3.476 \(\int x^2 (c+d x+e x^2) (a+b x^3)^p \, dx\)

Optimal. Leaf size=107 \[ \frac {c \left (a+b x^3\right )^{p+1}}{3 b (p+1)}+\frac {d x^4 \left (a+b x^3\right )^{p+1} \, _2F_1\left (1,p+\frac {7}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{4 a}+\frac {e x^5 \left (a+b x^3\right )^{p+1} \, _2F_1\left (1,p+\frac {8}{3};\frac {8}{3};-\frac {b x^3}{a}\right )}{5 a} \]

[Out]

1/3*c*(b*x^3+a)^(1+p)/b/(1+p)+1/4*d*x^4*(b*x^3+a)^(1+p)*hypergeom([1, 7/3+p],[7/3],-b*x^3/a)/a+1/5*e*x^5*(b*x^
3+a)^(1+p)*hypergeom([1, 8/3+p],[8/3],-b*x^3/a)/a

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Rubi [A]  time = 0.11, antiderivative size = 125, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1893, 261, 365, 364} \[ \frac {c \left (a+b x^3\right )^{p+1}}{3 b (p+1)}+\frac {1}{4} d x^4 \left (a+b x^3\right )^p \left (\frac {b x^3}{a}+1\right )^{-p} \, _2F_1\left (\frac {4}{3},-p;\frac {7}{3};-\frac {b x^3}{a}\right )+\frac {1}{5} e x^5 \left (a+b x^3\right )^p \left (\frac {b x^3}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{3},-p;\frac {8}{3};-\frac {b x^3}{a}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(a + b*x^3)^(1 + p))/(3*b*(1 + p)) + (d*x^4*(a + b*x^3)^p*Hypergeometric2F1[4/3, -p, 7/3, -((b*x^3)/a)])/(4
*(1 + (b*x^3)/a)^p) + (e*x^5*(a + b*x^3)^p*Hypergeometric2F1[5/3, -p, 8/3, -((b*x^3)/a)])/(5*(1 + (b*x^3)/a)^p
)

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 1893

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

Rubi steps

\begin {align*} \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^p \, dx &=\int \left (c x^2 \left (a+b x^3\right )^p+d x^3 \left (a+b x^3\right )^p+e x^4 \left (a+b x^3\right )^p\right ) \, dx\\ &=c \int x^2 \left (a+b x^3\right )^p \, dx+d \int x^3 \left (a+b x^3\right )^p \, dx+e \int x^4 \left (a+b x^3\right )^p \, dx\\ &=\frac {c \left (a+b x^3\right )^{1+p}}{3 b (1+p)}+\left (d \left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p}\right ) \int x^3 \left (1+\frac {b x^3}{a}\right )^p \, dx+\left (e \left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p}\right ) \int x^4 \left (1+\frac {b x^3}{a}\right )^p \, dx\\ &=\frac {c \left (a+b x^3\right )^{1+p}}{3 b (1+p)}+\frac {1}{4} d x^4 \left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p} \, _2F_1\left (\frac {4}{3},-p;\frac {7}{3};-\frac {b x^3}{a}\right )+\frac {1}{5} e x^5 \left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p} \, _2F_1\left (\frac {5}{3},-p;\frac {8}{3};-\frac {b x^3}{a}\right )\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 116, normalized size = 1.08 \[ \frac {\left (a+b x^3\right )^p \left (\frac {b x^3}{a}+1\right )^{-p} \left (20 c \left (a+b x^3\right ) \left (\frac {b x^3}{a}+1\right )^p+15 b d (p+1) x^4 \, _2F_1\left (\frac {4}{3},-p;\frac {7}{3};-\frac {b x^3}{a}\right )+12 b e (p+1) x^5 \, _2F_1\left (\frac {5}{3},-p;\frac {8}{3};-\frac {b x^3}{a}\right )\right )}{60 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^p*(20*c*(a + b*x^3)*(1 + (b*x^3)/a)^p + 15*b*d*(1 + p)*x^4*Hypergeometric2F1[4/3, -p, 7/3, -((b*x
^3)/a)] + 12*b*e*(1 + p)*x^5*Hypergeometric2F1[5/3, -p, 8/3, -((b*x^3)/a)]))/(60*b*(1 + p)*(1 + (b*x^3)/a)^p)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x^{4} + d x^{3} + c x^{2}\right )} {\left (b x^{3} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x^4 + d*x^3 + c*x^2)*(b*x^3 + a)^p, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{2} + d x + c\right )} {\left (b x^{3} + a\right )}^{p} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int \left (e \,x^{2}+d x +c \right ) x^{2} \left (b \,x^{3}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x^2+d*x+c)*(b*x^3+a)^p,x)

[Out]

int(x^2*(e*x^2+d*x+c)*(b*x^3+a)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b x^{3} + a\right )}^{p + 1} c}{3 \, b {\left (p + 1\right )}} + \int {\left (e x^{4} + d x^{3}\right )} {\left (b x^{3} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^3 + a)^(p + 1)*c/(b*(p + 1)) + integrate((e*x^4 + d*x^3)*(b*x^3 + a)^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\left (b\,x^3+a\right )}^p\,\left (e\,x^2+d\,x+c\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^3)^p*(c + d*x + e*x^2),x)

[Out]

int(x^2*(a + b*x^3)^p*(c + d*x + e*x^2), x)

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sympy [A]  time = 124.19, size = 114, normalized size = 1.07 \[ \frac {a^{p} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, - p \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{p} e x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{3}, - p \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} + c \left (\begin {cases} \frac {a^{p} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{3}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{3} \right )} & \text {otherwise} \end {cases}}{3 b} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x**2+d*x+c)*(b*x**3+a)**p,x)

[Out]

a**p*d*x**4*gamma(4/3)*hyper((4/3, -p), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + a**p*e*x**5*gamma(5
/3)*hyper((5/3, -p), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(8/3)) + c*Piecewise((a**p*x**3/3, Eq(b, 0)), (
Piecewise(((a + b*x**3)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + b*x**3), True))/(3*b), True))

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