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

Optimal. Leaf size=19 \[ \left (a+b x+c x^2+d x^3\right )^{p+1} \]

[Out]

(a + b*x + c*x^2 + d*x^3)^(1 + p)

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Rubi [A]  time = 0.0427075, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {1585, 1588} \[ \left (a+b x+c x^2+d x^3\right )^{p+1} \]

Antiderivative was successfully verified.

[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 1585

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 1588

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*} \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 &=\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]  time = 0.00958, size = 17, normalized size = 0.89 \[ (a+x (b+x (c+d x)))^{p+1} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.004, size = 20, normalized size = 1.1 \begin{align*} \left ( d{x}^{3}+c{x}^{2}+bx+a \right ) ^{1+p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

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

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Maxima [A]  time = 1.29372, size = 45, normalized size = 2.37 \begin{align*}{\left (d x^{3} + c x^{2} + b x + a\right )}{\left (d x^{3} + c x^{2} + b x + a\right )}^{p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.41991, size = 74, normalized size = 3.89 \begin{align*}{\left (d x^{3} + c x^{2} + b x + a\right )}{\left (d x^{3} + c x^{2} + b x + a\right )}^{p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (3 \, d{\left (p + 1\right )} x^{3} + 2 \, c{\left (p + 1\right )} x^{2} + b{\left (p + 1\right )} x\right )}{\left (d x^{3} + c x^{2} + b x + a\right )}^{p}}{x}\,{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*(b*(1+p)*x+c*(2+2*p)*x^2+d*(3+3*p)*x^3)/x,x, algorithm="giac")

[Out]

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