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

Optimal. Leaf size=43 \[ \frac{1}{3} x^3 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )-\frac{1}{6} b c^3 \log \left (c^2+x^2\right )+\frac{1}{6} b c x^2 \]

[Out]

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

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Rubi [A]  time = 0.029785, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5033, 263, 266, 43} \[ \frac{1}{3} x^3 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )-\frac{1}{6} b c^3 \log \left (c^2+x^2\right )+\frac{1}{6} b c x^2 \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTan[c/x]),x]

[Out]

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

Rule 5033

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

Rule 263

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

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

Mathematica [A]  time = 0.0073186, size = 48, normalized size = 1.12 \[ \frac{a x^3}{3}-\frac{1}{6} b c^3 \log \left (c^2+x^2\right )+\frac{1}{6} b c x^2+\frac{1}{3} b x^3 \tan ^{-1}\left (\frac{c}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTan[c/x]),x]

[Out]

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

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Maple [A]  time = 0.033, size = 55, normalized size = 1.3 \begin{align*}{\frac{{x}^{3}a}{3}}+{\frac{b{x}^{3}}{3}\arctan \left ({\frac{c}{x}} \right ) }-{\frac{{c}^{3}b}{6}\ln \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) }+{\frac{bc{x}^{2}}{6}}+{\frac{{c}^{3}b}{3}\ln \left ({\frac{c}{x}} \right ) } \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*c^3*b*ln(1+c^2/x^2)+1/6*b*c*x^2+1/3*c^3*b*ln(c/x)

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Maxima [A]  time = 0.988883, size = 58, normalized size = 1.35 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{6} \,{\left (2 \, x^{3} \arctan \left (\frac{c}{x}\right ) -{\left (c^{2} \log \left (c^{2} + x^{2}\right ) - x^{2}\right )} c\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^2*log(c^2 + x^2) - x^2)*c)*b

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Fricas [A]  time = 2.1818, size = 103, normalized size = 2.4 \begin{align*} \frac{1}{3} \, b x^{3} \arctan \left (\frac{c}{x}\right ) - \frac{1}{6} \, b c^{3} \log \left (c^{2} + x^{2}\right ) + \frac{1}{6} \, b c x^{2} + \frac{1}{3} \, a x^{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]

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

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Sympy [A]  time = 0.5446, size = 41, normalized size = 0.95 \begin{align*} \frac{a x^{3}}{3} - \frac{b c^{3} \log{\left (c^{2} + x^{2} \right )}}{6} + \frac{b c x^{2}}{6} + \frac{b x^{3} \operatorname{atan}{\left (\frac{c}{x} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/3 - b*c**3*log(c**2 + x**2)/6 + b*c*x**2/6 + b*x**3*atan(c/x)/3

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Giac [A]  time = 1.12529, size = 54, normalized size = 1.26 \begin{align*} \frac{1}{3} \, b x^{3} \arctan \left (\frac{c}{x}\right ) - \frac{1}{6} \, b c^{3} \log \left (c^{2} + x^{2}\right ) + \frac{1}{6} \, b c x^{2} + \frac{1}{3} \, a x^{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]

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