3.395 \(\int (d+e x)^2 \left (a+b x^2\right )^p \, dx\)

Optimal. Leaf size=133 \[ -\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a e^2-b d^2 (2 p+3)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b (2 p+3)}+\frac{e (d+e x) \left (a+b x^2\right )^{p+1}}{b (2 p+3)}+\frac{d e (p+2) \left (a+b x^2\right )^{p+1}}{b (p+1) (2 p+3)} \]

[Out]

(d*e*(2 + p)*(a + b*x^2)^(1 + p))/(b*(1 + p)*(3 + 2*p)) + (e*(d + e*x)*(a + b*x^
2)^(1 + p))/(b*(3 + 2*p)) - ((a*e^2 - b*d^2*(3 + 2*p))*x*(a + b*x^2)^p*Hypergeom
etric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(b*(3 + 2*p)*(1 + (b*x^2)/a)^p)

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Rubi [A]  time = 0.172484, antiderivative size = 125, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235 \[ x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d^2-\frac{a e^2}{2 b p+3 b}\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+\frac{e (d+e x) \left (a+b x^2\right )^{p+1}}{b (2 p+3)}+\frac{d e (p+2) \left (a+b x^2\right )^{p+1}}{b (p+1) (2 p+3)} \]

Antiderivative was successfully verified.

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

[Out]

(d*e*(2 + p)*(a + b*x^2)^(1 + p))/(b*(1 + p)*(3 + 2*p)) + (e*(d + e*x)*(a + b*x^
2)^(1 + p))/(b*(3 + 2*p)) + ((d^2 - (a*e^2)/(3*b + 2*b*p))*x*(a + b*x^2)^p*Hyper
geometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p

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Rubi in Sympy [A]  time = 28.0547, size = 107, normalized size = 0.8 \[ \frac{d e \left (a + b x^{2}\right )^{p + 1} \left (p + 2\right )}{b \left (p + 1\right ) \left (2 p + 3\right )} + \frac{e \left (a + b x^{2}\right )^{p + 1} \left (d + e x\right )}{b \left (2 p + 3\right )} - \frac{x \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p} \left (a e^{2} - b d^{2} \left (2 p + 3\right )\right ){{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{b \left (2 p + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.154338, size = 133, normalized size = 1. \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 b d^2 (p+1) x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+e \left (3 d \left (b x^2 \left (\frac{b x^2}{a}+1\right )^p+a \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )\right )+b e (p+1) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )\right )\right )}{3 b (p+1)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \[ \int \left ( ex+d \right ) ^{2} \left ( b{x}^{2}+a \right ) ^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x+d)^2*(b*x^2+a)^p,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 40.8541, size = 97, normalized size = 0.73 \[ a^{p} d^{2} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} + \frac{a^{p} e^{2} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3} + 2 d e \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\begin{cases} \frac{\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + b x^{2} \right )} & \text{otherwise} \end{cases}}{2 b} & \text{otherwise} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x+d)**2*(b*x**2+a)**p,x)

[Out]

a**p*d**2*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*e**2*x**3*
hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + 2*d*e*Piecewise((a**p*x**
2/2, Eq(b, 0)), (Piecewise(((a + b*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(a +
b*x**2), True))/(2*b), True))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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