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

Optimal. Leaf size=118 \[ \frac{(e x)^{m+1} (A b-a B) (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a b^2 e (m+1)}+\frac{(e x)^{m+1} (-a B d+A b d+b B c)}{b^2 e (m+1)}+\frac{B d (e x)^{m+3}}{b e^3 (m+3)} \]

[Out]

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

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Rubi [A]  time = 0.260416, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069 \[ \frac{(e x)^{m+1} (A b-a B) (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a b^2 e (m+1)}+\frac{(e x)^{m+1} (-a B d+A b d+b B c)}{b^2 e (m+1)}+\frac{B d (e x)^{m+3}}{b e^3 (m+3)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 39.7143, size = 99, normalized size = 0.84 \[ \frac{B d \left (e x\right )^{m + 3}}{b e^{3} \left (m + 3\right )} + \frac{\left (e x\right )^{m + 1} \left (A b d - B a d + B b c\right )}{b^{2} e \left (m + 1\right )} - \frac{\left (e x\right )^{m + 1} \left (A b - B a\right ) \left (a d - b c\right ){{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{a b^{2} e \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.207669, size = 121, normalized size = 1.03 \[ \frac{x (e x)^m \left (\frac{x^2 (A d+B c) \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};-\frac{b x^2}{a}\right )}{m+3}+\frac{A c \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{m+1}+\frac{B d x^4 \, _2F_1\left (1,\frac{m+5}{2};\frac{m+7}{2};-\frac{b x^2}{a}\right )}{m+5}\right )}{a} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex \right ) ^{m} \left ( B{x}^{2}+A \right ) \left ( d{x}^{2}+c \right ) }{b{x}^{2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 21.8469, size = 428, normalized size = 3.63 \[ \frac{A c e^{m} m x x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{A c e^{m} x x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{A d e^{m} m x^{3} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} + \frac{3 A d e^{m} x^{3} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} + \frac{B c e^{m} m x^{3} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} + \frac{3 B c e^{m} x^{3} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} + \frac{B d e^{m} m x^{5} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{5}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{7}{2}\right )} + \frac{5 B d e^{m} x^{5} x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{5}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{7}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

A*c*e**m*m*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1
/2)/(4*a*gamma(m/2 + 3/2)) + A*c*e**m*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a,
1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*a*gamma(m/2 + 3/2)) + A*d*e**m*m*x**3*x**m*ler
chphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*a*gamma(m/2 +
5/2)) + 3*A*d*e**m*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*ga
mma(m/2 + 3/2)/(4*a*gamma(m/2 + 5/2)) + B*c*e**m*m*x**3*x**m*lerchphi(b*x**2*exp
_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*a*gamma(m/2 + 5/2)) + 3*B*c*e*
*m*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(
4*a*gamma(m/2 + 5/2)) + B*d*e**m*m*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a,
1, m/2 + 5/2)*gamma(m/2 + 5/2)/(4*a*gamma(m/2 + 7/2)) + 5*B*d*e**m*x**5*x**m*ler
chphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(4*a*gamma(m/2 +
7/2))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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