\(\int x^m (a+b x^{2+2 m}) \, dx\) [2742]

Optimal result

Integrand size = 15, antiderivative size = 30 \[ \int x^m \left (a+b x^{2+2 m}\right ) \, dx=\frac {a x^{1+m}}{1+m}+\frac {b x^{3 (1+m)}}{3 (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14} \[ \int x^m \left (a+b x^{2+2 m}\right ) \, dx=\frac {a x^{m+1}}{m+1}+\frac {b x^{3 (m+1)}}{3 (m+1)} \]

[In]

Int[x^m*(a + b*x^(2 + 2*m)),x]

[Out]

(a*x^(1 + m))/(1 + m) + (b*x^(3*(1 + m)))/(3*(1 + 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^{2+3 m}\right ) \, dx \\ & = \frac {a x^{1+m}}{1+m}+\frac {b x^{3 (1+m)}}{3 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int x^m \left (a+b x^{2+2 m}\right ) \, dx=\frac {3 a x^{1+m}+b x^{3+3 m}}{3+3 m} \]

[In]

Integrate[x^m*(a + b*x^(2 + 2*m)),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
risch	\(\frac {a x \,x^{m}}{1+m}+\frac {b \,x^{3} x^{3 m}}{3+3 m}\)	\(29\)
parallelrisch	\(\frac {x \,x^{m} x^{2+2 m} b +3 x \,x^{m} a}{3+3 m}\)	\(29\)
norman	\(\frac {a x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}+\frac {b \,x^{3} {\mathrm e}^{3 m \ln \left (x \right )}}{3+3 m}\)	\(33\)

[In]

int(x^m*(a+b*x^(2+2*m)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int x^m \left (a+b x^{2+2 m}\right ) \, dx=\frac {b x^{3} x^{3 \, m} + 3 \, a x x^{m}}{3 \, {\left (m + 1\right )}} \]

[In]

integrate(x^m*(a+b*x^(2+2*m)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int x^m \left (a+b x^{2+2 m}\right ) \, dx=\begin {cases} \frac {3 a x x^{m}}{3 m + 3} + \frac {b x x^{m} x^{2 m + 2}}{3 m + 3} & \text {for}\: m \neq -1 \\\left (a + b\right ) \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**m*(a+b*x**(2+2*m)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int x^m \left (a+b x^{2+2 m}\right ) \, dx=\frac {b x^{3 \, m + 3}}{3 \, {\left (m + 1\right )}} + \frac {a x^{m + 1}}{m + 1} \]

[In]

integrate(x^m*(a+b*x^(2+2*m)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int x^m \left (a+b x^{2+2 m}\right ) \, dx=\frac {b x^{3} x^{3 \, m} + 3 \, a x x^{m}}{3 \, {\left (m + 1\right )}} \]

[In]

integrate(x^m*(a+b*x^(2+2*m)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.76 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int x^m \left (a+b x^{2+2 m}\right ) \, dx=\frac {x^{m+1}\,\left (a+\frac {b\,x^{2\,m+2}}{3}\right )}{m+1} \]

[In]

int(x^m*(a + b*x^(2*m + 2)),x)

[Out]

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