3.5.75 \(\int x (c+d x+e x^2) (a+b x^3)^p \, dx\) [475]

Optimal. Leaf size=107 \[ \frac {d \left (a+b x^3\right )^{1+p}}{3 b (1+p)}+\frac {c x^2 \left (a+b x^3\right )^{1+p} \, _2F_1\left (1,\frac {5}{3}+p;\frac {5}{3};-\frac {b x^3}{a}\right )}{2 a}+\frac {e x^4 \left (a+b x^3\right )^{1+p} \, _2F_1\left (1,\frac {7}{3}+p;\frac {7}{3};-\frac {b x^3}{a}\right )}{4 a} \]

[Out]

1/3*d*(b*x^3+a)^(1+p)/b/(1+p)+1/2*c*x^2*(b*x^3+a)^(1+p)*hypergeom([1, 5/3+p],[5/3],-b*x^3/a)/a+1/4*e*x^4*(b*x^
3+a)^(1+p)*hypergeom([1, 7/3+p],[7/3],-b*x^3/a)/a

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Rubi [A]
time = 0.06, antiderivative size = 125, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1907, 372, 371, 267} \begin {gather*} \frac {1}{2} c x^2 \left (a+b x^3\right )^p \left (\frac {b x^3}{a}+1\right )^{-p} \, _2F_1\left (\frac {2}{3},-p;\frac {5}{3};-\frac {b x^3}{a}\right )+\frac {d \left (a+b x^3\right )^{p+1}}{3 b (p+1)}+\frac {1}{4} e 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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(a + b*x^3)^(1 + p))/(3*b*(1 + p)) + (c*x^2*(a + b*x^3)^p*Hypergeometric2F1[2/3, -p, 5/3, -((b*x^3)/a)])/(2
*(1 + (b*x^3)/a)^p) + (e*x^4*(a + b*x^3)^p*Hypergeometric2F1[4/3, -p, 7/3, -((b*x^3)/a)])/(4*(1 + (b*x^3)/a)^p
)

Rule 267

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 1907

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

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Mathematica [A]
time = 0.59, size = 116, normalized size = 1.08 \begin {gather*} \frac {\left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p} \left (4 d \left (a+b x^3\right ) \left (1+\frac {b x^3}{a}\right )^p+6 b c (1+p) x^2 \, _2F_1\left (\frac {2}{3},-p;\frac {5}{3};-\frac {b x^3}{a}\right )+3 b e (1+p) x^4 \, _2F_1\left (\frac {4}{3},-p;\frac {7}{3};-\frac {b x^3}{a}\right )\right )}{12 b (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^p*(4*d*(a + b*x^3)*(1 + (b*x^3)/a)^p + 6*b*c*(1 + p)*x^2*Hypergeometric2F1[2/3, -p, 5/3, -((b*x^3
)/a)] + 3*b*e*(1 + p)*x^4*Hypergeometric2F1[4/3, -p, 7/3, -((b*x^3)/a)]))/(12*b*(1 + p)*(1 + (b*x^3)/a)^p)

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int x \left (e \,x^{2}+d x +c \right ) \left (b \,x^{3}+a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.39, size = 26, normalized size = 0.24 \begin {gather*} {\rm integral}\left ({\left (e x^{3} + d x^{2} + c x\right )} {\left (b x^{3} + a\right )}^{p}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 44.05, size = 114, normalized size = 1.07 \begin {gather*} \frac {a^{p} c x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, - p \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + \frac {a^{p} e 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 )} + d \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*c*x**2*gamma(2/3)*hyper((2/3, -p), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3)) + a**p*e*x**4*gamma(4
/3)*hyper((4/3, -p), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + d*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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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