3.324 \(\int \frac{x^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx\)

Optimal. Leaf size=93 \[ \frac{x^{m+1} (a B (m+1)+A b (3-m)) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{4 a^3 b (m+1)}+\frac{x^{m+1} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

[Out]

((A*b - a*B)*x^(1 + m))/(4*a*b*(a + b*x^2)^2) + ((A*b*(3 - m) + a*B*(1 + m))*x^(
1 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(4*a^3*b*(1 + m
))

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Rubi [A]  time = 0.119772, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1 \[ \frac{x^{m+1} (a B (m+1)+A b (3-m)) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{4 a^3 b (m+1)}+\frac{x^{m+1} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]  Int[(x^m*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

((A*b - a*B)*x^(1 + m))/(4*a*b*(a + b*x^2)^2) + ((A*b*(3 - m) + a*B*(1 + m))*x^(
1 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(4*a^3*b*(1 + m
))

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Rubi in Sympy [A]  time = 14.7396, size = 73, normalized size = 0.78 \[ \frac{x^{m + 1} \left (A b - B a\right )}{4 a b \left (a + b x^{2}\right )^{2}} + \frac{x^{m + 1} \left (A b \left (- m + 3\right ) + B a \left (m + 1\right )\right ){{}_{2}F_{1}\left (\begin{matrix} 2, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{4 a^{3} b \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**m*(B*x**2+A)/(b*x**2+a)**3,x)

[Out]

x**(m + 1)*(A*b - B*a)/(4*a*b*(a + b*x**2)**2) + x**(m + 1)*(A*b*(-m + 3) + B*a*
(m + 1))*hyper((2, m/2 + 1/2), (m/2 + 3/2,), -b*x**2/a)/(4*a**3*b*(m + 1))

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Mathematica [A]  time = 0.0801516, size = 80, normalized size = 0.86 \[ \frac{x^{m+1} \left ((A b-a B) \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )+a B \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )\right )}{a^3 b (m+1)} \]

Antiderivative was successfully verified.

[In]  Integrate[(x^m*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(x^(1 + m)*(a*B*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + (A*b
- a*B)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]))/(a^3*b*(1 + m)
)

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Maple [F]  time = 0.079, size = 0, normalized size = 0. \[ \int{\frac{{x}^{m} \left ( B{x}^{2}+A \right ) }{ \left ( b{x}^{2}+a \right ) ^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^m*(B*x^2+A)/(b*x^2+a)^3,x)

[Out]

int(x^m*(B*x^2+A)/(b*x^2+a)^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*x^m/(b*x^2 + a)^3,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^m/(b*x^2 + a)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (B x^{2} + A\right )} x^{m}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*x^m/(b*x^2 + a)^3,x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*x^m/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), 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(x**m*(B*x**2+A)/(b*x**2+a)**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*x^m/(b*x^2 + a)^3,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^m/(b*x^2 + a)^3, x)