3.62 \(\int x^3 (a+b \tan ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=43 \[ \frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{b \tan ^{-1}\left (c x^2\right )}{4 c^2}-\frac{b x^2}{4 c} \]

[Out]

-(b*x^2)/(4*c) + (b*ArcTan[c*x^2])/(4*c^2) + (x^4*(a + b*ArcTan[c*x^2]))/4

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Rubi [A]  time = 0.026978, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5033, 275, 321, 203} \[ \frac{1}{4} x^4 \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{b \tan ^{-1}\left (c x^2\right )}{4 c^2}-\frac{b x^2}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTan[c*x^2]),x]

[Out]

-(b*x^2)/(4*c) + (b*ArcTan[c*x^2])/(4*c^2) + (x^4*(a + b*ArcTan[c*x^2]))/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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [A]  time = 0.0053972, size = 48, normalized size = 1.12 \[ \frac{a x^4}{4}+\frac{b \tan ^{-1}\left (c x^2\right )}{4 c^2}-\frac{b x^2}{4 c}+\frac{1}{4} b x^4 \tan ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTan[c*x^2]),x]

[Out]

-(b*x^2)/(4*c) + (a*x^4)/4 + (b*ArcTan[c*x^2])/(4*c^2) + (b*x^4*ArcTan[c*x^2])/4

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Maple [A]  time = 0.025, size = 41, normalized size = 1. \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}\arctan \left ( c{x}^{2} \right ) }{4}}-{\frac{b{x}^{2}}{4\,c}}+{\frac{b\arctan \left ( c{x}^{2} \right ) }{4\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctan(c*x^2)),x)

[Out]

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

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Maxima [A]  time = 1.51948, size = 58, normalized size = 1.35 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{4} \,{\left (x^{4} \arctan \left (c x^{2}\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\arctan \left (c x^{2}\right )}{c^{3}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x^2)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.60519, size = 85, normalized size = 1.98 \begin{align*} \frac{a c^{2} x^{4} - b c x^{2} +{\left (b c^{2} x^{4} + b\right )} \arctan \left (c x^{2}\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x^2)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 40.6319, size = 48, normalized size = 1.12 \begin{align*} \begin{cases} \frac{a x^{4}}{4} + \frac{b x^{4} \operatorname{atan}{\left (c x^{2} \right )}}{4} - \frac{b x^{2}}{4 c} + \frac{b \operatorname{atan}{\left (c x^{2} \right )}}{4 c^{2}} & \text{for}\: c \neq 0 \\\frac{a x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atan(c*x**2)),x)

[Out]

Piecewise((a*x**4/4 + b*x**4*atan(c*x**2)/4 - b*x**2/(4*c) + b*atan(c*x**2)/(4*c**2), Ne(c, 0)), (a*x**4/4, Tr
ue))

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Giac [A]  time = 1.15399, size = 58, normalized size = 1.35 \begin{align*} \frac{a c x^{4} + \frac{{\left (c^{2} x^{4} \arctan \left (c x^{2}\right ) - c x^{2} + \arctan \left (c x^{2}\right )\right )} b}{c}}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x^2)),x, algorithm="giac")

[Out]

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