text {ArcTan}(a+b x)\right )}\right )\right )+(a c-b d) \left (\pi \left (\log \left (1+e^{-2 i \text {ArcTan}(a+b x)}\right )-\log \left (\frac {1}{\sqrt {1+(a+b x)^2}}\right )\right )-2 \text {ArcTan}\left (a-\frac {b d}{c}\right ) \left (\log \left (1-e^{2 i \left (-\text {ArcTan}\left (a-\frac {b d}{c}\right )+\text {ArcTan}(a+b x)\right )}\right )-\log \left (-\sin \left (\text {ArcTan}\left (a-\frac {b d}{c}\right )-\text {ArcTan}(a+b x)\right )\right )\right )\right )-i (a c-b d) \text {PolyLog}\left (2,e^{2 i \left (-\text {ArcTan}\left (a-\frac {b d}{c}\right )+\text {ArcTan}(a+b x)\right )}\right )\right )}{-a c+b d}}{2 b c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*(a + b*x)*ArcTan[a + b*x] + (b*c*d*ArcTan[a + b*x]^2)/(a*c - b*d) + 2*c*Log[1/Sqrt[1 + (a + b*x)^2]] - I*
b*d*(ArcTan[a + b*x]*(ArcTan[a + b*x] + (2*I)*Log[1 + E^((2*I)*ArcTan[a + b*x])]) + PolyLog[2, -E^((2*I)*ArcTa
n[a + b*x])]) + (b*d*((c*Sqrt[1 + a^2 - (2*a*b*d)/c + (b^2*d^2)/c^2]*ArcTan[a + b*x]^2)/E^(I*ArcTan[a - (b*d)/
c]) + (a*c - b*d)*ArcTan[a + b*x]*(I*Pi + (2*I)*ArcTan[a - (b*d)/c] + 2*Log[1 - E^((2*I)*(-ArcTan[a - (b*d)/c]
+ ArcTan[a + b*x]))]) + (a*c - b*d)*(Pi*(Log[1 + E^((-2*I)*ArcTan[a + b*x])] - Log[1/Sqrt[1 + (a + b*x)^2]])
- 2*ArcTan[a - (b*d)/c]*(Log[1 - E^((2*I)*(-ArcTan[a - (b*d)/c] + ArcTan[a + b*x]))] - Log[-Sin[ArcTan[a - (b*
d)/c] - ArcTan[a + b*x]]])) - I*(a*c - b*d)*PolyLog[2, E^((2*I)*(-ArcTan[a - (b*d)/c] + ArcTan[a + b*x]))]))/(
-(a*c) + b*d))/(2*b*c^2)

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Maple [A]
time = 0.06, size = 321, normalized size = 1.32

method	result	size
risch	\(\frac {i \ln \left (-i b x -i a +1\right ) x}{2 c}+\frac {a \arctan \left (b x +a \right )}{b c}-\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b c}+\frac {1}{b c}-\frac {i d \dilog \left (\frac {i a c -i b d +\left (-i b x -i a +1\right ) c -c}{i a c -i b d -c}\right )}{2 c^{2}}-\frac {i d \ln \left (-i b x -i a +1\right ) \ln \left (\frac {i a c -i b d +\left (-i b x -i a +1\right ) c -c}{i a c -i b d -c}\right )}{2 c^{2}}-\frac {i \ln \left (i b x +i a +1\right ) x}{2 c}+\frac {i d \dilog \left (\frac {-i a c +i b d +\left (i b x +i a +1\right ) c -c}{-i a c +i b d -c}\right )}{2 c^{2}}+\frac {i d \ln \left (i b x +i a +1\right ) \ln \left (\frac {-i a c +i b d +\left (i b x +i a +1\right ) c -c}{-i a c +i b d -c}\right )}{2 c^{2}}\)	\(319\)
derivativedivides	\(\frac {\frac {\arctan \left (b x +a \right ) \left (b x +a \right )}{c}-\frac {\arctan \left (b x +a \right ) d b \ln \left (a c -b d -c \left (b x +a \right )\right )}{c^{2}}+\frac {-\frac {\ln \left (a^{2} c^{2}-2 a b c d +b^{2} d^{2}-2 a c \left (a c -b d -c \left (b x +a \right )\right )+2 b d \left (a c -b d -c \left (b x +a \right )\right )+c^{2}+\left (a c -b d -c \left (b x +a \right )\right )^{2}\right )}{2}+\frac {i b d \ln \left (a c -b d -c \left (b x +a \right )\right ) \ln \left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )}{2 c}-\frac {i b d \ln \left (a c -b d -c \left (b x +a \right )\right ) \ln \left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )}{2 c}+\frac {i b d \dilog \left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )}{2 c}-\frac {i b d \dilog \left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )}{2 c}}{c}}{b}\)	\(321\)
default	\(\frac {\frac {\arctan \left (b x +a \right ) \left (b x +a \right )}{c}-\frac {\arctan \left (b x +a \right ) d b \ln \left (a c -b d -c \left (b x +a \right )\right )}{c^{2}}+\frac {-\frac {\ln \left (a^{2} c^{2}-2 a b c d +b^{2} d^{2}-2 a c \left (a c -b d -c \left (b x +a \right )\right )+2 b d \left (a c -b d -c \left (b x +a \right )\right )+c^{2}+\left (a c -b d -c \left (b x +a \right )\right )^{2}\right )}{2}+\frac {i b d \ln \left (a c -b d -c \left (b x +a \right )\right ) \ln \left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )}{2 c}-\frac {i b d \ln \left (a c -b d -c \left (b x +a \right )\right ) \ln \left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )}{2 c}+\frac {i b d \dilog \left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )}{2 c}-\frac {i b d \dilog \left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )}{2 c}}{c}}{b}\)	\(321\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(b*x+a)/(c+d/x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.57, size = 284, normalized size = 1.16 \begin {gather*} -\frac {b d \arctan \left (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 ) + i \, b d {\rm Li}_2\left (-\frac {i \, b c x + {\left (i \, a - 1\right )} c}{{\left (-i \, a + 1\right )} c + i \, b d}\right ) - i \, b d {\rm Li}_2\left (-\frac {i \, b c x + {\left (i \, a + 1\right )} c}{{\left (-i \, a - 1\right )} c + i \, b d}\right ) - 2 \, {\left (b c x + a c\right )} \arctan \left (b x + a\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*d*arctan(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)) +
I*b*d*dilog(-(I*b*c*x + (I*a - 1)*c)/((-I*a + 1)*c + I*b*d)) - I*b*d*dilog(-(I*b*c*x + (I*a + 1)*c)/((-I*a - 1
)*c + I*b*d)) - 2*(b*c*x + a*c)*arctan(b*x + a) - (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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\mathrm {atan}\left (a+b\,x\right )}{c+\frac {d}{x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a + b*x)/(c + d/x),x)

[Out]

int(atan(a + b*x)/(c + d/x), x)

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