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

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

[Out]

-((1 + I*a + I*b*x)*Log[1 + I*a + I*b*x])/(2*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

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Rubi [A]  time = 0.239131, antiderivative size = 244, normalized size of antiderivative = 1., 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 = {5051, 2409, 2389, 2295, 2394, 2393, 2391} \[ \frac{i d \text{PolyLog}\left (2,\frac{c (-a-b x+i)}{-a c+b d+i c}\right )}{2 c^2}-\frac{i d \text{PolyLog}\left (2,\frac{c (a+b x+i)}{-b d+(a+i) c}\right )}{2 c^2}+\frac{i d \log (i a+i b x+1) \log \left (\frac{b (c x+d)}{b d+(-a+i) c}\right )}{2 c^2}-\frac{i d \log (-i a-i b x+1) \log \left (-\frac{b (c x+d)}{-b d+(a+i) c}\right )}{2 c^2}-\frac{(i a+i b x+1) \log (i a+i b x+1)}{2 b c}-\frac{(-i a-i b x+1) \log (-i (a+b x+i))}{2 b c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/(c + d/x),x]

[Out]

-((1 + I*a + I*b*x)*Log[1 + I*a + I*b*x])/(2*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 5051

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]

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 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 2295

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

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 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 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]

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{\operatorname{Subst}(\int \log (x) \, dx,x,1-i a-i b x)}{2 b c}-\frac{\operatorname{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) \operatorname{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) \operatorname{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*}

Mathematica [B]  time = 11.0746, size = 771, normalized size = 3.16 \[ \frac{i b d (b d-a c) \text{PolyLog}\left (2,\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )+i b d (a c-b d) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(a+b x)}\right )+b c d \sqrt{a^2-\frac{2 a b d}{c}+\frac{b^2 d^2}{c^2}+1} \tan ^{-1}(a+b x)^2 e^{-i \tan ^{-1}\left (a-\frac{b d}{c}\right )}-2 a^2 c^2 \tan ^{-1}(a+b x)+2 b^2 d^2 \tan ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (1-\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )-2 b^2 d^2 \tan ^{-1}(a+b x) \log \left (1-\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )-2 i b^2 d^2 \tan ^{-1}(a+b x) \tan ^{-1}\left (a-\frac{b d}{c}\right )-2 b^2 d^2 \tan ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (-\sin \left (\tan ^{-1}\left (a-\frac{b d}{c}\right )-\tan ^{-1}(a+b x)\right )\right )+2 b^2 c d x \tan ^{-1}(a+b x)+\pi b^2 d^2 \log \left (\frac{1}{\sqrt{(a+b x)^2+1}}\right )-i b^2 d^2 \tan ^{-1}(a+b x)^2-i \pi b^2 d^2 \tan ^{-1}(a+b x)-\pi b^2 d^2 \log \left (1+e^{-2 i \tan ^{-1}(a+b x)}\right )+2 b^2 d^2 \tan ^{-1}(a+b x) \log \left (1+e^{2 i \tan ^{-1}(a+b x)}\right )-2 a c^2 \log \left (\frac{1}{\sqrt{(a+b x)^2+1}}\right )-2 a b c^2 x \tan ^{-1}(a+b x)-2 a b c d \tan ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (1-\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )+2 a b c d \tan ^{-1}(a+b x) \log \left (1-\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )+2 b c d \log \left (\frac{1}{\sqrt{(a+b x)^2+1}}\right )-\pi a b c d \log \left (\frac{1}{\sqrt{(a+b x)^2+1}}\right )+i a b c d \tan ^{-1}(a+b x)^2-b c d \tan ^{-1}(a+b x)^2+2 a b c d \tan ^{-1}(a+b x)+2 i a b c d \tan ^{-1}(a+b x) \tan ^{-1}\left (a-\frac{b d}{c}\right )+i \pi a b c d \tan ^{-1}(a+b x)+\pi a b c d \log \left (1+e^{-2 i \tan ^{-1}(a+b x)}\right )-2 a b c d \tan ^{-1}(a+b x) \log \left (1+e^{2 i \tan ^{-1}(a+b x)}\right )+2 a b c d \tan ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (-\sin \left (\tan ^{-1}\left (a-\frac{b d}{c}\right )-\tan ^{-1}(a+b x)\right )\right )}{b c^2 (2 b d-2 a c)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.06, size = 317, normalized size = 1.3 \begin{align*}{\frac{x\arctan \left ( bx+a \right ) }{c}}+{\frac{\arctan \left ( bx+a \right ) a}{bc}}-{\frac{\arctan \left ( bx+a \right ) d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{{c}^{2}}}-{\frac{\ln \left ({a}^{2}{c}^{2}-2\,abcd+{b}^{2}{d}^{2}+2\, \left ( c \left ( bx+a \right ) -ac+bd \right ) ac-2\, \left ( c \left ( bx+a \right ) -ac+bd \right ) bd+ \left ( c \left ( bx+a \right ) -ac+bd \right ) ^{2}+{c}^{2} \right ) }{2\,bc}}-{\frac{{\frac{i}{2}}d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{{c}^{2}}\ln \left ({\frac{ic-c \left ( bx+a \right ) }{ic-ac+bd}} \right ) }+{\frac{{\frac{i}{2}}d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{{c}^{2}}\ln \left ({\frac{ic+c \left ( bx+a \right ) }{ic+ac-bd}} \right ) }-{\frac{{\frac{i}{2}}d}{{c}^{2}}{\it dilog} \left ({\frac{ic-c \left ( bx+a \right ) }{ic-ac+bd}} \right ) }+{\frac{{\frac{i}{2}}d}{{c}^{2}}{\it dilog} \left ({\frac{ic+c \left ( bx+a \right ) }{ic+ac-bd}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(b*x+a)/c*x+1/b*arctan(b*x+a)/c*a-arctan(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*(c*(b*x+a)-a*c+b*d)*a*c-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 = 2.01758, size = 383, normalized size = 1.57 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \arctan \left (b x + a\right )}{c x + d}, x\right ) \end{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (b x + a\right )}{c + \frac{d}{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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