3.2520 \(\int x^{-1+2 n} (a+b x^n) \, dx\)

Optimal. Leaf size=27 \[ \frac {a x^{2 n}}{2 n}+\frac {b x^{3 n}}{3 n} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 27, 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} \[ \frac {a x^{2 n}}{2 n}+\frac {b x^{3 n}}{3 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)*(a + b*x^n),x]

[Out]

(a*x^(2*n))/(2*n) + (b*x^(3*n))/(3*n)

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^{-1+2 n} \left (a+b x^n\right ) \, dx &=\int \left (a x^{-1+2 n}+b x^{-1+3 n}\right ) \, dx\\ &=\frac {a x^{2 n}}{2 n}+\frac {b x^{3 n}}{3 n}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.81 \[ \frac {x^{2 n} \left (3 a+2 b x^n\right )}{6 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*n)*(a + b*x^n),x]

[Out]

(x^(2*n)*(3*a + 2*b*x^n))/(6*n)

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fricas [A]  time = 0.50, size = 22, normalized size = 0.81 \[ \frac {2 \, b x^{3 \, n} + 3 \, a x^{2 \, n}}{6 \, n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n),x, algorithm="fricas")

[Out]

1/6*(2*b*x^(3*n) + 3*a*x^(2*n))/n

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{n} + a\right )} x^{2 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n),x, algorithm="giac")

[Out]

integrate((b*x^n + a)*x^(2*n - 1), x)

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maple [A]  time = 0.02, size = 28, normalized size = 1.04 \[ \frac {a \,{\mathrm e}^{2 n \ln \relax (x )}}{2 n}+\frac {b \,{\mathrm e}^{3 n \ln \relax (x )}}{3 n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*n-1)*(b*x^n+a),x)

[Out]

1/2*a/n*exp(n*ln(x))^2+1/3*b/n*exp(n*ln(x))^3

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maxima [A]  time = 0.52, size = 23, normalized size = 0.85 \[ \frac {b x^{3 \, n}}{3 \, n} + \frac {a x^{2 \, n}}{2 \, n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n),x, algorithm="maxima")

[Out]

1/3*b*x^(3*n)/n + 1/2*a*x^(2*n)/n

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mupad [B]  time = 1.26, size = 20, normalized size = 0.74 \[ \frac {x^{2\,n}\,\left (3\,a+2\,b\,x^n\right )}{6\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*n - 1)*(a + b*x^n),x)

[Out]

(x^(2*n)*(3*a + 2*b*x^n))/(6*n)

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sympy [A]  time = 4.74, size = 26, normalized size = 0.96 \[ \begin {cases} \frac {a x^{2 n}}{2 n} + \frac {b x^{3 n}}{3 n} & \text {for}\: n \neq 0 \\\left (a + b\right ) \log {\relax (x )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+2*n)*(a+b*x**n),x)

[Out]

Piecewise((a*x**(2*n)/(2*n) + b*x**(3*n)/(3*n), Ne(n, 0)), ((a + b)*log(x), True))

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