3.1228 \(\int x^m (d+e x^2)^3 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=378 \[ \frac{b x^{m+2} \left (3 c^4 d^2 e \left (m^3+13 m^2+47 m+35\right )+c^6 \left (-d^3\right ) \left (m^3+15 m^2+71 m+105\right )-3 c^2 d e^2 \left (m^3+11 m^2+31 m+21\right )+e^3 \left (m^3+9 m^2+23 m+15\right )\right ) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-c^2 x^2\right )}{c^5 (m+1) (m+2) (m+3) (m+5) (m+7)}+\frac{3 d^2 e x^{m+3} \left (a+b \tan ^{-1}(c x)\right )}{m+3}+\frac{d^3 x^{m+1} \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac{3 d e^2 x^{m+5} \left (a+b \tan ^{-1}(c x)\right )}{m+5}+\frac{e^3 x^{m+7} \left (a+b \tan ^{-1}(c x)\right )}{m+7}-\frac{b e x^{m+2} \left (3 c^4 d^2 \left (m^2+12 m+35\right )-3 c^2 d e \left (m^2+10 m+21\right )+e^2 \left (m^2+8 m+15\right )\right )}{c^5 (m+2) (m+3) (m+5) (m+7)}+\frac{b e^2 x^{m+4} \left (e (m+5)-3 c^2 d (m+7)\right )}{c^3 (m+4) (m+5) (m+7)}-\frac{b e^3 x^{m+6}}{c (m+6) (m+7)} \]

[Out]

-((b*e*(e^2*(15 + 8*m + m^2) - 3*c^2*d*e*(21 + 10*m + m^2) + 3*c^4*d^2*(35 + 12*m + m^2))*x^(2 + m))/(c^5*(2 +
m)*(3 + m)*(5 + m)*(7 + m))) + (b*e^2*(e*(5 + m) - 3*c^2*d*(7 + m))*x^(4 + m))/(c^3*(4 + m)*(5 + m)*(7 + m))
- (b*e^3*x^(6 + m))/(c*(6 + m)*(7 + m)) + (d^3*x^(1 + m)*(a + b*ArcTan[c*x]))/(1 + m) + (3*d^2*e*x^(3 + m)*(a
+ b*ArcTan[c*x]))/(3 + m) + (3*d*e^2*x^(5 + m)*(a + b*ArcTan[c*x]))/(5 + m) + (e^3*x^(7 + m)*(a + b*ArcTan[c*x
]))/(7 + m) + (b*(e^3*(15 + 23*m + 9*m^2 + m^3) - 3*c^2*d*e^2*(21 + 31*m + 11*m^2 + m^3) + 3*c^4*d^2*e*(35 + 4
7*m + 13*m^2 + m^3) - c^6*d^3*(105 + 71*m + 15*m^2 + m^3))*x^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2
, -(c^2*x^2)])/(c^5*(1 + m)*(2 + m)*(3 + m)*(5 + m)*(7 + m))

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Rubi [A]  time = 1.98322, antiderivative size = 374, normalized size of antiderivative = 0.99, 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, 1802, 364} \[ \frac{3 d^2 e x^{m+3} \left (a+b \tan ^{-1}(c x)\right )}{m+3}+\frac{d^3 x^{m+1} \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac{3 d e^2 x^{m+5} \left (a+b \tan ^{-1}(c x)\right )}{m+5}+\frac{e^3 x^{m+7} \left (a+b \tan ^{-1}(c x)\right )}{m+7}+\frac{b x^{m+2} \left (3 c^4 d^2 e \left (m^3+13 m^2+47 m+35\right )+c^6 \left (-d^3\right ) \left (m^3+15 m^2+71 m+105\right )-3 c^2 d e^2 \left (m^3+11 m^2+31 m+21\right )+e^3 \left (m^3+9 m^2+23 m+15\right )\right ) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-c^2 x^2\right )}{c^5 (m+1) (m+2) (m+3) (m+5) (m+7)}-\frac{b e x^{m+2} \left (3 c^4 d^2 \left (m^2+12 m+35\right )-3 c^2 d e \left (m^2+10 m+21\right )+e^2 \left (m^2+8 m+15\right )\right )}{c^5 (m+2) (m+3) (m+5) (m+7)}-\frac{b e^2 x^{m+4} \left (\frac{3 c^2 d}{m+5}-\frac{e}{m+7}\right )}{c^3 (m+4)}-\frac{b e^3 x^{m+6}}{c (m+6) (m+7)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e*(e^2*(15 + 8*m + m^2) - 3*c^2*d*e*(21 + 10*m + m^2) + 3*c^4*d^2*(35 + 12*m + m^2))*x^(2 + m))/(c^5*(2 +
m)*(3 + m)*(5 + m)*(7 + m))) - (b*e^2*((3*c^2*d)/(5 + m) - e/(7 + m))*x^(4 + m))/(c^3*(4 + m)) - (b*e^3*x^(6
+ m))/(c*(6 + m)*(7 + m)) + (d^3*x^(1 + m)*(a + b*ArcTan[c*x]))/(1 + m) + (3*d^2*e*x^(3 + m)*(a + b*ArcTan[c*x
]))/(3 + m) + (3*d*e^2*x^(5 + m)*(a + b*ArcTan[c*x]))/(5 + m) + (e^3*x^(7 + m)*(a + b*ArcTan[c*x]))/(7 + m) +
(b*(e^3*(15 + 23*m + 9*m^2 + m^3) - 3*c^2*d*e^2*(21 + 31*m + 11*m^2 + m^3) + 3*c^4*d^2*e*(35 + 47*m + 13*m^2 +
m^3) - c^6*d^3*(105 + 71*m + 15*m^2 + m^3))*x^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])
/(c^5*(1 + m)*(2 + m)*(3 + m)*(5 + m)*(7 + 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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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 )^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{d^3 x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac{3 d^2 e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\frac{3 d e^2 x^{5+m} \left (a+b \tan ^{-1}(c x)\right )}{5+m}+\frac{e^3 x^{7+m} \left (a+b \tan ^{-1}(c x)\right )}{7+m}-(b c) \int \frac{x^{1+m} \left (\frac{d^3}{1+m}+\frac{3 d^2 e x^2}{3+m}+\frac{3 d e^2 x^4}{5+m}+\frac{e^3 x^6}{7+m}\right )}{1+c^2 x^2} \, dx\\ &=\frac{d^3 x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac{3 d^2 e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\frac{3 d e^2 x^{5+m} \left (a+b \tan ^{-1}(c x)\right )}{5+m}+\frac{e^3 x^{7+m} \left (a+b \tan ^{-1}(c x)\right )}{7+m}-(b c) \int \left (\frac{e \left (e^2 \left (15+8 m+m^2\right )-3 c^2 d e \left (21+10 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )\right ) x^{1+m}}{c^6 (3+m) (5+m) (7+m)}+\frac{e^2 \left (\frac{3 c^2 d}{5+m}-\frac{e}{7+m}\right ) x^{3+m}}{c^4}+\frac{e^3 x^{5+m}}{c^2 (7+m)}+\frac{\left (105 c^6 d^3-105 c^4 d^2 e+63 c^2 d e^2-15 e^3+71 c^6 d^3 m-141 c^4 d^2 e m+93 c^2 d e^2 m-23 e^3 m+15 c^6 d^3 m^2-39 c^4 d^2 e m^2+33 c^2 d e^2 m^2-9 e^3 m^2+c^6 d^3 m^3-3 c^4 d^2 e m^3+3 c^2 d e^2 m^3-e^3 m^3\right ) x^{1+m}}{c^6 (1+m) (3+m) (5+m) (7+m) \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{b e \left (e^2 \left (15+8 m+m^2\right )-3 c^2 d e \left (21+10 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )\right ) x^{2+m}}{c^5 (2+m) (3+m) (5+m) (7+m)}-\frac{b e^2 \left (\frac{3 c^2 d}{5+m}-\frac{e}{7+m}\right ) x^{4+m}}{c^3 (4+m)}-\frac{b e^3 x^{6+m}}{c (6+m) (7+m)}+\frac{d^3 x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac{3 d^2 e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\frac{3 d e^2 x^{5+m} \left (a+b \tan ^{-1}(c x)\right )}{5+m}+\frac{e^3 x^{7+m} \left (a+b \tan ^{-1}(c x)\right )}{7+m}+\frac{\left (b \left (e^3 \left (15+23 m+9 m^2+m^3\right )-3 c^2 d e^2 \left (21+31 m+11 m^2+m^3\right )+3 c^4 d^2 e \left (35+47 m+13 m^2+m^3\right )-c^6 d^3 \left (105+71 m+15 m^2+m^3\right )\right )\right ) \int \frac{x^{1+m}}{1+c^2 x^2} \, dx}{c^5 (1+m) (3+m) (5+m) (7+m)}\\ &=-\frac{b e \left (e^2 \left (15+8 m+m^2\right )-3 c^2 d e \left (21+10 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )\right ) x^{2+m}}{c^5 (2+m) (3+m) (5+m) (7+m)}-\frac{b e^2 \left (\frac{3 c^2 d}{5+m}-\frac{e}{7+m}\right ) x^{4+m}}{c^3 (4+m)}-\frac{b e^3 x^{6+m}}{c (6+m) (7+m)}+\frac{d^3 x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac{3 d^2 e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\frac{3 d e^2 x^{5+m} \left (a+b \tan ^{-1}(c x)\right )}{5+m}+\frac{e^3 x^{7+m} \left (a+b \tan ^{-1}(c x)\right )}{7+m}+\frac{b \left (e^3 \left (15+23 m+9 m^2+m^3\right )-3 c^2 d e^2 \left (21+31 m+11 m^2+m^3\right )+3 c^4 d^2 e \left (35+47 m+13 m^2+m^3\right )-c^6 d^3 \left (105+71 m+15 m^2+m^3\right )\right ) x^{2+m} \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-c^2 x^2\right )}{c^5 (1+m) (2+m) (3+m) (5+m) (7+m)}\\ \end{align*}

Mathematica [A]  time = 0.581661, size = 264, normalized size = 0.7 \[ x^{m+1} \left (-\frac{3 b c d^2 e x^3 \text{Hypergeometric2F1}\left (1,\frac{m+4}{2},\frac{m+6}{2},-c^2 x^2\right )}{m^2+7 m+12}-\frac{b c d^3 x \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-c^2 x^2\right )}{m^2+3 m+2}-\frac{3 b c d e^2 x^5 \text{Hypergeometric2F1}\left (1,\frac{m+6}{2},\frac{m+8}{2},-c^2 x^2\right )}{(m+5) (m+6)}-\frac{b c e^3 x^7 \text{Hypergeometric2F1}\left (1,\frac{m}{2}+4,\frac{m}{2}+5,-c^2 x^2\right )}{(m+7) (m+8)}+\frac{3 d^2 e x^2 \left (a+b \tan ^{-1}(c x)\right )}{m+3}+\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac{3 d e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{m+5}+\frac{e^3 x^6 \left (a+b \tan ^{-1}(c x)\right )}{m+7}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.137, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( e{x}^{2}+d \right ) ^{3} \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)^3*(a+b*arctan(c*x)),x)

[Out]

int(x^m*(e*x^2+d)^3*(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)^3*(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^{3} x^{6} + 3 \, a d e^{2} x^{4} + 3 \, a d^{2} e x^{2} + a d^{3} +{\left (b e^{3} x^{6} + 3 \, b d e^{2} x^{4} + 3 \, b d^{2} e x^{2} + b d^{3}\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)^3*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

integral((a*e^3*x^6 + 3*a*d*e^2*x^4 + 3*a*d^2*e*x^2 + a*d^3 + (b*e^3*x^6 + 3*b*d*e^2*x^4 + 3*b*d^2*e*x^2 + b*d
^3)*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 )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}^{3}{\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)^3*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

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