\(\int x^3 (c+a^2 c x^2)^2 \arctan (a x) \, dx\) [157]

Optimal result

Integrand size = 20, antiderivative size = 111 \[ \int x^3 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\frac {c^2 x}{24 a^3}-\frac {c^2 x^3}{72 a}-\frac {1}{24} a c^2 x^5-\frac {1}{56} a^3 c^2 x^7-\frac {c^2 \arctan (a x)}{24 a^4}+\frac {1}{4} c^2 x^4 \arctan (a x)+\frac {1}{3} a^2 c^2 x^6 \arctan (a x)+\frac {1}{8} a^4 c^2 x^8 \arctan (a x) \]

[Out]

1/24*c^2*x/a^3-1/72*c^2*x^3/a-1/24*a*c^2*x^5-1/56*a^3*c^2*x^7-1/24*c^2*arctan(a*x)/a^4+1/4*c^2*x^4*arctan(a*x)
+1/3*a^2*c^2*x^6*arctan(a*x)+1/8*a^4*c^2*x^8*arctan(a*x)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 111, 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, 308, 209} \[ \int x^3 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\frac {1}{8} a^4 c^2 x^8 \arctan (a x)-\frac {c^2 \arctan (a x)}{24 a^4}-\frac {1}{56} a^3 c^2 x^7+\frac {c^2 x}{24 a^3}+\frac {1}{3} a^2 c^2 x^6 \arctan (a x)+\frac {1}{4} c^2 x^4 \arctan (a x)-\frac {1}{24} a c^2 x^5-\frac {c^2 x^3}{72 a} \]

[In]

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

[Out]

(c^2*x)/(24*a^3) - (c^2*x^3)/(72*a) - (a*c^2*x^5)/24 - (a^3*c^2*x^7)/56 - (c^2*ArcTan[a*x])/(24*a^4) + (c^2*x^
4*ArcTan[a*x])/4 + (a^2*c^2*x^6*ArcTan[a*x])/3 + (a^4*c^2*x^8*ArcTan[a*x])/8

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^3 \arctan (a x)+2 a^2 c^2 x^5 \arctan (a x)+a^4 c^2 x^7 \arctan (a x)\right ) \, dx \\ & = c^2 \int x^3 \arctan (a x) \, dx+\left (2 a^2 c^2\right ) \int x^5 \arctan (a x) \, dx+\left (a^4 c^2\right ) \int x^7 \arctan (a x) \, dx \\ & = \frac {1}{4} c^2 x^4 \arctan (a x)+\frac {1}{3} a^2 c^2 x^6 \arctan (a x)+\frac {1}{8} a^4 c^2 x^8 \arctan (a x)-\frac {1}{4} \left (a c^2\right ) \int \frac {x^4}{1+a^2 x^2} \, dx-\frac {1}{3} \left (a^3 c^2\right ) \int \frac {x^6}{1+a^2 x^2} \, dx-\frac {1}{8} \left (a^5 c^2\right ) \int \frac {x^8}{1+a^2 x^2} \, dx \\ & = \frac {1}{4} c^2 x^4 \arctan (a x)+\frac {1}{3} a^2 c^2 x^6 \arctan (a x)+\frac {1}{8} a^4 c^2 x^8 \arctan (a x)-\frac {1}{4} \left (a c^2\right ) \int \left (-\frac {1}{a^4}+\frac {x^2}{a^2}+\frac {1}{a^4 \left (1+a^2 x^2\right )}\right ) \, dx-\frac {1}{3} \left (a^3 c^2\right ) \int \left (\frac {1}{a^6}-\frac {x^2}{a^4}+\frac {x^4}{a^2}-\frac {1}{a^6 \left (1+a^2 x^2\right )}\right ) \, dx-\frac {1}{8} \left (a^5 c^2\right ) \int \left (-\frac {1}{a^8}+\frac {x^2}{a^6}-\frac {x^4}{a^4}+\frac {x^6}{a^2}+\frac {1}{a^8 \left (1+a^2 x^2\right )}\right ) \, dx \\ & = \frac {c^2 x}{24 a^3}-\frac {c^2 x^3}{72 a}-\frac {1}{24} a c^2 x^5-\frac {1}{56} a^3 c^2 x^7+\frac {1}{4} c^2 x^4 \arctan (a x)+\frac {1}{3} a^2 c^2 x^6 \arctan (a x)+\frac {1}{8} a^4 c^2 x^8 \arctan (a x)-\frac {c^2 \int \frac {1}{1+a^2 x^2} \, dx}{8 a^3}-\frac {c^2 \int \frac {1}{1+a^2 x^2} \, dx}{4 a^3}+\frac {c^2 \int \frac {1}{1+a^2 x^2} \, dx}{3 a^3} \\ & = \frac {c^2 x}{24 a^3}-\frac {c^2 x^3}{72 a}-\frac {1}{24} a c^2 x^5-\frac {1}{56} a^3 c^2 x^7-\frac {c^2 \arctan (a x)}{24 a^4}+\frac {1}{4} c^2 x^4 \arctan (a x)+\frac {1}{3} a^2 c^2 x^6 \arctan (a x)+\frac {1}{8} a^4 c^2 x^8 \arctan (a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00 \[ \int x^3 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\frac {c^2 x}{24 a^3}-\frac {c^2 x^3}{72 a}-\frac {1}{24} a c^2 x^5-\frac {1}{56} a^3 c^2 x^7-\frac {c^2 \arctan (a x)}{24 a^4}+\frac {1}{4} c^2 x^4 \arctan (a x)+\frac {1}{3} a^2 c^2 x^6 \arctan (a x)+\frac {1}{8} a^4 c^2 x^8 \arctan (a x) \]

[In]

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

[Out]

(c^2*x)/(24*a^3) - (c^2*x^3)/(72*a) - (a*c^2*x^5)/24 - (a^3*c^2*x^7)/56 - (c^2*ArcTan[a*x])/(24*a^4) + (c^2*x^
4*ArcTan[a*x])/4 + (a^2*c^2*x^6*ArcTan[a*x])/3 + (a^4*c^2*x^8*ArcTan[a*x])/8

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {c^{2} \arctan \left (a x \right ) a^{8} x^{8}}{8}+\frac {c^{2} \arctan \left (a x \right ) a^{6} x^{6}}{3}+\frac {a^{4} c^{2} x^{4} \arctan \left (a x \right )}{4}-\frac {c^{2} \left (\frac {3 a^{7} x^{7}}{7}+a^{5} x^{5}+\frac {a^{3} x^{3}}{3}-a x +\arctan \left (a x \right )\right )}{24}}{a^{4}}\)	\(88\)
default	\(\frac {\frac {c^{2} \arctan \left (a x \right ) a^{8} x^{8}}{8}+\frac {c^{2} \arctan \left (a x \right ) a^{6} x^{6}}{3}+\frac {a^{4} c^{2} x^{4} \arctan \left (a x \right )}{4}-\frac {c^{2} \left (\frac {3 a^{7} x^{7}}{7}+a^{5} x^{5}+\frac {a^{3} x^{3}}{3}-a x +\arctan \left (a x \right )\right )}{24}}{a^{4}}\)	\(88\)
parts	\(\frac {a^{4} c^{2} x^{8} \arctan \left (a x \right )}{8}+\frac {a^{2} c^{2} x^{6} \arctan \left (a x \right )}{3}+\frac {c^{2} x^{4} \arctan \left (a x \right )}{4}-\frac {c^{2} a \left (\frac {\frac {3}{7} a^{6} x^{7}+a^{4} x^{5}+\frac {1}{3} a^{2} x^{3}-x}{a^{4}}+\frac {\arctan \left (a x \right )}{a^{5}}\right )}{24}\)	\(90\)
parallelrisch	\(\frac {63 c^{2} \arctan \left (a x \right ) a^{8} x^{8}-9 a^{7} c^{2} x^{7}+168 c^{2} \arctan \left (a x \right ) a^{6} x^{6}-21 a^{5} c^{2} x^{5}+126 a^{4} c^{2} x^{4} \arctan \left (a x \right )-7 a^{3} c^{2} x^{3}+21 a \,c^{2} x -21 c^{2} \arctan \left (a x \right )}{504 a^{4}}\)	\(101\)
risch	\(-\frac {i c^{2} x^{4} \left (3 a^{4} x^{4}+8 a^{2} x^{2}+6\right ) \ln \left (i a x +1\right )}{48}+\frac {i c^{2} a^{4} x^{8} \ln \left (-i a x +1\right )}{16}-\frac {a^{3} c^{2} x^{7}}{56}+\frac {i c^{2} a^{2} x^{6} \ln \left (-i a x +1\right )}{6}-\frac {a \,c^{2} x^{5}}{24}+\frac {i c^{2} x^{4} \ln \left (-i a x +1\right )}{8}-\frac {c^{2} x^{3}}{72 a}+\frac {c^{2} x}{24 a^{3}}-\frac {c^{2} \arctan \left (a x \right )}{24 a^{4}}\)	\(146\)
meijerg	\(\frac {c^{2} \left (\frac {x a \left (-45 a^{6} x^{6}+63 a^{4} x^{4}-105 a^{2} x^{2}+315\right )}{630}-\frac {x a \left (-9 a^{8} x^{8}+9\right ) \arctan \left (\sqrt {a^{2} x^{2}}\right )}{18 \sqrt {a^{2} x^{2}}}\right )}{4 a^{4}}+\frac {c^{2} \left (-\frac {2 a x \left (21 a^{4} x^{4}-35 a^{2} x^{2}+105\right )}{315}+\frac {2 a x \left (7 a^{6} x^{6}+7\right ) \arctan \left (\sqrt {a^{2} x^{2}}\right )}{21 \sqrt {a^{2} x^{2}}}\right )}{2 a^{4}}+\frac {c^{2} \left (\frac {a x \left (-5 a^{2} x^{2}+15\right )}{15}-\frac {a x \left (-5 a^{4} x^{4}+5\right ) \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}\right )}{4 a^{4}}\)	\(194\)

[In]

int(x^3*(a^2*c*x^2+c)^2*arctan(a*x),x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/8*c^2*arctan(a*x)*a^8*x^8+1/3*c^2*arctan(a*x)*a^6*x^6+1/4*a^4*c^2*x^4*arctan(a*x)-1/24*c^2*(3/7*a^7*x
^7+a^5*x^5+1/3*a^3*x^3-a*x+arctan(a*x)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82 \[ \int x^3 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=-\frac {9 \, a^{7} c^{2} x^{7} + 21 \, a^{5} c^{2} x^{5} + 7 \, a^{3} c^{2} x^{3} - 21 \, a c^{2} x - 21 \, {\left (3 \, a^{8} c^{2} x^{8} + 8 \, a^{6} c^{2} x^{6} + 6 \, a^{4} c^{2} x^{4} - c^{2}\right )} \arctan \left (a x\right )}{504 \, a^{4}} \]

[In]

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

[Out]

-1/504*(9*a^7*c^2*x^7 + 21*a^5*c^2*x^5 + 7*a^3*c^2*x^3 - 21*a*c^2*x - 21*(3*a^8*c^2*x^8 + 8*a^6*c^2*x^6 + 6*a^
4*c^2*x^4 - c^2)*arctan(a*x))/a^4

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.94 \[ \int x^3 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\begin {cases} \frac {a^{4} c^{2} x^{8} \operatorname {atan}{\left (a x \right )}}{8} - \frac {a^{3} c^{2} x^{7}}{56} + \frac {a^{2} c^{2} x^{6} \operatorname {atan}{\left (a x \right )}}{3} - \frac {a c^{2} x^{5}}{24} + \frac {c^{2} x^{4} \operatorname {atan}{\left (a x \right )}}{4} - \frac {c^{2} x^{3}}{72 a} + \frac {c^{2} x}{24 a^{3}} - \frac {c^{2} \operatorname {atan}{\left (a x \right )}}{24 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**4*c**2*x**8*atan(a*x)/8 - a**3*c**2*x**7/56 + a**2*c**2*x**6*atan(a*x)/3 - a*c**2*x**5/24 + c**2
*x**4*atan(a*x)/4 - c**2*x**3/(72*a) + c**2*x/(24*a**3) - c**2*atan(a*x)/(24*a**4), Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int x^3 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=-\frac {1}{504} \, a {\left (\frac {21 \, c^{2} \arctan \left (a x\right )}{a^{5}} + \frac {9 \, a^{6} c^{2} x^{7} + 21 \, a^{4} c^{2} x^{5} + 7 \, a^{2} c^{2} x^{3} - 21 \, c^{2} x}{a^{4}}\right )} + \frac {1}{24} \, {\left (3 \, a^{4} c^{2} x^{8} + 8 \, a^{2} c^{2} x^{6} + 6 \, c^{2} x^{4}\right )} \arctan \left (a x\right ) \]

[In]

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

[Out]

-1/504*a*(21*c^2*arctan(a*x)/a^5 + (9*a^6*c^2*x^7 + 21*a^4*c^2*x^5 + 7*a^2*c^2*x^3 - 21*c^2*x)/a^4) + 1/24*(3*
a^4*c^2*x^8 + 8*a^2*c^2*x^6 + 6*c^2*x^4)*arctan(a*x)

Giac [F]

\[ \int x^3 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )}^{2} x^{3} \arctan \left (a x\right ) \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.80 \[ \int x^3 \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\mathrm {atan}\left (a\,x\right )\,\left (\frac {a^4\,c^2\,x^8}{8}+\frac {a^2\,c^2\,x^6}{3}+\frac {c^2\,x^4}{4}\right )+\frac {c^2\,x}{24\,a^3}-\frac {a\,c^2\,x^5}{24}-\frac {c^2\,\mathrm {atan}\left (a\,x\right )}{24\,a^4}-\frac {c^2\,x^3}{72\,a}-\frac {a^3\,c^2\,x^7}{56} \]

[In]

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

[Out]

atan(a*x)*((c^2*x^4)/4 + (a^2*c^2*x^6)/3 + (a^4*c^2*x^8)/8) + (c^2*x)/(24*a^3) - (a*c^2*x^5)/24 - (c^2*atan(a*
x))/(24*a^4) - (c^2*x^3)/(72*a) - (a^3*c^2*x^7)/56