3.1.97 \(\int x (A+B x^2) (a+b x^2+c x^4)^3 \, dx\) [97]

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

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 = {1261, 645} \begin {gather*} \frac {1}{2} a^3 A x^2+\frac {1}{4} a^2 x^4 (a B+3 A b)+\frac {1}{4} c x^{12} \left (a B c+A b c+b^2 B\right )+\frac {1}{2} a x^6 \left (A \left (a c+b^2\right )+a b B\right )+\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}{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}{14} c^2 x^{14} (A c+3 b B)+\frac {1}{16} B c^3 x^{16} \end {gather*}

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 645

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])

Rule 1261

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]

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} \text {Subst}\left (\int (A+B x) \left (a+b x+c x^2\right )^3 \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {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*}

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Mathematica [A]
time = 0.04, size = 154, normalized size = 0.93 \begin {gather*} \frac {1}{560} x^2 \left (280 a^3 A+140 a^2 (3 A b+a B) x^2+280 a \left (a b B+A \left (b^2+a c\right )\right ) x^4+70 \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^6+56 \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^8+140 c \left (b^2 B+A b c+a B c\right ) x^{10}+40 c^2 (3 b B+A c) x^{12}+35 B c^3 x^{14}\right ) \end {gather*}

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 = 0.09, size = 226, normalized size = 1.36

method	result	size
norman	\(\frac {a^{3} A \,x^{2}}{2}+\left (\frac {3}{4} a^{2} b A +\frac {1}{4} a^{3} B \right ) x^{4}+\left (\frac {1}{2} a^{2} c A +\frac {1}{2} A a \,b^{2}+\frac {1}{2} B \,a^{2} b \right ) x^{6}+\left (\frac {3}{4} A a b c +\frac {1}{8} A \,b^{3}+\frac {3}{8} a^{2} c B +\frac {3}{8} B a \,b^{2}\right ) x^{8}+\left (\frac {3}{10} c^{2} a A +\frac {3}{10} A \,b^{2} c +\frac {3}{5} a b B c +\frac {1}{10} b^{3} B \right ) x^{10}+\left (\frac {1}{4} b \,c^{2} A +\frac {1}{4} c^{2} a B +\frac {1}{4} B \,b^{2} c \right ) x^{12}+\left (\frac {1}{14} c^{3} A +\frac {3}{14} B b \,c^{2}\right ) x^{14}+\frac {B \,c^{3} x^{16}}{16}\)	\(170\)
gosper	\(\frac {1}{2} a^{3} A \,x^{2}+\frac {3}{4} x^{4} a^{2} b A +\frac {1}{4} x^{4} a^{3} B +\frac {1}{2} x^{6} a^{2} c A +\frac {1}{2} x^{6} A a \,b^{2}+\frac {1}{2} x^{6} B \,a^{2} b +\frac {3}{4} x^{8} A a b c +\frac {1}{8} x^{8} A \,b^{3}+\frac {3}{8} x^{8} a^{2} c B +\frac {3}{8} x^{8} B a \,b^{2}+\frac {3}{10} x^{10} c^{2} a A +\frac {3}{10} x^{10} A \,b^{2} c +\frac {3}{5} x^{10} a b B c +\frac {1}{10} x^{10} b^{3} B +\frac {1}{4} x^{12} b \,c^{2} A +\frac {1}{4} x^{12} c^{2} a B +\frac {1}{4} x^{12} B \,b^{2} c +\frac {1}{14} x^{14} c^{3} A +\frac {3}{14} x^{14} B b \,c^{2}+\frac {1}{16} B \,c^{3} x^{16}\)	\(194\)
risch	\(\frac {1}{2} a^{3} A \,x^{2}+\frac {3}{4} x^{4} a^{2} b A +\frac {1}{4} x^{4} a^{3} B +\frac {1}{2} x^{6} a^{2} c A +\frac {1}{2} x^{6} A a \,b^{2}+\frac {1}{2} x^{6} B \,a^{2} b +\frac {3}{4} x^{8} A a b c +\frac {1}{8} x^{8} A \,b^{3}+\frac {3}{8} x^{8} a^{2} c B +\frac {3}{8} x^{8} B a \,b^{2}+\frac {3}{10} x^{10} c^{2} a A +\frac {3}{10} x^{10} A \,b^{2} c +\frac {3}{5} x^{10} a b B c +\frac {1}{10} x^{10} b^{3} B +\frac {1}{4} x^{12} b \,c^{2} A +\frac {1}{4} x^{12} c^{2} a B +\frac {1}{4} x^{12} B \,b^{2} c +\frac {1}{14} x^{14} c^{3} A +\frac {3}{14} x^{14} B b \,c^{2}+\frac {1}{16} B \,c^{3} x^{16}\)	\(194\)
default	\(\frac {B \,c^{3} x^{16}}{16}+\frac {\left (c^{3} A +3 B b \,c^{2}\right ) x^{14}}{14}+\frac {\left (3 b \,c^{2} A +B \left (c^{2} a +2 b^{2} c +c \left (2 a c +b^{2}\right )\right )\right ) x^{12}}{12}+\frac {\left (\left (c^{2} a +2 b^{2} c +c \left (2 a c +b^{2}\right )\right ) A +B \left (4 a b c +b \left (2 a c +b^{2}\right )\right )\right ) x^{10}}{10}+\frac {\left (\left (4 a b c +b \left (2 a c +b^{2}\right )\right ) A +B \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )\right ) x^{8}}{8}+\frac {\left (\left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right ) A +3 B \,a^{2} b \right ) x^{6}}{6}+\frac {\left (3 a^{2} b A +a^{3} B \right ) x^{4}}{4}+\frac {a^{3} A \,x^{2}}{2}\)	\(226\)

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,method=_RETURNVERBOSE)

[Out]

1/16*B*c^3*x^16+1/14*(A*c^3+3*B*b*c^2)*x^14+1/12*(3*b*c^2*A+B*(c^2*a+2*b^2*c+c*(2*a*c+b^2)))*x^12+1/10*((c^2*a
+2*b^2*c+c*(2*a*c+b^2))*A+B*(4*a*b*c+b*(2*a*c+b^2)))*x^10+1/8*((4*a*b*c+b*(2*a*c+b^2))*A+B*(a*(2*a*c+b^2)+2*a*
b^2+a^2*c))*x^8+1/6*((a*(2*a*c+b^2)+2*a*b^2+a^2*c)*A+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 = 0.28, size = 166, normalized size = 1.00 \begin {gather*} \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 {gather*}

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 = 0.33, size = 166, normalized size = 1.00 \begin {gather*} \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 {gather*}

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*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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Sympy [A]
time = 0.02, size = 199, normalized size = 1.20 \begin {gather*} \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} \cdot \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} \cdot \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} \cdot \left (\frac {3 A a^{2} b}{4} + \frac {B a^{3}}{4}\right ) \end {gather*}

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 = 3.05, size = 193, normalized size = 1.16 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.05, size = 169, normalized size = 1.02 \begin {gather*} x^8\,\left (\frac {3\,B\,c\,a^2}{8}+\frac {3\,B\,a\,b^2}{8}+\frac {3\,A\,c\,a\,b}{4}+\frac {A\,b^3}{8}\right )+x^{10}\,\left (\frac {B\,b^3}{10}+\frac {3\,A\,b^2\,c}{10}+\frac {3\,B\,a\,b\,c}{5}+\frac {3\,A\,a\,c^2}{10}\right )+x^4\,\left (\frac {B\,a^3}{4}+\frac {3\,A\,b\,a^2}{4}\right )+x^{14}\,\left (\frac {A\,c^3}{14}+\frac {3\,B\,b\,c^2}{14}\right )+x^6\,\left (\frac {B\,a^2\,b}{2}+\frac {A\,c\,a^2}{2}+\frac {A\,a\,b^2}{2}\right )+x^{12}\,\left (\frac {B\,b^2\,c}{4}+\frac {A\,b\,c^2}{4}+\frac {B\,a\,c^2}{4}\right )+\frac {A\,a^3\,x^2}{2}+\frac {B\,c^3\,x^{16}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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