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

Optimal. Leaf size=41 \[ \frac{i \text{PolyLog}(2,-i (a+b x))}{2 d}-\frac{i \text{PolyLog}(2,i (a+b x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0452878, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {5043, 12, 4848, 2391} \[ \frac{i \text{PolyLog}(2,-i (a+b x))}{2 d}-\frac{i \text{PolyLog}(2,i (a+b x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5043

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4848

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

Rubi steps

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

Mathematica [A]  time = 0.0074928, size = 34, normalized size = 0.83 \[ \frac{i (\text{PolyLog}(2,-i (a+b x))-\text{PolyLog}(2,i (a+b x)))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.046, size = 98, normalized size = 2.4 \begin{align*}{\frac{\ln \left ( bx+a \right ) \arctan \left ( bx+a \right ) }{d}}+{\frac{{\frac{i}{2}}\ln \left ( bx+a \right ) \ln \left ( 1+i \left ( bx+a \right ) \right ) }{d}}-{\frac{{\frac{i}{2}}\ln \left ( bx+a \right ) \ln \left ( 1-i \left ( bx+a \right ) \right ) }{d}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+i \left ( bx+a \right ) \right ) }{d}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( 1-i \left ( bx+a \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.68232, size = 166, normalized size = 4.05 \begin{align*} \frac{\arctan \left (b x + a\right ) \log \left (d x + \frac{a d}{b}\right )}{d} - \frac{\arctan \left (\frac{b^{2} x + a b}{b}\right ) \log \left (d x + \frac{a d}{b}\right )}{d} - \frac{\arctan \left (b x + a, 0\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, \arctan \left (b x + a\right ) \log \left ({\left | b x + a \right |}\right ) + i \,{\rm Li}_2\left (i \, b x + i \, a + 1\right ) - i \,{\rm Li}_2\left (-i \, b x - i \, a + 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(b*x + a)*log(d*x + a*d/b)/d - arctan((b^2*x + a*b)/b)*log(d*x + a*d/b)/d - 1/2*(arctan2(b*x + a, 0)*log
(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*arctan(b*x + a)*log(abs(b*x + a)) + I*dilog(I*b*x + I*a + 1) - I*dilog(-I*b*
x - I*a + 1))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{b \int \frac{\operatorname{atan}{\left (a + b x \right )}}{a + b x}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*Integral(atan(a + b*x)/(a + b*x), x)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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