3.400 \(\int x^4 (d+e x)^3 \left (a+b x^2\right )^p \, dx\)

Optimal. Leaf size=249 \[ \frac{a^2 e \left (3 b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^4 (p+1)}-\frac{3 a e \left (2 b d^2-a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^4 (p+2)}+\frac{3 e \left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+3}}{2 b^4 (p+3)}+\frac{e^3 \left (a+b x^2\right )^{p+4}}{2 b^4 (p+4)}-\frac{d x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (15 a e^2-b d^2 (2 p+7)\right ) \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )}{5 b (2 p+7)}+\frac{3 d e^2 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \]

[Out]

(a^2*e*(3*b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^4*(1 + p)) + (3*d*e^2*x^5*(a
+ b*x^2)^(1 + p))/(b*(7 + 2*p)) - (3*a*e*(2*b*d^2 - a*e^2)*(a + b*x^2)^(2 + p))/
(2*b^4*(2 + p)) + (3*e*(b*d^2 - a*e^2)*(a + b*x^2)^(3 + p))/(2*b^4*(3 + p)) + (e
^3*(a + b*x^2)^(4 + p))/(2*b^4*(4 + p)) - (d*(15*a*e^2 - b*d^2*(7 + 2*p))*x^5*(a
+ b*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*b*(7 + 2*p)*(1 + (
b*x^2)/a)^p)

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Rubi [A]  time = 0.50207, antiderivative size = 241, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3 \[ \frac{a^2 e \left (3 b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^4 (p+1)}-\frac{3 a e \left (2 b d^2-a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^4 (p+2)}+\frac{3 e \left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+3}}{2 b^4 (p+3)}+\frac{e^3 \left (a+b x^2\right )^{p+4}}{2 b^4 (p+4)}+\frac{1}{5} d x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d^2-\frac{15 a e^2}{2 b p+7 b}\right ) \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )+\frac{3 d e^2 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \]

Antiderivative was successfully verified.

[In]  Int[x^4*(d + e*x)^3*(a + b*x^2)^p,x]

[Out]

(a^2*e*(3*b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^4*(1 + p)) + (3*d*e^2*x^5*(a
+ b*x^2)^(1 + p))/(b*(7 + 2*p)) - (3*a*e*(2*b*d^2 - a*e^2)*(a + b*x^2)^(2 + p))/
(2*b^4*(2 + p)) + (3*e*(b*d^2 - a*e^2)*(a + b*x^2)^(3 + p))/(2*b^4*(3 + p)) + (e
^3*(a + b*x^2)^(4 + p))/(2*b^4*(4 + p)) + (d*(d^2 - (15*a*e^2)/(7*b + 2*b*p))*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 = 75.3312, size = 267, normalized size = 1.07 \[ - \frac{a^{3} e^{3} \left (a + b x^{2}\right )^{p + 1}}{2 b^{4} \left (p + 1\right )} + \frac{3 a^{2} d^{2} e \left (a + b x^{2}\right )^{p + 1}}{2 b^{3} \left (p + 1\right )} + \frac{3 a^{2} e^{3} \left (a + b x^{2}\right )^{p + 2}}{2 b^{4} \left (p + 2\right )} - \frac{3 a d^{2} e \left (a + b x^{2}\right )^{p + 2}}{b^{3} \left (p + 2\right )} - \frac{3 a e^{3} \left (a + b x^{2}\right )^{p + 3}}{2 b^{4} \left (p + 3\right )} + \frac{d^{3} x^{5} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{5} + \frac{3 d e^{2} x^{7} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{7}{2} \\ \frac{9}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{7} + \frac{3 d^{2} e \left (a + b x^{2}\right )^{p + 3}}{2 b^{3} \left (p + 3\right )} + \frac{e^{3} \left (a + b x^{2}\right )^{p + 4}}{2 b^{4} \left (p + 4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**4*(e*x+d)**3*(b*x**2+a)**p,x)

[Out]

-a**3*e**3*(a + b*x**2)**(p + 1)/(2*b**4*(p + 1)) + 3*a**2*d**2*e*(a + b*x**2)**
(p + 1)/(2*b**3*(p + 1)) + 3*a**2*e**3*(a + b*x**2)**(p + 2)/(2*b**4*(p + 2)) -
3*a*d**2*e*(a + b*x**2)**(p + 2)/(b**3*(p + 2)) - 3*a*e**3*(a + b*x**2)**(p + 3)
/(2*b**4*(p + 3)) + d**3*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*e**2*x**7*(1 + b*x**2/a)**(-p)*(a + b*x**2)**p*hy
per((-p, 7/2), (9/2,), -b*x**2/a)/7 + 3*d**2*e*(a + b*x**2)**(p + 3)/(2*b**3*(p
+ 3)) + e**3*(a + b*x**2)**(p + 4)/(2*b**4*(p + 4))

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Mathematica [A]  time = 0.69797, size = 357, normalized size = 1.43 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (14 b^4 d^3 \left (p^4+10 p^3+35 p^2+50 p+24\right ) x^5 \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )-5 e \left (7 \left (6 a^4 e^2 \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )-6 a^3 b \left (d^2 (p+4) \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+e^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p\right )+3 a^2 b^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p \left (2 d^2 (p+4)+e^2 (p+1) x^2\right )-b^4 \left (p^2+3 p+2\right ) x^6 \left (\frac{b x^2}{a}+1\right )^p \left (3 d^2 (p+4)+e^2 (p+3) x^2\right )-a b^3 p (p+1) x^4 \left (\frac{b x^2}{a}+1\right )^p \left (3 d^2 (p+4)+e^2 (p+2) x^2\right )\right )-6 b^4 d e \left (p^4+10 p^3+35 p^2+50 p+24\right ) x^7 \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )\right )\right )}{70 b^4 (p+1) (p+2) (p+3) (p+4)} \]

Antiderivative was successfully verified.

[In]  Integrate[x^4*(d + e*x)^3*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^p*(14*b^4*d^3*(24 + 50*p + 35*p^2 + 10*p^3 + p^4)*x^5*Hypergeometri
c2F1[5/2, -p, 7/2, -((b*x^2)/a)] - 5*e*(7*(3*a^2*b^2*p*x^2*(1 + (b*x^2)/a)^p*(2*
d^2*(4 + p) + e^2*(1 + p)*x^2) - a*b^3*p*(1 + p)*x^4*(1 + (b*x^2)/a)^p*(3*d^2*(4
+ p) + e^2*(2 + p)*x^2) - b^4*(2 + 3*p + p^2)*x^6*(1 + (b*x^2)/a)^p*(3*d^2*(4 +
p) + e^2*(3 + p)*x^2) + 6*a^4*e^2*(-1 + (1 + (b*x^2)/a)^p) - 6*a^3*b*(e^2*p*x^2
*(1 + (b*x^2)/a)^p + d^2*(4 + p)*(-1 + (1 + (b*x^2)/a)^p))) - 6*b^4*d*e*(24 + 50
*p + 35*p^2 + 10*p^3 + p^4)*x^7*Hypergeometric2F1[7/2, -p, 9/2, -((b*x^2)/a)])))
/(70*b^4*(1 + p)*(2 + p)*(3 + p)*(4 + p)*(1 + (b*x^2)/a)^p)

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Maple [F]  time = 0.101, size = 0, normalized size = 0. \[ \int{x}^{4} \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^4*(e*x+d)^3*(b*x^2+a)^p,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x + d)^3*(b*x^2 + a)^p*x^4,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (e^{3} x^{7} + 3 \, d e^{2} x^{6} + 3 \, d^{2} e x^{5} + d^{3} x^{4}\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x + d)^3*(b*x^2 + a)^p*x^4,x, algorithm="fricas")

[Out]

integral((e^3*x^7 + 3*d*e^2*x^6 + 3*d^2*e*x^5 + d^3*x^4)*(b*x^2 + a)^p, 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**4*(e*x+d)**3*(b*x**2+a)**p,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x + d)^3*(b*x^2 + a)^p*x^4,x, algorithm="giac")

[Out]

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