3.427 \(\int \frac{\left (a+b x^2\right )^p}{(d+e x)^3} \, dx\)

Optimal. Leaf size=322 \[ \frac{e^2 x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3}-\frac{3 b^2 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}+\frac{b e \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{4 (p+1) \left (a e^2+b d^2\right )^3}-\frac{d^2 e \left (a+b x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+b d^2\right )} \]

[Out]

-(d^2*e*(a + b*x^2)^(1 + p))/(4*(b*d^2 + a*e^2)*(d^2 - e^2*x^2)^2) + (x*(a + b*x
^2)^p*AppellF1[1/2, -p, 3, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^3*(1 + (b*x^2)/
a)^p) + (e^2*x^3*(a + b*x^2)^p*AppellF1[3/2, -p, 3, 5/2, -((b*x^2)/a), (e^2*x^2)
/d^2])/(d^5*(1 + (b*x^2)/a)^p) + (b*e*(2*a*e^2 + b*d^2*(1 + p))*(a + b*x^2)^(1 +
p)*Hypergeometric2F1[2, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(4*(b
*d^2 + a*e^2)^3*(1 + p)) - (3*b^2*d^2*e*(a + b*x^2)^(1 + p)*Hypergeometric2F1[3,
1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(2*(b*d^2 + a*e^2)^3*(1 + p))

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Rubi [A]  time = 0.723602, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529 \[ \frac{e^2 x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3}-\frac{3 b^2 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}+\frac{b e \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{4 (p+1) \left (a e^2+b d^2\right )^3}-\frac{d^2 e \left (a+b x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+b d^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

-(d^2*e*(a + b*x^2)^(1 + p))/(4*(b*d^2 + a*e^2)*(d^2 - e^2*x^2)^2) + (x*(a + b*x
^2)^p*AppellF1[1/2, -p, 3, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^3*(1 + (b*x^2)/
a)^p) + (e^2*x^3*(a + b*x^2)^p*AppellF1[3/2, -p, 3, 5/2, -((b*x^2)/a), (e^2*x^2)
/d^2])/(d^5*(1 + (b*x^2)/a)^p) + (b*e*(2*a*e^2 + b*d^2*(1 + p))*(a + b*x^2)^(1 +
p)*Hypergeometric2F1[2, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(4*(b
*d^2 + a*e^2)^3*(1 + p)) - (3*b^2*d^2*e*(a + b*x^2)^(1 + p)*Hypergeometric2F1[3,
1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(2*(b*d^2 + a*e^2)^3*(1 + p))

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Rubi in Sympy [A]  time = 18.4177, size = 144, normalized size = 0.45 \[ - \frac{\left (\frac{e \left (\sqrt{b} x + \sqrt{- a}\right )}{\sqrt{b} \left (d + e x\right )}\right )^{- p} \left (- \frac{e \left (- \sqrt{b} x + \sqrt{- a}\right )}{\sqrt{b} \left (d + e x\right )}\right )^{- p} \left (a + b x^{2}\right )^{p} \left (\frac{1}{d + e x}\right )^{2 p} \left (\frac{1}{d + e x}\right )^{- 2 p + 2} \operatorname{appellf_{1}}{\left (- 2 p + 2,- p,- p,- 2 p + 3,\frac{d - \frac{e \sqrt{- a}}{\sqrt{b}}}{d + e x},\frac{d + \frac{e \sqrt{- a}}{\sqrt{b}}}{d + e x} \right )}}{2 e \left (- p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-(e*(sqrt(b)*x + sqrt(-a))/(sqrt(b)*(d + e*x)))**(-p)*(-e*(-sqrt(b)*x + sqrt(-a)
)/(sqrt(b)*(d + e*x)))**(-p)*(a + b*x**2)**p*(1/(d + e*x))**(2*p)*(1/(d + e*x))*
*(-2*p + 2)*appellf1(-2*p + 2, -p, -p, -2*p + 3, (d - e*sqrt(-a)/sqrt(b))/(d + e
*x), (d + e*sqrt(-a)/sqrt(b))/(d + e*x))/(2*e*(-p + 1))

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Mathematica [A]  time = 0.181312, size = 142, normalized size = 0.44 \[ \frac{\left (a+b x^2\right )^p \left (\frac{e \left (x-\sqrt{-\frac{a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (2-2 p;-p,-p;3-2 p;\frac{d-\sqrt{-\frac{a}{b}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{b}} e}{d+e x}\right )}{2 e (p-1) (d+e x)^2} \]

Warning: Unable to verify antiderivative.

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

[Out]

((a + b*x^2)^p*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x)
, (d + Sqrt[-(a/b)]*e)/(d + e*x)])/(2*e*(-1 + p)*((e*(-Sqrt[-(a/b)] + x))/(d + e
*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p*(d + e*x)^2)

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Maple [F]  time = 0.087, size = 0, normalized size = 0. \[ \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p}}{ \left ( ex+d \right ) ^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x^2+a)^p/(e*x+d)^3,x)

[Out]

int((b*x^2+a)^p/(e*x+d)^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((b*x^2 + a)^p/(e*x + d)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((b*x^2 + a)^p/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^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((b*x**2+a)**p/(e*x+d)**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((b*x^2 + a)^p/(e*x + d)^3, x)