\(\int \frac {(e x)^m (a+b x^2) (A+B x^2)}{c+d x^2} \, dx\) [25]

Optimal result

Integrand size = 29, antiderivative size = 120 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=-\frac {(b B c-A b d-a B d) (e x)^{1+m}}{d^2 e (1+m)}+\frac {b B (e x)^{3+m}}{d e^3 (3+m)}+\frac {(b c-a d) (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c d^2 e (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {584, 371} \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {(e x)^{m+1} (b c-a d) (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c d^2 e (m+1)}-\frac {(e x)^{m+1} (-a B d-A b d+b B c)}{d^2 e (m+1)}+\frac {b B (e x)^{m+3}}{d e^3 (m+3)} \]

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 584

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]

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {x (e x)^m \left (\frac {-b B c+A b d+a B d}{1+m}+\frac {b B d x^2}{3+m}+\frac {(b c-a d) (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (1+m)}\right )}{d^2} \]

[In]

Integrate[((e*x)^m*(a + b*x^2)*(A + B*x^2))/(c + d*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) + ((b*c - a*d)*(B*c - A*d)*Hypergeometric
2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(1 + m))))/d^2

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.50 (sec) , antiderivative size = 418, normalized size of antiderivative = 3.48 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {A a e^{m} m x^{m + 1} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A a e^{m} x^{m + 1} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A b e^{m} m x^{m + 3} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 A b e^{m} x^{m + 3} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B a e^{m} m x^{m + 3} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 B a e^{m} x^{m + 3} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B b e^{m} m x^{m + 5} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 B b e^{m} x^{m + 5} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} \]

[In]

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

[Out]

A*a*e**m*m*x**(m + 1)*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*e**m*x**(m + 1)*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*b*e**m*m*x**(m + 3)*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*A*b*e**m*x**(m + 3)*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*e**m*m*x**(m + 3)*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*e**m*x**(m + 3)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamm
a(m/2 + 5/2)) + B*b*e**m*m*x**(m + 5)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(4*c*g
amma(m/2 + 7/2)) + 5*B*b*e**m*x**(m + 5)*lerchphi(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))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,\left (b\,x^2+a\right )}{d\,x^2+c} \,d x \]

[In]

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

[Out]

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