\(\int x^3 (a+b x^4) \, dx\) [606]

Optimal result

Integrand size = 11, antiderivative size = 17 \[ \int x^3 \left (a+b x^4\right ) \, dx=\frac {a x^4}{4}+\frac {b x^8}{8} \]

[Out]

1/4*a*x^4+1/8*b*x^8

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14} \[ \int x^3 \left (a+b x^4\right ) \, dx=\frac {a x^4}{4}+\frac {b x^8}{8} \]

[In]

Int[x^3*(a + b*x^4),x]

[Out]

(a*x^4)/4 + (b*x^8)/8

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^7\right ) \, dx \\ & = \frac {a x^4}{4}+\frac {b x^8}{8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b x^4\right ) \, dx=\frac {a x^4}{4}+\frac {b x^8}{8} \]

[In]

Integrate[x^3*(a + b*x^4),x]

[Out]

(a*x^4)/4 + (b*x^8)/8

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {1}{4} a \,x^{4}+\frac {1}{8} b \,x^{8}\)	\(14\)
norman	\(\frac {1}{4} a \,x^{4}+\frac {1}{8} b \,x^{8}\)	\(14\)
parallelrisch	\(\frac {1}{4} a \,x^{4}+\frac {1}{8} b \,x^{8}\)	\(14\)
default	\(\frac {\left (b \,x^{4}+a \right )^{2}}{8 b}\)	\(15\)
risch	\(\frac {b \,x^{8}}{8}+\frac {a \,x^{4}}{4}+\frac {a^{2}}{8 b}\)	\(22\)

[In]

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

[Out]

1/4*a*x^4+1/8*b*x^8

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int x^3 \left (a+b x^4\right ) \, dx=\frac {1}{8} \, b x^{8} + \frac {1}{4} \, a x^{4} \]

[In]

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

[Out]

1/8*b*x^8 + 1/4*a*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int x^3 \left (a+b x^4\right ) \, dx=\frac {a x^{4}}{4} + \frac {b x^{8}}{8} \]

[In]

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

[Out]

a*x**4/4 + b*x**8/8

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int x^3 \left (a+b x^4\right ) \, dx=\frac {{\left (b x^{4} + a\right )}^{2}}{8 \, b} \]

[In]

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

[Out]

1/8*(b*x^4 + a)^2/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int x^3 \left (a+b x^4\right ) \, dx=\frac {1}{8} \, b x^{8} + \frac {1}{4} \, a x^{4} \]

[In]

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

[Out]

1/8*b*x^8 + 1/4*a*x^4

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int x^3 \left (a+b x^4\right ) \, dx=\frac {b\,x^8}{8}+\frac {a\,x^4}{4} \]

[In]

int(x^3*(a + b*x^4),x)

[Out]

(a*x^4)/4 + (b*x^8)/8