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]
_______________________________________________________________________________________
Rubi [A] time = 1.60278, 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 \[ -\frac{a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{-1} a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{(-1)^{2/3} a^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}-\frac{e (e+f x)^{n+1}}{b f^2 (n+1)}+\frac{(e+f x)^{n+2}}{b f^2 (n+2)} \]
Antiderivative was successfully verified.
[In] Int[(x^4*(e + f*x)^n)/(a + b*x^3),x]
[Out]
_______________________________________________________________________________________
Rubi in Sympy [A] time = 167.999, size = 287, normalized size = 0.86 \[ - \frac{\sqrt [3]{-1} a^{\frac{2}{3}} \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{\left (-1\right )^{\frac{2}{3}} \sqrt [3]{b} \left (e + f x\right )}{- \sqrt [3]{a} f + \left (-1\right )^{\frac{2}{3}} \sqrt [3]{b} e}} \right )}}{3 b^{\frac{4}{3}} \left (n + 1\right ) \left (\sqrt [3]{a} f - \left (-1\right )^{\frac{2}{3}} \sqrt [3]{b} e\right )} + \frac{\left (-1\right )^{\frac{2}{3}} a^{\frac{2}{3}} \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{\sqrt [3]{-1} \sqrt [3]{b} \left (e + f x\right )}{\sqrt [3]{a} f + \sqrt [3]{-1} \sqrt [3]{b} e}} \right )}}{3 b^{\frac{4}{3}} \left (n + 1\right ) \left (\sqrt [3]{a} f + \sqrt [3]{-1} \sqrt [3]{b} e\right )} + \frac{a^{\frac{2}{3}} \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{\sqrt [3]{b} \left (e + f x\right )}{- \sqrt [3]{a} f + \sqrt [3]{b} e}} \right )}}{3 b^{\frac{4}{3}} \left (n + 1\right ) \left (\sqrt [3]{a} f - \sqrt [3]{b} e\right )} - \frac{e \left (e + f x\right )^{n + 1}}{b f^{2} \left (n + 1\right )} + \frac{\left (e + f x\right )^{n + 2}}{b f^{2} \left (n + 2\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate(x**4*(f*x+e)**n/(b*x**3+a),x)
[Out]
_______________________________________________________________________________________
Mathematica [C] time = 0.691891, size = 298, normalized size = 0.9 \[ \frac{(e+f x)^n \left (\frac{a e f^3 \text{RootSum}\left [-\text{$\#$1}^3 b+3 \text{$\#$1}^2 b e-3 \text{$\#$1} b e^2-a f^3+b e^3\&,\frac{\left (\frac{e+f x}{-\text{$\#$1}+e+f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;-\frac{\text{$\#$1}}{e+f x-\text{$\#$1}}\right )}{\text{$\#$1}^2-2 \text{$\#$1} e+e^2}\&\right ]}{b n}-\frac{a f^3 \text{RootSum}\left [-\text{$\#$1}^3 b+3 \text{$\#$1}^2 b e-3 \text{$\#$1} b e^2-a f^3+b e^3\&,\frac{\text{$\#$1} \left (\frac{e+f x}{-\text{$\#$1}+e+f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;-\frac{\text{$\#$1}}{e+f x-\text{$\#$1}}\right )}{\text{$\#$1}^2-2 \text{$\#$1} e+e^2}\&\right ]}{b n}-\frac{3 \left (e^2 \left (1-\left (\frac{f x}{e}+1\right )^{-n}\right )-e f n x-f^2 (n+1) x^2\right )}{n^2+3 n+2}\right )}{3 b f^2} \]
Warning: Unable to verify antiderivative.
[In] Integrate[(x^4*(e + f*x)^n)/(a + b*x^3),x]
[Out]
_______________________________________________________________________________________
Maple [F] time = 0.083, size = 0, normalized size = 0. \[ \int{\frac{{x}^{4} \left ( fx+e \right ) ^{n}}{b{x}^{3}+a}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int(x^4*(f*x+e)^n/(b*x^3+a),x)
[Out]
_______________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((f*x + e)^n*x^4/(b*x^3 + a),x, algorithm="maxima")
[Out]
_______________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((f*x + e)^n*x^4/(b*x^3 + a),x, algorithm="fricas")
[Out]
_______________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(x**4*(f*x+e)**n/(b*x**3+a),x)
[Out]
_______________________________________________________________________________________
GIAC/XCAS [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((f*x + e)^n*x^4/(b*x^3 + a),x, algorithm="giac")
[Out]