\(\int x^2 (a x^2+b x^3+c x^4)^2 \, dx\) [6]

Optimal result

Integrand size = 22, antiderivative size = 54 \[ \int x^2 \left (a x^2+b x^3+c x^4\right )^2 \, dx=\frac {a^2 x^7}{7}+\frac {1}{4} a b x^8+\frac {1}{9} \left (b^2+2 a c\right ) x^9+\frac {1}{5} b c x^{10}+\frac {c^2 x^{11}}{11} \]

[Out]

1/7*a^2*x^7+1/4*a*b*x^8+1/9*(2*a*c+b^2)*x^9+1/5*b*c*x^10+1/11*c^2*x^11

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1599, 712} \[ \int x^2 \left (a x^2+b x^3+c x^4\right )^2 \, dx=\frac {a^2 x^7}{7}+\frac {1}{9} x^9 \left (2 a c+b^2\right )+\frac {1}{4} a b x^8+\frac {1}{5} b c x^{10}+\frac {c^2 x^{11}}{11} \]

[In]

Int[x^2*(a*x^2 + b*x^3 + c*x^4)^2,x]

[Out]

(a^2*x^7)/7 + (a*b*x^8)/4 + ((b^2 + 2*a*c)*x^9)/9 + (b*c*x^10)/5 + (c^2*x^11)/11

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1599

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rubi steps \begin{align*} \text {integral}& = \int x^6 \left (a+b x+c x^2\right )^2 \, dx \\ & = \int \left (a^2 x^6+2 a b x^7+\left (b^2+2 a c\right ) x^8+2 b c x^9+c^2 x^{10}\right ) \, dx \\ & = \frac {a^2 x^7}{7}+\frac {1}{4} a b x^8+\frac {1}{9} \left (b^2+2 a c\right ) x^9+\frac {1}{5} b c x^{10}+\frac {c^2 x^{11}}{11} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a x^2+b x^3+c x^4\right )^2 \, dx=\frac {a^2 x^7}{7}+\frac {1}{4} a b x^8+\frac {1}{9} \left (b^2+2 a c\right ) x^9+\frac {1}{5} b c x^{10}+\frac {c^2 x^{11}}{11} \]

[In]

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

[Out]

(a^2*x^7)/7 + (a*b*x^8)/4 + ((b^2 + 2*a*c)*x^9)/9 + (b*c*x^10)/5 + (c^2*x^11)/11

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {a^{2} x^{7}}{7}+\frac {a b \,x^{8}}{4}+\frac {\left (2 a c +b^{2}\right ) x^{9}}{9}+\frac {b c \,x^{10}}{5}+\frac {c^{2} x^{11}}{11}\)	\(45\)
norman	\(\frac {c^{2} x^{11}}{11}+\frac {b c \,x^{10}}{5}+\left (\frac {2 a c}{9}+\frac {b^{2}}{9}\right ) x^{9}+\frac {a b \,x^{8}}{4}+\frac {a^{2} x^{7}}{7}\)	\(46\)
gosper	\(\frac {x^{7} \left (1260 c^{2} x^{4}+2772 b c \,x^{3}+3080 a c \,x^{2}+1540 b^{2} x^{2}+3465 a b x +1980 a^{2}\right )}{13860}\)	\(47\)
risch	\(\frac {1}{7} a^{2} x^{7}+\frac {1}{4} a b \,x^{8}+\frac {2}{9} x^{9} a c +\frac {1}{9} b^{2} x^{9}+\frac {1}{5} b c \,x^{10}+\frac {1}{11} c^{2} x^{11}\)	\(47\)
parallelrisch	\(\frac {1}{7} a^{2} x^{7}+\frac {1}{4} a b \,x^{8}+\frac {2}{9} x^{9} a c +\frac {1}{9} b^{2} x^{9}+\frac {1}{5} b c \,x^{10}+\frac {1}{11} c^{2} x^{11}\)	\(47\)

[In]

int(x^2*(c*x^4+b*x^3+a*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/7*a^2*x^7+1/4*a*b*x^8+1/9*(2*a*c+b^2)*x^9+1/5*b*c*x^10+1/11*c^2*x^11

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int x^2 \left (a x^2+b x^3+c x^4\right )^2 \, dx=\frac {1}{11} \, c^{2} x^{11} + \frac {1}{5} \, b c x^{10} + \frac {1}{4} \, a b x^{8} + \frac {1}{9} \, {\left (b^{2} + 2 \, a c\right )} x^{9} + \frac {1}{7} \, a^{2} x^{7} \]

[In]

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

[Out]

1/11*c^2*x^11 + 1/5*b*c*x^10 + 1/4*a*b*x^8 + 1/9*(b^2 + 2*a*c)*x^9 + 1/7*a^2*x^7

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int x^2 \left (a x^2+b x^3+c x^4\right )^2 \, dx=\frac {a^{2} x^{7}}{7} + \frac {a b x^{8}}{4} + \frac {b c x^{10}}{5} + \frac {c^{2} x^{11}}{11} + x^{9} \cdot \left (\frac {2 a c}{9} + \frac {b^{2}}{9}\right ) \]

[In]

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

[Out]

a**2*x**7/7 + a*b*x**8/4 + b*c*x**10/5 + c**2*x**11/11 + x**9*(2*a*c/9 + b**2/9)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int x^2 \left (a x^2+b x^3+c x^4\right )^2 \, dx=\frac {1}{11} \, c^{2} x^{11} + \frac {1}{5} \, b c x^{10} + \frac {1}{4} \, a b x^{8} + \frac {1}{9} \, {\left (b^{2} + 2 \, a c\right )} x^{9} + \frac {1}{7} \, a^{2} x^{7} \]

[In]

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

[Out]

1/11*c^2*x^11 + 1/5*b*c*x^10 + 1/4*a*b*x^8 + 1/9*(b^2 + 2*a*c)*x^9 + 1/7*a^2*x^7

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int x^2 \left (a x^2+b x^3+c x^4\right )^2 \, dx=\frac {1}{11} \, c^{2} x^{11} + \frac {1}{5} \, b c x^{10} + \frac {1}{9} \, b^{2} x^{9} + \frac {2}{9} \, a c x^{9} + \frac {1}{4} \, a b x^{8} + \frac {1}{7} \, a^{2} x^{7} \]

[In]

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

[Out]

1/11*c^2*x^11 + 1/5*b*c*x^10 + 1/9*b^2*x^9 + 2/9*a*c*x^9 + 1/4*a*b*x^8 + 1/7*a^2*x^7

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int x^2 \left (a x^2+b x^3+c x^4\right )^2 \, dx=x^9\,\left (\frac {b^2}{9}+\frac {2\,a\,c}{9}\right )+\frac {a^2\,x^7}{7}+\frac {c^2\,x^{11}}{11}+\frac {a\,b\,x^8}{4}+\frac {b\,c\,x^{10}}{5} \]

[In]

int(x^2*(a*x^2 + b*x^3 + c*x^4)^2,x)

[Out]

x^9*((2*a*c)/9 + b^2/9) + (a^2*x^7)/7 + (c^2*x^11)/11 + (a*b*x^8)/4 + (b*c*x^10)/5