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

Optimal. Leaf size=433 \[ -\frac{b (e x)^{m+1} \left (2 a^2 d^2 (m+1) (A d (3-m)+B c (m+1))+3 a b c d (m+3) (A d (m+1)-B c (m+5))-b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^2 d^4 e (m+1)}-\frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (-m^2-2 m+3\right )+B c \left (m^2+6 m+5\right )\right )+b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^3 d^4 e (m+1)}-\frac{b^2 (e x)^{m+3} (a d (m+3) (A d (3-m)+B c (m+1))+b c (m+5) (A d (m+3)-B c (m+7)))}{8 c^2 d^3 e^3 (m+3)}+\frac{\left (a+b x^2\right )^2 (e x)^{m+1} (a d (A d (3-m)+B c (m+1))+b c (A d (m+3)-B c (m+7)))}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac{\left (a+b x^2\right )^3 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]

[Out]

-(b*(2*a^2*d^2*(1 + m)*(A*d*(3 - m) + B*c*(1 + m)) + 3*a*b*c*d*(3 + m)*(A*d*(1 +
m) - B*c*(5 + m)) - b^2*c^2*(5 + m)*(A*d*(3 + m) - B*c*(7 + m)))*(e*x)^(1 + m))
/(8*c^2*d^4*e*(1 + m)) - (b^2*(a*d*(3 + m)*(A*d*(3 - m) + B*c*(1 + m)) + b*c*(5
+ m)*(A*d*(3 + m) - B*c*(7 + m)))*(e*x)^(3 + m))/(8*c^2*d^3*e^3*(3 + m)) - ((B*c
- A*d)*(e*x)^(1 + m)*(a + b*x^2)^3)/(4*c*d*e*(c + d*x^2)^2) + ((a*d*(A*d*(3 - m
) + B*c*(1 + m)) + b*c*(A*d*(3 + m) - B*c*(7 + m)))*(e*x)^(1 + m)*(a + b*x^2)^2)
/(8*c^2*d^2*e*(c + d*x^2)) - ((b*c - a*d)*(a^2*d^2*(1 - m)*(A*d*(3 - m) + B*c*(1
+ m)) + b^2*c^2*(5 + m)*(A*d*(3 + m) - B*c*(7 + m)) + 2*a*b*c*d*(A*d*(3 - 2*m -
m^2) + B*c*(5 + 6*m + m^2)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 +
m)/2, -((d*x^2)/c)])/(8*c^3*d^4*e*(1 + m))

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Rubi [A]  time = 2.90605, antiderivative size = 433, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097 \[ -\frac{b (e x)^{m+1} \left (2 a^2 d^2 (m+1) (A d (3-m)+B c (m+1))+3 a b c d (m+3) (A d (m+1)-B c (m+5))-b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^2 d^4 e (m+1)}-\frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (-m^2-2 m+3\right )+B c \left (m^2+6 m+5\right )\right )+b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^3 d^4 e (m+1)}-\frac{b^2 (e x)^{m+3} (a d (m+3) (A d (3-m)+B c (m+1))+b c (m+5) (A d (m+3)-B c (m+7)))}{8 c^2 d^3 e^3 (m+3)}+\frac{\left (a+b x^2\right )^2 (e x)^{m+1} (a d (A d (3-m)+B c (m+1))+b c (A d (m+3)-B c (m+7)))}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac{\left (a+b x^2\right )^3 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

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

[Out]

-(b*(2*a^2*d^2*(1 + m)*(A*d*(3 - m) + B*c*(1 + m)) + 3*a*b*c*d*(3 + m)*(A*d*(1 +
m) - B*c*(5 + m)) - b^2*c^2*(5 + m)*(A*d*(3 + m) - B*c*(7 + m)))*(e*x)^(1 + m))
/(8*c^2*d^4*e*(1 + m)) - (b^2*(a*d*(3 + m)*(A*d*(3 - m) + B*c*(1 + m)) + b*c*(5
+ m)*(A*d*(3 + m) - B*c*(7 + m)))*(e*x)^(3 + m))/(8*c^2*d^3*e^3*(3 + m)) - ((B*c
- A*d)*(e*x)^(1 + m)*(a + b*x^2)^3)/(4*c*d*e*(c + d*x^2)^2) + ((a*d*(A*d*(3 - m
) + B*c*(1 + m)) + b*c*(A*d*(3 + m) - B*c*(7 + m)))*(e*x)^(1 + m)*(a + b*x^2)^2)
/(8*c^2*d^2*e*(c + d*x^2)) - ((b*c - a*d)*(a^2*d^2*(1 - m)*(A*d*(3 - m) + B*c*(1
+ m)) + b^2*c^2*(5 + m)*(A*d*(3 + m) - B*c*(7 + m)) + 2*a*b*c*d*(A*d*(3 - 2*m -
m^2) + B*c*(5 + 6*m + m^2)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 +
m)/2, -((d*x^2)/c)])/(8*c^3*d^4*e*(1 + m))

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

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Mathematica [A]  time = 0.943397, size = 218, normalized size = 0.5 \[ \frac{x (e x)^m \left (\frac{a^3 A \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{m+1}+\frac{a^2 x^2 (a B+3 A b) \, _2F_1\left (3,\frac{m+3}{2};\frac{m+5}{2};-\frac{d x^2}{c}\right )}{m+3}+b x^4 \left (b x^2 \left (\frac{(3 a B+A b) \, _2F_1\left (3,\frac{m+7}{2};\frac{m+9}{2};-\frac{d x^2}{c}\right )}{m+7}+\frac{b B x^2 \, _2F_1\left (3,\frac{m+9}{2};\frac{m+11}{2};-\frac{d x^2}{c}\right )}{m+9}\right )+\frac{3 a (a B+A b) \, _2F_1\left (3,\frac{m+5}{2};\frac{m+7}{2};-\frac{d x^2}{c}\right )}{m+5}\right )\right )}{c^3} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex \right ) ^{m} \left ( b{x}^{2}+a \right ) ^{3} \left ( B{x}^{2}+A \right ) }{ \left ( d{x}^{2}+c \right ) ^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (B b^{3} x^{8} +{\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} x^{4} + A a^{3} +{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2}\right )} \left (e x\right )^{m}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*b^3*x^8 + (3*B*a*b^2 + A*b^3)*x^6 + 3*(B*a^2*b + A*a*b^2)*x^4 + A*a^
3 + (B*a^3 + 3*A*a^2*b)*x^2)*(e*x)^m/(d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^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)**3*(B*x**2+A)/(d*x**2+c)**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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