Optimal. Leaf size=324 \[ \frac {e^2 (e+f x)^{1+n}}{b f^3 (1+n)}-\frac {2 e (e+f x)^{2+n}}{b f^3 (2+n)}+\frac {(e+f x)^{3+n}}{b f^3 (3+n)}+\frac {a (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {a (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac {a (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)} \]
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Rubi [A]
time = 0.63, 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 = {6857, 70}
\begin {gather*} \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)} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 6857
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 = 0.49, size = 284, normalized size = 0.88 \begin {gather*} \frac {(e+f x)^{1+n} \left (\frac {3 b^{2/3} e^2}{f^3 (1+n)}-\frac {6 b^{2/3} e (e+f x)}{f^3 (2+n)}+\frac {3 b^{2/3} (e+f x)^2}{f^3 (3+n)}+\frac {a \, _2F_1\left (1,1+n;2+n;\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {a \, _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 )}{\left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac {a \, _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 )}{\left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)}\right )}{3 b^{5/3}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.06, 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.
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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.
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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.
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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.
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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.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^5\,{\left (e+f\,x\right )}^n}{b\,x^3+a} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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