3.2.89 \(\int \frac {x^2 (e+f x)^n}{a+b x^3} \, dx\) [189]

Optimal. Leaf size=253 \[ -\frac {(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{2/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}-\frac {(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{2/3} \left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}-\frac {(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{2/3} \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)} \]

[Out]

-1/3*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3)*e-a^(1/3)*f))/b^(2/3)/(b^(1/3)*e-a^(1/3)*
f)/(1+n)-1/3*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3)*e+(-1)^(1/3)*a^(1/3)*f))/b^(2/3)/
(b^(1/3)*e+(-1)^(1/3)*a^(1/3)*f)/(1+n)-1/3*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3)*e-(
-1)^(2/3)*a^(1/3)*f))/b^(2/3)/(b^(1/3)*e-(-1)^(2/3)*a^(1/3)*f)/(1+n)

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Rubi [A]
time = 0.26, antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6857, 70} \begin {gather*} -\frac {(e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{2/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}-\frac {(e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{2/3} (n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}-\frac {(e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{2/3} (n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(e + f*x)^n)/(a + b*x^3),x]

[Out]

-1/3*((e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)])/(b^(2
/3)*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) - ((e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x
))/(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f)])/(3*b^(2/3)*(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f)*(1 + n)) - ((e + f*x)^(1
+ n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - (-1)^(2/3)*a^(1/3)*f)])/(3*b^(2/3)*(
b^(1/3)*e - (-1)^(2/3)*a^(1/3)*f)*(1 + n))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {x^2 (e+f x)^n}{a+b x^3} \, dx &=\int \left (\frac {(e+f x)^n}{3 b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac {(e+f x)^n}{3 b^{2/3} \left (-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac {(e+f x)^n}{3 b^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}\right ) \, dx\\ &=\frac {\int \frac {(e+f x)^n}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{2/3}}+\frac {\int \frac {(e+f x)^n}{-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{2/3}}+\frac {\int \frac {(e+f x)^n}{(-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{2/3}}\\ &=-\frac {(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{2/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}-\frac {(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{2/3} \left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}-\frac {(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{2/3} \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 213, normalized size = 0.84 \begin {gather*} \frac {(e+f x)^{1+n} \left (-\frac {\, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-\sqrt [3]{a} f}-\frac {\, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}-\frac {\, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{2/3} (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(e + f*x)^n)/(a + b*x^3),x]

[Out]

((e + f*x)^(1 + n)*(-(Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)]/(b^(1/3)
*e - a^(1/3)*f)) - Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f)]/
(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f) - Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - (-1)^
(2/3)*a^(1/3)*f)]/(b^(1/3)*e - (-1)^(2/3)*a^(1/3)*f)))/(3*b^(2/3)*(1 + n))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(f*x+e)^n/(b*x^3+a),x)

[Out]

int(x^2*(f*x+e)^n/(b*x^3+a),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^2*(f*x+e)^n/(b*x^3+a),x, algorithm="maxima")

[Out]

integrate((f*x + e)^n*x^2/(b*x^3 + a), x)

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Fricas [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^2*(f*x+e)^n/(b*x^3+a),x, algorithm="fricas")

[Out]

integral((f*x + e)^n*x^2/(b*x^3 + a), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(f*x+e)**n/(b*x**3+a),x)

[Out]

Timed out

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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^2*(f*x+e)^n/(b*x^3+a),x, algorithm="giac")

[Out]

integrate((f*x + e)^n*x^2/(b*x^3 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(e + f*x)^n)/(a + b*x^3),x)

[Out]

int((x^2*(e + f*x)^n)/(a + b*x^3), x)

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