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

Optimal. Leaf size=210 \[ -\frac{a e \left (3 b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac{e \left (3 b d^2-2 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac{e^3 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}-\frac{d x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (9 a e^2-b d^2 (2 p+5)\right ) \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )}{3 b (2 p+5)}+\frac{3 d e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

[Out]

-(a*e*(3*b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) + (3*d*e^2*x^3*(a +
b*x^2)^(1 + p))/(b*(5 + 2*p)) + (e*(3*b*d^2 - 2*a*e^2)*(a + b*x^2)^(2 + p))/(2*
b^3*(2 + p)) + (e^3*(a + b*x^2)^(3 + p))/(2*b^3*(3 + p)) - (d*(9*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)])/(3*b*
(5 + 2*p)*(1 + (b*x^2)/a)^p)

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Rubi [A]  time = 0.420691, antiderivative size = 202, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3 \[ -\frac{a e \left (3 b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac{e \left (3 b d^2-2 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac{e^3 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac{1}{3} d x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d^2-\frac{9 a e^2}{2 b p+5 b}\right ) \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+\frac{3 d e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

Antiderivative was successfully verified.

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

[Out]

-(a*e*(3*b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) + (3*d*e^2*x^3*(a +
b*x^2)^(1 + p))/(b*(5 + 2*p)) + (e*(3*b*d^2 - 2*a*e^2)*(a + b*x^2)^(2 + p))/(2*
b^3*(2 + p)) + (e^3*(a + b*x^2)^(3 + p))/(2*b^3*(3 + p)) + (d*(d^2 - (9*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)])/(
3*(1 + (b*x^2)/a)^p)

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Rubi in Sympy [A]  time = 55.4724, size = 209, normalized size = 1. \[ \frac{a^{2} e^{3} \left (a + b x^{2}\right )^{p + 1}}{2 b^{3} \left (p + 1\right )} - \frac{3 a d^{2} e \left (a + b x^{2}\right )^{p + 1}}{2 b^{2} \left (p + 1\right )} - \frac{a e^{3} \left (a + b x^{2}\right )^{p + 2}}{b^{3} \left (p + 2\right )} + \frac{d^{3} x^{3} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{3} + \frac{3 d e^{2} x^{5} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{5} + \frac{3 d^{2} e \left (a + b x^{2}\right )^{p + 2}}{2 b^{2} \left (p + 2\right )} + \frac{e^{3} \left (a + b x^{2}\right )^{p + 3}}{2 b^{3} \left (p + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.474125, size = 291, normalized size = 1.39 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 e \left (5 \left (2 a^3 e^2 \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )-a^2 b \left (3 d^2 (p+3) \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+2 e^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p\right )+b^3 (p+1) x^4 \left (\frac{b x^2}{a}+1\right )^p \left (3 d^2 (p+3)+e^2 (p+2) x^2\right )+a b^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p \left (3 d^2 (p+3)+e^2 (p+1) x^2\right )\right )+6 b^3 d e \left (p^3+6 p^2+11 p+6\right ) x^5 \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )\right )+10 b^3 d^3 \left (p^3+6 p^2+11 p+6\right ) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )\right )}{30 b^3 (p+1) (p+2) (p+3)} \]

Antiderivative was successfully verified.

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

[Out]

((a + b*x^2)^p*(10*b^3*d^3*(6 + 11*p + 6*p^2 + p^3)*x^3*Hypergeometric2F1[3/2, -
p, 5/2, -((b*x^2)/a)] + 3*e*(5*(a*b^2*p*x^2*(1 + (b*x^2)/a)^p*(3*d^2*(3 + p) + e
^2*(1 + p)*x^2) + b^3*(1 + p)*x^4*(1 + (b*x^2)/a)^p*(3*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 + 3*d^2*(3 + p)*(-1 + (1 + (b*x^2)/a)^p))) + 6*b^3*d*e*(6 + 11*p + 6*p^2 + p^3
)*x^5*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])))/(30*b^3*(1 + p)*(2 + p)*(
3 + p)*(1 + (b*x^2)/a)^p)

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Maple [F]  time = 0.082, size = 0, normalized size = 0. \[ \int{x}^{2} \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 114.196, size = 1421, normalized size = 6.77 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

a**p*d**3*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + 3*a**p*d*e
**2*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + 3*d**2*e*Piecewi
se((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**3*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*sqr
t(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)*sqrt
(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*sqrt(
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*sqrt(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*s
qrt(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 + 1
2*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. \[ \int{\left (e x + d\right )}^{3}{\left (b x^{2} + a\right )}^{p} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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