3.132 \(\int x^3 (a+b \tan ^{-1}(\frac{c}{x})) \, dx\)

Optimal. Leaf size=50 \[ \frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )-\frac{1}{4} b c^3 x+\frac{1}{4} b c^4 \tan ^{-1}\left (\frac{x}{c}\right )+\frac{1}{12} b c x^3 \]

[Out]

-(b*c^3*x)/4 + (b*c*x^3)/12 + (x^4*(a + b*ArcTan[c/x]))/4 + (b*c^4*ArcTan[x/c])/4

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Rubi [A]  time = 0.0309079, antiderivative size = 50, 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, 302, 203} \[ \frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )-\frac{1}{4} b c^3 x+\frac{1}{4} b c^4 \tan ^{-1}\left (\frac{x}{c}\right )+\frac{1}{12} b c x^3 \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTan[c/x]),x]

[Out]

-(b*c^3*x)/4 + (b*c*x^3)/12 + (x^4*(a + b*ArcTan[c/x]))/4 + (b*c^4*ArcTan[x/c])/4

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^3 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{4} (b c) \int \frac{x^2}{1+\frac{c^2}{x^2}} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{4} (b c) \int \frac{x^4}{c^2+x^2} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{4} (b c) \int \left (-c^2+x^2+\frac{c^4}{c^2+x^2}\right ) \, dx\\ &=-\frac{1}{4} b c^3 x+\frac{1}{12} b c x^3+\frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{4} \left (b c^5\right ) \int \frac{1}{c^2+x^2} \, dx\\ &=-\frac{1}{4} b c^3 x+\frac{1}{12} b c x^3+\frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{4} b c^4 \tan ^{-1}\left (\frac{x}{c}\right )\\ \end{align*}

Mathematica [A]  time = 0.0103591, size = 55, normalized size = 1.1 \[ \frac{a x^4}{4}-\frac{1}{4} b c^3 x-\frac{1}{4} b c^4 \tan ^{-1}\left (\frac{c}{x}\right )+\frac{1}{12} b c x^3+\frac{1}{4} b x^4 \tan ^{-1}\left (\frac{c}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTan[c/x]),x]

[Out]

-(b*c^3*x)/4 + (b*c*x^3)/12 + (a*x^4)/4 - (b*c^4*ArcTan[c/x])/4 + (b*x^4*ArcTan[c/x])/4

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Maple [A]  time = 0.032, size = 46, normalized size = 0.9 \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}}{4}\arctan \left ({\frac{c}{x}} \right ) }+{\frac{b{c}^{4}}{4}\arctan \left ({\frac{x}{c}} \right ) }+{\frac{bc{x}^{3}}{12}}-{\frac{b{c}^{3}x}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctan(c/x)),x)

[Out]

1/4*x^4*a+1/4*b*x^4*arctan(c/x)+1/4*b*c^4*arctan(x/c)+1/12*b*c*x^3-1/4*b*c^3*x

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Maxima [A]  time = 1.47929, size = 61, normalized size = 1.22 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{12} \,{\left (3 \, x^{4} \arctan \left (\frac{c}{x}\right ) +{\left (3 \, c^{3} \arctan \left (\frac{x}{c}\right ) - 3 \, c^{2} x + x^{3}\right )} c\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c/x)),x, algorithm="maxima")

[Out]

1/4*a*x^4 + 1/12*(3*x^4*arctan(c/x) + (3*c^3*arctan(x/c) - 3*c^2*x + x^3)*c)*b

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Fricas [A]  time = 2.20965, size = 101, normalized size = 2.02 \begin{align*} -\frac{1}{4} \, b c^{3} x + \frac{1}{12} \, b c x^{3} + \frac{1}{4} \, a x^{4} - \frac{1}{4} \,{\left (b c^{4} - b x^{4}\right )} \arctan \left (\frac{c}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c/x)),x, algorithm="fricas")

[Out]

-1/4*b*c^3*x + 1/12*b*c*x^3 + 1/4*a*x^4 - 1/4*(b*c^4 - b*x^4)*arctan(c/x)

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Sympy [A]  time = 0.827307, size = 46, normalized size = 0.92 \begin{align*} \frac{a x^{4}}{4} - \frac{b c^{4} \operatorname{atan}{\left (\frac{c}{x} \right )}}{4} - \frac{b c^{3} x}{4} + \frac{b c x^{3}}{12} + \frac{b x^{4} \operatorname{atan}{\left (\frac{c}{x} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atan(c/x)),x)

[Out]

a*x**4/4 - b*c**4*atan(c/x)/4 - b*c**3*x/4 + b*c*x**3/12 + b*x**4*atan(c/x)/4

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Giac [A]  time = 1.16115, size = 81, normalized size = 1.62 \begin{align*} -\frac{1}{8} \, b c^{4} i \log \left (i x + c\right ) + \frac{1}{8} \, b c^{4} i \log \left (-i x + c\right ) + \frac{1}{4} \, b x^{4} \arctan \left (\frac{c}{x}\right ) - \frac{1}{4} \, b c^{3} x + \frac{1}{12} \, b c x^{3} + \frac{1}{4} \, a x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c/x)),x, algorithm="giac")

[Out]

-1/8*b*c^4*i*log(i*x + c) + 1/8*b*c^4*i*log(-i*x + c) + 1/4*b*x^4*arctan(c/x) - 1/4*b*c^3*x + 1/12*b*c*x^3 + 1
/4*a*x^4