3.1115 \(\int x^2 (d+e x^2) (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=94 \[ \frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {b \left (5 c^2 d-3 e\right ) \log \left (c^2 x^2+1\right )}{30 c^5}-\frac {b x^2 \left (5 c^2 d-3 e\right )}{30 c^3}-\frac {b e x^4}{20 c} \]

[Out]

-1/30*b*(5*c^2*d-3*e)*x^2/c^3-1/20*b*e*x^4/c+1/3*d*x^3*(a+b*arctan(c*x))+1/5*e*x^5*(a+b*arctan(c*x))+1/30*b*(5
*c^2*d-3*e)*ln(c^2*x^2+1)/c^5

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Rubi [A]  time = 0.14, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {14, 4976, 446, 77} \[ \frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {b x^2 \left (5 c^2 d-3 e\right )}{30 c^3}+\frac {b \left (5 c^2 d-3 e\right ) \log \left (c^2 x^2+1\right )}{30 c^5}-\frac {b e x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(5*c^2*d - 3*e)*x^2)/(30*c^3) - (b*e*x^4)/(20*c) + (d*x^3*(a + b*ArcTan[c*x]))/3 + (e*x^5*(a + b*ArcTan[c*
x]))/5 + (b*(5*c^2*d - 3*e)*Log[1 + c^2*x^2])/(30*c^5)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

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

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]))

Rubi steps

\begin {align*} \int x^2 \left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac {x^3 \left (5 d+3 e x^2\right )}{15+15 c^2 x^2} \, dx\\ &=\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{2} (b c) \operatorname {Subst}\left (\int \frac {x (5 d+3 e x)}{15+15 c^2 x} \, dx,x,x^2\right )\\ &=\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{2} (b c) \operatorname {Subst}\left (\int \left (\frac {5 c^2 d-3 e}{15 c^4}+\frac {e x}{5 c^2}+\frac {-5 c^2 d+3 e}{15 c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {b \left (5 c^2 d-3 e\right ) x^2}{30 c^3}-\frac {b e x^4}{20 c}+\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {b \left (5 c^2 d-3 e\right ) \log \left (1+c^2 x^2\right )}{30 c^5}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 119, normalized size = 1.27 \[ \frac {1}{3} a d x^3+\frac {1}{5} a e x^5+\frac {b e x^2}{10 c^3}-\frac {b e \log \left (c^2 x^2+1\right )}{10 c^5}+\frac {b d \log \left (c^2 x^2+1\right )}{6 c^3}+\frac {1}{3} b d x^3 \tan ^{-1}(c x)-\frac {b d x^2}{6 c}+\frac {1}{5} b e x^5 \tan ^{-1}(c x)-\frac {b e x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*d*x^2)/c + (b*e*x^2)/(10*c^3) + (a*d*x^3)/3 - (b*e*x^4)/(20*c) + (a*e*x^5)/5 + (b*d*x^3*ArcTan[c*x])/3
+ (b*e*x^5*ArcTan[c*x])/5 + (b*d*Log[1 + c^2*x^2])/(6*c^3) - (b*e*Log[1 + c^2*x^2])/(10*c^5)

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fricas [A]  time = 0.48, size = 107, normalized size = 1.14 \[ \frac {12 \, a c^{5} e x^{5} + 20 \, a c^{5} d x^{3} - 3 \, b c^{4} e x^{4} - 2 \, {\left (5 \, b c^{4} d - 3 \, b c^{2} e\right )} x^{2} + 4 \, {\left (3 \, b c^{5} e x^{5} + 5 \, b c^{5} d x^{3}\right )} \arctan \left (c x\right ) + 2 \, {\left (5 \, b c^{2} d - 3 \, b e\right )} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a*c^5*e*x^5 + 20*a*c^5*d*x^3 - 3*b*c^4*e*x^4 - 2*(5*b*c^4*d - 3*b*c^2*e)*x^2 + 4*(3*b*c^5*e*x^5 + 5*b
*c^5*d*x^3)*arctan(c*x) + 2*(5*b*c^2*d - 3*b*e)*log(c^2*x^2 + 1))/c^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 102, normalized size = 1.09 \[ \frac {a e \,x^{5}}{5}+\frac {a d \,x^{3}}{3}+\frac {b e \,x^{5} \arctan \left (c x \right )}{5}+\frac {b \arctan \left (c x \right ) d \,x^{3}}{3}-\frac {b d \,x^{2}}{6 c}-\frac {b e \,x^{4}}{20 c}+\frac {b e \,x^{2}}{10 c^{3}}+\frac {b d \ln \left (c^{2} x^{2}+1\right )}{6 c^{3}}-\frac {b e \ln \left (c^{2} x^{2}+1\right )}{10 c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e*x^5+1/3*a*d*x^3+1/5*b*e*x^5*arctan(c*x)+1/3*b*arctan(c*x)*d*x^3-1/6*b*d*x^2/c-1/20*b*e*x^4/c+1/10*b*e*
x^2/c^3+1/6*b*d*ln(c^2*x^2+1)/c^3-1/10*b*e*ln(c^2*x^2+1)/c^5

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maxima [A]  time = 0.31, size = 105, normalized size = 1.12 \[ \frac {1}{5} \, a e x^{5} + \frac {1}{3} \, a d x^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e*x^5 + 1/3*a*d*x^3 + 1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d + 1/20*(4*x^5*arc
tan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*e

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mupad [B]  time = 0.57, size = 101, normalized size = 1.07 \[ \frac {a\,d\,x^3}{3}+\frac {a\,e\,x^5}{5}+\frac {b\,d\,x^3\,\mathrm {atan}\left (c\,x\right )}{3}+\frac {b\,e\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}+\frac {b\,d\,\ln \left (c^2\,x^2+1\right )}{6\,c^3}-\frac {b\,e\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}-\frac {b\,d\,x^2}{6\,c}-\frac {b\,e\,x^4}{20\,c}+\frac {b\,e\,x^2}{10\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*d*x^3)/3 + (a*e*x^5)/5 + (b*d*x^3*atan(c*x))/3 + (b*e*x^5*atan(c*x))/5 + (b*d*log(c^2*x^2 + 1))/(6*c^3) - (
b*e*log(c^2*x^2 + 1))/(10*c^5) - (b*d*x^2)/(6*c) - (b*e*x^4)/(20*c) + (b*e*x^2)/(10*c^3)

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sympy [A]  time = 1.38, size = 128, normalized size = 1.36 \[ \begin {cases} \frac {a d x^{3}}{3} + \frac {a e x^{5}}{5} + \frac {b d x^{3} \operatorname {atan}{\left (c x \right )}}{3} + \frac {b e x^{5} \operatorname {atan}{\left (c x \right )}}{5} - \frac {b d x^{2}}{6 c} - \frac {b e x^{4}}{20 c} + \frac {b d \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6 c^{3}} + \frac {b e x^{2}}{10 c^{3}} - \frac {b e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10 c^{5}} & \text {for}\: c \neq 0 \\a \left (\frac {d x^{3}}{3} + \frac {e x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d*x**3/3 + a*e*x**5/5 + b*d*x**3*atan(c*x)/3 + b*e*x**5*atan(c*x)/5 - b*d*x**2/(6*c) - b*e*x**4/(
20*c) + b*d*log(x**2 + c**(-2))/(6*c**3) + b*e*x**2/(10*c**3) - b*e*log(x**2 + c**(-2))/(10*c**5), Ne(c, 0)),
(a*(d*x**3/3 + e*x**5/5), True))

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