3.399 \(\int x^5 (d+e x)^3 \left (a+b x^2\right )^p \, dx\)

Optimal. Leaf size=247 \[ \frac{a^2 d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^4 (p+1)}-\frac{a d \left (2 b d^2-9 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^4 (p+2)}+\frac{d \left (b d^2-9 a e^2\right ) \left (a+b x^2\right )^{p+3}}{2 b^4 (p+3)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+4}}{2 b^4 (p+4)}-\frac{e x^7 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (7 a e^2-3 b d^2 (2 p+9)\right ) \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )}{7 b (2 p+9)}+\frac{e^3 x^7 \left (a+b x^2\right )^{p+1}}{b (2 p+9)} \]

[Out]

(a^2*d*(b*d^2 - 3*a*e^2)*(a + b*x^2)^(1 + p))/(2*b^4*(1 + p)) + (e^3*x^7*(a + b*
x^2)^(1 + p))/(b*(9 + 2*p)) - (a*d*(2*b*d^2 - 9*a*e^2)*(a + b*x^2)^(2 + p))/(2*b
^4*(2 + p)) + (d*(b*d^2 - 9*a*e^2)*(a + b*x^2)^(3 + p))/(2*b^4*(3 + p)) + (3*d*e
^2*(a + b*x^2)^(4 + p))/(2*b^4*(4 + p)) - (e*(7*a*e^2 - 3*b*d^2*(9 + 2*p))*x^7*(
a + b*x^2)^p*Hypergeometric2F1[7/2, -p, 9/2, -((b*x^2)/a)])/(7*b*(9 + 2*p)*(1 +
(b*x^2)/a)^p)

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Rubi [A]  time = 0.502993, antiderivative size = 241, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3 \[ \frac{a^2 d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^4 (p+1)}-\frac{a d \left (2 b d^2-9 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^4 (p+2)}+\frac{d \left (b d^2-9 a e^2\right ) \left (a+b x^2\right )^{p+3}}{2 b^4 (p+3)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+4}}{2 b^4 (p+4)}+\frac{1}{7} e x^7 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac{7 a e^2}{2 b p+9 b}\right ) \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )+\frac{e^3 x^7 \left (a+b x^2\right )^{p+1}}{b (2 p+9)} \]

Antiderivative was successfully verified.

[In]  Int[x^5*(d + e*x)^3*(a + b*x^2)^p,x]

[Out]

(a^2*d*(b*d^2 - 3*a*e^2)*(a + b*x^2)^(1 + p))/(2*b^4*(1 + p)) + (e^3*x^7*(a + b*
x^2)^(1 + p))/(b*(9 + 2*p)) - (a*d*(2*b*d^2 - 9*a*e^2)*(a + b*x^2)^(2 + p))/(2*b
^4*(2 + p)) + (d*(b*d^2 - 9*a*e^2)*(a + b*x^2)^(3 + p))/(2*b^4*(3 + p)) + (3*d*e
^2*(a + b*x^2)^(4 + p))/(2*b^4*(4 + p)) + (e*(3*d^2 - (7*a*e^2)/(9*b + 2*b*p))*x
^7*(a + b*x^2)^p*Hypergeometric2F1[7/2, -p, 9/2, -((b*x^2)/a)])/(7*(1 + (b*x^2)/
a)^p)

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Rubi in Sympy [A]  time = 78.5849, size = 267, normalized size = 1.08 \[ - \frac{3 a^{3} d e^{2} \left (a + b x^{2}\right )^{p + 1}}{2 b^{4} \left (p + 1\right )} + \frac{a^{2} d^{3} \left (a + b x^{2}\right )^{p + 1}}{2 b^{3} \left (p + 1\right )} + \frac{9 a^{2} d e^{2} \left (a + b x^{2}\right )^{p + 2}}{2 b^{4} \left (p + 2\right )} - \frac{a d^{3} \left (a + b x^{2}\right )^{p + 2}}{b^{3} \left (p + 2\right )} - \frac{9 a d e^{2} \left (a + b x^{2}\right )^{p + 3}}{2 b^{4} \left (p + 3\right )} + \frac{3 d^{2} e x^{7} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{7}{2} \\ \frac{9}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{7} + \frac{e^{3} x^{9} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{9}{2} \\ \frac{11}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{9} + \frac{d^{3} \left (a + b x^{2}\right )^{p + 3}}{2 b^{3} \left (p + 3\right )} + \frac{3 d e^{2} \left (a + b x^{2}\right )^{p + 4}}{2 b^{4} \left (p + 4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**5*(e*x+d)**3*(b*x**2+a)**p,x)

[Out]

-3*a**3*d*e**2*(a + b*x**2)**(p + 1)/(2*b**4*(p + 1)) + a**2*d**3*(a + b*x**2)**
(p + 1)/(2*b**3*(p + 1)) + 9*a**2*d*e**2*(a + b*x**2)**(p + 2)/(2*b**4*(p + 2))
- a*d**3*(a + b*x**2)**(p + 2)/(b**3*(p + 2)) - 9*a*d*e**2*(a + b*x**2)**(p + 3)
/(2*b**4*(p + 3)) + 3*d**2*e*x**7*(1 + b*x**2/a)**(-p)*(a + b*x**2)**p*hyper((-p
, 7/2), (9/2,), -b*x**2/a)/7 + e**3*x**9*(1 + b*x**2/a)**(-p)*(a + b*x**2)**p*hy
per((-p, 9/2), (11/2,), -b*x**2/a)/9 + d**3*(a + b*x**2)**(p + 3)/(2*b**3*(p + 3
)) + 3*d*e**2*(a + b*x**2)**(p + 4)/(2*b**4*(p + 4))

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Mathematica [A]  time = 0.701644, size = 356, normalized size = 1.44 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (63 d \left (-18 a^4 e^2 \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+2 a^3 b \left (d^2 (p+4) \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+9 e^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p\right )-a^2 b^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p \left (2 d^2 (p+4)+9 e^2 (p+1) x^2\right )+b^4 \left (p^2+3 p+2\right ) x^6 \left (\frac{b x^2}{a}+1\right )^p \left (d^2 (p+4)+3 e^2 (p+3) x^2\right )+a b^3 p (p+1) x^4 \left (\frac{b x^2}{a}+1\right )^p \left (d^2 (p+4)+3 e^2 (p+2) x^2\right )\right )+54 b^4 d^2 e \left (p^4+10 p^3+35 p^2+50 p+24\right ) x^7 \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )+14 b^4 e^3 \left (p^4+10 p^3+35 p^2+50 p+24\right ) x^9 \, _2F_1\left (\frac{9}{2},-p;\frac{11}{2};-\frac{b x^2}{a}\right )\right )}{126 b^4 (p+1) (p+2) (p+3) (p+4)} \]

Antiderivative was successfully verified.

[In]  Integrate[x^5*(d + e*x)^3*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^p*(63*d*(-(a^2*b^2*p*x^2*(1 + (b*x^2)/a)^p*(2*d^2*(4 + p) + 9*e^2*(
1 + p)*x^2)) + a*b^3*p*(1 + p)*x^4*(1 + (b*x^2)/a)^p*(d^2*(4 + p) + 3*e^2*(2 + p
)*x^2) + b^4*(2 + 3*p + p^2)*x^6*(1 + (b*x^2)/a)^p*(d^2*(4 + p) + 3*e^2*(3 + p)*
x^2) - 18*a^4*e^2*(-1 + (1 + (b*x^2)/a)^p) + 2*a^3*b*(9*e^2*p*x^2*(1 + (b*x^2)/a
)^p + d^2*(4 + p)*(-1 + (1 + (b*x^2)/a)^p))) + 54*b^4*d^2*e*(24 + 50*p + 35*p^2
+ 10*p^3 + p^4)*x^7*Hypergeometric2F1[7/2, -p, 9/2, -((b*x^2)/a)] + 14*b^4*e^3*(
24 + 50*p + 35*p^2 + 10*p^3 + p^4)*x^9*Hypergeometric2F1[9/2, -p, 11/2, -((b*x^2
)/a)]))/(126*b^4*(1 + p)*(2 + p)*(3 + p)*(4 + p)*(1 + (b*x^2)/a)^p)

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Maple [F]  time = 0.096, size = 0, normalized size = 0. \[ \int{x}^{5} \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^5*(e*x+d)^3*(b*x^2+a)^p,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \frac{{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{6} +{\left (p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + 2 \, a^{3}\right )}{\left (b x^{2} + a\right )}^{p} d^{3}}{2 \,{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} + \int{\left (e^{3} x^{8} + 3 \, d e^{2} x^{7} + 3 \, d^{2} e x^{6}\right )}{\left (b x^{2} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x + d)^3*(b*x^2 + a)^p*x^5,x, algorithm="maxima")

[Out]

1/2*((p^2 + 3*p + 2)*b^3*x^6 + (p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + 2*a^3)*(b*x
^2 + a)^p*d^3/((p^3 + 6*p^2 + 11*p + 6)*b^3) + integrate((e^3*x^8 + 3*d*e^2*x^7
+ 3*d^2*e*x^6)*(b*x^2 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (e^{3} x^{8} + 3 \, d e^{2} x^{7} + 3 \, d^{2} e x^{6} + d^{3} x^{5}\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x + d)^3*(b*x^2 + a)^p*x^5,x, algorithm="fricas")

[Out]

integral((e^3*x^8 + 3*d*e^2*x^7 + 3*d^2*e*x^6 + d^3*x^5)*(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**5*(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. \[ \int{\left (e x + d\right )}^{3}{\left (b x^{2} + a\right )}^{p} x^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x + d)^3*(b*x^2 + a)^p*x^5,x, algorithm="giac")

[Out]

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