3.2739 \(\int x^m (a+b x^{1+m})^n \, dx\)

Optimal. Leaf size=27 \[ \frac {\left (a+b x^{m+1}\right )^{n+1}}{b (m+1) (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {261} \[ \frac {\left (a+b x^{m+1}\right )^{n+1}}{b (m+1) (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x^(1 + m))^n,x]

[Out]

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {\left (a+b x^{m+1}\right )^{n+1}}{b (m+1) (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x^(1 + m))^n,x]

[Out]

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

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fricas [A]  time = 0.58, size = 35, normalized size = 1.30 \[ \frac {{\left (b x^{m + 1} + a\right )} {\left (b x^{m + 1} + a\right )}^{n}}{b m + {\left (b m + b\right )} n + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 26, normalized size = 0.96 \[ \frac {{\left (b x^{m + 1} + a\right )}^{n + 1}}{{\left (b m + b\right )} {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 60, normalized size = 2.22 \[ \frac {x \,{\mathrm e}^{m \ln \relax (x )} {\mathrm e}^{n \ln \left (b x \,{\mathrm e}^{m \ln \relax (x )}+a \right )}}{m n +m +n +1}+\frac {a \,{\mathrm e}^{n \ln \left (b x \,{\mathrm e}^{m \ln \relax (x )}+a \right )}}{\left (m n +m +n +1\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 27, normalized size = 1.00 \[ \frac {{\left (b x^{m + 1} + a\right )}^{n + 1}}{b {\left (m + 1\right )} {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.49, size = 27, normalized size = 1.00 \[ \frac {{\left (a+b\,x^{m+1}\right )}^{n+1}}{b\,\left (m+1\right )\,\left (n+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a + b*x^(m + 1))^n,x)

[Out]

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

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sympy [A]  time = 134.23, size = 100, normalized size = 3.70 \[ \begin {cases} \frac {\log {\relax (x )}}{a} & \text {for}\: b = 0 \wedge m = -1 \wedge n = -1 \\\frac {a^{n} x x^{m}}{m + 1} & \text {for}\: b = 0 \\\left (a + b\right )^{n} \log {\relax (x )} & \text {for}\: m = -1 \\\frac {\log {\left (\frac {a}{b} + x x^{m} \right )}}{b m + b} & \text {for}\: n = -1 \\\frac {a \left (a + b x x^{m}\right )^{n}}{b m n + b m + b n + b} + \frac {b x x^{m} \left (a + b x x^{m}\right )^{n}}{b m n + b m + b n + b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(a+b*x**(1+m))**n,x)

[Out]

Piecewise((log(x)/a, Eq(b, 0) & Eq(m, -1) & Eq(n, -1)), (a**n*x*x**m/(m + 1), Eq(b, 0)), ((a + b)**n*log(x), E
q(m, -1)), (log(a/b + x*x**m)/(b*m + b), Eq(n, -1)), (a*(a + b*x*x**m)**n/(b*m*n + b*m + b*n + b) + b*x*x**m*(
a + b*x*x**m)**n/(b*m*n + b*m + b*n + b), True))

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