\(\int x^3 (c+a^2 c x^2) \arctan (a x) \, dx\) [149]

Optimal result

Integrand size = 18, antiderivative size = 69 \[ \int x^3 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\frac {c x}{12 a^3}-\frac {c x^3}{36 a}-\frac {1}{30} a c x^5-\frac {c \arctan (a x)}{12 a^4}+\frac {1}{4} c x^4 \arctan (a x)+\frac {1}{6} a^2 c x^6 \arctan (a x) \]

[Out]

1/12*c*x/a^3-1/36*c*x^3/a-1/30*a*c*x^5-1/12*c*arctan(a*x)/a^4+1/4*c*x^4*arctan(a*x)+1/6*a^2*c*x^6*arctan(a*x)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 69, 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, 308, 209} \[ \int x^3 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {c \arctan (a x)}{12 a^4}+\frac {c x}{12 a^3}+\frac {1}{6} a^2 c x^6 \arctan (a x)+\frac {1}{4} c x^4 \arctan (a x)-\frac {1}{30} a c x^5-\frac {c x^3}{36 a} \]

[In]

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

[Out]

(c*x)/(12*a^3) - (c*x^3)/(36*a) - (a*c*x^5)/30 - (c*ArcTan[a*x])/(12*a^4) + (c*x^4*ArcTan[a*x])/4 + (a^2*c*x^6
*ArcTan[a*x])/6

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 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^3 \arctan (a x) \, dx+\left (a^2 c\right ) \int x^5 \arctan (a x) \, dx \\ & = \frac {1}{4} c x^4 \arctan (a x)+\frac {1}{6} a^2 c x^6 \arctan (a x)-\frac {1}{4} (a c) \int \frac {x^4}{1+a^2 x^2} \, dx-\frac {1}{6} \left (a^3 c\right ) \int \frac {x^6}{1+a^2 x^2} \, dx \\ & = \frac {1}{4} c x^4 \arctan (a x)+\frac {1}{6} a^2 c x^6 \arctan (a x)-\frac {1}{4} (a c) \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}{6} \left (a^3 c\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 {c x}{12 a^3}-\frac {c x^3}{36 a}-\frac {1}{30} a c x^5+\frac {1}{4} c x^4 \arctan (a x)+\frac {1}{6} a^2 c x^6 \arctan (a x)+\frac {c \int \frac {1}{1+a^2 x^2} \, dx}{6 a^3}-\frac {c \int \frac {1}{1+a^2 x^2} \, dx}{4 a^3} \\ & = \frac {c x}{12 a^3}-\frac {c x^3}{36 a}-\frac {1}{30} a c x^5-\frac {c \arctan (a x)}{12 a^4}+\frac {1}{4} c x^4 \arctan (a x)+\frac {1}{6} a^2 c x^6 \arctan (a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int x^3 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\frac {c x}{12 a^3}-\frac {c x^3}{36 a}-\frac {1}{30} a c x^5-\frac {c \arctan (a x)}{12 a^4}+\frac {1}{4} c x^4 \arctan (a x)+\frac {1}{6} a^2 c x^6 \arctan (a x) \]

[In]

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

[Out]

(c*x)/(12*a^3) - (c*x^3)/(36*a) - (a*c*x^5)/30 - (c*ArcTan[a*x])/(12*a^4) + (c*x^4*ArcTan[a*x])/4 + (a^2*c*x^6
*ArcTan[a*x])/6

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\frac {c \arctan \left (a x \right ) a^{6} x^{6}}{6}+\frac {c \arctan \left (a x \right ) a^{4} x^{4}}{4}-\frac {c \left (\frac {2 a^{5} x^{5}}{5}+\frac {a^{3} x^{3}}{3}-a x +\arctan \left (a x \right )\right )}{12}}{a^{4}}\)	\(60\)
default	\(\frac {\frac {c \arctan \left (a x \right ) a^{6} x^{6}}{6}+\frac {c \arctan \left (a x \right ) a^{4} x^{4}}{4}-\frac {c \left (\frac {2 a^{5} x^{5}}{5}+\frac {a^{3} x^{3}}{3}-a x +\arctan \left (a x \right )\right )}{12}}{a^{4}}\)	\(60\)
parts	\(\frac {a^{2} c \,x^{6} \arctan \left (a x \right )}{6}+\frac {c \,x^{4} \arctan \left (a x \right )}{4}-\frac {c a \left (\frac {\frac {2}{5} a^{4} x^{5}+\frac {1}{3} a^{2} x^{3}-x}{a^{4}}+\frac {\arctan \left (a x \right )}{a^{5}}\right )}{12}\)	\(62\)
parallelrisch	\(\frac {30 c \arctan \left (a x \right ) a^{6} x^{6}-6 a^{5} c \,x^{5}+45 c \arctan \left (a x \right ) a^{4} x^{4}-5 a^{3} c \,x^{3}+15 a c x -15 c \arctan \left (a x \right )}{180 a^{4}}\)	\(63\)
risch	\(-\frac {i c \,x^{4} \left (2 a^{2} x^{2}+3\right ) \ln \left (i a x +1\right )}{24}+\frac {i c \,a^{2} x^{6} \ln \left (-i a x +1\right )}{12}-\frac {a c \,x^{5}}{30}+\frac {i c \,x^{4} \ln \left (-i a x +1\right )}{8}-\frac {c \,x^{3}}{36 a}+\frac {c x}{12 a^{3}}-\frac {c \arctan \left (a x \right )}{12 a^{4}}\)	\(93\)
meijerg	\(\frac {c \left (-\frac {2 x a \left (21 a^{4} x^{4}-35 a^{2} x^{2}+105\right )}{315}+\frac {2 x a \left (7 a^{6} x^{6}+7\right ) \arctan \left (\sqrt {a^{2} x^{2}}\right )}{21 \sqrt {a^{2} x^{2}}}\right )}{4 a^{4}}+\frac {c \left (\frac {a x \left (-5 a^{2} x^{2}+15\right )}{15}-\frac {a x \left (-5 a^{4} x^{4}+5\right ) \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}\right )}{4 a^{4}}\)	\(118\)

[In]

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

[Out]

1/a^4*(1/6*c*arctan(a*x)*a^6*x^6+1/4*c*arctan(a*x)*a^4*x^4-1/12*c*(2/5*a^5*x^5+1/3*a^3*x^3-a*x+arctan(a*x)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int x^3 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {6 \, a^{5} c x^{5} + 5 \, a^{3} c x^{3} - 15 \, a c x - 15 \, {\left (2 \, a^{6} c x^{6} + 3 \, a^{4} c x^{4} - c\right )} \arctan \left (a x\right )}{180 \, a^{4}} \]

[In]

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

[Out]

-1/180*(6*a^5*c*x^5 + 5*a^3*c*x^3 - 15*a*c*x - 15*(2*a^6*c*x^6 + 3*a^4*c*x^4 - c)*arctan(a*x))/a^4

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94 \[ \int x^3 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\begin {cases} \frac {a^{2} c x^{6} \operatorname {atan}{\left (a x \right )}}{6} - \frac {a c x^{5}}{30} + \frac {c x^{4} \operatorname {atan}{\left (a x \right )}}{4} - \frac {c x^{3}}{36 a} + \frac {c x}{12 a^{3}} - \frac {c \operatorname {atan}{\left (a x \right )}}{12 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*c*x**6*atan(a*x)/6 - a*c*x**5/30 + c*x**4*atan(a*x)/4 - c*x**3/(36*a) + c*x/(12*a**3) - c*atan
(a*x)/(12*a**4), Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int x^3 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {1}{180} \, a {\left (\frac {6 \, a^{4} c x^{5} + 5 \, a^{2} c x^{3} - 15 \, c x}{a^{4}} + \frac {15 \, c \arctan \left (a x\right )}{a^{5}}\right )} + \frac {1}{12} \, {\left (2 \, a^{2} c x^{6} + 3 \, c x^{4}\right )} \arctan \left (a x\right ) \]

[In]

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

[Out]

-1/180*a*((6*a^4*c*x^5 + 5*a^2*c*x^3 - 15*c*x)/a^4 + 15*c*arctan(a*x)/a^5) + 1/12*(2*a^2*c*x^6 + 3*c*x^4)*arct
an(a*x)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int x^3 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {c\,\left (15\,\mathrm {atan}\left (a\,x\right )-15\,a\,x+5\,a^3\,x^3+6\,a^5\,x^5-45\,a^4\,x^4\,\mathrm {atan}\left (a\,x\right )-30\,a^6\,x^6\,\mathrm {atan}\left (a\,x\right )\right )}{180\,a^4} \]

[In]

int(x^3*atan(a*x)*(c + a^2*c*x^2),x)

[Out]

-(c*(15*atan(a*x) - 15*a*x + 5*a^3*x^3 + 6*a^5*x^5 - 45*a^4*x^4*atan(a*x) - 30*a^6*x^6*atan(a*x)))/(180*a^4)