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

Optimal. Leaf size=33 \[ \frac{(a+b x) \tan ^{-1}(a+b x)}{b}-\frac{\log \left ((a+b x)^2+1\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0113888, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5039, 4846, 260} \[ \frac{(a+b x) \tan ^{-1}(a+b x)}{b}-\frac{\log \left ((a+b x)^2+1\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x],x]

[Out]

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

Rule 5039

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcTan[x])^p, x],
 x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rule 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.0138611, size = 39, normalized size = 1.18 \[ -\frac{\log \left (a^2+2 a b x+b^2 x^2+1\right )-2 (a+b x) \tan ^{-1}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.035, size = 36, normalized size = 1.1 \begin{align*} x\arctan \left ( bx+a \right ) +{\frac{\arctan \left ( bx+a \right ) a}{b}}-{\frac{\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(b*x+a)+1/b*arctan(b*x+a)*a-1/2*ln(1+(b*x+a)^2)/b

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Maxima [A]  time = 1.00836, size = 42, normalized size = 1.27 \begin{align*} \frac{2 \,{\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left ({\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b*x + a)*arctan(b*x + a) - log((b*x + a)^2 + 1))/b

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Fricas [A]  time = 1.74759, size = 97, normalized size = 2.94 \begin{align*} \frac{2 \,{\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b*x + a)*arctan(b*x + a) - log(b^2*x^2 + 2*a*b*x + a^2 + 1))/b

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Sympy [A]  time = 0.558806, size = 46, normalized size = 1.39 \begin{align*} \begin{cases} \frac{a \operatorname{atan}{\left (a + b x \right )}}{b} + x \operatorname{atan}{\left (a + b x \right )} - \frac{\log{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b} & \text{for}\: b \neq 0 \\x \operatorname{atan}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*atan(a + b*x)/b + x*atan(a + b*x) - log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*b), Ne(b, 0)), (x*atan
(a), True))

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Giac [A]  time = 1.11763, size = 42, normalized size = 1.27 \begin{align*} \frac{2 \,{\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left ({\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b*x + a)*arctan(b*x + a) - log((b*x + a)^2 + 1))/b