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

Optimal. Leaf size=180 \[ \frac{(e x)^{m+1} \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e (m+1)}-\frac{(e x)^{m+1} (b c-a d)^2 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d^3 e (m+1)}-\frac{b (e x)^{m+3} (-2 a B d-A b d+b B c)}{d^2 e^3 (m+3)}+\frac{b^2 B (e x)^{m+5}}{d e^5 (m+5)} \]

[Out]

((a^2*B*d^2 + b^2*c*(B*c - A*d) - 2*a*b*d*(B*c - A*d))*(e*x)^(1 + m))/(d^3*e*(1
+ m)) - (b*(b*B*c - A*b*d - 2*a*B*d)*(e*x)^(3 + m))/(d^2*e^3*(3 + m)) + (b^2*B*(
e*x)^(5 + m))/(d*e^5*(5 + m)) - ((b*c - a*d)^2*(B*c - A*d)*(e*x)^(1 + m)*Hyperge
ometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*d^3*e*(1 + m))

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Rubi [A]  time = 0.441483, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065 \[ \frac{(e x)^{m+1} \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e (m+1)}-\frac{(e x)^{m+1} (b c-a d)^2 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d^3 e (m+1)}-\frac{b (e x)^{m+3} (-2 a B d-A b d+b B c)}{d^2 e^3 (m+3)}+\frac{b^2 B (e x)^{m+5}}{d e^5 (m+5)} \]

Antiderivative was successfully verified.

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

[Out]

((a^2*B*d^2 + b^2*c*(B*c - A*d) - 2*a*b*d*(B*c - A*d))*(e*x)^(1 + m))/(d^3*e*(1
+ m)) - (b*(b*B*c - A*b*d - 2*a*B*d)*(e*x)^(3 + m))/(d^2*e^3*(3 + m)) + (b^2*B*(
e*x)^(5 + m))/(d*e^5*(5 + m)) - ((b*c - a*d)^2*(B*c - A*d)*(e*x)^(1 + m)*Hyperge
ometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*d^3*e*(1 + m))

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Rubi in Sympy [A]  time = 83.0784, size = 168, normalized size = 0.93 \[ \frac{B b^{2} \left (e x\right )^{m + 5}}{d e^{5} \left (m + 5\right )} + \frac{b \left (e x\right )^{m + 3} \left (A b d + 2 B a d - B b c\right )}{d^{2} e^{3} \left (m + 3\right )} + \frac{\left (e x\right )^{m + 1} \left (2 A a b d^{2} - A b^{2} c d + B a^{2} d^{2} - 2 B a b c d + B b^{2} c^{2}\right )}{d^{3} e \left (m + 1\right )} + \frac{\left (e x\right )^{m + 1} \left (A d - B c\right ) \left (a d - b c\right )^{2}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{d x^{2}}{c}} \right )}}{c d^{3} e \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.481526, size = 170, normalized size = 0.94 \[ \frac{x (e x)^m \left (\frac{a^2 A \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{m+1}+\frac{a x^2 (a B+2 A b) \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{d x^2}{c}\right )}{m+3}+b x^4 \left (\frac{(2 a B+A b) \, _2F_1\left (1,\frac{m+5}{2};\frac{m+7}{2};-\frac{d x^2}{c}\right )}{m+5}+\frac{b B x^2 \, _2F_1\left (1,\frac{m+7}{2};\frac{m+9}{2};-\frac{d x^2}{c}\right )}{m+7}\right )\right )}{c} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.068, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex \right ) ^{m} \left ( b{x}^{2}+a \right ) ^{2} \left ( B{x}^{2}+A \right ) }{d{x}^{2}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 47.9764, size = 666, normalized size = 3.7 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

A*a**2*e**m*m*x*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2
+ 1/2)/(4*c*gamma(m/2 + 3/2)) + A*a**2*e**m*x*x**m*lerchphi(d*x**2*exp_polar(I*p
i)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*c*gamma(m/2 + 3/2)) + A*a*b*e**m*m*x**3*
x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(2*c*gamm
a(m/2 + 5/2)) + 3*A*a*b*e**m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2
+ 3/2)*gamma(m/2 + 3/2)/(2*c*gamma(m/2 + 5/2)) + A*b**2*e**m*m*x**5*x**m*lerchp
hi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(4*c*gamma(m/2 + 7/2
)) + 5*A*b**2*e**m*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*ga
mma(m/2 + 5/2)/(4*c*gamma(m/2 + 7/2)) + B*a**2*e**m*m*x**3*x**m*lerchphi(d*x**2*
exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2)) + 3*B*a
**2*e**m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 +
3/2)/(4*c*gamma(m/2 + 5/2)) + B*a*b*e**m*m*x**5*x**m*lerchphi(d*x**2*exp_polar(I
*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(2*c*gamma(m/2 + 7/2)) + 5*B*a*b*e**m*x**
5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(2*c*ga
mma(m/2 + 7/2)) + B*b**2*e**m*m*x**7*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1,
m/2 + 7/2)*gamma(m/2 + 7/2)/(4*c*gamma(m/2 + 9/2)) + 7*B*b**2*e**m*x**7*x**m*ler
chphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 7/2)*gamma(m/2 + 7/2)/(4*c*gamma(m/2 +
9/2))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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