3.1229 \(\int x^m (d+e x^2)^2 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=230 \[ -\frac{b x^{m+2} \left (c^4 d^2 \left (m^2+8 m+15\right )-2 c^2 d e \left (m^2+6 m+5\right )+e^2 \left (m^2+4 m+3\right )\right ) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-c^2 x^2\right )}{c^3 (m+1) (m+2) (m+3) (m+5)}+\frac{d^2 x^{m+1} \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac{2 d e x^{m+3} \left (a+b \tan ^{-1}(c x)\right )}{m+3}+\frac{e^2 x^{m+5} \left (a+b \tan ^{-1}(c x)\right )}{m+5}+\frac{b e x^{m+2} \left (e (m+3)-2 c^2 d (m+5)\right )}{c^3 (m+2) (m+3) (m+5)}-\frac{b e^2 x^{m+4}}{c (m+4) (m+5)} \]

[Out]

(b*e*(e*(3 + m) - 2*c^2*d*(5 + m))*x^(2 + m))/(c^3*(2 + m)*(3 + m)*(5 + m)) - (b*e^2*x^(4 + m))/(c*(4 + m)*(5
+ m)) + (d^2*x^(1 + m)*(a + b*ArcTan[c*x]))/(1 + m) + (2*d*e*x^(3 + m)*(a + b*ArcTan[c*x]))/(3 + m) + (e^2*x^(
5 + m)*(a + b*ArcTan[c*x]))/(5 + m) - (b*(e^2*(3 + 4*m + m^2) - 2*c^2*d*e*(5 + 6*m + m^2) + c^4*d^2*(15 + 8*m
+ m^2))*x^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(c^3*(1 + m)*(2 + m)*(3 + m)*(5 + m)
)

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Rubi [A]  time = 0.293679, antiderivative size = 226, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {270, 4976, 1261, 364} \[ \frac{d^2 x^{m+1} \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac{2 d e x^{m+3} \left (a+b \tan ^{-1}(c x)\right )}{m+3}+\frac{e^2 x^{m+5} \left (a+b \tan ^{-1}(c x)\right )}{m+5}-\frac{b x^{m+2} \left (c^4 d^2 \left (m^2+8 m+15\right )-2 c^2 d e \left (m^2+6 m+5\right )+e^2 \left (m^2+4 m+3\right )\right ) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-c^2 x^2\right )}{c^3 (m+1) (m+2) (m+3) (m+5)}-\frac{b e x^{m+2} \left (\frac{2 c^2 d}{m+3}-\frac{e}{m+5}\right )}{c^3 (m+2)}-\frac{b e^2 x^{m+4}}{c (m+4) (m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e*((2*c^2*d)/(3 + m) - e/(5 + m))*x^(2 + m))/(c^3*(2 + m))) - (b*e^2*x^(4 + m))/(c*(4 + m)*(5 + m)) + (d^
2*x^(1 + m)*(a + b*ArcTan[c*x]))/(1 + m) + (2*d*e*x^(3 + m)*(a + b*ArcTan[c*x]))/(3 + m) + (e^2*x^(5 + m)*(a +
b*ArcTan[c*x]))/(5 + m) - (b*(e^2*(3 + 4*m + m^2) - 2*c^2*d*e*(5 + 6*m + m^2) + c^4*d^2*(15 + 8*m + m^2))*x^(
2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(c^3*(1 + m)*(2 + m)*(3 + m)*(5 + m))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int x^m \left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{d^2 x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac{2 d e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\frac{e^2 x^{5+m} \left (a+b \tan ^{-1}(c x)\right )}{5+m}-(b c) \int \frac{x^{1+m} \left (\frac{d^2}{1+m}+\frac{2 d e x^2}{3+m}+\frac{e^2 x^4}{5+m}\right )}{1+c^2 x^2} \, dx\\ &=\frac{d^2 x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac{2 d e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\frac{e^2 x^{5+m} \left (a+b \tan ^{-1}(c x)\right )}{5+m}-(b c) \int \left (\frac{e \left (\frac{2 c^2 d}{3+m}-\frac{e}{5+m}\right ) x^{1+m}}{c^4}+\frac{e^2 x^{3+m}}{c^2 (5+m)}+\frac{\left (15 c^4 d^2-10 c^2 d e+3 e^2+8 c^4 d^2 m-12 c^2 d e m+4 e^2 m+c^4 d^2 m^2-2 c^2 d e m^2+e^2 m^2\right ) x^{1+m}}{c^4 (1+m) (3+m) (5+m) \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{b e \left (\frac{2 c^2 d}{3+m}-\frac{e}{5+m}\right ) x^{2+m}}{c^3 (2+m)}-\frac{b e^2 x^{4+m}}{c (4+m) (5+m)}+\frac{d^2 x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac{2 d e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\frac{e^2 x^{5+m} \left (a+b \tan ^{-1}(c x)\right )}{5+m}-\frac{\left (b \left (e^2 \left (3+4 m+m^2\right )-2 c^2 d e \left (5+6 m+m^2\right )+c^4 d^2 \left (15+8 m+m^2\right )\right )\right ) \int \frac{x^{1+m}}{1+c^2 x^2} \, dx}{c^3 (1+m) (3+m) (5+m)}\\ &=-\frac{b e \left (\frac{2 c^2 d}{3+m}-\frac{e}{5+m}\right ) x^{2+m}}{c^3 (2+m)}-\frac{b e^2 x^{4+m}}{c (4+m) (5+m)}+\frac{d^2 x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac{2 d e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\frac{e^2 x^{5+m} \left (a+b \tan ^{-1}(c x)\right )}{5+m}-\frac{b \left (e^2 \left (3+4 m+m^2\right )-2 c^2 d e \left (5+6 m+m^2\right )+c^4 d^2 \left (15+8 m+m^2\right )\right ) x^{2+m} \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-c^2 x^2\right )}{c^3 (1+m) (2+m) (3+m) (5+m)}\\ \end{align*}

Mathematica [A]  time = 0.205338, size = 193, normalized size = 0.84 \[ x^{m+1} \left (-\frac{b c d^2 x \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-c^2 x^2\right )}{m^2+3 m+2}-\frac{2 b c d e x^3 \text{Hypergeometric2F1}\left (1,\frac{m+4}{2},\frac{m+6}{2},-c^2 x^2\right )}{m^2+7 m+12}-\frac{b c e^2 x^5 \text{Hypergeometric2F1}\left (1,\frac{m+6}{2},\frac{m+8}{2},-c^2 x^2\right )}{(m+5) (m+6)}+\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac{2 d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{m+3}+\frac{e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{m+5}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^(1 + m)*((d^2*(a + b*ArcTan[c*x]))/(1 + m) + (2*d*e*x^2*(a + b*ArcTan[c*x]))/(3 + m) + (e^2*x^4*(a + b*ArcTa
n[c*x]))/(5 + m) - (b*c*d^2*x*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(2 + 3*m + m^2) - (2*b*c
*d*e*x^3*Hypergeometric2F1[1, (4 + m)/2, (6 + m)/2, -(c^2*x^2)])/(12 + 7*m + m^2) - (b*c*e^2*x^5*Hypergeometri
c2F1[1, (6 + m)/2, (8 + m)/2, -(c^2*x^2)])/((5 + m)*(6 + m)))

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Maple [F]  time = 0.935, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( e{x}^{2}+d \right ) ^{2} \left ( a+b\arctan \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} +{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \arctan \left (c x\right )\right )} x^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arctan(c*x))*x^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \left (a + b \operatorname{atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}^{2}{\left (b \arctan \left (c x\right ) + a\right )} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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