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

Optimal. Leaf size=23 \[ \frac{\left (a+b x+c x^2+d x^3\right )^{p+1}}{x^2} \]

[Out]

(a + b*x + c*x^2 + d*x^3)^(1 + p)/x^2

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Rubi [A]  time = 0.0353046, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.021, Rules used = {1590} \[ \frac{\left (a+b x+c x^2+d x^3\right )^{p+1}}{x^2} \]

Antiderivative was successfully verified.

[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 1590

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*} \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}\\ \end{align*}

Mathematica [A]  time = 0.182003, size = 21, normalized size = 0.91 \[ \frac{(a+x (b+x (c+d x)))^{p+1}}{x^2} \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.007, size = 24, normalized size = 1. \begin{align*}{\frac{ \left ( d{x}^{3}+c{x}^{2}+bx+a \right ) ^{1+p}}{{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

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

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Maxima [A]  time = 1.31428, size = 49, normalized size = 2.13 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.82445, size = 80, normalized size = 3.48 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)