\(\int \frac {(a+b x+c x^2+d x^3)^p (b (1+p) x+c (2+2 p) x^2+d (3+3 p) x^3)}{x} \, dx\) [238]

Optimal result

Integrand size = 48, antiderivative size = 19 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (b (1+p) x+c (2+2 p) x^2+d (3+3 p) x^3\right )}{x} \, dx=\left (a+b x+c x^2+d x^3\right )^{1+p} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1599, 1602} \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (b (1+p) x+c (2+2 p) x^2+d (3+3 p) x^3\right )}{x} \, dx=\left (a+b x+c x^2+d x^3\right )^{p+1} \]

[In]

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

[Out]

(a + b*x + c*x^2 + d*x^3)^(1 + p)

Rule 1599

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

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (b (1+p) x+c (2+2 p) x^2+d (3+3 p) x^3\right )}{x} \, dx=(a+x (b+x (c+d x)))^{1+p} \]

[In]

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

[Out]

(a + x*(b + x*(c + d*x)))^(1 + p)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\left (x^{3} d +c \,x^{2}+b x +a \right )^{1+p}\)	\(20\)
risch	\(\left (x^{3} d +c \,x^{2}+b x +a \right )^{p} \left (x^{3} d +c \,x^{2}+b x +a \right )\)	\(34\)
norman	\(a \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+b x \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+c \,x^{2} {\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+x^{3} d \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}\)	\(93\)
parallelrisch	\(\frac {x^{3} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} d^{2}+x^{2} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} c d +x \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} b d +\left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a d}{d}\)	\(94\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (b (1+p) x+c (2+2 p) x^2+d (3+3 p) x^3\right )}{x} \, dx={\left (d x^{3} + c x^{2} + b x + a\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (b (1+p) x+c (2+2 p) x^2+d (3+3 p) x^3\right )}{x} \, dx={\left (d x^{3} + c x^{2} + b x + a\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.74 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (b (1+p) x+c (2+2 p) x^2+d (3+3 p) x^3\right )}{x} \, dx=\frac {{\left (d x^{3} + c x^{2} + b x + a\right )}^{p + 1} p}{p + 1} + \frac {{\left (d x^{3} + c x^{2} + b x + a\right )}^{p + 1}}{p + 1} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (b (1+p) x+c (2+2 p) x^2+d (3+3 p) x^3\right )}{x} \, dx={\left (d\,x^3+c\,x^2+b\,x+a\right )}^{p+1} \]

[In]

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

[Out]

(a + b*x + c*x^2 + d*x^3)^(p + 1)