3.1.22 \(\int \frac {(e x)^m (a+b x^2)^4 (A+B x^2)}{c+d x^2} \, dx\) [22]

3.1.22.1 Optimal result

Integrand size = 31, antiderivative size = 363 \[ \int \frac {(e x)^m \left (a+b x^2\right )^4 \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {\left (a^4 B d^4+b^4 c^3 (B c-A d)-4 a b^3 c^2 d (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a^3 b d^3 (B c-A d)\right ) (e x)^{1+m}}{d^5 e (1+m)}+\frac {b \left (4 a^3 B d^3-b^3 c^2 (B c-A d)+4 a b^2 c d (B c-A d)-6 a^2 b d^2 (B c-A d)\right ) (e x)^{3+m}}{d^4 e^3 (3+m)}+\frac {b^2 \left (6 a^2 B d^2+b^2 c (B c-A d)-4 a b d (B c-A d)\right ) (e x)^{5+m}}{d^3 e^5 (5+m)}-\frac {b^3 (b B c-A b d-4 a B d) (e x)^{7+m}}{d^2 e^7 (7+m)}+\frac {b^4 B (e x)^{9+m}}{d e^9 (9+m)}-\frac {(b c-a d)^4 (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c d^5 e (1+m)} \]

output

(a^4*B*d^4+b^4*c^3*(-A*d+B*c)-4*a*b^3*c^2*d*(-A*d+B*c)+6*a^2*b^2*c*d^2*(-A 
*d+B*c)-4*a^3*b*d^3*(-A*d+B*c))*(e*x)^(1+m)/d^5/e/(1+m)+b*(4*a^3*B*d^3-b^3 
*c^2*(-A*d+B*c)+4*a*b^2*c*d*(-A*d+B*c)-6*a^2*b*d^2*(-A*d+B*c))*(e*x)^(3+m) 
/d^4/e^3/(3+m)+b^2*(6*a^2*B*d^2+b^2*c*(-A*d+B*c)-4*a*b*d*(-A*d+B*c))*(e*x) 
^(5+m)/d^3/e^5/(5+m)-b^3*(-A*b*d-4*B*a*d+B*b*c)*(e*x)^(7+m)/d^2/e^7/(7+m)+ 
b^4*B*(e*x)^(9+m)/d/e^9/(9+m)-(-a*d+b*c)^4*(-A*d+B*c)*(e*x)^(1+m)*hypergeo 
m([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c/d^5/e/(1+m)
 

3.1.22.2 Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 315, normalized size of antiderivative = 0.87 \[ \int \frac {(e x)^m \left (a+b x^2\right )^4 \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {x (e x)^m \left (\frac {a^4 B d^4+b^4 c^3 (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)+4 a b^3 c^2 d (-B c+A d)+4 a^3 b d^3 (-B c+A d)}{1+m}+\frac {b d \left (4 a^3 B d^3+4 a b^2 c d (B c-A d)+b^3 c^2 (-B c+A d)+6 a^2 b d^2 (-B c+A d)\right ) x^2}{3+m}+\frac {b^2 d^2 \left (6 a^2 B d^2+b^2 c (B c-A d)+4 a b d (-B c+A d)\right ) x^4}{5+m}+\frac {b^3 d^3 (-b B c+A b d+4 a B d) x^6}{7+m}+\frac {b^4 B d^4 x^8}{9+m}-\frac {(b c-a d)^4 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (1+m)}\right )}{d^5} \]

input

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

output

(x*(e*x)^m*((a^4*B*d^4 + b^4*c^3*(B*c - A*d) + 6*a^2*b^2*c*d^2*(B*c - A*d) 
 + 4*a*b^3*c^2*d*(-(B*c) + A*d) + 4*a^3*b*d^3*(-(B*c) + A*d))/(1 + m) + (b 
*d*(4*a^3*B*d^3 + 4*a*b^2*c*d*(B*c - A*d) + b^3*c^2*(-(B*c) + A*d) + 6*a^2 
*b*d^2*(-(B*c) + A*d))*x^2)/(3 + m) + (b^2*d^2*(6*a^2*B*d^2 + b^2*c*(B*c - 
 A*d) + 4*a*b*d*(-(B*c) + A*d))*x^4)/(5 + m) + (b^3*d^3*(-(b*B*c) + A*b*d 
+ 4*a*B*d)*x^6)/(7 + m) + (b^4*B*d^4*x^8)/(9 + m) - ((b*c - a*d)^4*(B*c - 
A*d)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(1 + m)) 
))/d^5
 

3.1.22.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 363, 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.

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

\(\Big \downarrow \) 437

\(\displaystyle \int \left (\frac {b^2 (e x)^{m+4} \left (6 a^2 B d^2-4 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e^4}+\frac {b (e x)^{m+2} \left (4 a^3 B d^3-6 a^2 b d^2 (B c-A d)+4 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e^2}+\frac {(e x)^m \left (a^4 B d^4-4 a^3 b d^3 (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a b^3 c^2 d (B c-A d)+b^4 c^3 (B c-A d)\right )}{d^5}+\frac {(e x)^m \left (a^4 A d^5-a^4 B c d^4-4 a^3 A b c d^4+4 a^3 b B c^2 d^3+6 a^2 A b^2 c^2 d^3-6 a^2 b^2 B c^3 d^2-4 a A b^3 c^3 d^2+4 a b^3 B c^4 d+A b^4 c^4 d-b^4 B c^5\right )}{d^5 \left (c+d x^2\right )}-\frac {b^3 (e x)^{m+6} (-4 a B d-A b d+b B c)}{d^2 e^6}+\frac {b^4 B (e x)^{m+8}}{d e^8}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 (e x)^{m+5} \left (6 a^2 B d^2-4 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e^5 (m+5)}+\frac {b (e x)^{m+3} \left (4 a^3 B d^3-6 a^2 b d^2 (B c-A d)+4 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e^3 (m+3)}+\frac {(e x)^{m+1} \left (a^4 B d^4-4 a^3 b d^3 (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a b^3 c^2 d (B c-A d)+b^4 c^3 (B c-A d)\right )}{d^5 e (m+1)}-\frac {b^3 (e x)^{m+7} (-4 a B d-A b d+b B c)}{d^2 e^7 (m+7)}-\frac {(e x)^{m+1} (b c-a d)^4 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c d^5 e (m+1)}+\frac {b^4 B (e x)^{m+9}}{d e^9 (m+9)}\)

input

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

output

((a^4*B*d^4 + b^4*c^3*(B*c - A*d) - 4*a*b^3*c^2*d*(B*c - A*d) + 6*a^2*b^2* 
c*d^2*(B*c - A*d) - 4*a^3*b*d^3*(B*c - A*d))*(e*x)^(1 + m))/(d^5*e*(1 + m) 
) + (b*(4*a^3*B*d^3 - b^3*c^2*(B*c - A*d) + 4*a*b^2*c*d*(B*c - A*d) - 6*a^ 
2*b*d^2*(B*c - A*d))*(e*x)^(3 + m))/(d^4*e^3*(3 + m)) + (b^2*(6*a^2*B*d^2 
+ b^2*c*(B*c - A*d) - 4*a*b*d*(B*c - A*d))*(e*x)^(5 + m))/(d^3*e^5*(5 + m) 
) - (b^3*(b*B*c - A*b*d - 4*a*B*d)*(e*x)^(7 + m))/(d^2*e^7*(7 + m)) + (b^4 
*B*(e*x)^(9 + m))/(d*e^9*(9 + m)) - ((b*c - a*d)^4*(B*c - A*d)*(e*x)^(1 + 
m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*d^5*e*(1 + 
 m))
 

3.1.22.3.1 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]
 

3.1.22.4 Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{4} \left (x^{2} B +A \right )}{d \,x^{2}+c}d x\]

input

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

output

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

3.1.22.5 Fricas [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^4 \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{4} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]

input

integrate((e*x)^m*(b*x^2+a)^4*(B*x^2+A)/(d*x^2+c),x, algorithm="fricas")
 

output

integral((B*b^4*x^10 + (4*B*a*b^3 + A*b^4)*x^8 + 2*(3*B*a^2*b^2 + 2*A*a*b^ 
3)*x^6 + A*a^4 + 2*(2*B*a^3*b + 3*A*a^2*b^2)*x^4 + (B*a^4 + 4*A*a^3*b)*x^2 
)*(e*x)^m/(d*x^2 + c), x)
 

3.1.22.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 13.39 (sec) , antiderivative size = 1102, normalized size of antiderivative = 3.04 \[ \int \frac {(e x)^m \left (a+b x^2\right )^4 \left (A+B x^2\right )}{c+d x^2} \, dx=\text {Too large to display} \]

input

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

output

A*a**4*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**4*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**3*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)/(c*gamma(m/2 + 5/2)) + 3*A*a**3*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)/(c*ga 
mma(m/2 + 5/2)) + 3*A*a**2*b**2*e**m*m*x**(m + 5)*lerchphi(d*x**2*exp_pola 
r(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(2*c*gamma(m/2 + 7/2)) + 15*A*a* 
*2*b**2*e**m*x**(m + 5)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*g 
amma(m/2 + 5/2)/(2*c*gamma(m/2 + 7/2)) + A*a*b**3*e**m*m*x**(m + 7)*lerchp 
hi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 7/2)*gamma(m/2 + 7/2)/(c*gamma(m/2 + 
 9/2)) + 7*A*a*b**3*e**m*x**(m + 7)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, 
m/2 + 7/2)*gamma(m/2 + 7/2)/(c*gamma(m/2 + 9/2)) + A*b**4*e**m*m*x**(m + 9 
)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 9/2)*gamma(m/2 + 9/2)/(4*c*g 
amma(m/2 + 11/2)) + 9*A*b**4*e**m*x**(m + 9)*lerchphi(d*x**2*exp_polar(I*p 
i)/c, 1, m/2 + 9/2)*gamma(m/2 + 9/2)/(4*c*gamma(m/2 + 11/2)) + B*a**4*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**4*e**m*x**(m + 3)*lerchphi(d*x**2*ex 
p_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2)) + B 
*a**3*b*e**m*m*x**(m + 5)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5...
 

3.1.22.7 Maxima [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^4 \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{4} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]

input

integrate((e*x)^m*(b*x^2+a)^4*(B*x^2+A)/(d*x^2+c),x, algorithm="maxima")
 

output

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

3.1.22.8 Giac [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^4 \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{4} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]

input

integrate((e*x)^m*(b*x^2+a)^4*(B*x^2+A)/(d*x^2+c),x, algorithm="giac")
 

output

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

3.1.22.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^4 \left (A+B x^2\right )}{c+d x^2} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^4}{d\,x^2+c} \,d x \]

input

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

output

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