3.340 \(\int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx\)

Optimal. Leaf size=176 \[ \frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a^2 d^2-2 a b c d (2 p+5)+b^2 c^2 \left (4 p^2+16 p+15\right )\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b^2 (2 p+3) (2 p+5)}-\frac{d x \left (a+b x^2\right )^{p+1} (3 a d-b c (2 p+7))}{b^2 (2 p+3) (2 p+5)}+\frac{d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

[Out]

-((d*(3*a*d - b*c*(7 + 2*p))*x*(a + b*x^2)^(1 + p))/(b^2*(3 + 2*p)*(5 + 2*p))) +
(d*x*(a + b*x^2)^(1 + p)*(c + d*x^2))/(b*(5 + 2*p)) + ((3*a^2*d^2 - 2*a*b*c*d*(
5 + 2*p) + b^2*c^2*(15 + 16*p + 4*p^2))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -
p, 3/2, -((b*x^2)/a)])/(b^2*(3 + 2*p)*(5 + 2*p)*(1 + (b*x^2)/a)^p)

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Rubi [A]  time = 0.26312, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21 \[ \frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a^2 d^2-2 a b c d (2 p+5)+b^2 c^2 \left (4 p^2+16 p+15\right )\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b^2 (2 p+3) (2 p+5)}-\frac{d x \left (a+b x^2\right )^{p+1} (3 a d-b c (2 p+7))}{b^2 (2 p+3) (2 p+5)}+\frac{d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

Antiderivative was successfully verified.

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

[Out]

-((d*(3*a*d - b*c*(7 + 2*p))*x*(a + b*x^2)^(1 + p))/(b^2*(3 + 2*p)*(5 + 2*p))) +
(d*x*(a + b*x^2)^(1 + p)*(c + d*x^2))/(b*(5 + 2*p)) + ((3*a^2*d^2 - 2*a*b*c*d*(
5 + 2*p) + b^2*c^2*(15 + 16*p + 4*p^2))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -
p, 3/2, -((b*x^2)/a)])/(b^2*(3 + 2*p)*(5 + 2*p)*(1 + (b*x^2)/a)^p)

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Rubi in Sympy [A]  time = 37.0355, size = 160, normalized size = 0.91 \[ \frac{d x \left (a + b x^{2}\right )^{p + 1} \left (c + d x^{2}\right )}{b \left (2 p + 5\right )} - \frac{d x \left (a + b x^{2}\right )^{p + 1} \left (3 a d - 2 b c p - 7 b c\right )}{b^{2} \left (2 p + 3\right ) \left (2 p + 5\right )} + \frac{x \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p} \left (a d \left (3 a d - 2 b c p - 7 b c\right ) - b c \left (2 p + 3\right ) \left (a d - b c \left (2 p + 5\right )\right )\right ){{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{b^{2} \left (2 p + 3\right ) \left (2 p + 5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.0637352, size = 106, normalized size = 0.6 \[ \frac{1}{15} x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (15 c^2 \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+d x^2 \left (10 c \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+3 d x^2 \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )\right )\right ) \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.061, size = 0, normalized size = 0. \[ \int \left ( b{x}^{2}+a \right ) ^{p} \left ( d{x}^{2}+c \right ) ^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x^2+a)^p*(d*x^2+c)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 88.4894, size = 88, normalized size = 0.5 \[ a^{p} c^{2} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} + \frac{2 a^{p} c d x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac{a^{p} d^{2} x^{5}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{2}, - p \\ \frac{7}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

a**p*c**2*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + 2*a**p*c*d*x**3
*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + a**p*d**2*x**5*hyper((5/
2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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