3.1114 \(\int x^3 (d+e x^2) (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=107 \[ \frac{1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{b x^3 \left (3 c^2 d-2 e\right )}{36 c^3}+\frac{b x \left (3 c^2 d-2 e\right )}{12 c^5}-\frac{b \left (3 c^2 d-2 e\right ) \tan ^{-1}(c x)}{12 c^6}-\frac{b e x^5}{30 c} \]

[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

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Rubi [A]  time = 0.110298, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {14, 4976, 459, 302, 203} \[ \frac{1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{b x^3 \left (3 c^2 d-2 e\right )}{36 c^3}+\frac{b x \left (3 c^2 d-2 e\right )}{12 c^5}-\frac{b \left (3 c^2 d-2 e\right ) \tan ^{-1}(c x)}{12 c^6}-\frac{b e x^5}{30 c} \]

Antiderivative was successfully verified.

[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 4976

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]))

Rule 459

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 302

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 203

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

Rubi steps

\begin{align*} \int x^3 \left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \tan ^{-1}(c x)\right )-(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 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \tan ^{-1}(c x)\right )+\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 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \tan ^{-1}(c x)\right )+\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 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \tan ^{-1}(c x)\right )+\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 ) \tan ^{-1}(c x)}{12 c^4}+\frac{1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \tan ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0054455, size = 127, normalized size = 1.19 \[ \frac{1}{4} a d x^4+\frac{1}{6} a e x^6+\frac{b d x}{4 c^3}-\frac{b d \tan ^{-1}(c x)}{4 c^4}+\frac{b e x^3}{18 c^3}-\frac{b e x}{6 c^5}+\frac{b e \tan ^{-1}(c x)}{6 c^6}-\frac{b d x^3}{12 c}+\frac{1}{4} b d x^4 \tan ^{-1}(c x)-\frac{b e x^5}{30 c}+\frac{1}{6} b e x^6 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.038, size = 106, normalized size = 1. \begin{align*}{\frac{ae{x}^{6}}{6}}+{\frac{a{x}^{4}d}{4}}+{\frac{b\arctan \left ( cx \right ) e{x}^{6}}{6}}+{\frac{b\arctan \left ( cx \right ){x}^{4}d}{4}}-{\frac{be{x}^{5}}{30\,c}}-{\frac{bd{x}^{3}}{12\,c}}+{\frac{b{x}^{3}e}{18\,{c}^{3}}}+{\frac{bdx}{4\,{c}^{3}}}-{\frac{bex}{6\,{c}^{5}}}-{\frac{\arctan \left ( cx \right ) bd}{4\,{c}^{4}}}+{\frac{b\arctan \left ( cx \right ) e}{6\,{c}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*e*x^6+1/4*a*x^4*d+1/6*b*arctan(c*x)*e*x^6+1/4*b*arctan(c*x)*x^4*d-1/30*b*e*x^5/c-1/12*b*d*x^3/c+1/18/c^3
*b*x^3*e+1/4*b*d*x/c^3-1/6/c^5*b*e*x-1/4*b*d*arctan(c*x)/c^4+1/6/c^6*b*arctan(c*x)*e

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Maxima [A]  time = 1.47799, size = 146, normalized size = 1.36 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.64274, size = 258, normalized size = 2.41 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 2.86752, size = 138, normalized size = 1.29 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.18206, size = 180, normalized size = 1.68 \begin{align*} \frac{30 \, b c^{6} x^{6} \arctan \left (c x\right ) e + 30 \, a c^{6} x^{6} e + 45 \, b c^{6} d x^{4} \arctan \left (c x\right ) + 45 \, a c^{6} d x^{4} - 6 \, b c^{5} x^{5} e - 15 \, b c^{5} d x^{3} + 10 \, b c^{3} x^{3} e + 45 \, b c^{3} d x - 45 \, b c^{2} d \arctan \left (c x\right ) - 30 \, \pi b e \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - 30 \, b c x e + 30 \, b \arctan \left (c x\right ) e}{180 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/180*(30*b*c^6*x^6*arctan(c*x)*e + 30*a*c^6*x^6*e + 45*b*c^6*d*x^4*arctan(c*x) + 45*a*c^6*d*x^4 - 6*b*c^5*x^5
*e - 15*b*c^5*d*x^3 + 10*b*c^3*x^3*e + 45*b*c^3*d*x - 45*b*c^2*d*arctan(c*x) - 30*pi*b*e*sgn(c)*sgn(x) - 30*b*
c*x*e + 30*b*arctan(c*x)*e)/c^6