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

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 = 0.0980765, 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, Rules used = {570, 364} \[ \frac{(e x)^{m+1} (A b-a B) (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a b^2 e (m+1)}+\frac{(e x)^{m+1} (-a B d+A b d+b B c)}{b^2 e (m+1)}+\frac{B d (e x)^{m+3}}{b e^3 (m+3)} \]

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))

Rule 570

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )}{a+b x^2} \, dx &=\int \left (\frac{(b B c+A b d-a B d) (e x)^m}{b^2}+\frac{B d (e x)^{2+m}}{b e^2}+\frac{\left (A b^2 c-a b B c-a A b d+a^2 B d\right ) (e x)^m}{b^2 \left (a+b x^2\right )}\right ) \, dx\\ &=\frac{(b B c+A b d-a B d) (e x)^{1+m}}{b^2 e (1+m)}+\frac{B d (e x)^{3+m}}{b e^3 (3+m)}+\frac{((A b-a B) (b c-a d)) \int \frac{(e x)^m}{a+b x^2} \, dx}{b^2}\\ &=\frac{(b B c+A b d-a B d) (e x)^{1+m}}{b^2 e (1+m)}+\frac{B d (e x)^{3+m}}{b e^3 (3+m)}+\frac{(A b-a B) (b c-a d) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{a b^2 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.106945, size = 93, normalized size = 0.79 \[ \frac{x (e x)^m \left (\frac{(a B-A b) (a d-b c) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a (m+1)}+\frac{-a B d+A b d+b B c}{m+1}+\frac{b B d x^2}{m+3}\right )}{b^2} \]

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*((b*B*c + A*b*d - a*B*d)/(1 + m) + (b*B*d*x^2)/(3 + m) + ((-(A*b) + a*B)*(-(b*c) + a*d)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(1 + m))))/b^2

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( B{x}^{2}+A \right ) \left ( d{x}^{2}+c \right ) }{b{x}^{2}+a}}\, dx \end{align*}

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. \begin{align*} \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} \end{align*}

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, 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. \begin{align*}{\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 ) \end{align*}

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, 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 [C]  time = 12.6763, size = 428, normalized size = 3.63 \begin{align*} \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 )} \end{align*}

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*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*
A*d*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*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*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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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, algorithm="giac")

[Out]

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