3.9.67 \(\int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx\)

Optimal. Leaf size=22 \[ \frac {x^{2 (n+1)} (a+b x)^{n+1}}{n+1} \]

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Rubi [A]  time = 0.00, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {74} \begin {gather*} \frac {x^{2 (n+1)} (a+b x)^{n+1}}{n+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(1 + 2*n)*(a + b*x)^n*(2*a + 3*b*x),x]

[Out]

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

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {x^{2 n+2} (a+b x)^{n+1}}{n+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(1 + 2*n)*(a + b*x)^n*(2*a + 3*b*x),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.05, size = 0, normalized size = 0.00 \begin {gather*} \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^(1 + 2*n)*(a + b*x)^n*(2*a + 3*b*x),x]

[Out]

Defer[IntegrateAlgebraic][x^(1 + 2*n)*(a + b*x)^n*(2*a + 3*b*x), x]

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fricas [A]  time = 1.60, size = 29, normalized size = 1.32 \begin {gather*} \frac {{\left (b x^{2} + a x\right )} {\left (b x + a\right )}^{n} x^{2 \, n + 1}}{n + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+2*n)*(b*x+a)^n*(3*b*x+2*a),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.19, size = 47, normalized size = 2.14 \begin {gather*} \frac {{\left (b x + a\right )}^{n} b x^{2} e^{\left (2 \, n \log \relax (x) + \log \relax (x)\right )} + {\left (b x + a\right )}^{n} a x e^{\left (2 \, n \log \relax (x) + \log \relax (x)\right )}}{n + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+2*n)*(b*x+a)^n*(3*b*x+2*a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 23, normalized size = 1.05 \begin {gather*} \frac {x^{2 n +2} \left (b x +a \right )^{n +1}}{n +1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.83, size = 32, normalized size = 1.45 \begin {gather*} \frac {{\left (b x^{3} + a x^{2}\right )} e^{\left (n \log \left (b x + a\right ) + 2 \, n \log \relax (x)\right )}}{n + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+2*n)*(b*x+a)^n*(3*b*x+2*a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.17, size = 41, normalized size = 1.86 \begin {gather*} \left (\frac {a\,x\,x^{2\,n+1}}{n+1}+\frac {b\,x^{2\,n+1}\,x^2}{n+1}\right )\,{\left (a+b\,x\right )}^n \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 49.85, size = 53, normalized size = 2.41 \begin {gather*} \begin {cases} \frac {a x^{2} x^{2 n} \left (a + b x\right )^{n}}{n + 1} + \frac {b x^{3} x^{2 n} \left (a + b x\right )^{n}}{n + 1} & \text {for}\: n \neq -1 \\2 \log {\relax (x )} + \log {\left (\frac {a}{b} + x \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1+2*n)*(b*x+a)**n*(3*b*x+2*a),x)

[Out]

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

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