3.188 \(\int x^n (c x+d x^2)^n (2 c x+3 d x^2) \, dx\)

Optimal. Leaf size=24 \[ \frac {x^{n+1} \left (c x+d x^2\right )^{n+1}}{n+1} \]

[Out]

x^(1+n)*(d*x^2+c*x)^(1+n)/(1+n)

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1584, 763} \[ \frac {x^{n+1} \left (c x+d x^2\right )^{n+1}}{n+1} \]

Antiderivative was successfully verified.

[In]

Int[x^n*(c*x + d*x^2)^n*(2*c*x + 3*d*x^2),x]

[Out]

(x^(1 + n)*(c*x + d*x^2)^(1 + n))/(1 + n)

Rule 763

Int[((e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(e*x)^m*(b*
x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] /; FreeQ[{b, c, e, f, g, m, p}, x] && EqQ[b*g*(m + p + 1) - c*f*(m +
2*p + 2), 0] && NeQ[m + 2*p + 2, 0]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int x^n \left (c x+d x^2\right )^n \left (2 c x+3 d x^2\right ) \, dx &=\int x^{1+n} (2 c+3 d x) \left (c x+d x^2\right )^n \, dx\\ &=\frac {x^{1+n} \left (c x+d x^2\right )^{1+n}}{1+n}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.92 \[ \frac {x^{n+1} (x (c+d x))^{n+1}}{n+1} \]

Antiderivative was successfully verified.

[In]

Integrate[x^n*(c*x + d*x^2)^n*(2*c*x + 3*d*x^2),x]

[Out]

(x^(1 + n)*(x*(c + d*x))^(1 + n))/(1 + n)

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fricas [A]  time = 0.91, size = 31, normalized size = 1.29 \[ \frac {{\left (d x^{3} + c x^{2}\right )} {\left (d x^{2} + c x\right )}^{n} x^{n}}{n + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^n*(d*x^2+c*x)^n*(3*d*x^2+2*c*x),x, algorithm="fricas")

[Out]

(d*x^3 + c*x^2)*(d*x^2 + c*x)^n*x^n/(n + 1)

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giac [B]  time = 0.46, size = 51, normalized size = 2.12 \[ \frac {d x^{3} x^{n} e^{\left (n \log \left (d x + c\right ) + n \log \relax (x)\right )} + c x^{2} x^{n} e^{\left (n \log \left (d x + c\right ) + n \log \relax (x)\right )}}{n + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^n*(d*x^2+c*x)^n*(3*d*x^2+2*c*x),x, algorithm="giac")

[Out]

(d*x^3*x^n*e^(n*log(d*x + c) + n*log(x)) + c*x^2*x^n*e^(n*log(d*x + c) + n*log(x)))/(n + 1)

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maple [A]  time = 0.00, size = 28, normalized size = 1.17 \[ \frac {\left (d x +c \right ) x^{n +2} \left (d \,x^{2}+c x \right )^{n}}{n +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^n*(d*x^2+c*x)^n*(3*d*x^2+2*c*x),x)

[Out]

(d*x^2+c*x)^n*x^(2+n)*(d*x+c)/(n+1)

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maxima [A]  time = 0.83, size = 32, normalized size = 1.33 \[ \frac {{\left (d x^{3} + c x^{2}\right )} e^{\left (n \log \left (d x + c\right ) + 2 \, n \log \relax (x)\right )}}{n + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^n*(d*x^2+c*x)^n*(3*d*x^2+2*c*x),x, algorithm="maxima")

[Out]

(d*x^3 + c*x^2)*e^(n*log(d*x + c) + 2*n*log(x))/(n + 1)

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mupad [B]  time = 2.20, size = 28, normalized size = 1.17 \[ \frac {x^n\,x^2\,{\left (d\,x^2+c\,x\right )}^n\,\left (c+d\,x\right )}{n+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^n*(c*x + d*x^2)^n*(2*c*x + 3*d*x^2),x)

[Out]

(x^n*x^2*(c*x + d*x^2)^n*(c + d*x))/(n + 1)

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sympy [A]  time = 6.17, size = 56, normalized size = 2.33 \[ \begin {cases} \frac {c x^{2} x^{n} \left (c x + d x^{2}\right )^{n}}{n + 1} + \frac {d x^{3} x^{n} \left (c x + d x^{2}\right )^{n}}{n + 1} & \text {for}\: n \neq -1 \\2 \log {\relax (x )} + \log {\left (\frac {c}{d} + x \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**n*(d*x**2+c*x)**n*(3*d*x**2+2*c*x),x)

[Out]

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

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