3.403 \(\int x (d+e x)^3 \left (a+b x^2\right )^p \, dx\)

Optimal. Leaf size=167 \[ \frac{d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}-\frac{e x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a e^2-b d^2 (2 p+5)\right ) \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )}{b (2 p+5)}+\frac{e^3 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

[Out]

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

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Rubi [A]  time = 0.325148, antiderivative size = 159, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333 \[ \frac{d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+e x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d^2-\frac{a e^2}{2 b p+5 b}\right ) \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+\frac{e^3 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 64.9472, size = 212, normalized size = 1.27 \[ \frac{3 a e x \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p} \left (a e^{2} - 2 b d^{2} p - 5 b d^{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{b^{2} \left (2 p + 3\right ) \left (2 p + 5\right )} + \frac{3 d \left (a + b x^{2}\right )^{p + 1} \left (d + e x\right )^{2}}{2 b \left (p + 2\right ) \left (2 p + 5\right )} + \frac{\left (a + b x^{2}\right )^{p + 1} \left (d + e x\right )^{3}}{b \left (2 p + 5\right )} - \frac{\left (a + b x^{2}\right )^{p + 1} \left (6 d \left (2 p + 3\right ) \left (2 a e^{2} p + 5 a e^{2} - b d^{2}\right ) - 12 e x \left (p + 1\right ) \left (- a e^{2} \left (p + 2\right ) + b d^{2}\right )\right )}{4 b^{2} \left (p + 1\right ) \left (p + 2\right ) \left (2 p + 3\right ) \left (2 p + 5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

3*a*e*x*(1 + b*x**2/a)**(-p)*(a + b*x**2)**p*(a*e**2 - 2*b*d**2*p - 5*b*d**2)*hy
per((-p, 1/2), (3/2,), -b*x**2/a)/(b**2*(2*p + 3)*(2*p + 5)) + 3*d*(a + b*x**2)*
*(p + 1)*(d + e*x)**2/(2*b*(p + 2)*(2*p + 5)) + (a + b*x**2)**(p + 1)*(d + e*x)*
*3/(b*(2*p + 5)) - (a + b*x**2)**(p + 1)*(6*d*(2*p + 3)*(2*a*e**2*p + 5*a*e**2 -
b*d**2) - 12*e*x*(p + 1)*(-a*e**2*(p + 2) + b*d**2))/(4*b**2*(p + 1)*(p + 2)*(2
*p + 3)*(2*p + 5))

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Mathematica [A]  time = 0.367588, size = 228, normalized size = 1.37 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (5 d \left (-3 a^2 e^2 \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+b^2 x^2 \left (\frac{b x^2}{a}+1\right )^p \left (d^2 (p+2)+3 e^2 (p+1) x^2\right )+a b \left (d^2 (p+2) \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+3 e^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p\right )\right )+10 b^2 d^2 e \left (p^2+3 p+2\right ) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+2 b^2 e^3 \left (p^2+3 p+2\right ) x^5 \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )\right )}{10 b^2 (p+1) (p+2)} \]

Antiderivative was successfully verified.

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

[Out]

((a + b*x^2)^p*(5*d*(b^2*x^2*(1 + (b*x^2)/a)^p*(d^2*(2 + p) + 3*e^2*(1 + p)*x^2)
- 3*a^2*e^2*(-1 + (1 + (b*x^2)/a)^p) + a*b*(3*e^2*p*x^2*(1 + (b*x^2)/a)^p + d^2
*(2 + p)*(-1 + (1 + (b*x^2)/a)^p))) + 10*b^2*d^2*e*(2 + 3*p + p^2)*x^3*Hypergeom
etric2F1[3/2, -p, 5/2, -((b*x^2)/a)] + 2*b^2*e^3*(2 + 3*p + p^2)*x^5*Hypergeomet
ric2F1[5/2, -p, 7/2, -((b*x^2)/a)]))/(10*b^2*(1 + p)*(2 + p)*(1 + (b*x^2)/a)^p)

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Maple [F]  time = 0.074, size = 0, normalized size = 0. \[ \int x \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x*(e*x+d)^3*(b*x^2+a)^p,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((e^3*x^4 + 3*d*e^2*x^3 + 3*d^2*e*x^2 + d^3*x)*(b*x^2 + a)^p, x)

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Sympy [A]  time = 85.6439, size = 471, normalized size = 2.82 \[ a^{p} d^{2} e x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} + \frac{a^{p} e^{3} x^{5}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{2}, - p \\ \frac{7}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{5} + d^{3} \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\begin{cases} \frac{\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + b x^{2} \right )} & \text{otherwise} \end{cases}}{2 b} & \text{otherwise} \end{cases}\right ) + 3 d e^{2} \left (\begin{cases} \frac{a^{p} x^{4}}{4} & \text{for}\: b = 0 \\\frac{a \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{b x^{2} \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{b x^{2} \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text{for}\: p = -2 \\- \frac{a \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 b^{2}} - \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 b^{2}} + \frac{x^{2}}{2 b} & \text{for}\: p = -1 \\- \frac{a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac{a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac{b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac{b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text{otherwise} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

a**p*d**2*e*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a) + a**p*e**3*
x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + d**3*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)) + 3*d*e**2*Piecewise((a**p*x**4/4, Eq(b, 0)),
(a*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**2 + 2*b**3*x**2) + a*log(I*sqrt(a)*sqrt
(1/b) + x)/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(-I
*sqrt(a)*sqrt(1/b) + x)/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(I*sqrt(a)*sqrt(1/b
) + x)/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*log(-I*sqrt(a)*sqrt(1/b) + x)/(
2*b**2) - a*log(I*sqrt(a)*sqrt(1/b) + x)/(2*b**2) + x**2/(2*b), Eq(p, -1)), (-a*
*2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*p*x**2*(a + b*x**2)**
p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*p*x**4*(a + b*x**2)**p/(2*b**2*p**2 +
6*b**2*p + 4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2
), True))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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