3.4 \(\int x^2 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=45 \[ \frac{1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \log \left (c^2 x^2+1\right )}{6 c^3}-\frac{b x^2}{6 c} \]

[Out]

-(b*x^2)/(6*c) + (x^3*(a + b*ArcTan[c*x]))/3 + (b*Log[1 + c^2*x^2])/(6*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.0318549, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4852, 266, 43} \[ \frac{1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \log \left (c^2 x^2+1\right )}{6 c^3}-\frac{b x^2}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTan[c*x]),x]

[Out]

-(b*x^2)/(6*c) + (x^3*(a + b*ArcTan[c*x]))/3 + (b*Log[1 + c^2*x^2])/(6*c^3)

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{3} (b c) \int \frac{x^3}{1+c^2 x^2} \, dx\\ &=\frac{1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )\\ &=\frac{1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{b x^2}{6 c}+\frac{1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \log \left (1+c^2 x^2\right )}{6 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0082896, size = 50, normalized size = 1.11 \[ \frac{a x^3}{3}+\frac{b \log \left (c^2 x^2+1\right )}{6 c^3}-\frac{b x^2}{6 c}+\frac{1}{3} b x^3 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTan[c*x]),x]

[Out]

-(b*x^2)/(6*c) + (a*x^3)/3 + (b*x^3*ArcTan[c*x])/3 + (b*Log[1 + c^2*x^2])/(6*c^3)

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 43, normalized size = 1. \begin{align*}{\frac{{x}^{3}a}{3}}+{\frac{b{x}^{3}\arctan \left ( cx \right ) }{3}}-{\frac{b{x}^{2}}{6\,c}}+{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{6\,{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctan(c*x)),x)

[Out]

1/3*x^3*a+1/3*b*x^3*arctan(c*x)-1/6*b*x^2/c+1/6*b*ln(c^2*x^2+1)/c^3

________________________________________________________________________________________

Maxima [A]  time = 0.982194, size = 62, normalized size = 1.38 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{6} \,{\left (2 \, x^{3} \arctan \left (c x\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

1/3*a*x^3 + 1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b

________________________________________________________________________________________

Fricas [A]  time = 2.30553, size = 111, normalized size = 2.47 \begin{align*} \frac{2 \, b c^{3} x^{3} \arctan \left (c x\right ) + 2 \, a c^{3} x^{3} - b c^{2} x^{2} + b \log \left (c^{2} x^{2} + 1\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

1/6*(2*b*c^3*x^3*arctan(c*x) + 2*a*c^3*x^3 - b*c^2*x^2 + b*log(c^2*x^2 + 1))/c^3

________________________________________________________________________________________

Sympy [A]  time = 0.765492, size = 49, normalized size = 1.09 \begin{align*} \begin{cases} \frac{a x^{3}}{3} + \frac{b x^{3} \operatorname{atan}{\left (c x \right )}}{3} - \frac{b x^{2}}{6 c} + \frac{b \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6 c^{3}} & \text{for}\: c \neq 0 \\\frac{a x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*atan(c*x)),x)

[Out]

Piecewise((a*x**3/3 + b*x**3*atan(c*x)/3 - b*x**2/(6*c) + b*log(x**2 + c**(-2))/(6*c**3), Ne(c, 0)), (a*x**3/3
, True))

________________________________________________________________________________________

Giac [A]  time = 1.20281, size = 66, normalized size = 1.47 \begin{align*} \frac{2 \, b c^{3} x^{3} \arctan \left (c x\right ) + 2 \, a c^{3} x^{3} - b c^{2} x^{2} + b \log \left (c^{2} x^{2} + 1\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

1/6*(2*b*c^3*x^3*arctan(c*x) + 2*a*c^3*x^3 - b*c^2*x^2 + b*log(c^2*x^2 + 1))/c^3