\(\int x^{2 (1+p)} (b+2 c x^3) (b x+c x^4)^p \, dx\) [175]

Optimal result

Integrand size = 27, antiderivative size = 29 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {x^{2 (1+p)} \left (b x+c x^4\right )^{1+p}}{3 (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {1604} \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {x^{2 (p+1)} \left (b x+c x^4\right )^{p+1}}{3 (p+1)} \]

[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 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 {x^{2 (1+p)} \left (b x+c x^4\right )^{1+p}}{3 (1+p)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.41 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {x^{3+2 p} \left (x \left (b+c x^3\right )\right )^p \left (1+\frac {c x^3}{b}\right )^{-p} \left (b (2+p) \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,-\frac {c x^3}{b}\right )+2 c (1+p) x^3 \operatorname {Hypergeometric2F1}\left (-p,2+p,3+p,-\frac {c x^3}{b}\right )\right )}{3 (1+p) (2+p)} \]

[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)

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14

method	result	size
gosper	\(\frac {x^{3+2 p} \left (c \,x^{3}+b \right ) \left (c \,x^{4}+b x \right )^{p}}{3+3 p}\)	\(33\)
parallelrisch	\(\frac {x^{4} x^{2+2 p} {\left (x \left (c \,x^{3}+b \right )\right )}^{p} c^{2}+x \,x^{2+2 p} {\left (x \left (c \,x^{3}+b \right )\right )}^{p} b c}{3 c \left (1+p \right )}\)	\(59\)
risch	\(\frac {\left (c \,x^{3}+b \right ) x \,x^{2+2 p} \left (c \,x^{3}+b \right )^{p} x^{p} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i x \left (c \,x^{3}+b \right )\right ) p \left (-\operatorname {csgn}\left (i x \left (c \,x^{3}+b \right )\right )+\operatorname {csgn}\left (i \left (c \,x^{3}+b \right )\right )\right ) \left (-\operatorname {csgn}\left (i x \left (c \,x^{3}+b \right )\right )+\operatorname {csgn}\left (i x \right )\right )}{2}}}{3+3 p}\)	\(99\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {{\left (c x^{4} + b x\right )} {\left (c x^{4} + b x\right )}^{p} x^{2 \, p + 2}}{3 \, {\left (p + 1\right )}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {{\left (c x^{6} + b x^{3}\right )} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right )\right )}}{3 \, {\left (p + 1\right )}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.00 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {c x^{4} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right ) + 2 \, \log \left (x\right )\right )} + b x e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right ) + 2 \, \log \left (x\right )\right )}}{3 \, {\left (p + 1\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx={\left (c\,x^4+b\,x\right )}^p\,\left (\frac {c\,x^{2\,p+2}\,x^4}{3\,p+3}+\frac {b\,x\,x^{2\,p+2}}{3\,p+3}\right ) \]

[In]

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

[Out]

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