\(\int x^2 (c+a^2 c x^2)^2 \arctan (a x) \, dx\) [158]

Optimal result

Integrand size = 20, antiderivative size = 106 \[ \int x^2 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=-\frac {4 c^2 x^2}{105 a}-\frac {9}{140} a c^2 x^4-\frac {1}{42} a^3 c^2 x^6+\frac {1}{3} c^2 x^3 \arctan (a x)+\frac {2}{5} a^2 c^2 x^5 \arctan (a x)+\frac {1}{7} a^4 c^2 x^7 \arctan (a x)+\frac {4 c^2 \log \left (1+a^2 x^2\right )}{105 a^3} \]

[Out]

-4/105*c^2*x^2/a-9/140*a*c^2*x^4-1/42*a^3*c^2*x^6+1/3*c^2*x^3*arctan(a*x)+2/5*a^2*c^2*x^5*arctan(a*x)+1/7*a^4*
c^2*x^7*arctan(a*x)+4/105*c^2*ln(a^2*x^2+1)/a^3

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5068, 4946, 272, 45} \[ \int x^2 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\frac {1}{7} a^4 c^2 x^7 \arctan (a x)-\frac {1}{42} a^3 c^2 x^6+\frac {2}{5} a^2 c^2 x^5 \arctan (a x)+\frac {4 c^2 \log \left (a^2 x^2+1\right )}{105 a^3}+\frac {1}{3} c^2 x^3 \arctan (a x)-\frac {9}{140} a c^2 x^4-\frac {4 c^2 x^2}{105 a} \]

[In]

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

[Out]

(-4*c^2*x^2)/(105*a) - (9*a*c^2*x^4)/140 - (a^3*c^2*x^6)/42 + (c^2*x^3*ArcTan[a*x])/3 + (2*a^2*c^2*x^5*ArcTan[
a*x])/5 + (a^4*c^2*x^7*ArcTan[a*x])/7 + (4*c^2*Log[1 + a^2*x^2])/(105*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 5068

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

Rubi steps \begin{align*} \text {integral}& = \int \left (c^2 x^2 \arctan (a x)+2 a^2 c^2 x^4 \arctan (a x)+a^4 c^2 x^6 \arctan (a x)\right ) \, dx \\ & = c^2 \int x^2 \arctan (a x) \, dx+\left (2 a^2 c^2\right ) \int x^4 \arctan (a x) \, dx+\left (a^4 c^2\right ) \int x^6 \arctan (a x) \, dx \\ & = \frac {1}{3} c^2 x^3 \arctan (a x)+\frac {2}{5} a^2 c^2 x^5 \arctan (a x)+\frac {1}{7} a^4 c^2 x^7 \arctan (a x)-\frac {1}{3} \left (a c^2\right ) \int \frac {x^3}{1+a^2 x^2} \, dx-\frac {1}{5} \left (2 a^3 c^2\right ) \int \frac {x^5}{1+a^2 x^2} \, dx-\frac {1}{7} \left (a^5 c^2\right ) \int \frac {x^7}{1+a^2 x^2} \, dx \\ & = \frac {1}{3} c^2 x^3 \arctan (a x)+\frac {2}{5} a^2 c^2 x^5 \arctan (a x)+\frac {1}{7} a^4 c^2 x^7 \arctan (a x)-\frac {1}{6} \left (a c^2\right ) \text {Subst}\left (\int \frac {x}{1+a^2 x} \, dx,x,x^2\right )-\frac {1}{5} \left (a^3 c^2\right ) \text {Subst}\left (\int \frac {x^2}{1+a^2 x} \, dx,x,x^2\right )-\frac {1}{14} \left (a^5 c^2\right ) \text {Subst}\left (\int \frac {x^3}{1+a^2 x} \, dx,x,x^2\right ) \\ & = \frac {1}{3} c^2 x^3 \arctan (a x)+\frac {2}{5} a^2 c^2 x^5 \arctan (a x)+\frac {1}{7} a^4 c^2 x^7 \arctan (a x)-\frac {1}{6} \left (a c^2\right ) \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}{5} \left (a^3 c^2\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 {1}{14} \left (a^5 c^2\right ) \text {Subst}\left (\int \left (\frac {1}{a^6}-\frac {x}{a^4}+\frac {x^2}{a^2}-\frac {1}{a^6 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right ) \\ & = -\frac {4 c^2 x^2}{105 a}-\frac {9}{140} a c^2 x^4-\frac {1}{42} a^3 c^2 x^6+\frac {1}{3} c^2 x^3 \arctan (a x)+\frac {2}{5} a^2 c^2 x^5 \arctan (a x)+\frac {1}{7} a^4 c^2 x^7 \arctan (a x)+\frac {4 c^2 \log \left (1+a^2 x^2\right )}{105 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00 \[ \int x^2 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=-\frac {4 c^2 x^2}{105 a}-\frac {9}{140} a c^2 x^4-\frac {1}{42} a^3 c^2 x^6+\frac {1}{3} c^2 x^3 \arctan (a x)+\frac {2}{5} a^2 c^2 x^5 \arctan (a x)+\frac {1}{7} a^4 c^2 x^7 \arctan (a x)+\frac {4 c^2 \log \left (1+a^2 x^2\right )}{105 a^3} \]

[In]

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

[Out]

(-4*c^2*x^2)/(105*a) - (9*a*c^2*x^4)/140 - (a^3*c^2*x^6)/42 + (c^2*x^3*ArcTan[a*x])/3 + (2*a^2*c^2*x^5*ArcTan[
a*x])/5 + (a^4*c^2*x^7*ArcTan[a*x])/7 + (4*c^2*Log[1 + a^2*x^2])/(105*a^3)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\frac {c^{2} \arctan \left (a x \right ) a^{7} x^{7}}{7}+\frac {2 c^{2} \arctan \left (a x \right ) a^{5} x^{5}}{5}+\frac {a^{3} c^{2} x^{3} \arctan \left (a x \right )}{3}-\frac {c^{2} \left (\frac {5 a^{6} x^{6}}{2}+\frac {27 a^{4} x^{4}}{4}+4 a^{2} x^{2}-4 \ln \left (a^{2} x^{2}+1\right )\right )}{105}}{a^{3}}\)	\(93\)
default	\(\frac {\frac {c^{2} \arctan \left (a x \right ) a^{7} x^{7}}{7}+\frac {2 c^{2} \arctan \left (a x \right ) a^{5} x^{5}}{5}+\frac {a^{3} c^{2} x^{3} \arctan \left (a x \right )}{3}-\frac {c^{2} \left (\frac {5 a^{6} x^{6}}{2}+\frac {27 a^{4} x^{4}}{4}+4 a^{2} x^{2}-4 \ln \left (a^{2} x^{2}+1\right )\right )}{105}}{a^{3}}\)	\(93\)
parts	\(\frac {a^{4} c^{2} x^{7} \arctan \left (a x \right )}{7}+\frac {2 a^{2} c^{2} x^{5} \arctan \left (a x \right )}{5}+\frac {c^{2} x^{3} \arctan \left (a x \right )}{3}-\frac {c^{2} a \left (\frac {5 a^{4} x^{6}+\frac {27}{2} a^{2} x^{4}+8 x^{2}}{2 a^{2}}-\frac {4 \ln \left (a^{2} x^{2}+1\right )}{a^{4}}\right )}{105}\)	\(93\)
parallelrisch	\(\frac {60 c^{2} \arctan \left (a x \right ) a^{7} x^{7}-10 a^{6} c^{2} x^{6}+168 c^{2} \arctan \left (a x \right ) a^{5} x^{5}-27 a^{4} c^{2} x^{4}+140 a^{3} c^{2} x^{3} \arctan \left (a x \right )-16 a^{2} c^{2} x^{2}+16 c^{2} \ln \left (a^{2} x^{2}+1\right )}{420 a^{3}}\)	\(100\)
risch	\(-\frac {i c^{2} x^{3} \left (15 a^{4} x^{4}+42 a^{2} x^{2}+35\right ) \ln \left (i a x +1\right )}{210}+\frac {i c^{2} a^{4} x^{7} \ln \left (-i a x +1\right )}{14}-\frac {a^{3} c^{2} x^{6}}{42}+\frac {i c^{2} a^{2} x^{5} \ln \left (-i a x +1\right )}{5}-\frac {9 a \,c^{2} x^{4}}{140}+\frac {i c^{2} x^{3} \ln \left (-i a x +1\right )}{6}-\frac {4 c^{2} x^{2}}{105 a}+\frac {4 c^{2} \ln \left (-a^{2} x^{2}-1\right )}{105 a^{3}}\)	\(144\)
meijerg	\(\frac {c^{2} \left (-\frac {a^{2} x^{2} \left (4 a^{4} x^{4}-6 a^{2} x^{2}+12\right )}{42}+\frac {4 a^{8} x^{8} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{7 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{7}\right )}{4 a^{3}}+\frac {c^{2} \left (\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+6\right )}{15}+\frac {4 a^{6} x^{6} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (a^{2} x^{2}+1\right )}{5}\right )}{2 a^{3}}+\frac {c^{2} \left (-\frac {2 a^{2} x^{2}}{3}+\frac {4 a^{4} x^{4} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4 a^{3}}\)	\(198\)

[In]

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

[Out]

1/a^3*(1/7*c^2*arctan(a*x)*a^7*x^7+2/5*c^2*arctan(a*x)*a^5*x^5+1/3*a^3*c^2*x^3*arctan(a*x)-1/105*c^2*(5/2*a^6*
x^6+27/4*a^4*x^4+4*a^2*x^2-4*ln(a^2*x^2+1)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89 \[ \int x^2 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=-\frac {10 \, a^{6} c^{2} x^{6} + 27 \, a^{4} c^{2} x^{4} + 16 \, a^{2} c^{2} x^{2} - 16 \, c^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, {\left (15 \, a^{7} c^{2} x^{7} + 42 \, a^{5} c^{2} x^{5} + 35 \, a^{3} c^{2} x^{3}\right )} \arctan \left (a x\right )}{420 \, a^{3}} \]

[In]

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

[Out]

-1/420*(10*a^6*c^2*x^6 + 27*a^4*c^2*x^4 + 16*a^2*c^2*x^2 - 16*c^2*log(a^2*x^2 + 1) - 4*(15*a^7*c^2*x^7 + 42*a^
5*c^2*x^5 + 35*a^3*c^2*x^3)*arctan(a*x))/a^3

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int x^2 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\begin {cases} \frac {a^{4} c^{2} x^{7} \operatorname {atan}{\left (a x \right )}}{7} - \frac {a^{3} c^{2} x^{6}}{42} + \frac {2 a^{2} c^{2} x^{5} \operatorname {atan}{\left (a x \right )}}{5} - \frac {9 a c^{2} x^{4}}{140} + \frac {c^{2} x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {4 c^{2} x^{2}}{105 a} + \frac {4 c^{2} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{105 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**4*c**2*x**7*atan(a*x)/7 - a**3*c**2*x**6/42 + 2*a**2*c**2*x**5*atan(a*x)/5 - 9*a*c**2*x**4/140 +
c**2*x**3*atan(a*x)/3 - 4*c**2*x**2/(105*a) + 4*c**2*log(x**2 + a**(-2))/(105*a**3), Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int x^2 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=-\frac {1}{420} \, a {\left (\frac {10 \, a^{4} c^{2} x^{6} + 27 \, a^{2} c^{2} x^{4} + 16 \, c^{2} x^{2}}{a^{2}} - \frac {16 \, c^{2} \log \left (a^{2} x^{2} + 1\right )}{a^{4}}\right )} + \frac {1}{105} \, {\left (15 \, a^{4} c^{2} x^{7} + 42 \, a^{2} c^{2} x^{5} + 35 \, c^{2} x^{3}\right )} \arctan \left (a x\right ) \]

[In]

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

[Out]

-1/420*a*((10*a^4*c^2*x^6 + 27*a^2*c^2*x^4 + 16*c^2*x^2)/a^2 - 16*c^2*log(a^2*x^2 + 1)/a^4) + 1/105*(15*a^4*c^
2*x^7 + 42*a^2*c^2*x^5 + 35*c^2*x^3)*arctan(a*x)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.76 \[ \int x^2 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\frac {c^2\,\left (16\,\ln \left (a^2\,x^2+1\right )-16\,a^2\,x^2-27\,a^4\,x^4-10\,a^6\,x^6+140\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )+168\,a^5\,x^5\,\mathrm {atan}\left (a\,x\right )+60\,a^7\,x^7\,\mathrm {atan}\left (a\,x\right )\right )}{420\,a^3} \]

[In]

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

[Out]

(c^2*(16*log(a^2*x^2 + 1) - 16*a^2*x^2 - 27*a^4*x^4 - 10*a^6*x^6 + 140*a^3*x^3*atan(a*x) + 168*a^5*x^5*atan(a*
x) + 60*a^7*x^7*atan(a*x)))/(420*a^3)