3.1.12 \(\int \frac {(e x)^m (A+B x^2) (c+d x^2)^2}{a+b x^2} \, dx\) [12]

3.1.12.1 Optimal result

Integrand size = 31, antiderivative size = 178 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {\left (a^2 B d^2-a b d (2 B c+A d)+b^2 c (B c+2 A d)\right ) (e x)^{1+m}}{b^3 e (1+m)}+\frac {d (2 b B c+A b d-a B d) (e x)^{3+m}}{b^2 e^3 (3+m)}+\frac {B d^2 (e x)^{5+m}}{b e^5 (5+m)}+\frac {(A b-a B) (b c-a d)^2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a b^3 e (1+m)} \]

output

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

3.1.12.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.82 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {x (e x)^m \left (\frac {a^2 B d^2-a b d (2 B c+A d)+b^2 c (B c+2 A d)}{1+m}+\frac {b d (2 b B c+A b d-a B d) x^2}{3+m}+\frac {b^2 B d^2 x^4}{5+m}+\frac {(A b-a B) (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a (1+m)}\right )}{b^3} \]

input

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

output

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

3.1.12.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 178, 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.

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

\(\Big \downarrow \) 437

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.1.12.3.1 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]
 

3.1.12.4 Maple [F]

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

input

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

output

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

3.1.12.5 Fricas [F]

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

input

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

output

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

3.1.12.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.86 (sec) , antiderivative size = 649, normalized size of antiderivative = 3.65 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {A c^{2} e^{m} m x^{m + 1} \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^{2} e^{m} x^{m + 1} \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 d e^{m} m x^{m + 3} \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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 A c d e^{m} x^{m + 3} \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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {A d^{2} e^{m} m x^{m + 5} \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 A d^{2} e^{m} x^{m + 5} \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 {B c^{2} e^{m} m x^{m + 3} \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^{2} e^{m} x^{m + 3} \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 d e^{m} m x^{m + 5} \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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 B c d e^{m} x^{m + 5} \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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {B d^{2} e^{m} m x^{m + 7} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} + \frac {7 B d^{2} e^{m} x^{m + 7} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} \]

input

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

output

A*c**2*e**m*m*x**(m + 1)*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**2*e**m*x**(m + 1)*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*d*e**m*m*x**(m + 3)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 
+ 3/2)*gamma(m/2 + 3/2)/(2*a*gamma(m/2 + 5/2)) + 3*A*c*d*e**m*x**(m + 3)*l 
erchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(2*a*gamm 
a(m/2 + 5/2)) + A*d**2*e**m*m*x**(m + 5)*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*A*d**2*e**m*x* 
*(m + 5)*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)) + B*c**2*e**m*m*x**(m + 3)*lerchphi(b*x**2*exp_pol 
ar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*a*gamma(m/2 + 5/2)) + 3*B*c* 
*2*e**m*x**(m + 3)*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*d*e**m*m*x**(m + 5)*lerchphi(b*x** 
2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(2*a*gamma(m/2 + 7/2)) 
 + 5*B*c*d*e**m*x**(m + 5)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2 
)*gamma(m/2 + 5/2)/(2*a*gamma(m/2 + 7/2)) + B*d**2*e**m*m*x**(m + 7)*lerch 
phi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 7/2)*gamma(m/2 + 7/2)/(4*a*gamma(m/ 
2 + 9/2)) + 7*B*d**2*e**m*x**(m + 7)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, 
 m/2 + 7/2)*gamma(m/2 + 7/2)/(4*a*gamma(m/2 + 9/2))
 

3.1.12.7 Maxima [F]

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

input

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

output

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

3.1.12.8 Giac [F]

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

input

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

output

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

3.1.12.9 Mupad [F(-1)]

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

input

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

output

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