3.170 \(\int \frac{(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx\)

Optimal. Leaf size=326 \[ -\frac{b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{-1} b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{(-1)^{2/3} b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a e^2 (n+1)} \]

[Out]

-(b^(2/3)*(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)])/(3*a^(4/3)*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)
^(1/3)*b^(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*a^(4/3)*((-1)^(2/3)*b
^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(2/3)*b^(2/3)*(e + f*x)^(1 + n)*Hypergeom
etric2F1[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*a^(4/3)*((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)*(1 + n)) + (f*(e + f
*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/(a*e^2*(1 + n))

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Rubi [A]  time = 1.28852, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15 \[ -\frac{b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{-1} b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{(-1)^{2/3} b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a e^2 (n+1)} \]

Antiderivative was successfully verified.

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

[Out]

-(b^(2/3)*(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)])/(3*a^(4/3)*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)
^(1/3)*b^(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*a^(4/3)*((-1)^(2/3)*b
^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(2/3)*b^(2/3)*(e + f*x)^(1 + n)*Hypergeom
etric2F1[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*a^(4/3)*((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)*(1 + n)) + (f*(e + f
*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/(a*e^2*(1 + n))

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Rubi in Sympy [A]  time = 163.288, size = 282, normalized size = 0.87 \[ \frac{f \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 2, n + 1 \\ n + 2 \end{matrix}\middle |{1 + \frac{f x}{e}} \right )}}{a e^{2} \left (n + 1\right )} - \frac{\sqrt [3]{-1} b^{\frac{2}{3}} \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{\left (-1\right )^{\frac{2}{3}} \sqrt [3]{b} \left (e + f x\right )}{- \sqrt [3]{a} f + \left (-1\right )^{\frac{2}{3}} \sqrt [3]{b} e}} \right )}}{3 a^{\frac{4}{3}} \left (n + 1\right ) \left (\sqrt [3]{a} f - \left (-1\right )^{\frac{2}{3}} \sqrt [3]{b} e\right )} + \frac{\left (-1\right )^{\frac{2}{3}} b^{\frac{2}{3}} \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{\sqrt [3]{-1} \sqrt [3]{b} \left (e + f x\right )}{\sqrt [3]{a} f + \sqrt [3]{-1} \sqrt [3]{b} e}} \right )}}{3 a^{\frac{4}{3}} \left (n + 1\right ) \left (\sqrt [3]{a} f + \sqrt [3]{-1} \sqrt [3]{b} e\right )} + \frac{b^{\frac{2}{3}} \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{\sqrt [3]{b} \left (e + f x\right )}{- \sqrt [3]{a} f + \sqrt [3]{b} e}} \right )}}{3 a^{\frac{4}{3}} \left (n + 1\right ) \left (\sqrt [3]{a} f - \sqrt [3]{b} e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

f*(e + f*x)**(n + 1)*hyper((2, n + 1), (n + 2,), 1 + f*x/e)/(a*e**2*(n + 1)) - (
-1)**(1/3)*b**(2/3)*(e + f*x)**(n + 1)*hyper((1, n + 1), (n + 2,), (-1)**(2/3)*b
**(1/3)*(e + f*x)/(-a**(1/3)*f + (-1)**(2/3)*b**(1/3)*e))/(3*a**(4/3)*(n + 1)*(a
**(1/3)*f - (-1)**(2/3)*b**(1/3)*e)) + (-1)**(2/3)*b**(2/3)*(e + f*x)**(n + 1)*h
yper((1, n + 1), (n + 2,), (-1)**(1/3)*b**(1/3)*(e + f*x)/(a**(1/3)*f + (-1)**(1
/3)*b**(1/3)*e))/(3*a**(4/3)*(n + 1)*(a**(1/3)*f + (-1)**(1/3)*b**(1/3)*e)) + b*
*(2/3)*(e + f*x)**(n + 1)*hyper((1, n + 1), (n + 2,), b**(1/3)*(e + f*x)/(-a**(1
/3)*f + b**(1/3)*e))/(3*a**(4/3)*(n + 1)*(a**(1/3)*f - b**(1/3)*e))

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Mathematica [C]  time = 0.247376, size = 280, normalized size = 0.86 \[ \frac{(e+f x)^n \left (\frac{e f \text{RootSum}\left [-\text{$\#$1}^3 b+3 \text{$\#$1}^2 b e-3 \text{$\#$1} b e^2-a f^3+b e^3\&,\frac{\left (\frac{e+f x}{-\text{$\#$1}+e+f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;-\frac{\text{$\#$1}}{e+f x-\text{$\#$1}}\right )}{\text{$\#$1}^2-2 \text{$\#$1} e+e^2}\&\right ]}{n}-\frac{f \text{RootSum}\left [-\text{$\#$1}^3 b+3 \text{$\#$1}^2 b e-3 \text{$\#$1} b e^2-a f^3+b e^3\&,\frac{\text{$\#$1} \left (\frac{e+f x}{-\text{$\#$1}+e+f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;-\frac{\text{$\#$1}}{e+f x-\text{$\#$1}}\right )}{\text{$\#$1}^2-2 \text{$\#$1} e+e^2}\&\right ]}{n}+\frac{3 \left (\frac{e}{f x}+1\right )^{-n} \, _2F_1\left (1-n,-n;2-n;-\frac{e}{f x}\right )}{(n-1) x}\right )}{3 a} \]

Warning: Unable to verify antiderivative.

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

[Out]

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

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Maple [F]  time = 0.09, size = 0, normalized size = 0. \[ \int{\frac{ \left ( fx+e \right ) ^{n}}{{x}^{2} \left ( b{x}^{3}+a \right ) }}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((f*x+e)^n/x^2/(b*x^3+a),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x + e)^n/((b*x^3 + a)*x^2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (f x + e\right )}^{n}}{b x^{5} + a x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((f*x + e)^n/(b*x^5 + a*x^2), x)

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

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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