3.14 \(\int \frac{(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{\left (a+b x^2\right )^3} \, dx\)

Optimal. Leaf size=292 \[ -\frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (a d (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))-b c (a B (m+1)+A b (3-m)) (a d (m+1)+b (c-c m)))}{8 a^3 b^3 e (m+1)}+\frac{d (e x)^{m+1} (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))}{8 a^2 b^3 e (m+1)}+\frac{(e x)^{m+1} (b c-a d) \left (c (a B (m+1)+A b (3-m))-d x^2 (A b (m+1)-a B (m+5))\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac{\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \]

[Out]

(d*(b*c*(1 + m) - a*d*(3 + m))*(A*b*(1 + m) - a*B*(5 + m))*(e*x)^(1 + m))/(8*a^2
*b^3*e*(1 + m)) + ((A*b - a*B)*(e*x)^(1 + m)*(c + d*x^2)^2)/(4*a*b*e*(a + b*x^2)
^2) + ((b*c - a*d)*(e*x)^(1 + m)*(c*(A*b*(3 - m) + a*B*(1 + m)) - d*(A*b*(1 + m)
- a*B*(5 + m))*x^2))/(8*a^2*b^2*e*(a + b*x^2)) - ((a*d*(b*c*(1 + m) - a*d*(3 +
m))*(A*b*(1 + m) - a*B*(5 + m)) - b*c*(A*b*(3 - m) + a*B*(1 + m))*(a*d*(1 + m) +
b*(c - c*m)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2
)/a)])/(8*a^3*b^3*e*(1 + m))

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Rubi [A]  time = 1.12674, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097 \[ -\frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (a d (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))-b c (a B (m+1)+A b (3-m)) (a d (m+1)+b (c-c m)))}{8 a^3 b^3 e (m+1)}+\frac{d (e x)^{m+1} (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))}{8 a^2 b^3 e (m+1)}+\frac{(e x)^{m+1} (b c-a d) \left (c (a B (m+1)+A b (3-m))-d x^2 (A b (m+1)-a B (m+5))\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac{\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]  Int[((e*x)^m*(A + B*x^2)*(c + d*x^2)^2)/(a + b*x^2)^3,x]

[Out]

(d*(b*c*(1 + m) - a*d*(3 + m))*(A*b*(1 + m) - a*B*(5 + m))*(e*x)^(1 + m))/(8*a^2
*b^3*e*(1 + m)) + ((A*b - a*B)*(e*x)^(1 + m)*(c + d*x^2)^2)/(4*a*b*e*(a + b*x^2)
^2) + ((b*c - a*d)*(e*x)^(1 + m)*(c*(A*b*(3 - m) + a*B*(1 + m)) - d*(A*b*(1 + m)
- a*B*(5 + m))*x^2))/(8*a^2*b^2*e*(a + b*x^2)) - ((a*d*(b*c*(1 + m) - a*d*(3 +
m))*(A*b*(1 + m) - a*B*(5 + m)) - b*c*(A*b*(3 - m) + a*B*(1 + m))*(a*d*(1 + m) +
b*(c - c*m)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2
)/a)])/(8*a^3*b^3*e*(1 + m))

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Rubi in Sympy [A]  time = 106.696, size = 274, normalized size = 0.94 \[ \frac{\left (e x\right )^{m + 1} \left (c + d x^{2}\right )^{2} \left (A b - B a\right )}{4 a b e \left (a + b x^{2}\right )^{2}} - \frac{\left (e x\right )^{m + 1} \left (a d - b c\right ) \left (c \left (- A b m + 3 A b + B a m + B a\right ) + d x^{2} \left (4 A b - \left (m + 5\right ) \left (A b - B a\right )\right )\right )}{8 a^{2} b^{2} e \left (a + b x^{2}\right )} + \frac{d \left (e x\right )^{m + 1} \left (4 A b - \left (m + 5\right ) \left (A b - B a\right )\right ) \left (2 b c + \left (m + 3\right ) \left (a d - b c\right )\right )}{8 a^{2} b^{3} e \left (m + 1\right )} - \frac{\left (e x\right )^{m + 1} \left (a d \left (4 A b - \left (m + 5\right ) \left (A b - B a\right )\right ) \left (2 b c + \left (m + 3\right ) \left (a d - b c\right )\right ) - b c \left (- A b m + 3 A b + B a m + B a\right ) \left (a d m + a d - b c m + b c\right )\right ){{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{8 a^{3} b^{3} e \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x)**m*(B*x**2+A)*(d*x**2+c)**2/(b*x**2+a)**3,x)

[Out]

(e*x)**(m + 1)*(c + d*x**2)**2*(A*b - B*a)/(4*a*b*e*(a + b*x**2)**2) - (e*x)**(m
+ 1)*(a*d - b*c)*(c*(-A*b*m + 3*A*b + B*a*m + B*a) + d*x**2*(4*A*b - (m + 5)*(A
*b - B*a)))/(8*a**2*b**2*e*(a + b*x**2)) + d*(e*x)**(m + 1)*(4*A*b - (m + 5)*(A*
b - B*a))*(2*b*c + (m + 3)*(a*d - b*c))/(8*a**2*b**3*e*(m + 1)) - (e*x)**(m + 1)
*(a*d*(4*A*b - (m + 5)*(A*b - B*a))*(2*b*c + (m + 3)*(a*d - b*c)) - b*c*(-A*b*m
+ 3*A*b + B*a*m + B*a)*(a*d*m + a*d - b*c*m + b*c))*hyper((1, m/2 + 1/2), (m/2 +
3/2,), -b*x**2/a)/(8*a**3*b**3*e*(m + 1))

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Mathematica [A]  time = 0.479671, size = 170, normalized size = 0.58 \[ \frac{x (e x)^m \left (\frac{c x^2 (2 A d+B c) \, _2F_1\left (3,\frac{m+3}{2};\frac{m+5}{2};-\frac{b x^2}{a}\right )}{m+3}+d x^4 \left (\frac{(A d+2 B c) \, _2F_1\left (3,\frac{m+5}{2};\frac{m+7}{2};-\frac{b x^2}{a}\right )}{m+5}+\frac{B d x^2 \, _2F_1\left (3,\frac{m+7}{2};\frac{m+9}{2};-\frac{b x^2}{a}\right )}{m+7}\right )+\frac{A c^2 \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{m+1}\right )}{a^3} \]

Antiderivative was successfully verified.

[In]  Integrate[((e*x)^m*(A + B*x^2)*(c + d*x^2)^2)/(a + b*x^2)^3,x]

[Out]

(x*(e*x)^m*((A*c^2*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(1
+ m) + (c*(B*c + 2*A*d)*x^2*Hypergeometric2F1[3, (3 + m)/2, (5 + m)/2, -((b*x^2)
/a)])/(3 + m) + d*x^4*(((2*B*c + A*d)*Hypergeometric2F1[3, (5 + m)/2, (7 + m)/2,
-((b*x^2)/a)])/(5 + m) + (B*d*x^2*Hypergeometric2F1[3, (7 + m)/2, (9 + m)/2, -(
(b*x^2)/a)])/(7 + m))))/a^3

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Maple [F]  time = 0.1, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex \right ) ^{m} \left ( B{x}^{2}+A \right ) \left ( d{x}^{2}+c \right ) ^{2}}{ \left ( b{x}^{2}+a \right ) ^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^3,x)

[Out]

int((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(d*x^2 + c)^2*(e*x)^m/(b*x^2 + a)^3,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(d*x^2 + c)^2*(e*x)^m/(b*x^2 + a)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (B d^{2} x^{6} +{\left (2 \, B c d + A d^{2}\right )} x^{4} + A c^{2} +{\left (B c^{2} + 2 \, A c d\right )} x^{2}\right )} \left (e x\right )^{m}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(d*x^2 + c)^2*(e*x)^m/(b*x^2 + a)^3,x, algorithm="fricas")

[Out]

integral((B*d^2*x^6 + (2*B*c*d + A*d^2)*x^4 + A*c^2 + (B*c^2 + 2*A*c*d)*x^2)*(e*
x)^m/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), 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((e*x)**m*(B*x**2+A)*(d*x**2+c)**2/(b*x**2+a)**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(d*x^2 + c)^2*(e*x)^m/(b*x^2 + a)^3,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(d*x^2 + c)^2*(e*x)^m/(b*x^2 + a)^3, x)