3.5.74 \(\int (c+d x+e x^2) (a+b x^3)^p \, dx\) [474]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 120, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1900, 267, 1907, 252, 251, 372, 371} \begin {gather*} 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)} \end {gather*}

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 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 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 1900

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 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 \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*}

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Mathematica [A]
time = 0.60, size = 114, normalized size = 1.12 \begin {gather*} \frac {\left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p} \left (2 e \left (a+b x^3\right ) \left (1+\frac {b x^3}{a}\right )^p+6 b c (1+p) x \, _2F_1\left (\frac {1}{3},-p;\frac {4}{3};-\frac {b x^3}{a}\right )+3 b d (1+p) x^2 \, _2F_1\left (\frac {2}{3},-p;\frac {5}{3};-\frac {b x^3}{a}\right )\right )}{6 b (1+p)} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \left (e \,x^{2}+d x +c \right ) \left (b \,x^{3}+a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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 = 0.00, 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((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)

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Fricas [F]
time = 0.38, size = 22, normalized size = 0.22 \begin {gather*} {\rm integral}\left ({\left (e x^{2} + d x + c\right )} {\left (b x^{3} + a\right )}^{p}, x\right ) \end {gather*}

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 = 29.24, size = 112, normalized size = 1.10 \begin {gather*} \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 {gather*}

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 = 0.00, 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((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)

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

[Out]

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

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