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

Optimal. Leaf size=321 \[ \frac{d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+3)\right ) F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^3 \left (a e^2+b d^2\right )}-\frac{d^2 \left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 (2 p+3)\right ) \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^2 (p+1) \left (a e^2+b d^2\right )^2}-\frac{d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (2 a e^2+b d^2 (2 p+3)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{e^3 \left (a e^2+b d^2\right )}+\frac{d^3 \left (a+b x^2\right )^{p+1}}{e^2 (d+e x) \left (a e^2+b d^2\right )}+\frac{\left (a+b x^2\right )^{p+1}}{2 b e^2 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.999818, antiderivative size = 321, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5 \[ \frac{d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+3)\right ) F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^3 \left (a e^2+b d^2\right )}-\frac{d^2 \left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 (2 p+3)\right ) \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^2 (p+1) \left (a e^2+b d^2\right )^2}-\frac{d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (2 a e^2+b d^2 (2 p+3)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{e^3 \left (a e^2+b d^2\right )}+\frac{d^3 \left (a+b x^2\right )^{p+1}}{e^2 (d+e x) \left (a e^2+b d^2\right )}+\frac{\left (a+b x^2\right )^{p+1}}{2 b e^2 (p+1)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 64.7212, size = 338, normalized size = 1.05 \[ \frac{d^{3} \left (\frac{e \left (\sqrt{b} x + \sqrt{- a}\right )}{\sqrt{b} \left (d + e x\right )}\right )^{- p} \left (- \frac{e \left (- \sqrt{b} x + \sqrt{- a}\right )}{\sqrt{b} \left (d + e x\right )}\right )^{- p} \left (a + b x^{2}\right )^{p} \left (\frac{1}{d + e x}\right )^{2 p} \left (\frac{1}{d + e x}\right )^{- 2 p + 1} \operatorname{appellf_{1}}{\left (- 2 p + 1,- p,- p,- 2 p + 2,\frac{d - \frac{e \sqrt{- a}}{\sqrt{b}}}{d + e x},\frac{d + \frac{e \sqrt{- a}}{\sqrt{b}}}{d + e x} \right )}}{e^{4} \left (- 2 p + 1\right )} + \frac{3 d^{2} \left (\frac{e \left (\sqrt{b} x + \sqrt{- a}\right )}{\sqrt{b} \left (d + e x\right )}\right )^{- p} \left (- \frac{e \left (- \sqrt{b} x + \sqrt{- a}\right )}{\sqrt{b} \left (d + e x\right )}\right )^{- p} \left (a + b x^{2}\right )^{p} \operatorname{appellf_{1}}{\left (- 2 p,- p,- p,- 2 p + 1,\frac{d - \frac{e \sqrt{- a}}{\sqrt{b}}}{d + e x},\frac{d + \frac{e \sqrt{- a}}{\sqrt{b}}}{d + e x} \right )}}{2 e^{4} p} - \frac{2 d x \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{e^{3}} + \frac{\left (a + b x^{2}\right )^{p + 1}}{2 b e^{2} \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

d**3*(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 + 1)*appellf1(-2*p + 1, -p, -p, -2*p + 2, (d - e*sqrt(-a)/sqrt(b))/(d
+ e*x), (d + e*sqrt(-a)/sqrt(b))/(d + e*x))/(e**4*(-2*p + 1)) + 3*d**2*(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*appellf1(-2*p, -p, -p, -2*p + 1, (d - e*sqrt
(-a)/sqrt(b))/(d + e*x), (d + e*sqrt(-a)/sqrt(b))/(d + e*x))/(2*e**4*p) - 2*d*x*
(1 + b*x**2/a)**(-p)*(a + b*x**2)**p*hyper((-p, 1/2), (3/2,), -b*x**2/a)/e**3 +
(a + b*x**2)**(p + 1)/(2*b*e**2*(p + 1))

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Mathematica [A]  time = 1.47693, size = 343, normalized size = 1.07 \[ \frac{\left (a+b x^2\right )^p \left (-\frac{2 d^3 \left (\frac{e \left (x-\sqrt{-\frac{a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac{d-\sqrt{-\frac{a}{b}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{b}} e}{d+e x}\right )}{(2 p-1) (d+e x)}+\frac{3 d^2 \left (\frac{e \left (x-\sqrt{-\frac{a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac{d-\sqrt{-\frac{a}{b}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{b}} e}{d+e x}\right )}{p}+e \left (\frac{e \left (-a \left (\frac{b x^2}{a}+1\right )^{-p}+a+b x^2\right )}{b p+b}-4 d x \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )\right )\right )}{2 e^4} \]

Warning: Unable to verify antiderivative.

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

[Out]

((a + b*x^2)^p*((-2*d^3*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (d - Sqrt[-(a/b)]*e)/
(d + e*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)])/((-1 + 2*p)*((e*(-Sqrt[-(a/b)] + x))
/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p*(d + e*x)) + (3*d^2*AppellF1[
-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(d +
e*x)])/(p*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*
x))^p) + e*((e*(a + b*x^2 - a/(1 + (b*x^2)/a)^p))/(b + b*p) - (4*d*x*Hypergeomet
ric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p)))/(2*e^4)

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Maple [F]  time = 0.088, size = 0, normalized size = 0. \[ \int{\frac{{x}^{3} \left ( b{x}^{2}+a \right ) ^{p}}{ \left ( ex+d \right ) ^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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