### 3.1449 $$\int (1+x)^m (1+2 x+x^2)^n \, dx$$

Optimal. Leaf size=26 $\frac{(x+1)^{m+1} \left (x^2+2 x+1\right )^n}{m+2 n+1}$

[Out]

((1 + x)^(1 + m)*(1 + 2*x + x^2)^n)/(1 + m + 2*n)

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Rubi [A]  time = 0.0108419, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {644, 32} $\frac{(x+1)^{m+1} \left (x^2+2 x+1\right )^n}{m+2 n+1}$

Antiderivative was successfully veriﬁed.

[In]

Int[(1 + x)^m*(1 + 2*x + x^2)^n,x]

[Out]

((1 + x)^(1 + m)*(1 + 2*x + x^2)^n)/(1 + m + 2*n)

Rule 644

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^p/(d
+ e*x)^(2*p), Int[(d + e*x)^(m + 2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !
IntegerQ[p] && EqQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0107284, size = 23, normalized size = 0.88 $\frac{(x+1)^{m+1} \left ((x+1)^2\right )^n}{m+2 n+1}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(1 + x)^m*(1 + 2*x + x^2)^n,x]

[Out]

((1 + x)^(1 + m)*((1 + x)^2)^n)/(1 + m + 2*n)

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Maple [A]  time = 0.04, size = 27, normalized size = 1. \begin{align*}{\frac{ \left ( 1+x \right ) ^{1+m} \left ({x}^{2}+2\,x+1 \right ) ^{n}}{1+m+2\,n}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^m*(x^2+2*x+1)^n,x)

[Out]

(1+x)^(1+m)*(x^2+2*x+1)^n/(1+m+2*n)

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Maxima [A]  time = 1.31031, size = 36, normalized size = 1.38 \begin{align*} \frac{{\left (x + 1\right )} e^{\left (m \log \left (x + 1\right ) + 2 \, n \log \left (x + 1\right )\right )}}{m + 2 \, n + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^m*(x^2+2*x+1)^n,x, algorithm="maxima")

[Out]

(x + 1)*e^(m*log(x + 1) + 2*n*log(x + 1))/(m + 2*n + 1)

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Fricas [A]  time = 2.18157, size = 63, normalized size = 2.42 \begin{align*} \frac{{\left (x + 1\right )}^{m}{\left (x + 1\right )}^{2 \, n}{\left (x + 1\right )}}{m + 2 \, n + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^m*(x^2+2*x+1)^n,x, algorithm="fricas")

[Out]

(x + 1)^m*(x + 1)^(2*n)*(x + 1)/(m + 2*n + 1)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**m*(x**2+2*x+1)**n,x)

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.23872, size = 50, normalized size = 1.92 \begin{align*} \frac{{\left (x + 1\right )}^{m}{\left (x + 1\right )}^{2 \, n} x +{\left (x + 1\right )}^{m}{\left (x + 1\right )}^{2 \, n}}{m + 2 \, n + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^m*(x^2+2*x+1)^n,x, algorithm="giac")

[Out]

((x + 1)^m*(x + 1)^(2*n)*x + (x + 1)^m*(x + 1)^(2*n))/(m + 2*n + 1)