3.332 \(\int \frac{x^m \left (c+d x^2\right )^2}{a+b x^2} \, dx\)

Optimal. Leaf size=94 \[ \frac{x^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a b^2 (m+1)}+\frac{d x^{m+1} (2 b c-a d)}{b^2 (m+1)}+\frac{d^2 x^{m+3}}{b (m+3)} \]

[Out]

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

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Rubi [A]  time = 0.151754, antiderivative size = 94, 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 \[ \frac{x^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a b^2 (m+1)}+\frac{d x^{m+1} (2 b c-a d)}{b^2 (m+1)}+\frac{d^2 x^{m+3}}{b (m+3)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 26.7751, size = 76, normalized size = 0.81 \[ \frac{d^{2} x^{m + 3}}{b \left (m + 3\right )} - \frac{d x^{m + 1} \left (a d - 2 b c\right )}{b^{2} \left (m + 1\right )} + \frac{x^{m + 1} \left (a d - b c\right )^{2}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{a b^{2} \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

d**2*x**(m + 3)/(b*(m + 3)) - d*x**(m + 1)*(a*d - 2*b*c)/(b**2*(m + 1)) + x**(m
+ 1)*(a*d - b*c)**2*hyper((1, m/2 + 1/2), (m/2 + 3/2,), -b*x**2/a)/(a*b**2*(m +
1))

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Mathematica [A]  time = 0.14129, size = 118, normalized size = 1.26 \[ \frac{x^{m+1} \left (\frac{c^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{m+1}+d x^2 \left (\frac{2 c \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{b x^2}{a}\right )}{m+3}+\frac{d x^2 \, _2F_1\left (1,\frac{m+5}{2};\frac{m+7}{2};-\frac{b x^2}{a}\right )}{m+5}\right )\right )}{a} \]

Antiderivative was successfully verified.

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

[Out]

(x^(1 + m)*((c^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(1 +
m) + d*x^2*((2*c*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, -((b*x^2)/a)])/(3 +
m) + (d*x^2*Hypergeometric2F1[1, (5 + m)/2, (7 + m)/2, -((b*x^2)/a)])/(5 + m))))
/a

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((d^2*x^4 + 2*c*d*x^2 + c^2)*x^m/(b*x^2 + a), x)

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Sympy [A]  time = 34.3137, size = 299, normalized size = 3.18 \[ \frac{c^{2} m x x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{c^{2} x x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{c d m x^{3} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{2 a \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} + \frac{3 c d x^{3} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{2 a \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} + \frac{d^{2} m x^{5} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{5}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{7}{2}\right )} + \frac{5 d^{2} x^{5} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{5}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{7}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

c**2*m*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/
(4*a*gamma(m/2 + 3/2)) + c**2*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 +
1/2)*gamma(m/2 + 1/2)/(4*a*gamma(m/2 + 3/2)) + c*d*m*x**3*x**m*lerchphi(b*x**2*
exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(2*a*gamma(m/2 + 5/2)) + 3*c*d
*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(2*
a*gamma(m/2 + 5/2)) + d**2*m*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2
+ 5/2)*gamma(m/2 + 5/2)/(4*a*gamma(m/2 + 7/2)) + 5*d**2*x**5*x**m*lerchphi(b*x*
*2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(4*a*gamma(m/2 + 7/2))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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