3.323 \(\int \frac{x^m (A+B x^2)}{(a+b x^2)^2} \, dx\)
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 = 0.0439417, 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, Rules used =
{457, 364} \[ \frac{x^{m+1} (a B (m+1)+A (b-b m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 b (m+1)}+\frac{x^{m+1} (A b-a B)}{2 a b \left (a+b x^2\right )} \]
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))
Rule 457
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))
Rule 364
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])
Rubi steps
\begin{align*} \int \frac{x^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac{(A b-a B) x^{1+m}}{2 a b \left (a+b x^2\right )}+\frac{(a B (1+m)+A (b-b m)) \int \frac{x^m}{a+b x^2} \, dx}{2 a b}\\ &=\frac{(A b-a B) x^{1+m}}{2 a b \left (a+b x^2\right )}+\frac{(a B (1+m)+A (b-b m)) x^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{2 a^2 b (1+m)}\\ \end{align*}
Mathematica [A] time = 0.0558589, 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 = 0.038, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ){x}^{m}}{ \left ( b{x}^{2}+a \right ) ^{2}}}\, dx \end{align*}
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. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{m}}{{\left (b x^{2} + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^m*(B*x^2+A)/(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. \begin{align*}{\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^m*(B*x^2+A)/(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 [C] time = 35.6348, size = 906, normalized size = 9.74 \begin{align*} \text{result too large to display} \end{align*}
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*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamm
a(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*gamma(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*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)) + b*x**3*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))) + B*(-a*m**2*x**3*x**m*lerc
hphi(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)*gamma(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)*g
amma(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*e
xp_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)) -
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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{m}}{{\left (b x^{2} + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^m*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="giac")
[Out]
integrate((B*x^2 + A)*x^m/(b*x^2 + a)^2, x)