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

Optimal. Leaf size=230 \[ \frac {d^2 x^{m+1} \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac {2 d e x^{m+3} \left (a+b \tan ^{-1}(c x)\right )}{m+3}+\frac {e^2 x^{m+5} \left (a+b \tan ^{-1}(c x)\right )}{m+5}+\frac {b e x^{m+2} \left (e (m+3)-2 c^2 d (m+5)\right )}{c^3 (m+2) (m+3) (m+5)}-\frac {b x^{m+2} \left (c^4 d^2 \left (m^2+8 m+15\right )-2 c^2 d e \left (m^2+6 m+5\right )+e^2 \left (m^2+4 m+3\right )\right ) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{c^3 (m+1) (m+2) (m+3) (m+5)}-\frac {b e^2 x^{m+4}}{c (m+4) (m+5)} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 226, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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])

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 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^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*}

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Mathematica [A]  time = 0.22, size = 193, normalized size = 0.84 \[ x^{m+1} \left (\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac {2 d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{m+3}+\frac {e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{m+5}-\frac {b c d^2 x \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{m^2+3 m+2}-\frac {2 b c d e x^3 \, _2F_1\left (1,\frac {m+4}{2};\frac {m+6}{2};-c^2 x^2\right )}{m^2+7 m+12}-\frac {b c e^2 x^5 \, _2F_1\left (1,\frac {m+6}{2};\frac {m+8}{2};-c^2 x^2\right )}{(m+5) (m+6)}\right ) \]

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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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} + {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \arctan \left (c x\right )\right )} x^{m}, x\right ) \]

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

[Out]

sage0*x

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int x^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \arctan \left (c x \right )\right )\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a e^{2} x^{m + 5}}{m + 5} + \frac {2 \, a d e x^{m + 3}}{m + 3} + \frac {a d^{2} x^{m + 1}}{m + 1} + \frac {{\left ({\left (b e^{2} m^{2} + 4 \, b e^{2} m + 3 \, b e^{2}\right )} x^{5} + 2 \, {\left (b d e m^{2} + 6 \, b d e m + 5 \, b d e\right )} x^{3} + {\left (b d^{2} m^{2} + 8 \, b d^{2} m + 15 \, b d^{2}\right )} x\right )} x^{m} \arctan \left (c x\right ) - {\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} \int \frac {{\left ({\left (b c e^{2} m^{2} + 4 \, b c e^{2} m + 3 \, b c e^{2}\right )} x^{5} + 2 \, {\left (b c d e m^{2} + 6 \, b c d e m + 5 \, b c d e\right )} x^{3} + {\left (b c d^{2} m^{2} + 8 \, b c d^{2} m + 15 \, b c d^{2}\right )} x\right )} x^{m}}{m^{3} + {\left (c^{2} m^{3} + 9 \, c^{2} m^{2} + 23 \, c^{2} m + 15 \, c^{2}\right )} x^{2} + 9 \, m^{2} + 23 \, m + 15}\,{d x}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

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]

a*e^2*x^(m + 5)/(m + 5) + 2*a*d*e*x^(m + 3)/(m + 3) + a*d^2*x^(m + 1)/(m + 1) + (((b*e^2*m^2 + 4*b*e^2*m + 3*b
*e^2)*x^5 + 2*(b*d*e*m^2 + 6*b*d*e*m + 5*b*d*e)*x^3 + (b*d^2*m^2 + 8*b*d^2*m + 15*b*d^2)*x)*x^m*arctan(c*x) -
(m^3 + 9*m^2 + 23*m + 15)*integrate(((b*c*e^2*m^2 + 4*b*c*e^2*m + 3*b*c*e^2)*x^5 + 2*(b*c*d*e*m^2 + 6*b*c*d*e*
m + 5*b*c*d*e)*x^3 + (b*c*d^2*m^2 + 8*b*c*d^2*m + 15*b*c*d^2)*x)*x^m/(m^3 + (c^2*m^3 + 9*c^2*m^2 + 23*c^2*m +
15*c^2)*x^2 + 9*m^2 + 23*m + 15), x))/(m^3 + 9*m^2 + 23*m + 15)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^m\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \]

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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