3.599 \(\int \frac{(d x)^m}{a+b x^n+c x^{2 n}} \, dx\)

Optimal. Leaf size=175 \[ \frac{2 c (d x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{d (m+1) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{2 c (d x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{d (m+1) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )} \]

[Out]

(2*c*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b
- Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*(1 + m)) - (
2*c*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b +
Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*(1 + m))

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Rubi [A]  time = 0.406217, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091 \[ \frac{2 c (d x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{d (m+1) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{2 c (d x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{d (m+1) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )} \]

Antiderivative was successfully verified.

[In]  Int[(d*x)^m/(a + b*x^n + c*x^(2*n)),x]

[Out]

(2*c*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b
- Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*(1 + m)) - (
2*c*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b +
Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*(1 + m))

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Rubi in Sympy [A]  time = 29.5365, size = 144, normalized size = 0.82 \[ - \frac{2 c \left (d x\right )^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m + 1}{n} \\ \frac{m + n + 1}{n} \end{matrix}\middle |{- \frac{2 c x^{n}}{b + \sqrt{- 4 a c + b^{2}}}} \right )}}{d \left (b + \sqrt{- 4 a c + b^{2}}\right ) \left (m + 1\right ) \sqrt{- 4 a c + b^{2}}} + \frac{2 c \left (d x\right )^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m + 1}{n} \\ \frac{m + n + 1}{n} \end{matrix}\middle |{- \frac{2 c x^{n}}{b - \sqrt{- 4 a c + b^{2}}}} \right )}}{d \left (b - \sqrt{- 4 a c + b^{2}}\right ) \left (m + 1\right ) \sqrt{- 4 a c + b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((d*x)**m/(a+b*x**n+c*x**(2*n)),x)

[Out]

-2*c*(d*x)**(m + 1)*hyper((1, (m + 1)/n), ((m + n + 1)/n,), -2*c*x**n/(b + sqrt(
-4*a*c + b**2)))/(d*(b + sqrt(-4*a*c + b**2))*(m + 1)*sqrt(-4*a*c + b**2)) + 2*c
*(d*x)**(m + 1)*hyper((1, (m + 1)/n), ((m + n + 1)/n,), -2*c*x**n/(b - sqrt(-4*a
*c + b**2)))/(d*(b - sqrt(-4*a*c + b**2))*(m + 1)*sqrt(-4*a*c + b**2))

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Mathematica [A]  time = 1.91469, size = 307, normalized size = 1.75 \[ -\frac{x (d x)^m \left (\frac{2 c \left (1-2^{-\frac{m+1}{n}} \left (\frac{c x^n}{-\sqrt{b^2-4 a c}+b+2 c x^n}\right )^{-\frac{m+1}{n}} \, _2F_1\left (-\frac{m+1}{n},-\frac{m+1}{n};1-\frac{m+1}{n};\frac{b-\sqrt{b^2-4 a c}}{2 c x^n+b-\sqrt{b^2-4 a c}}\right )\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}+\frac{2 c \left (1-2^{-\frac{m+1}{n}} \left (\frac{c x^n}{\sqrt{b^2-4 a c}+b+2 c x^n}\right )^{-\frac{m+1}{n}} \, _2F_1\left (-\frac{m+1}{n},-\frac{m+1}{n};\frac{-m+n-1}{n};\frac{b+\sqrt{b^2-4 a c}}{2 c x^n+b+\sqrt{b^2-4 a c}}\right )\right )}{\sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]  Integrate[(d*x)^m/(a + b*x^n + c*x^(2*n)),x]

[Out]

-((x*(d*x)^m*((2*c*(1 - Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m
)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(2^((1 + m)/n)*(
(c*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n))))/(b^2 - 4*a*c - b*Sqrt[
b^2 - 4*a*c]) + (2*c*(1 - Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), (-1 - m
+ n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(2^((1 + m)/n
)*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n))))/(Sqrt[b^2 - 4*a*c]*
(b + Sqrt[b^2 - 4*a*c]))))/(1 + m))

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Maple [F]  time = 0.045, size = 0, normalized size = 0. \[ \int{\frac{ \left ( dx \right ) ^{m}}{a+b{x}^{n}+c{x}^{2\,n}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((d*x)^m/(a+b*x^n+c*x^(2*n)),x)

[Out]

int((d*x)^m/(a+b*x^n+c*x^(2*n)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x)^m/(c*x^(2*n) + b*x^n + a),x, algorithm="maxima")

[Out]

integrate((d*x)^m/(c*x^(2*n) + b*x^n + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x)^m/(c*x^(2*n) + b*x^n + a),x, algorithm="fricas")

[Out]

integral((d*x)^m/(c*x^(2*n) + b*x^n + a), 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((d*x)**m/(a+b*x**n+c*x**(2*n)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x)^m/(c*x^(2*n) + b*x^n + a),x, algorithm="giac")

[Out]

integrate((d*x)^m/(c*x^(2*n) + b*x^n + a), x)