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

Optimal. Leaf size=149 \[ -\frac{a \left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac{\left (b d^2-2 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac{e^2 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac{2}{5} d e x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right ) \]

[Out]

-(a*(b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) + ((b*d^2 - 2*a*e^2)*(a
+ b*x^2)^(2 + p))/(2*b^3*(2 + p)) + (e^2*(a + b*x^2)^(3 + p))/(2*b^3*(3 + p)) +
(2*d*e*x^5*(a + b*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*(1 +
(b*x^2)/a)^p)

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Rubi [A]  time = 0.288477, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3 \[ -\frac{a \left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac{\left (b d^2-2 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac{e^2 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac{2}{5} d e x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right ) \]

Antiderivative was successfully verified.

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

[Out]

-(a*(b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) + ((b*d^2 - 2*a*e^2)*(a
+ b*x^2)^(2 + p))/(2*b^3*(2 + p)) + (e^2*(a + b*x^2)^(3 + p))/(2*b^3*(3 + p)) +
(2*d*e*x^5*(a + b*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*(1 +
(b*x^2)/a)^p)

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Rubi in Sympy [A]  time = 45.4117, size = 158, normalized size = 1.06 \[ \frac{a^{2} e^{2} \left (a + b x^{2}\right )^{p + 1}}{2 b^{3} \left (p + 1\right )} - \frac{a d^{2} \left (a + b x^{2}\right )^{p + 1}}{2 b^{2} \left (p + 1\right )} - \frac{a e^{2} \left (a + b x^{2}\right )^{p + 2}}{b^{3} \left (p + 2\right )} + \frac{2 d e x^{5} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{5} + \frac{d^{2} \left (a + b x^{2}\right )^{p + 2}}{2 b^{2} \left (p + 2\right )} + \frac{e^{2} \left (a + b x^{2}\right )^{p + 3}}{2 b^{3} \left (p + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

a**2*e**2*(a + b*x**2)**(p + 1)/(2*b**3*(p + 1)) - a*d**2*(a + b*x**2)**(p + 1)/
(2*b**2*(p + 1)) - a*e**2*(a + b*x**2)**(p + 2)/(b**3*(p + 2)) + 2*d*e*x**5*(1 +
b*x**2/a)**(-p)*(a + b*x**2)**p*hyper((-p, 5/2), (7/2,), -b*x**2/a)/5 + d**2*(a
+ b*x**2)**(p + 2)/(2*b**2*(p + 2)) + e**2*(a + b*x**2)**(p + 3)/(2*b**3*(p + 3
))

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Mathematica [A]  time = 0.301101, size = 241, normalized size = 1.62 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (5 \left (2 a^3 e^2 \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )-a^2 b \left (d^2 (p+3) \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+2 e^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p\right )+b^3 (p+1) x^4 \left (\frac{b x^2}{a}+1\right )^p \left (d^2 (p+3)+e^2 (p+2) x^2\right )+a b^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p \left (d^2 (p+3)+e^2 (p+1) x^2\right )\right )+4 b^3 d e \left (p^3+6 p^2+11 p+6\right ) x^5 \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )\right )}{10 b^3 (p+1) (p+2) (p+3)} \]

Antiderivative was successfully verified.

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

[Out]

((a + b*x^2)^p*(5*(a*b^2*p*x^2*(1 + (b*x^2)/a)^p*(d^2*(3 + p) + e^2*(1 + p)*x^2)
+ b^3*(1 + p)*x^4*(1 + (b*x^2)/a)^p*(d^2*(3 + p) + e^2*(2 + p)*x^2) + 2*a^3*e^2
*(-1 + (1 + (b*x^2)/a)^p) - a^2*b*(2*e^2*p*x^2*(1 + (b*x^2)/a)^p + d^2*(3 + p)*(
-1 + (1 + (b*x^2)/a)^p))) + 4*b^3*d*e*(6 + 11*p + 6*p^2 + p^3)*x^5*Hypergeometri
c2F1[5/2, -p, 7/2, -((b*x^2)/a)]))/(10*b^3*(1 + p)*(2 + p)*(3 + p)*(1 + (b*x^2)/
a)^p)

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Maple [F]  time = 0.075, size = 0, normalized size = 0. \[ \int{x}^{3} \left ( ex+d \right ) ^{2} \left ( b{x}^{2}+a \right ) ^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \frac{{\left (b^{2}{\left (p + 1\right )} x^{4} + a b p x^{2} - a^{2}\right )}{\left (b x^{2} + a\right )}^{p} d^{2}}{2 \,{\left (p^{2} + 3 \, p + 2\right )} b^{2}} + \int{\left (e^{2} x^{5} + 2 \, d e x^{4}\right )}{\left (b x^{2} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/2*(b^2*(p + 1)*x^4 + a*b*p*x^2 - a^2)*(b*x^2 + a)^p*d^2/((p^2 + 3*p + 2)*b^2)
+ integrate((e^2*x^5 + 2*d*e*x^4)*(b*x^2 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (e^{2} x^{5} + 2 \, d e x^{4} + d^{2} x^{3}\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((e^2*x^5 + 2*d*e*x^4 + d^2*x^3)*(b*x^2 + a)^p, x)

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Sympy [A]  time = 93.3054, size = 1386, normalized size = 9.3 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2*a**p*d*e*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + d**2*Piec
ewise((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)) + e**2*Piecewise((a**p*x**6/6, Eq(b, 0)
), (2*a**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x
**4) + 2*a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5
*x**4) + 3*a**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(-I*
sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*
log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b
*x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(-I*sqrt(a)*s
qrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(I*sq
rt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4), Eq(p, -3)), (-
2*a**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(I*sqr
t(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a**2/(2*a*b**3 + 2*b**4*x**2) -
2*a*b*x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*
log(I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) + b**2*x**4/(2*a*b**3 + 2*
b**4*x**2), Eq(p, -2)), (a**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*b**3) + a**2*log(
I*sqrt(a)*sqrt(1/b) + x)/(2*b**3) - a*x**2/(2*b**2) + x**4/(4*b), Eq(p, -1)), (2
*a**3*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) - 2*a**
2*b*p*x**2*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) +
a*b**2*p**2*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b*
*3) + a*b**2*p*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12
*b**3) + b**3*p**2*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p
+ 12*b**3) + 3*b**3*p*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3
*p + 12*b**3) + 2*b**3*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**
3*p + 12*b**3), True))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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