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

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

[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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Rubi [A]  time = 0.09, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 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])

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 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 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]))

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*}

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Mathematica [A]  time = 0.02, size = 66, normalized size = 1.00 \[ \frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)+\frac {c \log \left (a^2 x^2+1\right )}{15 a^3}-\frac {1}{20} a c x^4+\frac {1}{3} c x^3 \tan ^{-1}(a x)-\frac {c x^2}{15 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(c*x^2)/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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fricas [A]  time = 0.68, size = 62, normalized size = 0.94 \[ -\frac {3 \, a^{4} c x^{4} + 4 \, a^{2} c x^{2} - 4 \, {\left (3 \, a^{5} c x^{5} + 5 \, a^{3} c x^{3}\right )} \arctan \left (a x\right ) - 4 \, c \log \left (a^{2} x^{2} + 1\right )}{60 \, a^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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]

sage0*x

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maple [A]  time = 0.03, size = 57, normalized size = 0.86 \[ -\frac {c \,x^{2}}{15 a}-\frac {a c \,x^{4}}{20}+\frac {c \,x^{3} \arctan \left (a x \right )}{3}+\frac {a^{2} c \,x^{5} \arctan \left (a x \right )}{5}+\frac {c \ln \left (a^{2} x^{2}+1\right )}{15 a^{3}} \]

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 = 0.32, size = 63, normalized size = 0.95 \[ -\frac {1}{60} \, a {\left (\frac {3 \, a^{2} c x^{4} + 4 \, c x^{2}}{a^{2}} - \frac {4 \, c \log \left (a^{2} x^{2} + 1\right )}{a^{4}}\right )} + \frac {1}{15} \, {\left (3 \, a^{2} c x^{5} + 5 \, c x^{3}\right )} \arctan \left (a x\right ) \]

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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mupad [B]  time = 0.23, size = 58, normalized size = 0.88 \[ \frac {\frac {c\,\ln \left (a^2\,x^2+1\right )}{15}-\frac {a^2\,c\,x^2}{15}}{a^3}+\frac {c\,x^3\,\mathrm {atan}\left (a\,x\right )}{3}-\frac {a\,c\,x^4}{20}+\frac {a^2\,c\,x^5\,\mathrm {atan}\left (a\,x\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atan(a*x)*(c + a^2*c*x^2),x)

[Out]

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

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sympy [A]  time = 1.00, size = 61, normalized size = 0.92 \[ \begin {cases} \frac {a^{2} c x^{5} \operatorname {atan}{\left (a x \right )}}{5} - \frac {a c x^{4}}{20} + \frac {c x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {c x^{2}}{15 a} + \frac {c \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{15 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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