\(\int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx\) [936]

Optimal result

Integrand size = 23, antiderivative size = 22 \[ \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx=\frac {x^{2 (1+n)} (a+b x)^{1+n}}{1+n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {75} \[ \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx=\frac {x^{2 (n+1)} (a+b x)^{n+1}}{n+1} \]

[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 75

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*} \text {integral}& = \frac {x^{2 (1+n)} (a+b x)^{1+n}}{1+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx=\frac {x^{2+2 n} (a+b x)^{1+n}}{1+n} \]

[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)

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\frac {x^{2+2 n} \left (b x +a \right )^{1+n}}{1+n}\)	\(23\)
risch	\(\frac {x \left (b x +a \right ) x^{1+2 n} \left (b x +a \right )^{n}}{1+n}\)	\(27\)
parallelrisch	\(\frac {x^{2} x^{1+2 n} \left (b x +a \right )^{n} a b +x \,x^{1+2 n} \left (b x +a \right )^{n} a^{2}}{a \left (1+n \right )}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx=\frac {{\left (b x^{2} + a x\right )} {\left (b x + a\right )}^{n} x^{2 \, n + 1}}{n + 1} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (17) = 34\).

Time = 0.88 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45 \[ \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx=\begin {cases} \frac {a x x^{2 n + 1} \left (a + b x\right )^{n}}{n + 1} + \frac {b x^{2} x^{2 n + 1} \left (a + b x\right )^{n}}{n + 1} & \text {for}\: n \neq -1 \\2 \log {\left (x \right )} + \log {\left (\frac {a}{b} + x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx=\frac {{\left (b x^{3} + a x^{2}\right )} e^{\left (n \log \left (b x + a\right ) + 2 \, n \log \left (x\right )\right )}}{n + 1} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.14 \[ \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx=\frac {{\left (b x + a\right )}^{n} b x^{2} e^{\left (2 \, n \log \left (x\right ) + \log \left (x\right )\right )} + {\left (b x + a\right )}^{n} a x e^{\left (2 \, n \log \left (x\right ) + \log \left (x\right )\right )}}{n + 1} \]

[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)

Mupad [B] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int x^{1+2 n} (a+b x)^n (2 a+3 b x) \, dx=\left (\frac {a\,x\,x^{2\,n+1}}{n+1}+\frac {b\,x^{2\,n+1}\,x^2}{n+1}\right )\,{\left (a+b\,x\right )}^n \]

[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