\(\int x^3 (d+e x^2) (a+b \arctan (c x)) \, dx\) [1114]

Optimal result

Integrand size = 19, antiderivative size = 107 \[ \int x^3 \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\frac {b \left (3 c^2 d-2 e\right ) x}{12 c^5}-\frac {b \left (3 c^2 d-2 e\right ) x^3}{36 c^3}-\frac {b e x^5}{30 c}-\frac {b \left (3 c^2 d-2 e\right ) \arctan (c x)}{12 c^6}+\frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{6} e x^6 (a+b \arctan (c x)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 5096, 470, 308, 209} \[ \int x^3 \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{6} e x^6 (a+b \arctan (c x))-\frac {b \arctan (c x) \left (3 c^2 d-2 e\right )}{12 c^6}+\frac {b x \left (3 c^2 d-2 e\right )}{12 c^5}-\frac {b x^3 \left (3 c^2 d-2 e\right )}{36 c^3}-\frac {b e x^5}{30 c} \]

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 5096

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.19 \[ \int x^3 \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\frac {b d x}{4 c^3}-\frac {b e x}{6 c^5}-\frac {b d x^3}{12 c}+\frac {b e x^3}{18 c^3}+\frac {1}{4} a d x^4-\frac {b e x^5}{30 c}+\frac {1}{6} a e x^6-\frac {b d \arctan (c x)}{4 c^4}+\frac {b e \arctan (c x)}{6 c^6}+\frac {1}{4} b d x^4 \arctan (c x)+\frac {1}{6} b e x^6 \arctan (c x) \]

[In]

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

[Out]

(b*d*x)/(4*c^3) - (b*e*x)/(6*c^5) - (b*d*x^3)/(12*c) + (b*e*x^3)/(18*c^3) + (a*d*x^4)/4 - (b*e*x^5)/(30*c) + (
a*e*x^6)/6 - (b*d*ArcTan[c*x])/(4*c^4) + (b*e*ArcTan[c*x])/(6*c^6) + (b*d*x^4*ArcTan[c*x])/4 + (b*e*x^6*ArcTan
[c*x])/6

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.01

method	result	size
parts	\(a \left (\frac {1}{6} e \,x^{6}+\frac {1}{4} d \,x^{4}\right )+\frac {b \left (\frac {c^{4} \arctan \left (c x \right ) e \,x^{6}}{6}+\frac {\arctan \left (c x \right ) d \,c^{4} x^{4}}{4}-\frac {\frac {2 e \,c^{5} x^{5}}{5}+d \,c^{5} x^{3}-\frac {2 e \,c^{3} x^{3}}{3}-3 c^{3} x d +2 e c x +\left (3 c^{2} d -2 e \right ) \arctan \left (c x \right )}{12 c^{2}}\right )}{c^{4}}\)	\(108\)
derivativedivides	\(\frac {\frac {a \left (\frac {1}{4} d \,c^{6} x^{4}+\frac {1}{6} e \,c^{6} x^{6}\right )}{c^{2}}+\frac {b \left (\frac {\arctan \left (c x \right ) d \,c^{6} x^{4}}{4}+\frac {\arctan \left (c x \right ) e \,c^{6} x^{6}}{6}-\frac {d \,c^{5} x^{3}}{12}-\frac {e \,c^{5} x^{5}}{30}+\frac {c^{3} x d}{4}+\frac {e \,c^{3} x^{3}}{18}-\frac {e c x}{6}-\frac {\left (3 c^{2} d -2 e \right ) \arctan \left (c x \right )}{12}\right )}{c^{2}}}{c^{4}}\)	\(117\)
default	\(\frac {\frac {a \left (\frac {1}{4} d \,c^{6} x^{4}+\frac {1}{6} e \,c^{6} x^{6}\right )}{c^{2}}+\frac {b \left (\frac {\arctan \left (c x \right ) d \,c^{6} x^{4}}{4}+\frac {\arctan \left (c x \right ) e \,c^{6} x^{6}}{6}-\frac {d \,c^{5} x^{3}}{12}-\frac {e \,c^{5} x^{5}}{30}+\frac {c^{3} x d}{4}+\frac {e \,c^{3} x^{3}}{18}-\frac {e c x}{6}-\frac {\left (3 c^{2} d -2 e \right ) \arctan \left (c x \right )}{12}\right )}{c^{2}}}{c^{4}}\)	\(117\)
parallelrisch	\(\frac {30 x^{6} \arctan \left (c x \right ) b \,c^{6} e +30 a \,c^{6} e \,x^{6}+45 d b \arctan \left (c x \right ) x^{4} c^{6}-6 b \,c^{5} e \,x^{5}+45 a \,c^{6} d \,x^{4}-15 b \,c^{5} d \,x^{3}+10 b \,c^{3} e \,x^{3}+45 b \,c^{3} d x -45 b \,c^{2} d \arctan \left (c x \right )-30 b c e x +30 e b \arctan \left (c x \right )}{180 c^{6}}\)	\(118\)
risch	\(-\frac {i b \left (2 e \,x^{6}+3 d \,x^{4}\right ) \ln \left (i c x +1\right )}{24}+\frac {x^{6} e a}{6}+\frac {i b e \,x^{6} \ln \left (-i c x +1\right )}{12}+\frac {x^{4} d a}{4}-\frac {b e \,x^{5}}{30 c}+\frac {i b d \,x^{4} \ln \left (-i c x +1\right )}{8}-\frac {b d \,x^{3}}{12 c}+\frac {b e \,x^{3}}{18 c^{3}}+\frac {b d x}{4 c^{3}}-\frac {b d \arctan \left (c x \right )}{4 c^{4}}-\frac {b e x}{6 c^{5}}+\frac {e b \arctan \left (c x \right )}{6 c^{6}}\)	\(141\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.03 \[ \int x^3 \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\frac {30 \, a c^{6} e x^{6} + 45 \, a c^{6} d x^{4} - 6 \, b c^{5} e x^{5} - 5 \, {\left (3 \, b c^{5} d - 2 \, b c^{3} e\right )} x^{3} + 15 \, {\left (3 \, b c^{3} d - 2 \, b c e\right )} x + 15 \, {\left (2 \, b c^{6} e x^{6} + 3 \, b c^{6} d x^{4} - 3 \, b c^{2} d + 2 \, b e\right )} \arctan \left (c x\right )}{180 \, c^{6}} \]

[In]

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

[Out]

1/180*(30*a*c^6*e*x^6 + 45*a*c^6*d*x^4 - 6*b*c^5*e*x^5 - 5*(3*b*c^5*d - 2*b*c^3*e)*x^3 + 15*(3*b*c^3*d - 2*b*c
*e)*x + 15*(2*b*c^6*e*x^6 + 3*b*c^6*d*x^4 - 3*b*c^2*d + 2*b*e)*arctan(c*x))/c^6

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.29 \[ \int x^3 \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\begin {cases} \frac {a d x^{4}}{4} + \frac {a e x^{6}}{6} + \frac {b d x^{4} \operatorname {atan}{\left (c x \right )}}{4} + \frac {b e x^{6} \operatorname {atan}{\left (c x \right )}}{6} - \frac {b d x^{3}}{12 c} - \frac {b e x^{5}}{30 c} + \frac {b d x}{4 c^{3}} + \frac {b e x^{3}}{18 c^{3}} - \frac {b d \operatorname {atan}{\left (c x \right )}}{4 c^{4}} - \frac {b e x}{6 c^{5}} + \frac {b e \operatorname {atan}{\left (c x \right )}}{6 c^{6}} & \text {for}\: c \neq 0 \\a \left (\frac {d x^{4}}{4} + \frac {e x^{6}}{6}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*d*x**4/4 + a*e*x**6/6 + b*d*x**4*atan(c*x)/4 + b*e*x**6*atan(c*x)/6 - b*d*x**3/(12*c) - b*e*x**5/
(30*c) + b*d*x/(4*c**3) + b*e*x**3/(18*c**3) - b*d*atan(c*x)/(4*c**4) - b*e*x/(6*c**5) + b*e*atan(c*x)/(6*c**6
), Ne(c, 0)), (a*(d*x**4/4 + e*x**6/6), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.01 \[ \int x^3 \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\frac {1}{6} \, a e x^{6} + \frac {1}{4} \, a d x^{4} + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d + \frac {1}{90} \, {\left (15 \, x^{6} \arctan \left (c x\right ) - c {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b e \]

[In]

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

[Out]

1/6*a*e*x^6 + 1/4*a*d*x^4 + 1/12*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*d + 1/90*
(15*x^6*arctan(c*x) - c*((3*c^4*x^5 - 5*c^2*x^3 + 15*x)/c^6 - 15*arctan(c*x)/c^7))*b*e

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98 \[ \int x^3 \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\frac {a\,d\,x^4}{4}+\frac {a\,e\,x^6}{6}+\frac {b\,d\,x}{4\,c^3}-\frac {b\,e\,x}{6\,c^5}-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{4\,c^4}+\frac {b\,e\,\mathrm {atan}\left (c\,x\right )}{6\,c^6}+\frac {b\,d\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}+\frac {b\,e\,x^6\,\mathrm {atan}\left (c\,x\right )}{6}-\frac {b\,d\,x^3}{12\,c}-\frac {b\,e\,x^5}{30\,c}+\frac {b\,e\,x^3}{18\,c^3} \]

[In]

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

[Out]

(a*d*x^4)/4 + (a*e*x^6)/6 + (b*d*x)/(4*c^3) - (b*e*x)/(6*c^5) - (b*d*atan(c*x))/(4*c^4) + (b*e*atan(c*x))/(6*c
^6) + (b*d*x^4*atan(c*x))/4 + (b*e*x^6*atan(c*x))/6 - (b*d*x^3)/(12*c) - (b*e*x^5)/(30*c) + (b*e*x^3)/(18*c^3)