3.96 \(\int x^2 (A+B x^2) (a+b x^2+c x^4)^3 \, dx\)

Optimal. Leaf size=166 \[ \frac{1}{5} a^2 x^5 (a B+3 A b)+\frac{1}{3} a^3 A x^3+\frac{1}{11} x^{11} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{3}{13} c x^{13} \left (a B c+A b c+b^2 B\right )+\frac{1}{9} x^9 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac{3}{7} a x^7 \left (A \left (a c+b^2\right )+a b B\right )+\frac{1}{15} c^2 x^{15} (A c+3 b B)+\frac{1}{17} B c^3 x^{17} \]

[Out]

(a^3*A*x^3)/3 + (a^2*(3*A*b + a*B)*x^5)/5 + (3*a*(a*b*B + A*(b^2 + a*c))*x^7)/7 + ((3*a*B*(b^2 + a*c) + A*(b^3
+ 6*a*b*c))*x^9)/9 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^11)/11 + (3*c*(b^2*B + A*b*c + a*B*c)*x^1
3)/13 + (c^2*(3*b*B + A*c)*x^15)/15 + (B*c^3*x^17)/17

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Rubi [A]  time = 0.153507, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {1261} \[ \frac{1}{5} a^2 x^5 (a B+3 A b)+\frac{1}{3} a^3 A x^3+\frac{1}{11} x^{11} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{3}{13} c x^{13} \left (a B c+A b c+b^2 B\right )+\frac{1}{9} x^9 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac{3}{7} a x^7 \left (A \left (a c+b^2\right )+a b B\right )+\frac{1}{15} c^2 x^{15} (A c+3 b B)+\frac{1}{17} B c^3 x^{17} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(A + B*x^2)*(a + b*x^2 + c*x^4)^3,x]

[Out]

(a^3*A*x^3)/3 + (a^2*(3*A*b + a*B)*x^5)/5 + (3*a*(a*b*B + A*(b^2 + a*c))*x^7)/7 + ((3*a*B*(b^2 + a*c) + A*(b^3
+ 6*a*b*c))*x^9)/9 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^11)/11 + (3*c*(b^2*B + A*b*c + a*B*c)*x^1
3)/13 + (c^2*(3*b*B + A*c)*x^15)/15 + (B*c^3*x^17)/17

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]

Rubi steps

\begin{align*} \int x^2 \left (A+B x^2\right ) \left (a+b x^2+c x^4\right )^3 \, dx &=\int \left (a^3 A x^2+a^2 (3 A b+a B) x^4+3 a \left (a b B+A \left (b^2+a c\right )\right ) x^6+\left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^8+\left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^{10}+3 c \left (b^2 B+A b c+a B c\right ) x^{12}+c^2 (3 b B+A c) x^{14}+B c^3 x^{16}\right ) \, dx\\ &=\frac{1}{3} a^3 A x^3+\frac{1}{5} a^2 (3 A b+a B) x^5+\frac{3}{7} a \left (a b B+A \left (b^2+a c\right )\right ) x^7+\frac{1}{9} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^9+\frac{1}{11} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^{11}+\frac{3}{13} c \left (b^2 B+A b c+a B c\right ) x^{13}+\frac{1}{15} c^2 (3 b B+A c) x^{15}+\frac{1}{17} B c^3 x^{17}\\ \end{align*}

Mathematica [A]  time = 0.05703, size = 166, normalized size = 1. \[ \frac{1}{5} a^2 x^5 (a B+3 A b)+\frac{1}{3} a^3 A x^3+\frac{1}{11} x^{11} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{3}{13} c x^{13} \left (a B c+A b c+b^2 B\right )+\frac{1}{9} x^9 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac{3}{7} a x^7 \left (A \left (a c+b^2\right )+a b B\right )+\frac{1}{15} c^2 x^{15} (A c+3 b B)+\frac{1}{17} B c^3 x^{17} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(A + B*x^2)*(a + b*x^2 + c*x^4)^3,x]

[Out]

(a^3*A*x^3)/3 + (a^2*(3*A*b + a*B)*x^5)/5 + (3*a*(a*b*B + A*(b^2 + a*c))*x^7)/7 + ((3*a*B*(b^2 + a*c) + A*(b^3
+ 6*a*b*c))*x^9)/9 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^11)/11 + (3*c*(b^2*B + A*b*c + a*B*c)*x^1
3)/13 + (c^2*(3*b*B + A*c)*x^15)/15 + (B*c^3*x^17)/17

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Maple [A]  time = 0., size = 226, normalized size = 1.4 \begin{align*}{\frac{B{c}^{3}{x}^{17}}{17}}+{\frac{ \left ( A{c}^{3}+3\,Bb{c}^{2} \right ){x}^{15}}{15}}+{\frac{ \left ( 3\,Ab{c}^{2}+B \left ( a{c}^{2}+2\,{b}^{2}c+c \left ( 2\,ac+{b}^{2} \right ) \right ) \right ){x}^{13}}{13}}+{\frac{ \left ( A \left ( a{c}^{2}+2\,{b}^{2}c+c \left ( 2\,ac+{b}^{2} \right ) \right ) +B \left ( 4\,abc+b \left ( 2\,ac+{b}^{2} \right ) \right ) \right ){x}^{11}}{11}}+{\frac{ \left ( A \left ( 4\,abc+b \left ( 2\,ac+{b}^{2} \right ) \right ) +B \left ( a \left ( 2\,ac+{b}^{2} \right ) +2\,{b}^{2}a+c{a}^{2} \right ) \right ){x}^{9}}{9}}+{\frac{ \left ( A \left ( a \left ( 2\,ac+{b}^{2} \right ) +2\,{b}^{2}a+c{a}^{2} \right ) +3\,B{a}^{2}b \right ){x}^{7}}{7}}+{\frac{ \left ( 3\,A{a}^{2}b+B{a}^{3} \right ){x}^{5}}{5}}+{\frac{{a}^{3}A{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x^2+A)*(c*x^4+b*x^2+a)^3,x)

[Out]

1/17*B*c^3*x^17+1/15*(A*c^3+3*B*b*c^2)*x^15+1/13*(3*A*b*c^2+B*(a*c^2+2*b^2*c+c*(2*a*c+b^2)))*x^13+1/11*(A*(a*c
^2+2*b^2*c+c*(2*a*c+b^2))+B*(4*a*b*c+b*(2*a*c+b^2)))*x^11+1/9*(A*(4*a*b*c+b*(2*a*c+b^2))+B*(a*(2*a*c+b^2)+2*b^
2*a+c*a^2))*x^9+1/7*(A*(a*(2*a*c+b^2)+2*b^2*a+c*a^2)+3*B*a^2*b)*x^7+1/5*(3*A*a^2*b+B*a^3)*x^5+1/3*a^3*A*x^3

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Maxima [A]  time = 0.988011, size = 224, normalized size = 1.35 \begin{align*} \frac{1}{17} \, B c^{3} x^{17} + \frac{1}{15} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{15} + \frac{3}{13} \,{\left (B b^{2} c +{\left (B a + A b\right )} c^{2}\right )} x^{13} + \frac{1}{11} \,{\left (B b^{3} + 3 \, A a c^{2} + 3 \,{\left (2 \, B a b + A b^{2}\right )} c\right )} x^{11} + \frac{1}{9} \,{\left (3 \, B a b^{2} + A b^{3} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{9} + \frac{3}{7} \,{\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{7} + \frac{1}{3} \, A a^{3} x^{3} + \frac{1}{5} \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/17*B*c^3*x^17 + 1/15*(3*B*b*c^2 + A*c^3)*x^15 + 3/13*(B*b^2*c + (B*a + A*b)*c^2)*x^13 + 1/11*(B*b^3 + 3*A*a*
c^2 + 3*(2*B*a*b + A*b^2)*c)*x^11 + 1/9*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^9 + 3/7*(B*a^2*b + A*a*b
^2 + A*a^2*c)*x^7 + 1/3*A*a^3*x^3 + 1/5*(B*a^3 + 3*A*a^2*b)*x^5

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Fricas [A]  time = 1.25964, size = 494, normalized size = 2.98 \begin{align*} \frac{1}{17} x^{17} c^{3} B + \frac{1}{5} x^{15} c^{2} b B + \frac{1}{15} x^{15} c^{3} A + \frac{3}{13} x^{13} c b^{2} B + \frac{3}{13} x^{13} c^{2} a B + \frac{3}{13} x^{13} c^{2} b A + \frac{1}{11} x^{11} b^{3} B + \frac{6}{11} x^{11} c b a B + \frac{3}{11} x^{11} c b^{2} A + \frac{3}{11} x^{11} c^{2} a A + \frac{1}{3} x^{9} b^{2} a B + \frac{1}{3} x^{9} c a^{2} B + \frac{1}{9} x^{9} b^{3} A + \frac{2}{3} x^{9} c b a A + \frac{3}{7} x^{7} b a^{2} B + \frac{3}{7} x^{7} b^{2} a A + \frac{3}{7} x^{7} c a^{2} A + \frac{1}{5} x^{5} a^{3} B + \frac{3}{5} x^{5} b a^{2} A + \frac{1}{3} x^{3} a^{3} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/17*x^17*c^3*B + 1/5*x^15*c^2*b*B + 1/15*x^15*c^3*A + 3/13*x^13*c*b^2*B + 3/13*x^13*c^2*a*B + 3/13*x^13*c^2*b
*A + 1/11*x^11*b^3*B + 6/11*x^11*c*b*a*B + 3/11*x^11*c*b^2*A + 3/11*x^11*c^2*a*A + 1/3*x^9*b^2*a*B + 1/3*x^9*c
*a^2*B + 1/9*x^9*b^3*A + 2/3*x^9*c*b*a*A + 3/7*x^7*b*a^2*B + 3/7*x^7*b^2*a*A + 3/7*x^7*c*a^2*A + 1/5*x^5*a^3*B
+ 3/5*x^5*b*a^2*A + 1/3*x^3*a^3*A

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Sympy [A]  time = 0.098198, size = 204, normalized size = 1.23 \begin{align*} \frac{A a^{3} x^{3}}{3} + \frac{B c^{3} x^{17}}{17} + x^{15} \left (\frac{A c^{3}}{15} + \frac{B b c^{2}}{5}\right ) + x^{13} \left (\frac{3 A b c^{2}}{13} + \frac{3 B a c^{2}}{13} + \frac{3 B b^{2} c}{13}\right ) + x^{11} \left (\frac{3 A a c^{2}}{11} + \frac{3 A b^{2} c}{11} + \frac{6 B a b c}{11} + \frac{B b^{3}}{11}\right ) + x^{9} \left (\frac{2 A a b c}{3} + \frac{A b^{3}}{9} + \frac{B a^{2} c}{3} + \frac{B a b^{2}}{3}\right ) + x^{7} \left (\frac{3 A a^{2} c}{7} + \frac{3 A a b^{2}}{7} + \frac{3 B a^{2} b}{7}\right ) + x^{5} \left (\frac{3 A a^{2} b}{5} + \frac{B a^{3}}{5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x**2+A)*(c*x**4+b*x**2+a)**3,x)

[Out]

A*a**3*x**3/3 + B*c**3*x**17/17 + x**15*(A*c**3/15 + B*b*c**2/5) + x**13*(3*A*b*c**2/13 + 3*B*a*c**2/13 + 3*B*
b**2*c/13) + x**11*(3*A*a*c**2/11 + 3*A*b**2*c/11 + 6*B*a*b*c/11 + B*b**3/11) + x**9*(2*A*a*b*c/3 + A*b**3/9 +
B*a**2*c/3 + B*a*b**2/3) + x**7*(3*A*a**2*c/7 + 3*A*a*b**2/7 + 3*B*a**2*b/7) + x**5*(3*A*a**2*b/5 + B*a**3/5)

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Giac [A]  time = 1.1507, size = 261, normalized size = 1.57 \begin{align*} \frac{1}{17} \, B c^{3} x^{17} + \frac{1}{5} \, B b c^{2} x^{15} + \frac{1}{15} \, A c^{3} x^{15} + \frac{3}{13} \, B b^{2} c x^{13} + \frac{3}{13} \, B a c^{2} x^{13} + \frac{3}{13} \, A b c^{2} x^{13} + \frac{1}{11} \, B b^{3} x^{11} + \frac{6}{11} \, B a b c x^{11} + \frac{3}{11} \, A b^{2} c x^{11} + \frac{3}{11} \, A a c^{2} x^{11} + \frac{1}{3} \, B a b^{2} x^{9} + \frac{1}{9} \, A b^{3} x^{9} + \frac{1}{3} \, B a^{2} c x^{9} + \frac{2}{3} \, A a b c x^{9} + \frac{3}{7} \, B a^{2} b x^{7} + \frac{3}{7} \, A a b^{2} x^{7} + \frac{3}{7} \, A a^{2} c x^{7} + \frac{1}{5} \, B a^{3} x^{5} + \frac{3}{5} \, A a^{2} b x^{5} + \frac{1}{3} \, A a^{3} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2+a)^3,x, algorithm="giac")

[Out]

1/17*B*c^3*x^17 + 1/5*B*b*c^2*x^15 + 1/15*A*c^3*x^15 + 3/13*B*b^2*c*x^13 + 3/13*B*a*c^2*x^13 + 3/13*A*b*c^2*x^
13 + 1/11*B*b^3*x^11 + 6/11*B*a*b*c*x^11 + 3/11*A*b^2*c*x^11 + 3/11*A*a*c^2*x^11 + 1/3*B*a*b^2*x^9 + 1/9*A*b^3
*x^9 + 1/3*B*a^2*c*x^9 + 2/3*A*a*b*c*x^9 + 3/7*B*a^2*b*x^7 + 3/7*A*a*b^2*x^7 + 3/7*A*a^2*c*x^7 + 1/5*B*a^3*x^5
+ 3/5*A*a^2*b*x^5 + 1/3*A*a^3*x^3