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

Optimal. Leaf size=37 \[ \frac{1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \tan ^{-1}(c x)}{2 c^2}-\frac{b x}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.0156177, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4852, 321, 203} \[ \frac{1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \tan ^{-1}(c x)}{2 c^2}-\frac{b x}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTan[c*x]),x]

[Out]

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

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0022397, size = 42, normalized size = 1.14 \[ \frac{a x^2}{2}+\frac{b \tan ^{-1}(c x)}{2 c^2}+\frac{1}{2} b x^2 \tan ^{-1}(c x)-\frac{b x}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTan[c*x]),x]

[Out]

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

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Maple [A]  time = 0.004, size = 35, normalized size = 1. \begin{align*}{\frac{a{x}^{2}}{2}}+{\frac{b{x}^{2}\arctan \left ( cx \right ) }{2}}-{\frac{bx}{2\,c}}+{\frac{b\arctan \left ( cx \right ) }{2\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.48385, size = 50, normalized size = 1.35 \begin{align*} \frac{1}{2} \, a x^{2} + \frac{1}{2} \,{\left (x^{2} \arctan \left (c x\right ) - c{\left (\frac{x}{c^{2}} - \frac{\arctan \left (c x\right )}{c^{3}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 1/2*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b

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Fricas [A]  time = 2.21952, size = 80, normalized size = 2.16 \begin{align*} \frac{a c^{2} x^{2} - b c x +{\left (b c^{2} x^{2} + b\right )} \arctan \left (c x\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.561452, size = 42, normalized size = 1.14 \begin{align*} \begin{cases} \frac{a x^{2}}{2} + \frac{b x^{2} \operatorname{atan}{\left (c x \right )}}{2} - \frac{b x}{2 c} + \frac{b \operatorname{atan}{\left (c x \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\\frac{a x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19385, size = 61, normalized size = 1.65 \begin{align*} \frac{b c^{2} x^{2} \arctan \left (c x\right ) + a c^{2} x^{2} - \pi b \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - b c x + b \arctan \left (c x\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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