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

Optimal. Leaf size=491 \[ -\frac{d (e x)^{m+1} \left (A \left (4 a^2 d^2+a b c d (11-m)-b^2 c^2 (3-m)\right )-a B c (a d (11-m)+b c (m+1))\right )}{8 a^2 c e \left (c+d x^2\right ) (b c-a d)^3}+\frac{(e x)^{m+1} (A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (m+1)))}{8 a^2 e \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)^2}+\frac{b (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2-12 m+35\right )-2 a b c d \left (m^2-8 m+7\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (-m^2+4 m+5\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 e (m+1) (b c-a d)^4}+\frac{d^2 (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (5-m)-A d (7-m)))}{2 c^2 e (m+1) (b c-a d)^4}+\frac{(e x)^{m+1} (A b-a B)}{4 a e \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)} \]

[Out]

-(d*(A*(4*a^2*d^2 - b^2*c^2*(3 - m) + a*b*c*d*(11 - m)) - a*B*c*(a*d*(11 - m) +
b*c*(1 + m)))*(e*x)^(1 + m))/(8*a^2*c*(b*c - a*d)^3*e*(c + d*x^2)) + ((A*b - a*B
)*(e*x)^(1 + m))/(4*a*(b*c - a*d)*e*(a + b*x^2)^2*(c + d*x^2)) + ((A*b*(b*c*(3 -
m) - a*d*(9 - m)) + a*B*(a*d*(5 - m) + b*c*(1 + m)))*(e*x)^(1 + m))/(8*a^2*(b*c
- a*d)^2*e*(a + b*x^2)*(c + d*x^2)) + (b*(a*B*(b^2*c^2*(1 - m^2) - 2*a*b*c*d*(5
+ 4*m - m^2) - a^2*d^2*(15 - 8*m + m^2)) + A*b*(a^2*d^2*(35 - 12*m + m^2) - 2*a
*b*c*d*(7 - 8*m + m^2) + b^2*c^2*(3 - 4*m + m^2)))*(e*x)^(1 + m)*Hypergeometric2
F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(8*a^3*(b*c - a*d)^4*e*(1 + m)) + (d^
2*(b*c*(B*c*(5 - m) - A*d*(7 - m)) + a*d*(A*d*(1 - m) + B*c*(1 + m)))*(e*x)^(1 +
m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(2*c^2*(b*c - a*d)
^4*e*(1 + m))

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Rubi [A]  time = 4.05301, antiderivative size = 491, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097 \[ -\frac{d (e x)^{m+1} \left (A \left (4 a^2 d^2+a b c d (11-m)-b^2 c^2 (3-m)\right )-a B c (a d (11-m)+b c (m+1))\right )}{8 a^2 c e \left (c+d x^2\right ) (b c-a d)^3}+\frac{(e x)^{m+1} (A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (m+1)))}{8 a^2 e \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)^2}+\frac{b (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2-12 m+35\right )-2 a b c d \left (m^2-8 m+7\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (-m^2+4 m+5\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 e (m+1) (b c-a d)^4}+\frac{d^2 (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (5-m)-A d (7-m)))}{2 c^2 e (m+1) (b c-a d)^4}+\frac{(e x)^{m+1} (A b-a B)}{4 a e \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

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

[Out]

-(d*(A*(4*a^2*d^2 - b^2*c^2*(3 - m) + a*b*c*d*(11 - m)) - a*B*c*(a*d*(11 - m) +
b*c*(1 + m)))*(e*x)^(1 + m))/(8*a^2*c*(b*c - a*d)^3*e*(c + d*x^2)) + ((A*b - a*B
)*(e*x)^(1 + m))/(4*a*(b*c - a*d)*e*(a + b*x^2)^2*(c + d*x^2)) + ((A*b*(b*c*(3 -
m) - a*d*(9 - m)) + a*B*(a*d*(5 - m) + b*c*(1 + m)))*(e*x)^(1 + m))/(8*a^2*(b*c
- a*d)^2*e*(a + b*x^2)*(c + d*x^2)) + (b*(a*B*(b^2*c^2*(1 - m^2) - 2*a*b*c*d*(5
+ 4*m - m^2) - a^2*d^2*(15 - 8*m + m^2)) + A*b*(a^2*d^2*(35 - 12*m + m^2) - 2*a
*b*c*d*(7 - 8*m + m^2) + b^2*c^2*(3 - 4*m + m^2)))*(e*x)^(1 + m)*Hypergeometric2
F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(8*a^3*(b*c - a*d)^4*e*(1 + m)) + (d^
2*(b*c*(B*c*(5 - m) - A*d*(7 - m)) + a*d*(A*d*(1 - m) + B*c*(1 + m)))*(e*x)^(1 +
m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(2*c^2*(b*c - a*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)/(b*x**2+a)**3/(d*x**2+c)**2,x)

[Out]

Timed out

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Mathematica [C]  time = 1.4808, size = 379, normalized size = 0.77 \[ \frac{a c x (e x)^m \left (\frac{A (m+3)^2 F_1\left (\frac{m+1}{2};3,2;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{(m+1) \left (a c (m+3) F_1\left (\frac{m+1}{2};3,2;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-2 x^2 \left (2 a d F_1\left (\frac{m+3}{2};3,3;\frac{m+5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )+3 b c F_1\left (\frac{m+3}{2};4,2;\frac{m+5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )\right )}+\frac{B (m+5) x^2 F_1\left (\frac{m+3}{2};3,2;\frac{m+5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{a c (m+5) F_1\left (\frac{m+3}{2};3,2;\frac{m+5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-2 x^2 \left (2 a d F_1\left (\frac{m+5}{2};3,3;\frac{m+7}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )+3 b c F_1\left (\frac{m+5}{2};4,2;\frac{m+7}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )}\right )}{(m+3) \left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \]

Warning: Unable to verify antiderivative.

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

[Out]

(a*c*x*(e*x)^m*((A*(3 + m)^2*AppellF1[(1 + m)/2, 3, 2, (3 + m)/2, -((b*x^2)/a),
-((d*x^2)/c)])/((1 + m)*(a*c*(3 + m)*AppellF1[(1 + m)/2, 3, 2, (3 + m)/2, -((b*x
^2)/a), -((d*x^2)/c)] - 2*x^2*(2*a*d*AppellF1[(3 + m)/2, 3, 3, (5 + m)/2, -((b*x
^2)/a), -((d*x^2)/c)] + 3*b*c*AppellF1[(3 + m)/2, 4, 2, (5 + m)/2, -((b*x^2)/a),
-((d*x^2)/c)]))) + (B*(5 + m)*x^2*AppellF1[(3 + m)/2, 3, 2, (5 + m)/2, -((b*x^2
)/a), -((d*x^2)/c)])/(a*c*(5 + m)*AppellF1[(3 + m)/2, 3, 2, (5 + m)/2, -((b*x^2)
/a), -((d*x^2)/c)] - 2*x^2*(2*a*d*AppellF1[(5 + m)/2, 3, 3, (7 + m)/2, -((b*x^2)
/a), -((d*x^2)/c)] + 3*b*c*AppellF1[(5 + m)/2, 4, 2, (7 + m)/2, -((b*x^2)/a), -(
(d*x^2)/c)]))))/((3 + m)*(a + b*x^2)^3*(c + d*x^2)^2)

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Maple [F]  time = 0.109, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex \right ) ^{m} \left ( B{x}^{2}+A \right ) }{ \left ( b{x}^{2}+a \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{b^{3} d^{2} x^{10} +{\left (2 \, b^{3} c d + 3 \, a b^{2} d^{2}\right )} x^{8} +{\left (b^{3} c^{2} + 6 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x^{6} + a^{3} c^{2} +{\left (3 \, a b^{2} c^{2} + 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{4} +{\left (3 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*x^2 + A)*(e*x)^m/(b^3*d^2*x^10 + (2*b^3*c*d + 3*a*b^2*d^2)*x^8 + (b^
3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*x^6 + a^3*c^2 + (3*a*b^2*c^2 + 6*a^2*b*c*d +
a^3*d^2)*x^4 + (3*a^2*b*c^2 + 2*a^3*c*d)*x^2), 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)/(b*x**2+a)**3/(d*x**2+c)**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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