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

Optimal. Leaf size=483 \[ \frac{(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))-2 a b c d (A d (1-m) (-m-2 p+3)+B c (m+1) (-m-2 p+1))+b^2 c^2 (-m-2 p+1) (A d (-m-2 p+3)+B c (m+2 p+1))\right ) F_1\left (\frac{m+1}{2};-p,1;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{8 c^3 d e (m+1) (b c-a d)^2}-\frac{b (m+2 p+1) (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{b x^2}{a}\right ) (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 d e (m+1) (b c-a d)^2}+\frac{(e x)^{m+1} \left (a+b x^2\right )^{p+1} (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 e \left (c+d x^2\right ) (b c-a d)^2}+\frac{(e x)^{m+1} \left (a+b x^2\right )^{p+1} (B c-A d)}{4 c e \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

((B*c - A*d)*(e*x)^(1 + m)*(a + b*x^2)^(1 + p))/(4*c*(b*c - a*d)*e*(c + d*x^2)^2
) + ((a*d*(A*d*(3 - m) + B*c*(1 + m)) + b*c*(B*c*(1 - m - 2*p) - A*d*(5 - m - 2*
p)))*(e*x)^(1 + m)*(a + b*x^2)^(1 + p))/(8*c^2*(b*c - a*d)^2*e*(c + d*x^2)) + ((
a^2*d^2*(1 - m)*(A*d*(3 - m) + B*c*(1 + m)) - 2*a*b*c*d*(B*c*(1 + m)*(1 - m - 2*
p) + A*d*(1 - m)*(3 - m - 2*p)) + b^2*c^2*(1 - m - 2*p)*(A*d*(3 - m - 2*p) + B*c
*(1 + m + 2*p)))*(e*x)^(1 + m)*(a + b*x^2)^p*AppellF1[(1 + m)/2, -p, 1, (3 + m)/
2, -((b*x^2)/a), -((d*x^2)/c)])/(8*c^3*d*(b*c - a*d)^2*e*(1 + m)*(1 + (b*x^2)/a)
^p) - (b*(a*d*(A*d*(3 - m) + B*c*(1 + m)) + b*c*(B*c*(1 - m - 2*p) - A*d*(5 - m
- 2*p)))*(1 + m + 2*p)*(e*x)^(1 + m)*(a + b*x^2)^p*Hypergeometric2F1[(1 + m)/2,
-p, (3 + m)/2, -((b*x^2)/a)])/(8*c^2*d*(b*c - a*d)^2*e*(1 + m)*(1 + (b*x^2)/a)^p
)

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Rubi [A]  time = 2.89441, antiderivative size = 483, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194 \[ \frac{(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))-2 a b c d (A d (1-m) (-m-2 p+3)+B c (m+1) (-m-2 p+1))+b^2 c^2 (-m-2 p+1) (A d (-m-2 p+3)+B c (m+2 p+1))\right ) F_1\left (\frac{m+1}{2};-p,1;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{8 c^3 d e (m+1) (b c-a d)^2}-\frac{b (m+2 p+1) (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{b x^2}{a}\right ) (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 d e (m+1) (b c-a d)^2}+\frac{(e x)^{m+1} \left (a+b x^2\right )^{p+1} (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 e \left (c+d x^2\right ) (b c-a d)^2}+\frac{(e x)^{m+1} \left (a+b x^2\right )^{p+1} (B c-A d)}{4 c e \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

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

[Out]

((B*c - A*d)*(e*x)^(1 + m)*(a + b*x^2)^(1 + p))/(4*c*(b*c - a*d)*e*(c + d*x^2)^2
) + ((a*d*(A*d*(3 - m) + B*c*(1 + m)) + b*c*(B*c*(1 - m - 2*p) - A*d*(5 - m - 2*
p)))*(e*x)^(1 + m)*(a + b*x^2)^(1 + p))/(8*c^2*(b*c - a*d)^2*e*(c + d*x^2)) + ((
a^2*d^2*(1 - m)*(A*d*(3 - m) + B*c*(1 + m)) - 2*a*b*c*d*(B*c*(1 + m)*(1 - m - 2*
p) + A*d*(1 - m)*(3 - m - 2*p)) + b^2*c^2*(1 - m - 2*p)*(A*d*(3 - m - 2*p) + B*c
*(1 + m + 2*p)))*(e*x)^(1 + m)*(a + b*x^2)^p*AppellF1[(1 + m)/2, -p, 1, (3 + m)/
2, -((b*x^2)/a), -((d*x^2)/c)])/(8*c^3*d*(b*c - a*d)^2*e*(1 + m)*(1 + (b*x^2)/a)
^p) - (b*(a*d*(A*d*(3 - m) + B*c*(1 + m)) + b*c*(B*c*(1 - m - 2*p) - A*d*(5 - m
- 2*p)))*(1 + m + 2*p)*(e*x)^(1 + m)*(a + b*x^2)^p*Hypergeometric2F1[(1 + m)/2,
-p, (3 + m)/2, -((b*x^2)/a)])/(8*c^2*d*(b*c - a*d)^2*e*(1 + m)*(1 + (b*x^2)/a)^p
)

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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)**p*(B*x**2+A)/(d*x**2+c)**3,x)

[Out]

Timed out

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Mathematica [A]  time = 1.70517, size = 403, normalized size = 0.83 \[ \frac{a c (m+3) x (e x)^m \left (a+b x^2\right )^p \left (\frac{(A d-B c) F_1\left (\frac{m+1}{2};-p,3;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{2 x^2 \left (b c p F_1\left (\frac{m+3}{2};1-p,3;\frac{m+5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-3 a d F_1\left (\frac{m+3}{2};-p,4;\frac{m+5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )+a c (m+3) F_1\left (\frac{m+1}{2};-p,3;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}+\frac{B \left (c+d x^2\right ) F_1\left (\frac{m+1}{2};-p,2;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{2 x^2 \left (b c p F_1\left (\frac{m+3}{2};1-p,2;\frac{m+5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-2 a d F_1\left (\frac{m+3}{2};-p,3;\frac{m+5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )+a c (m+3) F_1\left (\frac{m+1}{2};-p,2;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}\right )}{d (m+1) \left (c+d x^2\right )^3} \]

Warning: Unable to verify antiderivative.

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

[Out]

(a*c*(3 + m)*x*(e*x)^m*(a + b*x^2)^p*((B*(c + d*x^2)*AppellF1[(1 + m)/2, -p, 2,
(3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)])/(a*c*(3 + m)*AppellF1[(1 + m)/2, -p, 2,
(3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)] + 2*x^2*(b*c*p*AppellF1[(3 + m)/2, 1 - p,
2, (5 + m)/2, -((b*x^2)/a), -((d*x^2)/c)] - 2*a*d*AppellF1[(3 + m)/2, -p, 3, (5
+ m)/2, -((b*x^2)/a), -((d*x^2)/c)])) + ((-(B*c) + A*d)*AppellF1[(1 + m)/2, -p,
3, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)])/(a*c*(3 + m)*AppellF1[(1 + m)/2, -p,
3, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)] + 2*x^2*(b*c*p*AppellF1[(3 + m)/2, 1
- p, 3, (5 + m)/2, -((b*x^2)/a), -((d*x^2)/c)] - 3*a*d*AppellF1[(3 + m)/2, -p, 4
, (5 + m)/2, -((b*x^2)/a), -((d*x^2)/c)]))))/(d*(1 + m)*(c + d*x^2)^3)

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Maple [F]  time = 0.116, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex \right ) ^{m} \left ( b{x}^{2}+a \right ) ^{p} \left ( B{x}^{2}+A \right ) }{ \left ( d{x}^{2}+c \right ) ^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*x^2 + A)*(b*x^2 + a)^p*(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)**p*(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. \[ \int \frac{{\left (B x^{2} + A\right )}{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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