3.330 \(\int \frac{x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx\)

Optimal. Leaf size=171 \[ \frac{x^{m+1} \left (a^2 d^2 \left (m^2-4 m+3\right )+2 a b c d \left (1-m^2\right )+b^2 c^2 \left (m^2+4 m+3\right )\right ) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{8 c^3 d^2 (m+1)}-\frac{x^{m+1} (b c-a d) (a d (3-m)+b c (m+5))}{8 c^2 d^2 \left (c+d x^2\right )}+\frac{x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

[Out]

((b*c - a*d)^2*x^(1 + m))/(4*c*d^2*(c + d*x^2)^2) - ((b*c - a*d)*(a*d*(3 - m) +
b*c*(5 + m))*x^(1 + m))/(8*c^2*d^2*(c + d*x^2)) + ((2*a*b*c*d*(1 - m^2) + a^2*d^
2*(3 - 4*m + m^2) + b^2*c^2*(3 + 4*m + m^2))*x^(1 + m)*Hypergeometric2F1[1, (1 +
m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*d^2*(1 + m))

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Rubi [A]  time = 0.386817, antiderivative size = 166, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136 \[ \frac{x^{m+1} \left (\frac{(1-m) \left (4 a^2 d^2-(m+1) (b c-a d)^2\right )}{c^2 (m+1)}+4 b^2\right ) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{8 c d^2}-\frac{x^{m+1} (b c-a d) (a d (3-m)+b c (m+5))}{8 c^2 d^2 \left (c+d x^2\right )}+\frac{x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

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

[Out]

((b*c - a*d)^2*x^(1 + m))/(4*c*d^2*(c + d*x^2)^2) - ((b*c - a*d)*(a*d*(3 - m) +
b*c*(5 + m))*x^(1 + m))/(8*c^2*d^2*(c + d*x^2)) + ((4*b^2 + ((1 - m)*(4*a^2*d^2
- (b*c - a*d)^2*(1 + m)))/(c^2*(1 + m)))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/
2, (3 + m)/2, -((d*x^2)/c)])/(8*c*d^2)

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Rubi in Sympy [A]  time = 41.7323, size = 148, normalized size = 0.87 \[ \frac{x^{m + 1} \left (a d - b c\right )^{2}}{4 c d^{2} \left (c + d x^{2}\right )^{2}} + \frac{x^{m + 1} \left (a d - b c\right ) \left (- a d m + 3 a d + b c m + 5 b c\right )}{8 c^{2} d^{2} \left (c + d x^{2}\right )} + \frac{x^{m + 1} \left (4 b^{2} c^{2} \left (m + 1\right ) + \left (- m + 1\right ) \left (4 a^{2} d^{2} - \left (m + 1\right ) \left (a d - b c\right )^{2}\right )\right ){{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{d x^{2}}{c}} \right )}}{8 c^{3} d^{2} \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

x**(m + 1)*(a*d - b*c)**2/(4*c*d**2*(c + d*x**2)**2) + x**(m + 1)*(a*d - b*c)*(-
a*d*m + 3*a*d + b*c*m + 5*b*c)/(8*c**2*d**2*(c + d*x**2)) + x**(m + 1)*(4*b**2*c
**2*(m + 1) + (-m + 1)*(4*a**2*d**2 - (m + 1)*(a*d - b*c)**2))*hyper((1, m/2 + 1
/2), (m/2 + 3/2,), -d*x**2/c)/(8*c**3*d**2*(m + 1))

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Mathematica [A]  time = 0.164144, size = 118, normalized size = 0.69 \[ \frac{x^{m+1} \left (\frac{a^2 \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{m+1}+b x^2 \left (\frac{2 a \, _2F_1\left (3,\frac{m+3}{2};\frac{m+5}{2};-\frac{d x^2}{c}\right )}{m+3}+\frac{b x^2 \, _2F_1\left (3,\frac{m+5}{2};\frac{m+7}{2};-\frac{d x^2}{c}\right )}{m+5}\right )\right )}{c^3} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.086, size = 0, normalized size = 0. \[ \int{\frac{{x}^{m} \left ( b{x}^{2}+a \right ) ^{2}}{ \left ( d{x}^{2}+c \right ) ^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

int(x^m*(b*x^2+a)^2/(d*x^2+c)^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((b*x^2 + a)^2*x^m/(d*x^2 + c)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((b*x^2 + a)^2*x^m/(d*x^2 + c)^3, x)