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

Optimal. Leaf size=66 \[ \frac{c \log \left (a^2 x^2+1\right )}{15 a^3}+\frac{1}{5} a^2 c x^5 \tan ^{-1}(a x)-\frac{1}{20} a c x^4-\frac{c x^2}{15 a}+\frac{1}{3} c x^3 \tan ^{-1}(a x) \]

[Out]

-(c*x^2)/(15*a) - (a*c*x^4)/20 + (c*x^3*ArcTan[a*x])/3 + (a^2*c*x^5*ArcTan[a*x])/5 + (c*Log[1 + a^2*x^2])/(15*
a^3)

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Rubi [A]  time = 0.0948406, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4950, 4852, 266, 43} \[ \frac{c \log \left (a^2 x^2+1\right )}{15 a^3}+\frac{1}{5} a^2 c x^5 \tan ^{-1}(a x)-\frac{1}{20} a c x^4-\frac{c x^2}{15 a}+\frac{1}{3} c x^3 \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*(c + a^2*c*x^2)*ArcTan[a*x],x]

[Out]

-(c*x^2)/(15*a) - (a*c*x^4)/20 + (c*x^3*ArcTan[a*x])/3 + (a^2*c*x^5*ArcTan[a*x])/5 + (c*Log[1 + a^2*x^2])/(15*
a^3)

Rule 4950

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[
d, Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] + Dist[(c^2*d)/f^2, Int[(f*x)^(m + 2)*(d + e*
x^2)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[q, 0] &&
IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))

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 (c+a^2 c x^2\right ) \tan ^{-1}(a x) \, dx &=c \int x^2 \tan ^{-1}(a x) \, dx+\left (a^2 c\right ) \int x^4 \tan ^{-1}(a x) \, dx\\ &=\frac{1}{3} c x^3 \tan ^{-1}(a x)+\frac{1}{5} a^2 c x^5 \tan ^{-1}(a x)-\frac{1}{3} (a c) \int \frac{x^3}{1+a^2 x^2} \, dx-\frac{1}{5} \left (a^3 c\right ) \int \frac{x^5}{1+a^2 x^2} \, dx\\ &=\frac{1}{3} c x^3 \tan ^{-1}(a x)+\frac{1}{5} a^2 c x^5 \tan ^{-1}(a x)-\frac{1}{6} (a c) \operatorname{Subst}\left (\int \frac{x}{1+a^2 x} \, dx,x,x^2\right )-\frac{1}{10} \left (a^3 c\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+a^2 x} \, dx,x,x^2\right )\\ &=\frac{1}{3} c x^3 \tan ^{-1}(a x)+\frac{1}{5} a^2 c x^5 \tan ^{-1}(a x)-\frac{1}{6} (a c) \operatorname{Subst}\left (\int \left (\frac{1}{a^2}-\frac{1}{a^2 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right )-\frac{1}{10} \left (a^3 c\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{a^4}+\frac{x}{a^2}+\frac{1}{a^4 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{c x^2}{15 a}-\frac{1}{20} a c x^4+\frac{1}{3} c x^3 \tan ^{-1}(a x)+\frac{1}{5} a^2 c x^5 \tan ^{-1}(a x)+\frac{c \log \left (1+a^2 x^2\right )}{15 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0224866, size = 66, normalized size = 1. \[ \frac{c \log \left (a^2 x^2+1\right )}{15 a^3}+\frac{1}{5} a^2 c x^5 \tan ^{-1}(a x)-\frac{1}{20} a c x^4-\frac{c x^2}{15 a}+\frac{1}{3} c x^3 \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*x^2)/(15*a) - (a*c*x^4)/20 + (c*x^3*ArcTan[a*x])/3 + (a^2*c*x^5*ArcTan[a*x])/5 + (c*Log[1 + a^2*x^2])/(15*
a^3)

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Maple [A]  time = 0.023, size = 57, normalized size = 0.9 \begin{align*} -{\frac{c{x}^{2}}{15\,a}}-{\frac{ac{x}^{4}}{20}}+{\frac{c{x}^{3}\arctan \left ( ax \right ) }{3}}+{\frac{{a}^{2}c{x}^{5}\arctan \left ( ax \right ) }{5}}+{\frac{c\ln \left ({a}^{2}{x}^{2}+1 \right ) }{15\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*c*x^2/a-1/20*a*c*x^4+1/3*c*x^3*arctan(a*x)+1/5*a^2*c*x^5*arctan(a*x)+1/15*c*ln(a^2*x^2+1)/a^3

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Maxima [A]  time = 0.987573, size = 85, normalized size = 1.29 \begin{align*} -\frac{1}{60} \, a{\left (\frac{3 \, a^{2} c x^{4} + 4 \, c x^{2}}{a^{2}} - \frac{4 \, c \log \left (a^{2} x^{2} + 1\right )}{a^{4}}\right )} + \frac{1}{15} \,{\left (3 \, a^{2} c x^{5} + 5 \, c x^{3}\right )} \arctan \left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*a*((3*a^2*c*x^4 + 4*c*x^2)/a^2 - 4*c*log(a^2*x^2 + 1)/a^4) + 1/15*(3*a^2*c*x^5 + 5*c*x^3)*arctan(a*x)

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Fricas [A]  time = 1.63626, size = 143, normalized size = 2.17 \begin{align*} -\frac{3 \, a^{4} c x^{4} + 4 \, a^{2} c x^{2} - 4 \,{\left (3 \, a^{5} c x^{5} + 5 \, a^{3} c x^{3}\right )} \arctan \left (a x\right ) - 4 \, c \log \left (a^{2} x^{2} + 1\right )}{60 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(3*a^4*c*x^4 + 4*a^2*c*x^2 - 4*(3*a^5*c*x^5 + 5*a^3*c*x^3)*arctan(a*x) - 4*c*log(a^2*x^2 + 1))/a^3

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Sympy [A]  time = 1.36341, size = 61, normalized size = 0.92 \begin{align*} \begin{cases} \frac{a^{2} c x^{5} \operatorname{atan}{\left (a x \right )}}{5} - \frac{a c x^{4}}{20} + \frac{c x^{3} \operatorname{atan}{\left (a x \right )}}{3} - \frac{c x^{2}}{15 a} + \frac{c \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{15 a^{3}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c*x**5*atan(a*x)/5 - a*c*x**4/20 + c*x**3*atan(a*x)/3 - c*x**2/(15*a) + c*log(x**2 + a**(-2))/
(15*a**3), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.15673, size = 85, normalized size = 1.29 \begin{align*} \frac{1}{15} \,{\left (3 \, a^{2} c x^{5} + 5 \, c x^{3}\right )} \arctan \left (a x\right ) + \frac{c \log \left (a^{2} x^{2} + 1\right )}{15 \, a^{3}} - \frac{3 \, a^{5} c x^{4} + 4 \, a^{3} c x^{2}}{60 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a^2*c*x^5 + 5*c*x^3)*arctan(a*x) + 1/15*c*log(a^2*x^2 + 1)/a^3 - 1/60*(3*a^5*c*x^4 + 4*a^3*c*x^2)/a^4