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

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

[Out]

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

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Rubi [A]  time = 0.121905, 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 \[ \frac{x^{m+1} (a B (m+1)+A (b-b m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 b (m+1)}+\frac{x^{m+1} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 14.8714, size = 71, normalized size = 0.76 \[ \frac{x^{m + 1} \left (A b - B a\right )}{2 a b \left (a + b x^{2}\right )} + \frac{x^{m + 1} \left (A b \left (- m + 1\right ) + B a \left (m + 1\right )\right ){{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{2 a^{2} b \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.0708068, size = 80, normalized size = 0.86 \[ \frac{x^{m+1} \left ((A b-a B) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )+a B \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )\right )}{a^2 b (m+1)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.058, size = 0, normalized size = 0. \[ \int{\frac{{x}^{m} \left ( B{x}^{2}+A \right ) }{ \left ( b{x}^{2}+a \right ) ^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 137.154, size = 906, normalized size = 9.74 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

A*(-a*m**2*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1
/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) + 2*a*m*x*x**m*ga
mma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) + a*x*
x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**3*g
amma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) + 2*a*x*x**m*gamma(m/2 + 1/2)/
(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) - b*m**2*x**3*x**m*le
rchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/
2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) + b*x**3*x**m*lerchphi(b*x**2*exp_pol
ar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x
**2*gamma(m/2 + 3/2))) + B*(-a*m**2*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a,
1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m
/2 + 5/2)) - 4*a*m*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*ga
mma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) + 2*a*
m*x**3*x**m*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2
+ 5/2)) - 3*a*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m
/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) + 6*a*x**3*
x**m*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2))
- b*m**2*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 +
3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 4*b*m*x**5*x*
*m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gam
ma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 3*b*x**5*x**m*lerchphi(b*x**2*
exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a
**2*b*x**2*gamma(m/2 + 5/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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