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

Optimal result

Integrand size = 18, antiderivative size = 25 \[ \int x \left (a x^2+b x^3+c x^4\right ) \, dx=\frac {a x^4}{4}+\frac {b x^5}{5}+\frac {c x^6}{6} \]

[Out]

1/4*a*x^4+1/5*b*x^5+1/6*c*x^6

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14} \[ \int x \left (a x^2+b x^3+c x^4\right ) \, dx=\frac {a x^4}{4}+\frac {b x^5}{5}+\frac {c x^6}{6} \]

[In]

Int[x*(a*x^2 + b*x^3 + c*x^4),x]

[Out]

(a*x^4)/4 + (b*x^5)/5 + (c*x^6)/6

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (a x^3+b x^4+c x^5\right ) \, dx \\ & = \frac {a x^4}{4}+\frac {b x^5}{5}+\frac {c x^6}{6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int x \left (a x^2+b x^3+c x^4\right ) \, dx=\frac {a x^4}{4}+\frac {b x^5}{5}+\frac {c x^6}{6} \]

[In]

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

[Out]

(a*x^4)/4 + (b*x^5)/5 + (c*x^6)/6

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80

method	result	size
gosper	\(\frac {x^{4} \left (10 c \,x^{2}+12 b x +15 a \right )}{60}\)	\(20\)
default	\(\frac {1}{4} a \,x^{4}+\frac {1}{5} b \,x^{5}+\frac {1}{6} c \,x^{6}\)	\(20\)
norman	\(\frac {1}{4} a \,x^{4}+\frac {1}{5} b \,x^{5}+\frac {1}{6} c \,x^{6}\)	\(20\)
risch	\(\frac {1}{4} a \,x^{4}+\frac {1}{5} b \,x^{5}+\frac {1}{6} c \,x^{6}\)	\(20\)
parallelrisch	\(\frac {1}{4} a \,x^{4}+\frac {1}{5} b \,x^{5}+\frac {1}{6} c \,x^{6}\)	\(20\)

[In]

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

[Out]

1/60*x^4*(10*c*x^2+12*b*x+15*a)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \left (a x^2+b x^3+c x^4\right ) \, dx=\frac {1}{6} \, c x^{6} + \frac {1}{5} \, b x^{5} + \frac {1}{4} \, a x^{4} \]

[In]

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

[Out]

1/6*c*x^6 + 1/5*b*x^5 + 1/4*a*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \left (a x^2+b x^3+c x^4\right ) \, dx=\frac {a x^{4}}{4} + \frac {b x^{5}}{5} + \frac {c x^{6}}{6} \]

[In]

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

[Out]

a*x**4/4 + b*x**5/5 + c*x**6/6

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \left (a x^2+b x^3+c x^4\right ) \, dx=\frac {1}{6} \, c x^{6} + \frac {1}{5} \, b x^{5} + \frac {1}{4} \, a x^{4} \]

[In]

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

[Out]

1/6*c*x^6 + 1/5*b*x^5 + 1/4*a*x^4

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \left (a x^2+b x^3+c x^4\right ) \, dx=\frac {1}{6} \, c x^{6} + \frac {1}{5} \, b x^{5} + \frac {1}{4} \, a x^{4} \]

[In]

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

[Out]

1/6*c*x^6 + 1/5*b*x^5 + 1/4*a*x^4

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \left (a x^2+b x^3+c x^4\right ) \, dx=\frac {x^4\,\left (10\,c\,x^2+12\,b\,x+15\,a\right )}{60} \]

[In]

int(x*(a*x^2 + b*x^3 + c*x^4),x)

[Out]

(x^4*(15*a + 12*b*x + 10*c*x^2))/60