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

Optimal. Leaf size=102 \[ \frac{c x \left (a+b x^3\right )^{p+1} \, _2F_1\left (1,p+\frac{4}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{a}+\frac{d x^2 \left (a+b x^3\right )^{p+1} \, _2F_1\left (1,p+\frac{5}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{2 a}+\frac{e \left (a+b x^3\right )^{p+1}}{3 b (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0762832, antiderivative size = 120, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1886, 261, 1893, 246, 245, 365, 364} \[ c x \left (a+b x^3\right )^p \left (\frac{b x^3}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{3},-p;\frac{4}{3};-\frac{b x^3}{a}\right )+\frac{1}{2} d x^2 \left (a+b x^3\right )^p \left (\frac{b x^3}{a}+1\right )^{-p} \, _2F_1\left (\frac{2}{3},-p;\frac{5}{3};-\frac{b x^3}{a}\right )+\frac{e \left (a+b x^3\right )^{p+1}}{3 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1886

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[Coeff[Pq, x, n - 1], Int[x^(n - 1)*(a + b*x^n)^p, x
], x] + Int[ExpandToSum[Pq - Coeff[Pq, x, n - 1]*x^(n - 1), x]*(a + b*x^n)^p, x] /; FreeQ[{a, b, p}, x] && Pol
yQ[Pq, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1

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 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])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

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

Rubi steps

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

Mathematica [A]  time = 0.0616251, size = 114, normalized size = 1.12 \[ \frac{\left (a+b x^3\right )^p \left (\frac{b x^3}{a}+1\right )^{-p} \left (6 b c (p+1) x \, _2F_1\left (\frac{1}{3},-p;\frac{4}{3};-\frac{b x^3}{a}\right )+3 b d (p+1) x^2 \, _2F_1\left (\frac{2}{3},-p;\frac{5}{3};-\frac{b x^3}{a}\right )+2 e \left (a+b x^3\right ) \left (\frac{b x^3}{a}+1\right )^p\right )}{6 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.264, size = 0, normalized size = 0. \begin{align*} \int \left ( e{x}^{2}+dx+c \right ) \left ( b{x}^{3}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d x + c\right )}{\left (b x^{3} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e x^{2} + d x + c\right )}{\left (b x^{3} + a\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 91.6781, size = 112, normalized size = 1.1 \begin{align*} \frac{a^{p} c x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, - p \\ \frac{4}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{4}{3}\right )} + \frac{a^{p} d 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 )} + e \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*c*x*gamma(1/3)*hyper((1/3, -p), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + a**p*d*x**2*gamma(2/3)
*hyper((2/3, -p), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3)) + e*Piecewise((a**p*x**3/3, Eq(b, 0)), (Pie
cewise(((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., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d x + c\right )}{\left (b x^{3} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)