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

Optimal. Leaf size=166 \[ \frac{1}{4} a^2 x^4 (a B+3 A b)+\frac{1}{2} a^3 A x^2+\frac{1}{10} x^{10} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{1}{4} c x^{12} \left (a B c+A b c+b^2 B\right )+\frac{1}{8} x^8 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac{1}{2} a x^6 \left (A \left (a c+b^2\right )+a b B\right )+\frac{1}{14} c^2 x^{14} (A c+3 b B)+\frac{1}{16} B c^3 x^{16} \]

[Out]

(a^3*A*x^2)/2 + (a^2*(3*A*b + a*B)*x^4)/4 + (a*(a*b*B + A*(b^2 + a*c))*x^6)/2 + ((3*a*B*(b^2 + a*c) + A*(b^3 +
6*a*b*c))*x^8)/8 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^10)/10 + (c*(b^2*B + A*b*c + a*B*c)*x^12)/4
+ (c^2*(3*b*B + A*c)*x^14)/14 + (B*c^3*x^16)/16

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Rubi [A]  time = 0.286762, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {1247, 631} \[ \frac{1}{4} a^2 x^4 (a B+3 A b)+\frac{1}{2} a^3 A x^2+\frac{1}{10} x^{10} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{1}{4} c x^{12} \left (a B c+A b c+b^2 B\right )+\frac{1}{8} x^8 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac{1}{2} a x^6 \left (A \left (a c+b^2\right )+a b B\right )+\frac{1}{14} c^2 x^{14} (A c+3 b B)+\frac{1}{16} B c^3 x^{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*A*x^2)/2 + (a^2*(3*A*b + a*B)*x^4)/4 + (a*(a*b*B + A*(b^2 + a*c))*x^6)/2 + ((3*a*B*(b^2 + a*c) + A*(b^3 +
6*a*b*c))*x^8)/8 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^10)/10 + (c*(b^2*B + A*b*c + a*B*c)*x^12)/4
+ (c^2*(3*b*B + A*c)*x^14)/14 + (B*c^3*x^16)/16

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 631

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.0620843, size = 154, normalized size = 0.93 \[ \frac{1}{560} x^2 \left (140 a^2 x^2 (a B+3 A b)+280 a^3 A+56 x^8 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+140 c x^{10} \left (a B c+A b c+b^2 B\right )+70 x^6 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+280 a x^4 \left (A \left (a c+b^2\right )+a b B\right )+40 c^2 x^{12} (A c+3 b B)+35 B c^3 x^{14}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(280*a^3*A + 140*a^2*(3*A*b + a*B)*x^2 + 280*a*(a*b*B + A*(b^2 + a*c))*x^4 + 70*(3*a*B*(b^2 + a*c) + A*(b
^3 + 6*a*b*c))*x^6 + 56*(b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^8 + 140*c*(b^2*B + A*b*c + a*B*c)*x^10 +
40*c^2*(3*b*B + A*c)*x^12 + 35*B*c^3*x^14))/560

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Maple [A]  time = 0.001, size = 226, normalized size = 1.4 \begin{align*}{\frac{B{c}^{3}{x}^{16}}{16}}+{\frac{ \left ( A{c}^{3}+3\,Bb{c}^{2} \right ){x}^{14}}{14}}+{\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}^{12}}{12}}+{\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}^{10}}{10}}+{\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}^{8}}{8}}+{\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}^{6}}{6}}+{\frac{ \left ( 3\,A{a}^{2}b+B{a}^{3} \right ){x}^{4}}{4}}+{\frac{{a}^{3}A{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*B*c^3*x^16+1/14*(A*c^3+3*B*b*c^2)*x^14+1/12*(3*A*b*c^2+B*(a*c^2+2*b^2*c+c*(2*a*c+b^2)))*x^12+1/10*(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^10+1/8*(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^8+1/6*(A*(a*(2*a*c+b^2)+2*b^2*a+c*a^2)+3*B*a^2*b)*x^6+1/4*(3*A*a^2*b+B*a^3)*x^4+1/2*a^3*A*x^2

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Maxima [A]  time = 1.09262, size = 224, normalized size = 1.35 \begin{align*} \frac{1}{16} \, B c^{3} x^{16} + \frac{1}{14} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{14} + \frac{1}{4} \,{\left (B b^{2} c +{\left (B a + A b\right )} c^{2}\right )} x^{12} + \frac{1}{10} \,{\left (B b^{3} + 3 \, A a c^{2} + 3 \,{\left (2 \, B a b + A b^{2}\right )} c\right )} x^{10} + \frac{1}{8} \,{\left (3 \, B a b^{2} + A b^{3} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{8} + \frac{1}{2} \,{\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{6} + \frac{1}{2} \, A a^{3} x^{2} + \frac{1}{4} \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*B*c^3*x^16 + 1/14*(3*B*b*c^2 + A*c^3)*x^14 + 1/4*(B*b^2*c + (B*a + A*b)*c^2)*x^12 + 1/10*(B*b^3 + 3*A*a*c
^2 + 3*(2*B*a*b + A*b^2)*c)*x^10 + 1/8*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^8 + 1/2*(B*a^2*b + A*a*b^
2 + A*a^2*c)*x^6 + 1/2*A*a^3*x^2 + 1/4*(B*a^3 + 3*A*a^2*b)*x^4

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Fricas [A]  time = 1.27122, size = 490, normalized size = 2.95 \begin{align*} \frac{1}{16} x^{16} c^{3} B + \frac{3}{14} x^{14} c^{2} b B + \frac{1}{14} x^{14} c^{3} A + \frac{1}{4} x^{12} c b^{2} B + \frac{1}{4} x^{12} c^{2} a B + \frac{1}{4} x^{12} c^{2} b A + \frac{1}{10} x^{10} b^{3} B + \frac{3}{5} x^{10} c b a B + \frac{3}{10} x^{10} c b^{2} A + \frac{3}{10} x^{10} c^{2} a A + \frac{3}{8} x^{8} b^{2} a B + \frac{3}{8} x^{8} c a^{2} B + \frac{1}{8} x^{8} b^{3} A + \frac{3}{4} x^{8} c b a A + \frac{1}{2} x^{6} b a^{2} B + \frac{1}{2} x^{6} b^{2} a A + \frac{1}{2} x^{6} c a^{2} A + \frac{1}{4} x^{4} a^{3} B + \frac{3}{4} x^{4} b a^{2} A + \frac{1}{2} x^{2} a^{3} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^16*c^3*B + 3/14*x^14*c^2*b*B + 1/14*x^14*c^3*A + 1/4*x^12*c*b^2*B + 1/4*x^12*c^2*a*B + 1/4*x^12*c^2*b*A
+ 1/10*x^10*b^3*B + 3/5*x^10*c*b*a*B + 3/10*x^10*c*b^2*A + 3/10*x^10*c^2*a*A + 3/8*x^8*b^2*a*B + 3/8*x^8*c*a^
2*B + 1/8*x^8*b^3*A + 3/4*x^8*c*b*a*A + 1/2*x^6*b*a^2*B + 1/2*x^6*b^2*a*A + 1/2*x^6*c*a^2*A + 1/4*x^4*a^3*B +
3/4*x^4*b*a^2*A + 1/2*x^2*a^3*A

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Sympy [A]  time = 0.100483, size = 199, normalized size = 1.2 \begin{align*} \frac{A a^{3} x^{2}}{2} + \frac{B c^{3} x^{16}}{16} + x^{14} \left (\frac{A c^{3}}{14} + \frac{3 B b c^{2}}{14}\right ) + x^{12} \left (\frac{A b c^{2}}{4} + \frac{B a c^{2}}{4} + \frac{B b^{2} c}{4}\right ) + x^{10} \left (\frac{3 A a c^{2}}{10} + \frac{3 A b^{2} c}{10} + \frac{3 B a b c}{5} + \frac{B b^{3}}{10}\right ) + x^{8} \left (\frac{3 A a b c}{4} + \frac{A b^{3}}{8} + \frac{3 B a^{2} c}{8} + \frac{3 B a b^{2}}{8}\right ) + x^{6} \left (\frac{A a^{2} c}{2} + \frac{A a b^{2}}{2} + \frac{B a^{2} b}{2}\right ) + x^{4} \left (\frac{3 A a^{2} b}{4} + \frac{B a^{3}}{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**3*x**2/2 + B*c**3*x**16/16 + x**14*(A*c**3/14 + 3*B*b*c**2/14) + x**12*(A*b*c**2/4 + B*a*c**2/4 + B*b**2*
c/4) + x**10*(3*A*a*c**2/10 + 3*A*b**2*c/10 + 3*B*a*b*c/5 + B*b**3/10) + x**8*(3*A*a*b*c/4 + A*b**3/8 + 3*B*a*
*2*c/8 + 3*B*a*b**2/8) + x**6*(A*a**2*c/2 + A*a*b**2/2 + B*a**2*b/2) + x**4*(3*A*a**2*b/4 + B*a**3/4)

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Giac [A]  time = 1.11051, size = 261, normalized size = 1.57 \begin{align*} \frac{1}{16} \, B c^{3} x^{16} + \frac{3}{14} \, B b c^{2} x^{14} + \frac{1}{14} \, A c^{3} x^{14} + \frac{1}{4} \, B b^{2} c x^{12} + \frac{1}{4} \, B a c^{2} x^{12} + \frac{1}{4} \, A b c^{2} x^{12} + \frac{1}{10} \, B b^{3} x^{10} + \frac{3}{5} \, B a b c x^{10} + \frac{3}{10} \, A b^{2} c x^{10} + \frac{3}{10} \, A a c^{2} x^{10} + \frac{3}{8} \, B a b^{2} x^{8} + \frac{1}{8} \, A b^{3} x^{8} + \frac{3}{8} \, B a^{2} c x^{8} + \frac{3}{4} \, A a b c x^{8} + \frac{1}{2} \, B a^{2} b x^{6} + \frac{1}{2} \, A a b^{2} x^{6} + \frac{1}{2} \, A a^{2} c x^{6} + \frac{1}{4} \, B a^{3} x^{4} + \frac{3}{4} \, A a^{2} b x^{4} + \frac{1}{2} \, A a^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*B*c^3*x^16 + 3/14*B*b*c^2*x^14 + 1/14*A*c^3*x^14 + 1/4*B*b^2*c*x^12 + 1/4*B*a*c^2*x^12 + 1/4*A*b*c^2*x^12
+ 1/10*B*b^3*x^10 + 3/5*B*a*b*c*x^10 + 3/10*A*b^2*c*x^10 + 3/10*A*a*c^2*x^10 + 3/8*B*a*b^2*x^8 + 1/8*A*b^3*x^
8 + 3/8*B*a^2*c*x^8 + 3/4*A*a*b*c*x^8 + 1/2*B*a^2*b*x^6 + 1/2*A*a*b^2*x^6 + 1/2*A*a^2*c*x^6 + 1/4*B*a^3*x^4 +
3/4*A*a^2*b*x^4 + 1/2*A*a^3*x^2