3.91 \(\int \frac{(f x)^m \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\)

Optimal. Leaf size=154 \[ \frac{(f x)^{m+1} (b d-a e)}{4 a b f \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) (f x)^{m+1} (a e (m+1)+b d (3-m)) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{4 a^3 b f (m+1) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

((b*d - a*e)*(f*x)^(1 + m))/(4*a*b*f*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
) + ((b*d*(3 - m) + a*e*(1 + m))*(f*x)^(1 + m)*(a + b*x^2)*Hypergeometric2F1[2,
(1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(4*a^3*b*f*(1 + m)*Sqrt[a^2 + 2*a*b*x^2 + b
^2*x^4])

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Rubi [A]  time = 0.312783, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086 \[ \frac{(f x)^{m+1} (b d-a e)}{4 a b f \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) (f x)^{m+1} (a e (m+1)+b d (3-m)) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{4 a^3 b f (m+1) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]  Int[((f*x)^m*(d + e*x^2))/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

((b*d - a*e)*(f*x)^(1 + m))/(4*a*b*f*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
) + ((b*d*(3 - m) + a*e*(1 + m))*(f*x)^(1 + m)*(a + b*x^2)*Hypergeometric2F1[2,
(1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(4*a^3*b*f*(1 + m)*Sqrt[a^2 + 2*a*b*x^2 + b
^2*x^4])

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Rubi in Sympy [A]  time = 36.5839, size = 131, normalized size = 0.85 \[ - \frac{\left (f x\right )^{m + 1} \left (a e - b d\right ) \sqrt{a^{2} + 2 a b x^{2} + b^{2} x^{4}}}{4 a b f \left (a + b x^{2}\right )^{3}} + \frac{\left (f x\right )^{m + 1} \left (a e \left (m + 1\right ) + b d \left (- m + 3\right )\right ) \sqrt{a^{2} + 2 a b x^{2} + b^{2} x^{4}}{{}_{2}F_{1}\left (\begin{matrix} 2, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{4 a^{3} b f \left (a + b x^{2}\right ) \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((f*x)**m*(e*x**2+d)/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

-(f*x)**(m + 1)*(a*e - b*d)*sqrt(a**2 + 2*a*b*x**2 + b**2*x**4)/(4*a*b*f*(a + b*
x**2)**3) + (f*x)**(m + 1)*(a*e*(m + 1) + b*d*(-m + 3))*sqrt(a**2 + 2*a*b*x**2 +
b**2*x**4)*hyper((2, m/2 + 1/2), (m/2 + 3/2,), -b*x**2/a)/(4*a**3*b*f*(a + b*x*
*2)*(m + 1))

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Mathematica [A]  time = 0.127806, size = 101, normalized size = 0.66 \[ \frac{x \left (a+b x^2\right ) (f x)^m \left ((b d-a e) \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )+a e \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )\right )}{a^3 b (m+1) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]  Integrate[((f*x)^m*(d + e*x^2))/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

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

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \[ \int{ \left ( fx \right ) ^{m} \left ( e{x}^{2}+d \right ) \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{-{\frac{3}{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((f*x)^m*(e*x^2+d)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

int((f*x)^m*(e*x^2+d)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x^2 + d)*(f*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)*(f*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x^2 + d)*(f*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x^2 + d)*(f*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(3/2),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(f*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(3/2), x)