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

Optimal result

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

[Out]

(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)*hype
rgeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c/d^5/e/(1+m)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {584, 371} \[ \int \frac {(e x)^m \left (a+b x^2\right )^4 \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {b^2 (e x)^{m+5} \left (6 a^2 B d^2-4 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e^5 (m+5)}+\frac {b (e x)^{m+3} \left (4 a^3 B d^3-6 a^2 b d^2 (B c-A d)+4 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e^3 (m+3)}+\frac {(e x)^{m+1} \left (a^4 B d^4-4 a^3 b d^3 (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a b^3 c^2 d (B c-A d)+b^4 c^3 (B c-A d)\right )}{d^5 e (m+1)}-\frac {b^3 (e x)^{m+7} (-4 a B d-A b d+b B c)}{d^2 e^7 (m+7)}-\frac {(e x)^{m+1} (b c-a d)^4 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c d^5 e (m+1)}+\frac {b^4 B (e x)^{m+9}}{d e^9 (m+9)} \]

[In]

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

[Out]

((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))

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 {\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)^m}{d^5}+\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)^{2+m}}{d^4 e^2}+\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)^{4+m}}{d^3 e^4}-\frac {b^3 (b B c-A b d-4 a B d) (e x)^{6+m}}{d^2 e^6}+\frac {b^4 B (e x)^{8+m}}{d e^8}+\frac {\left (-b^4 B c^5+A b^4 c^4 d+4 a b^3 B c^4 d-4 a A b^3 c^3 d^2-6 a^2 b^2 B c^3 d^2+6 a^2 A b^2 c^2 d^3+4 a^3 b B c^2 d^3-4 a^3 A b c d^4-a^4 B c d^4+a^4 A d^5\right ) (e x)^m}{d^5 \left (c+d x^2\right )}\right ) \, 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 {\left ((b c-a d)^4 (B c-A d)\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{d^5} \\ & = \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} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{c d^5 e (1+m)} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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)

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

[In]

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

[Out]

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*gamm
a(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*gamma(m/2 + 5/2)) + 3*A*a**2*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)/(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)*gamma(m/2 + 5/2)/(2*c*gamma(m/2 + 7/2)) + A*a*b**3*e**m*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)) + 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*gamma(m/2 + 11/2)) + 9*A*b**4*e**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*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)*lerchp
hi(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**3*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)/(c*gamma(m/2 + 7/2)) + 5*B*a**3*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)/(c*gamma(m/2 + 7/2)) + 3*B*a**2*b**2
*e**m*m*x**(m + 7)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 7/2)*gamma(m/2 + 7/2)/(2*c*gamma(m/2 + 9/2)) +
21*B*a**2*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)/(2*c*gamma(m/
2 + 9/2)) + B*a*b**3*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)/(c*ga
mma(m/2 + 11/2)) + 9*B*a*b**3*e**m*x**(m + 9)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 9/2)*gamma(m/2 + 9/2
)/(c*gamma(m/2 + 11/2)) + B*b**4*e**m*m*x**(m + 11)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 11/2)*gamma(m/
2 + 11/2)/(4*c*gamma(m/2 + 13/2)) + 11*B*b**4*e**m*x**(m + 11)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 11/
2)*gamma(m/2 + 11/2)/(4*c*gamma(m/2 + 13/2))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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