\(\int x^m (a+b x+c x^2+d x^3)^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx\) [234]

Optimal result

Integrand size = 56, antiderivative size = 25 \[ \int x^m \left (a+b x+c x^2+d x^3\right )^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx=x^{1+m} \left (a+b x+c x^2+d x^3\right )^{1+p} \]

[Out]

x^(1+m)*(d*x^3+c*x^2+b*x+a)^(p+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {1604} \[ \int x^m \left (a+b x+c x^2+d x^3\right )^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx=x^{m+1} \left (a+b x+c x^2+d x^3\right )^{p+1} \]

[In]

Int[x^m*(a + b*x + c*x^2 + d*x^3)^p*(a*(1 + m) + x*(b*(2 + m + p) + x*(c*(3 + m + 2*p) + d*(4 + m + 3*p)*x))),
x]

[Out]

x^(1 + m)*(a + b*x + c*x^2 + d*x^3)^(1 + p)

Rule 1604

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*(Rr^(n + 1)/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x,
r])), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = x^{1+m} \left (a+b x+c x^2+d x^3\right )^{1+p} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int x^m \left (a+b x+c x^2+d x^3\right )^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx=x^{1+m} (a+x (b+x (c+d x)))^{1+p} \]

[In]

Integrate[x^m*(a + b*x + c*x^2 + d*x^3)^p*(a*(1 + m) + x*(b*(2 + m + p) + x*(c*(3 + m + 2*p) + d*(4 + m + 3*p)
*x))),x]

[Out]

x^(1 + m)*(a + x*(b + x*(c + d*x)))^(1 + p)

Maple [A] (verified)

Time = 4.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04

method	result	size
gosper	\(x^{1+m} \left (x^{3} d +c \,x^{2}+b x +a \right )^{1+p}\)	\(26\)
risch	\(\left (x^{3} d +c \,x^{2}+b x +a \right )^{p} x^{m} x \left (x^{3} d +c \,x^{2}+b x +a \right )\)	\(38\)
parallelrisch	\(\frac {x^{4} x^{m} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a d +x^{3} x^{m} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a c +x^{2} x^{m} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a b +x \,x^{m} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a^{2}}{a}\)	\(109\)

[In]

int(x^m*(d*x^3+c*x^2+b*x+a)^p*(a*(1+m)+x*(b*(2+m+p)+x*(c*(3+m+2*p)+d*(4+m+3*p)*x))),x,method=_RETURNVERBOSE)

[Out]

x^(1+m)*(d*x^3+c*x^2+b*x+a)^(1+p)

Fricas [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int x^m \left (a+b x+c x^2+d x^3\right )^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx={\left (d x^{4} + c x^{3} + b x^{2} + a x\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} x^{m} \]

[In]

integrate(x^m*(d*x^3+c*x^2+b*x+a)^p*(a*(1+m)+x*(b*(2+m+p)+x*(c*(3+m+2*p)+d*(4+m+3*p)*x))),x, algorithm="fricas
")

[Out]

(d*x^4 + c*x^3 + b*x^2 + a*x)*(d*x^3 + c*x^2 + b*x + a)^p*x^m

Sympy [F(-1)]

Timed out. \[ \int x^m \left (a+b x+c x^2+d x^3\right )^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx=\text {Timed out} \]

[In]

integrate(x**m*(d*x**3+c*x**2+b*x+a)**p*(a*(1+m)+x*(b*(2+m+p)+x*(c*(3+m+2*p)+d*(4+m+3*p)*x))),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int x^m \left (a+b x+c x^2+d x^3\right )^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx={\left (d x^{4} + c x^{3} + b x^{2} + a x\right )} e^{\left (p \log \left (d x^{3} + c x^{2} + b x + a\right ) + m \log \left (x\right )\right )} \]

[In]

integrate(x^m*(d*x^3+c*x^2+b*x+a)^p*(a*(1+m)+x*(b*(2+m+p)+x*(c*(3+m+2*p)+d*(4+m+3*p)*x))),x, algorithm="maxima
")

[Out]

(d*x^4 + c*x^3 + b*x^2 + a*x)*e^(p*log(d*x^3 + c*x^2 + b*x + a) + m*log(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (25) = 50\).

Time = 0.54 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.96 \[ \int x^m \left (a+b x+c x^2+d x^3\right )^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx={\left (d x^{3} + c x^{2} + b x + a\right )}^{p} d x^{4} x^{m} + {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} c x^{3} x^{m} + {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} b x^{2} x^{m} + {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} a x x^{m} \]

[In]

integrate(x^m*(d*x^3+c*x^2+b*x+a)^p*(a*(1+m)+x*(b*(2+m+p)+x*(c*(3+m+2*p)+d*(4+m+3*p)*x))),x, algorithm="giac")

[Out]

(d*x^3 + c*x^2 + b*x + a)^p*d*x^4*x^m + (d*x^3 + c*x^2 + b*x + a)^p*c*x^3*x^m + (d*x^3 + c*x^2 + b*x + a)^p*b*
x^2*x^m + (d*x^3 + c*x^2 + b*x + a)^p*a*x*x^m

Mupad [B] (verification not implemented)

Time = 9.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int x^m \left (a+b x+c x^2+d x^3\right )^p (a (1+m)+x (b (2+m+p)+x (c (3+m+2 p)+d (4+m+3 p) x))) \, dx={\left (d\,x^3+c\,x^2+b\,x+a\right )}^p\,\left (a\,x\,x^m+b\,x^m\,x^2+c\,x^m\,x^3+d\,x^m\,x^4\right ) \]

[In]

int(x^m*(a*(m + 1) + x*(x*(c*(m + 2*p + 3) + d*x*(m + 3*p + 4)) + b*(m + p + 2)))*(a + b*x + c*x^2 + d*x^3)^p,
x)

[Out]

(a + b*x + c*x^2 + d*x^3)^p*(a*x*x^m + b*x^m*x^2 + c*x^m*x^3 + d*x^m*x^4)