3.175 \(\int x^{2 (1+p)} (b+2 c x^3) (b x+c x^4)^p \, dx\)

Optimal. Leaf size=29 \[ \frac{x^{2 (p+1)} \left (b x+c x^4\right )^{p+1}}{3 (p+1)} \]

[Out]

(x^(2*(1 + p))*(b*x + c*x^4)^(1 + p))/(3*(1 + p))

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Rubi [A]  time = 0.0238577, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.037, Rules used = {1590} \[ \frac{x^{2 (p+1)} \left (b x+c x^4\right )^{p+1}}{3 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(2*(1 + p))*(b*x + c*x^4)^(1 + p))/(3*(1 + p))

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 x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx &=\frac{x^{2 (1+p)} \left (b x+c x^4\right )^{1+p}}{3 (1+p)}\\ \end{align*}

Mathematica [C]  time = 0.0764508, size = 99, normalized size = 3.41 \[ \frac{x^{2 p+3} \left (x \left (b+c x^3\right )\right )^p \left (\frac{c x^3}{b}+1\right )^{-p} \left (2 c (p+1) x^3 \, _2F_1\left (-p,p+2;p+3;-\frac{c x^3}{b}\right )+b (p+2) \, _2F_1\left (-p,p+1;p+2;-\frac{c x^3}{b}\right )\right )}{3 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(3 + 2*p)*(x*(b + c*x^3))^p*(b*(2 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, -((c*x^3)/b)] + 2*c*(1 + p)*x^3*
Hypergeometric2F1[-p, 2 + p, 3 + p, -((c*x^3)/b)]))/(3*(1 + p)*(2 + p)*(1 + (c*x^3)/b)^p)

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Maple [A]  time = 0.003, size = 33, normalized size = 1.1 \begin{align*}{\frac{{x}^{3+2\,p} \left ( c{x}^{3}+b \right ) \left ( c{x}^{4}+bx \right ) ^{p}}{3+3\,p}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2+2*p)*(2*c*x^3+b)*(c*x^4+b*x)^p,x)

[Out]

1/3*x^(3+2*p)*(c*x^3+b)/(1+p)*(c*x^4+b*x)^p

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Maxima [A]  time = 1.59304, size = 47, normalized size = 1.62 \begin{align*} \frac{{\left (c x^{6} + b x^{3}\right )} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right )\right )}}{3 \,{\left (p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(c*x^6 + b*x^3)*e^(p*log(c*x^3 + b) + 3*p*log(x))/(p + 1)

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Fricas [A]  time = 1.39162, size = 74, normalized size = 2.55 \begin{align*} \frac{{\left (c x^{4} + b x\right )}{\left (c x^{4} + b x\right )}^{p} x^{2 \, p + 2}}{3 \,{\left (p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(c*x^4 + b*x)*(c*x^4 + b*x)^p*x^(2*p + 2)/(p + 1)

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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(x**(2+2*p)*(2*c*x**3+b)*(c*x**4+b*x)**p,x)

[Out]

Timed out

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Giac [B]  time = 1.17391, size = 78, normalized size = 2.69 \begin{align*} \frac{c x^{4} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right ) + 2 \, \log \left (x\right )\right )} + b x e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right ) + 2 \, \log \left (x\right )\right )}}{3 \,{\left (p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(c*x^4*e^(p*log(c*x^3 + b) + 3*p*log(x) + 2*log(x)) + b*x*e^(p*log(c*x^3 + b) + 3*p*log(x) + 2*log(x)))/(p
+ 1)