3.1.24 \(\int \frac {(e x)^m (a+b x^2)^2 (A+B x^2)}{c+d x^2} \, dx\) [24]

3.1.24.1 Optimal result

Integrand size = 31, antiderivative size = 180 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {\left (a^2 B d^2+b^2 c (B c-A d)-2 a b d (B c-A d)\right ) (e x)^{1+m}}{d^3 e (1+m)}-\frac {b (b B c-A b d-2 a B d) (e x)^{3+m}}{d^2 e^3 (3+m)}+\frac {b^2 B (e x)^{5+m}}{d e^5 (5+m)}-\frac {(b c-a d)^2 (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c d^3 e (1+m)} \]

output

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

3.1.24.2 Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.82 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {x (e x)^m \left (\frac {a^2 B d^2+b^2 c (B c-A d)+2 a b d (-B c+A d)}{1+m}+\frac {b d (-b B c+A b d+2 a B d) x^2}{3+m}+\frac {b^2 B d^2 x^4}{5+m}-\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (1+m)}\right )}{d^3} \]

input

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

output

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

3.1.24.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {437, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 437

\(\displaystyle \int \left (\frac {(e x)^m \left (a^2 A d^3-a^2 B c d^2-2 a A b c d^2+2 a b B c^2 d+A b^2 c^2 d-b^2 B c^3\right )}{d^3 \left (c+d x^2\right )}+\frac {(e x)^m \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3}-\frac {b (e x)^{m+2} (-2 a B d-A b d+b B c)}{d^2 e^2}+\frac {b^2 B (e x)^{m+4}}{d e^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(e x)^{m+1} \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e (m+1)}-\frac {(e x)^{m+1} (b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c d^3 e (m+1)}-\frac {b (e x)^{m+3} (-2 a B d-A b d+b B c)}{d^2 e^3 (m+3)}+\frac {b^2 B (e x)^{m+5}}{d e^5 (m+5)}\)

input

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

output

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

3.1.24.3.1 Defintions of rubi rules used

rule 437

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

3.1.24.4 Maple [F]

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

input

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

output

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

3.1.24.5 Fricas [F]

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

input

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

output

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)
 

3.1.24.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.84 (sec) , antiderivative size = 649, normalized size of antiderivative = 3.61 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {A a^{2} 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^{2} 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 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 )}{2 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 A 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 )}{2 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {A b^{2} 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 A b^{2} 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 )} + \frac {B a^{2} 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^{2} 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 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 )}{2 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 B a 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 )}{2 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {B b^{2} e^{m} m x^{m + 7} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} + \frac {7 B b^{2} e^{m} x^{m + 7} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} \]

input

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

output

A*a**2*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**2*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*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)/(2*c*gamma(m/2 + 5/2)) + 3*A*a*b*e**m*x**(m + 3)*l 
erchphi(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)) + A*b**2*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*gamma(m/2 + 7/2)) + 5*A*b**2*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)) + B*a**2*e**m*m*x**(m + 3)*lerchphi(d*x**2*exp_pol 
ar(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**(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*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)/(2*c*gamma(m/2 + 7/2)) 
 + 5*B*a*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)/(2*c*gamma(m/2 + 7/2)) + B*b**2*e**m*m*x**(m + 7)*lerch 
phi(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**(m + 7)*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))
 

3.1.24.7 Maxima [F]

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

input

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

output

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

3.1.24.8 Giac [F]

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

input

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

output

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

3.1.24.9 Mupad [F(-1)]

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

input

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

output

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