\(\int \frac {(a+b x+c x^2+d x^3)^p (-2 a+b (-1+p) x+2 c p x^2+d (1+3 p) x^3)}{x^3} \, dx\) [240]

Optimal result

Integrand size = 48, antiderivative size = 23 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-2 a+b (-1+p) x+2 c p x^2+d (1+3 p) x^3\right )}{x^3} \, dx=\frac {\left (a+b x+c x^2+d x^3\right )^{1+p}}{x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {1604} \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-2 a+b (-1+p) x+2 c p x^2+d (1+3 p) x^3\right )}{x^3} \, dx=\frac {\left (a+b x+c x^2+d x^3\right )^{p+1}}{x^2} \]

[In]

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

[Out]

(a + b*x + c*x^2 + d*x^3)^(1 + p)/x^2

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}& = \frac {\left (a+b x+c x^2+d x^3\right )^{1+p}}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-2 a+b (-1+p) x+2 c p x^2+d (1+3 p) x^3\right )}{x^3} \, dx=\frac {(a+x (b+x (c+d x)))^{1+p}}{x^2} \]

[In]

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

[Out]

(a + x*(b + x*(c + d*x)))^(1 + p)/x^2

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
gosper	\(\frac {\left (x^{3} d +c \,x^{2}+b x +a \right )^{1+p}}{x^{2}}\)	\(24\)
risch	\(\frac {\left (x^{3} d +c \,x^{2}+b x +a \right ) \left (x^{3} d +c \,x^{2}+b x +a \right )^{p}}{x^{2}}\)	\(37\)
norman	\(\frac {a \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+b x \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+c \,x^{2} {\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+x^{3} d \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}}{x^{2}}\)	\(97\)
parallelrisch	\(\frac {x^{3} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} c d +x^{2} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} c^{2}+x \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} b c +\left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a c}{x^{2} c}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-2 a+b (-1+p) x+2 c p x^2+d (1+3 p) x^3\right )}{x^3} \, dx=\frac {{\left (d x^{3} + c x^{2} + b x + a\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p}}{x^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-2 a+b (-1+p) x+2 c p x^2+d (1+3 p) x^3\right )}{x^3} \, dx=\frac {{\left (d x^{3} + c x^{2} + b x + a\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p}}{x^{2}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-2 a+b (-1+p) x+2 c p x^2+d (1+3 p) x^3\right )}{x^3} \, dx=\frac {{\left (d\,x^3+c\,x^2+b\,x+a\right )}^{p+1}}{x^2} \]

[In]

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

[Out]

(a + b*x + c*x^2 + d*x^3)^(p + 1)/x^2