3.186 \(\int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx\)

Optimal. Leaf size=324 \[ \frac {a (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {a (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}+\frac {a (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )}+\frac {e^2 (e+f x)^{n+1}}{b f^3 (n+1)}-\frac {2 e (e+f x)^{n+2}}{b f^3 (n+2)}+\frac {(e+f x)^{n+3}}{b f^3 (n+3)} \]

[Out]

e^2*(f*x+e)^(1+n)/b/f^3/(1+n)-2*e*(f*x+e)^(2+n)/b/f^3/(2+n)+(f*x+e)^(3+n)/b/f^3/(3+n)+1/3*a*(f*x+e)^(1+n)*hype
rgeom([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3)*e-a^(1/3)*f))/b^(5/3)/(b^(1/3)*e-a^(1/3)*f)/(1+n)+1/3*a*(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^(5/3)/(b^(1/3)*e+(-1)^(1/3)
*a^(1/3)*f)/(1+n)+1/3*a*(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^(5/3)/(b^(1/3)*e-(-1)^(2/3)*a^(1/3)*f)/(1+n)

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Rubi [A]  time = 0.86, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6725, 68} \[ \frac {a (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {a (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}+\frac {a (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )}+\frac {e^2 (e+f x)^{n+1}}{b f^3 (n+1)}-\frac {2 e (e+f x)^{n+2}}{b f^3 (n+2)}+\frac {(e+f x)^{n+3}}{b f^3 (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(e + f*x)^n)/(a + b*x^3),x]

[Out]

(e^2*(e + f*x)^(1 + n))/(b*f^3*(1 + n)) - (2*e*(e + f*x)^(2 + n))/(b*f^3*(2 + n)) + (e + f*x)^(3 + n)/(b*f^3*(
3 + n)) + (a*(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)]
)/(3*b^(5/3)*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + (a*(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^(5/3)*(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f)*(1 + n)) + (a
*(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^(5/3)*(b^(1/3)*e - (-1)^(2/3)*a^(1/3)*f)*(1 + n))

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]

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]

Rubi steps

\begin {align*} \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx &=\int \left (\frac {e^2 (e+f x)^n}{b f^2}-\frac {2 e (e+f x)^{1+n}}{b f^2}+\frac {(e+f x)^{2+n}}{b f^2}-\frac {a x^2 (e+f x)^n}{b \left (a+b x^3\right )}\right ) \, dx\\ &=\frac {e^2 (e+f x)^{1+n}}{b f^3 (1+n)}-\frac {2 e (e+f x)^{2+n}}{b f^3 (2+n)}+\frac {(e+f x)^{3+n}}{b f^3 (3+n)}-\frac {a \int \frac {x^2 (e+f x)^n}{a+b x^3} \, dx}{b}\\ &=\frac {e^2 (e+f x)^{1+n}}{b f^3 (1+n)}-\frac {2 e (e+f x)^{2+n}}{b f^3 (2+n)}+\frac {(e+f x)^{3+n}}{b f^3 (3+n)}-\frac {a \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}{b}\\ &=\frac {e^2 (e+f x)^{1+n}}{b f^3 (1+n)}-\frac {2 e (e+f x)^{2+n}}{b f^3 (2+n)}+\frac {(e+f x)^{3+n}}{b f^3 (3+n)}-\frac {a \int \frac {(e+f x)^n}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{5/3}}-\frac {a \int \frac {(e+f x)^n}{-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{5/3}}-\frac {a \int \frac {(e+f x)^n}{(-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{5/3}}\\ &=\frac {e^2 (e+f x)^{1+n}}{b f^3 (1+n)}-\frac {2 e (e+f x)^{2+n}}{b f^3 (2+n)}+\frac {(e+f x)^{3+n}}{b f^3 (3+n)}+\frac {a (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^{5/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {a (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^{5/3} \left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac {a (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^{5/3} \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 284, normalized size = 0.88 \[ \frac {(e+f x)^{n+1} \left (\frac {a \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{(n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {a \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{(n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}+\frac {a \, _2F_1\left (1,n+1;n+2;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{(n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )}+\frac {3 b^{2/3} e^2}{f^3 (n+1)}-\frac {6 b^{2/3} e (e+f x)}{f^3 (n+2)}+\frac {3 b^{2/3} (e+f x)^2}{f^3 (n+3)}\right )}{3 b^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(e + f*x)^n)/(a + b*x^3),x]

[Out]

((e + f*x)^(1 + n)*((3*b^(2/3)*e^2)/(f^3*(1 + n)) - (6*b^(2/3)*e*(e + f*x))/(f^3*(2 + n)) + (3*b^(2/3)*(e + f*
x)^2)/(f^3*(3 + n)) + (a*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)*(1 + n)) + (a*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)*(1 + n)) + (a*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)*(1 + n))))/(3*b^(5/3))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(f*x+e)^n/(b*x^3+a),x, algorithm="fricas")

[Out]

integral((f*x + e)^n*x^5/(b*x^3 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(f*x+e)^n/(b*x^3+a),x, algorithm="giac")

[Out]

integrate((f*x + e)^n*x^5/(b*x^3 + a), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (f x +e \right )^{n}}{b \,x^{3}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(f*x+e)^n/(b*x^3+a),x)

[Out]

int(x^5*(f*x+e)^n/(b*x^3+a),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(f*x+e)^n/(b*x^3+a),x, algorithm="maxima")

[Out]

integrate((f*x + e)^n*x^5/(b*x^3 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,{\left (e+f\,x\right )}^n}{b\,x^3+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(e + f*x)^n)/(a + b*x^3),x)

[Out]

int((x^5*(e + f*x)^n)/(a + b*x^3), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(f*x+e)**n/(b*x**3+a),x)

[Out]

Timed out

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