\(\int x^2 (c+a^2 c x^2) \arctan (a x) \, dx\) [150]

Optimal result

Integrand size = 18, antiderivative size = 66 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {c x^2}{15 a}-\frac {1}{20} a c x^4+\frac {1}{3} c x^3 \arctan (a x)+\frac {1}{5} a^2 c x^5 \arctan (a x)+\frac {c \log \left (1+a^2 x^2\right )}{15 a^3} \]

[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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5070, 4946, 272, 45} \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\frac {1}{5} a^2 c x^5 \arctan (a x)+\frac {c \log \left (a^2 x^2+1\right )}{15 a^3}+\frac {1}{3} c x^3 \arctan (a x)-\frac {1}{20} a c x^4-\frac {c x^2}{15 a} \]

[In]

Int[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)

Rule 45

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 272

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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 5070

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {c x^2}{15 a}-\frac {1}{20} a c x^4+\frac {1}{3} c x^3 \arctan (a x)+\frac {1}{5} a^2 c x^5 \arctan (a x)+\frac {c \log \left (1+a^2 x^2\right )}{15 a^3} \]

[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)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {c \arctan \left (a x \right ) a^{5} x^{5}}{5}+\frac {c \arctan \left (a x \right ) a^{3} x^{3}}{3}-\frac {c \left (\frac {3 a^{4} x^{4}}{4}+a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )\right )}{15}}{a^{3}}\)	\(63\)
default	\(\frac {\frac {c \arctan \left (a x \right ) a^{5} x^{5}}{5}+\frac {c \arctan \left (a x \right ) a^{3} x^{3}}{3}-\frac {c \left (\frac {3 a^{4} x^{4}}{4}+a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )\right )}{15}}{a^{3}}\)	\(63\)
parallelrisch	\(\frac {12 c \arctan \left (a x \right ) a^{5} x^{5}-3 a^{4} c \,x^{4}+20 c \arctan \left (a x \right ) a^{3} x^{3}-4 a^{2} c \,x^{2}+4 c \ln \left (a^{2} x^{2}+1\right )}{60 a^{3}}\)	\(64\)
parts	\(\frac {a^{2} c \,x^{5} \arctan \left (a x \right )}{5}+\frac {c \,x^{3} \arctan \left (a x \right )}{3}-\frac {c a \left (\frac {\frac {3}{2} a^{2} x^{4}+2 x^{2}}{2 a^{2}}-\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{4}}\right )}{15}\)	\(64\)
risch	\(-\frac {i c \,x^{3} \left (3 a^{2} x^{2}+5\right ) \ln \left (i a x +1\right )}{30}+\frac {i c \,a^{2} x^{5} \ln \left (-i a x +1\right )}{10}-\frac {a c \,x^{4}}{20}+\frac {i c \,x^{3} \ln \left (-i a x +1\right )}{6}-\frac {c \,x^{2}}{15 a}+\frac {c \ln \left (-a^{2} x^{2}-1\right )}{15 a^{3}}-\frac {c}{45 a^{3}}\)	\(99\)
meijerg	\(\frac {c \left (\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+6\right )}{15}+\frac {4 a^{6} x^{6} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (a^{2} x^{2}+1\right )}{5}\right )}{4 a^{3}}+\frac {c \left (-\frac {2 a^{2} x^{2}}{3}+\frac {4 a^{4} x^{4} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4 a^{3}}\)	\(120\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {3 \, a^{4} c x^{4} + 4 \, a^{2} c x^{2} - 4 \, {\left (3 \, a^{5} c x^{5} + 5 \, a^{3} c x^{3}\right )} \arctan \left (a x\right ) - 4 \, c \log \left (a^{2} x^{2} + 1\right )}{60 \, a^{3}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\begin {cases} \frac {a^{2} c x^{5} \operatorname {atan}{\left (a x \right )}}{5} - \frac {a c x^{4}}{20} + \frac {c x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {c x^{2}}{15 a} + \frac {c \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{15 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {1}{60} \, a {\left (\frac {3 \, a^{2} c x^{4} + 4 \, c x^{2}}{a^{2}} - \frac {4 \, c \log \left (a^{2} x^{2} + 1\right )}{a^{4}}\right )} + \frac {1}{15} \, {\left (3 \, a^{2} c x^{5} + 5 \, c x^{3}\right )} \arctan \left (a x\right ) \]

[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)

Giac [F]

\[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )} x^{2} \arctan \left (a x\right ) \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\frac {\frac {c\,\ln \left (a^2\,x^2+1\right )}{15}-\frac {a^2\,c\,x^2}{15}}{a^3}+\frac {c\,x^3\,\mathrm {atan}\left (a\,x\right )}{3}-\frac {a\,c\,x^4}{20}+\frac {a^2\,c\,x^5\,\mathrm {atan}\left (a\,x\right )}{5} \]

[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