\(\int x^m (a+b x^3) \, dx\) [587]

Optimal result

Integrand size = 11, antiderivative size = 25 \[ \int x^m \left (a+b x^3\right ) \, dx=\frac {a x^{1+m}}{1+m}+\frac {b x^{4+m}}{4+m} \]

[Out]

a*x^(1+m)/(1+m)+b*x^(4+m)/(4+m)

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14} \[ \int x^m \left (a+b x^3\right ) \, dx=\frac {a x^{m+1}}{m+1}+\frac {b x^{m+4}}{m+4} \]

[In]

Int[x^m*(a + b*x^3),x]

[Out]

(a*x^(1 + m))/(1 + m) + (b*x^(4 + m))/(4 + m)

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^m+b x^{3+m}\right ) \, dx \\ & = \frac {a x^{1+m}}{1+m}+\frac {b x^{4+m}}{4+m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int x^m \left (a+b x^3\right ) \, dx=\frac {a x^{1+m}}{1+m}+\frac {b x^{4+m}}{4+m} \]

[In]

Integrate[x^m*(a + b*x^3),x]

[Out]

(a*x^(1 + m))/(1 + m) + (b*x^(4 + m))/(4 + m)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
norman	\(\frac {a x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}+\frac {b \,x^{4} {\mathrm e}^{m \ln \left (x \right )}}{4+m}\)	\(30\)
risch	\(\frac {x \left (b m \,x^{3}+b \,x^{3}+a m +4 a \right ) x^{m}}{\left (4+m \right ) \left (1+m \right )}\)	\(34\)
gosper	\(\frac {x^{1+m} \left (b m \,x^{3}+b \,x^{3}+a m +4 a \right )}{\left (1+m \right ) \left (4+m \right )}\)	\(35\)
parallelrisch	\(\frac {x^{4} x^{m} b m +x^{4} x^{m} b +x \,x^{m} a m +4 x \,x^{m} a}{\left (4+m \right ) \left (1+m \right )}\)	\(44\)

[In]

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

[Out]

a/(1+m)*x*exp(m*ln(x))+b/(4+m)*x^4*exp(m*ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int x^m \left (a+b x^3\right ) \, dx=\frac {{\left ({\left (b m + b\right )} x^{4} + {\left (a m + 4 \, a\right )} x\right )} x^{m}}{m^{2} + 5 \, m + 4} \]

[In]

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

[Out]

((b*m + b)*x^4 + (a*m + 4*a)*x)*x^m/(m^2 + 5*m + 4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (19) = 38\).

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.76 \[ \int x^m \left (a+b x^3\right ) \, dx=\begin {cases} - \frac {a}{3 x^{3}} + b \log {\left (x \right )} & \text {for}\: m = -4 \\a \log {\left (x \right )} + \frac {b x^{3}}{3} & \text {for}\: m = -1 \\\frac {a m x x^{m}}{m^{2} + 5 m + 4} + \frac {4 a x x^{m}}{m^{2} + 5 m + 4} + \frac {b m x^{4} x^{m}}{m^{2} + 5 m + 4} + \frac {b x^{4} x^{m}}{m^{2} + 5 m + 4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a/(3*x**3) + b*log(x), Eq(m, -4)), (a*log(x) + b*x**3/3, Eq(m, -1)), (a*m*x*x**m/(m**2 + 5*m + 4)
+ 4*a*x*x**m/(m**2 + 5*m + 4) + b*m*x**4*x**m/(m**2 + 5*m + 4) + b*x**4*x**m/(m**2 + 5*m + 4), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int x^m \left (a+b x^3\right ) \, dx=\frac {b x^{m + 4}}{m + 4} + \frac {a x^{m + 1}}{m + 1} \]

[In]

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

[Out]

b*x^(m + 4)/(m + 4) + a*x^(m + 1)/(m + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int x^m \left (a+b x^3\right ) \, dx=\frac {b m x^{4} x^{m} + b x^{4} x^{m} + a m x x^{m} + 4 \, a x x^{m}}{m^{2} + 5 \, m + 4} \]

[In]

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

[Out]

(b*m*x^4*x^m + b*x^4*x^m + a*m*x*x^m + 4*a*x*x^m)/(m^2 + 5*m + 4)

Mupad [B] (verification not implemented)

Time = 5.84 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int x^m \left (a+b x^3\right ) \, dx=\frac {x^{m+1}\,\left (4\,a+a\,m+b\,x^3+b\,m\,x^3\right )}{m^2+5\,m+4} \]

[In]

int(x^m*(a + b*x^3),x)

[Out]

(x^(m + 1)*(4*a + a*m + b*x^3 + b*m*x^3))/(5*m + m^2 + 4)