3.187 \(\int \frac{x^4 (e+f x)^n}{a+b x^3} \, dx\)

Optimal. Leaf size=332 \[ -\frac{a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{-1} a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{(-1)^{2/3} a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}-\frac{e (e+f x)^{n+1}}{b f^2 (n+1)}+\frac{(e+f x)^{n+2}}{b f^2 (n+2)} \]

[Out]

-((e*(e + f*x)^(1 + n))/(b*f^2*(1 + n))) + (e + f*x)^(2 + n)/(b*f^2*(2 + n)) - (a^(2/3)*(e + f*x)^(1 + n)*Hype
rgeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)])/(3*b^(4/3)*(b^(1/3)*e - a^(1/3)*f
)*(1 + n)) + ((-1)^(1/3)*a^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(2/3)*b^(1/3)*(e +
f*x))/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)])/(3*b^(4/3)*((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(2/
3)*a^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(1/3)*b^(1/3)*(e + f*x))/((-1)^(1/3)*b^(
1/3)*e + a^(1/3)*f)])/(3*b^(4/3)*((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)*(1 + n))

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Rubi [A]  time = 0.862376, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {6725, 68} \[ -\frac{a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{-1} a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{(-1)^{2/3} a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}-\frac{e (e+f x)^{n+1}}{b f^2 (n+1)}+\frac{(e+f x)^{n+2}}{b f^2 (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(e + f*x)^n)/(a + b*x^3),x]

[Out]

-((e*(e + f*x)^(1 + n))/(b*f^2*(1 + n))) + (e + f*x)^(2 + n)/(b*f^2*(2 + n)) - (a^(2/3)*(e + f*x)^(1 + n)*Hype
rgeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)])/(3*b^(4/3)*(b^(1/3)*e - a^(1/3)*f
)*(1 + n)) + ((-1)^(1/3)*a^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(2/3)*b^(1/3)*(e +
f*x))/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)])/(3*b^(4/3)*((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(2/
3)*a^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(1/3)*b^(1/3)*(e + f*x))/((-1)^(1/3)*b^(
1/3)*e + a^(1/3)*f)])/(3*b^(4/3)*((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)*(1 + n))

Rule 6725

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]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{x^4 (e+f x)^n}{a+b x^3} \, dx &=\int \left (-\frac{e (e+f x)^n}{b f}+\frac{(e+f x)^{1+n}}{b f}-\frac{a x (e+f x)^n}{b \left (a+b x^3\right )}\right ) \, dx\\ &=-\frac{e (e+f x)^{1+n}}{b f^2 (1+n)}+\frac{(e+f x)^{2+n}}{b f^2 (2+n)}-\frac{a \int \frac{x (e+f x)^n}{a+b x^3} \, dx}{b}\\ &=-\frac{e (e+f x)^{1+n}}{b f^2 (1+n)}+\frac{(e+f x)^{2+n}}{b f^2 (2+n)}-\frac{a \int \left (-\frac{(e+f x)^n}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac{(-1)^{2/3} (e+f x)^n}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}+\frac{\sqrt [3]{-1} (e+f x)^n}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}\right ) \, dx}{b}\\ &=-\frac{e (e+f x)^{1+n}}{b f^2 (1+n)}+\frac{(e+f x)^{2+n}}{b f^2 (2+n)}+\frac{a^{2/3} \int \frac{(e+f x)^n}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{4/3}}-\frac{\left (\sqrt [3]{-1} a^{2/3}\right ) \int \frac{(e+f x)^n}{\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x} \, dx}{3 b^{4/3}}+\frac{\left ((-1)^{2/3} a^{2/3}\right ) \int \frac{(e+f x)^n}{\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x} \, dx}{3 b^{4/3}}\\ &=-\frac{e (e+f x)^{1+n}}{b f^2 (1+n)}+\frac{(e+f x)^{2+n}}{b f^2 (2+n)}-\frac{a^{2/3} (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^{4/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac{\sqrt [3]{-1} a^{2/3} (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac{(-1)^{2/3} a^{2/3} (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 b^{4/3} \left (\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f\right ) (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.68365, size = 292, normalized size = 0.88 \[ \frac{(e+f x)^{n+1} \left (-\frac{a^{2/3} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{(n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{-1} a^{2/3} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{(n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{(-1)^{2/3} a^{2/3} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{(n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}+\frac{3 \sqrt [3]{b} (e+f x)}{f^2 (n+2)}-\frac{3 \sqrt [3]{b} e}{f^2 (n+1)}\right )}{3 b^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(e + f*x)^n)/(a + b*x^3),x]

[Out]

((e + f*x)^(1 + n)*((-3*b^(1/3)*e)/(f^2*(1 + n)) + (3*b^(1/3)*(e + f*x))/(f^2*(2 + n)) - (a^(2/3)*Hypergeometr
ic2F1[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)*(1 + n)) + ((-1)
^(1/3)*a^(2/3)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(2/3)*b^(1/3)*(e + f*x))/((-1)^(2/3)*b^(1/3)*e - a^(1/
3)*f)])/(((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(2/3)*a^(2/3)*Hypergeometric2F1[1, 1 + n, 2 + n,
((-1)^(1/3)*b^(1/3)*(e + f*x))/((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)])/(((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)*(1 + n
))))/(3*b^(4/3))

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Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{4} \left ( fx+e \right ) ^{n}}{b{x}^{3}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(f*x+e)^n/(b*x^3+a),x)

[Out]

int(x^4*(f*x+e)^n/(b*x^3+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(f*x+e)^n/(b*x^3+a),x, algorithm="maxima")

[Out]

integrate((f*x + e)^n*x^4/(b*x^3 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(f*x+e)^n/(b*x^3+a),x, algorithm="fricas")

[Out]

integral((f*x + e)^n*x^4/(b*x^3 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(f*x+e)**n/(b*x**3+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(f*x+e)^n/(b*x^3+a),x, algorithm="giac")

[Out]

integrate((f*x + e)^n*x^4/(b*x^3 + a), x)