3.2742 \(\int x^m (a+b x^{2+2 m}) \, dx\)

Optimal. Leaf size=30 \[ \frac {a x^{m+1}}{m+1}+\frac {b x^{3 (m+1)}}{3 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14} \[ \frac {a x^{m+1}}{m+1}+\frac {b x^{3 (m+1)}}{3 (m+1)} \]

Antiderivative was successfully verified.

[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*} \int x^m \left (a+b x^{2+2 m}\right ) \, dx &=\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*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.87 \[ \frac {3 a x^{m+1}+b x^{3 m+3}}{3 m+3} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.65, size = 25, normalized size = 0.83 \[ \frac {b x^{3} x^{3 \, m} + 3 \, a x x^{m}}{3 \, {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 25, normalized size = 0.83 \[ \frac {b x^{3} x^{3 \, m} + 3 \, a x x^{m}}{3 \, {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.02, size = 33, normalized size = 1.10 \[ \frac {b \,x^{3} {\mathrm e}^{3 m \ln \relax (x )}}{3 m +3}+\frac {a x \,{\mathrm e}^{m \ln \relax (x )}}{m +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 28, normalized size = 0.93 \[ \frac {b x^{3 \, m + 3}}{3 \, {\left (m + 1\right )}} + \frac {a x^{m + 1}}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.29, size = 23, normalized size = 0.77 \[ \frac {x^{m+1}\,\left (a+\frac {b\,x^{2\,m+2}}{3}\right )}{m+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a + b*x^(2*m + 2)),x)

[Out]

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

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sympy [A]  time = 9.91, size = 36, normalized size = 1.20 \[ \begin {cases} \frac {3 a x x^{m}}{3 m + 3} + \frac {b x^{3} x^{3 m}}{3 m + 3} & \text {for}\: m \neq -1 \\\left (a + b\right ) \log {\relax (x )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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