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

Optimal. Leaf size=29 \[ a x-\frac{b \log \left (c^2 x^2+1\right )}{2 c}+b x \tan ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.0113169, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4846, 260} \[ a x-\frac{b \log \left (c^2 x^2+1\right )}{2 c}+b x \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcTan[c*x],x]

[Out]

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

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

Mathematica [A]  time = 0.0029365, size = 29, normalized size = 1. \[ a x-\frac{b \log \left (c^2 x^2+1\right )}{2 c}+b x \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcTan[c*x],x]

[Out]

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

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Maple [A]  time = 0.003, size = 28, normalized size = 1. \begin{align*} ax+bx\arctan \left ( cx \right ) -{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.985285, size = 42, normalized size = 1.45 \begin{align*} a x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.23199, size = 81, normalized size = 2.79 \begin{align*} \frac{2 \, b c x \arctan \left (c x\right ) + 2 \, a c x - b \log \left (c^{2} x^{2} + 1\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.252711, size = 26, normalized size = 0.9 \begin{align*} a x + b \left (\begin{cases} x \operatorname{atan}{\left (c x \right )} - \frac{\log{\left (c^{2} x^{2} + 1 \right )}}{2 c} & \text{for}\: c \neq 0 \\0 & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.33481, size = 42, normalized size = 1.45 \begin{align*} a x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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