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]

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

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.02, 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]

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

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fricas [A]  time = 0.40, size = 39, normalized size = 1.18 \[ \frac {2 \, {\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} \]

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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giac [A]  time = 0.13, size = 31, normalized size = 0.94 \[ \frac {2 \, {\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left ({\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \]

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

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maple [A]  time = 0.04, size = 36, normalized size = 1.09 \[ x \arctan \left (b x +a \right )+\frac {\arctan \left (b x +a \right ) a}{b}-\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2 b} \]

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 = 0.31, size = 31, normalized size = 0.94 \[ \frac {2 \, {\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left ({\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \]

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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mupad [B]  time = 0.45, size = 42, normalized size = 1.27 \[ x\,\mathrm {atan}\left (a+b\,x\right )-\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )-2\,a\,\mathrm {atan}\left (a+b\,x\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a + b*x),x)

[Out]

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

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sympy [A]  time = 0.39, size = 46, normalized size = 1.39 \[ \begin {cases} \frac {a \operatorname {atan}{\left (a + b x \right )}}{b} + x \operatorname {atan}{\left (a + b x \right )} - \frac {\log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b} & \text {for}\: b \neq 0 \\x \operatorname {atan}{\relax (a )} & \text {otherwise} \end {cases} \]

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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