\(\int \frac {(e x)^m (a+b x^2)^2 (A+B x^2)}{(c+d x^2)^3} \, dx\) [38]

Optimal result

Integrand size = 31, antiderivative size = 292 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) (e x)^{1+m}}{8 c^2 d^3 e (1+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {(a d (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))-b c (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 d^3 e (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {591, 470, 371} \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right ) (a d (a d (1-m)+b c (m+1)) (A d (3-m)+B c (m+1))-b c (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5)))}{8 c^3 d^3 e (m+1)}+\frac {b (e x)^{m+1} (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5))}{8 c^2 d^3 e (m+1)}-\frac {(e x)^{m+1} (b c-a d) \left (a (A d (3-m)+B c (m+1))-b x^2 (A d (m+1)-B c (m+5))\right )}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac {\left (a+b x^2\right )^2 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]

[In]

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

[Out]

(b*(a*d*(1 + m) - b*c*(3 + m))*(A*d*(1 + m) - B*c*(5 + m))*(e*x)^(1 + m))/(8*c^2*d^3*e*(1 + m)) - ((B*c - A*d)
*(e*x)^(1 + m)*(a + b*x^2)^2)/(4*c*d*e*(c + d*x^2)^2) - ((b*c - a*d)*(e*x)^(1 + m)*(a*(A*d*(3 - m) + B*c*(1 +
m)) - b*(A*d*(1 + m) - B*c*(5 + m))*x^2))/(8*c^2*d^2*e*(c + d*x^2)) + ((a*d*(a*d*(1 - m) + b*c*(1 + m))*(A*d*(
3 - m) + B*c*(1 + m)) - b*c*(a*d*(1 + m) - b*c*(3 + m))*(A*d*(1 + m) - B*c*(5 + m)))*(e*x)^(1 + m)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*d^3*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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 591

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*g*n*(p + 1))), x] + Dis
t[1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(
m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

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

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.58 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {x (e x)^m \left (b^2 B-\frac {b (3 b B c-A b d-2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c}+\frac {(b c-a d) (3 b B c-2 A b d-a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^2}-\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^3}\right )}{d^3 (1+m)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{2} \left (x^{2} B +A \right )}{\left (d \,x^{2}+c \right )^{3}}d x\]

[In]

int((e*x)^m*(b*x^2+a)^2*(B*x^2+A)/(d*x^2+c)^3,x)

[Out]

int((e*x)^m*(b*x^2+a)^2*(B*x^2+A)/(d*x^2+c)^3,x)

Fricas [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]

[In]

integrate((e*x)^m*(b*x^2+a)^2*(B*x^2+A)/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

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^3*x^6 + 3*c*d^2*x^4 +
3*c^2*d*x^2 + c^3), x)

Sympy [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int \frac {\left (e x\right )^{m} \left (A + B x^{2}\right ) \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{3}}\, dx \]

[In]

integrate((e*x)**m*(b*x**2+a)**2*(B*x**2+A)/(d*x**2+c)**3,x)

[Out]

Integral((e*x)**m*(A + B*x**2)*(a + b*x**2)**2/(c + d*x**2)**3, x)

Maxima [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]

[In]

integrate((e*x)^m*(b*x^2+a)^2*(B*x^2+A)/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^2*(e*x)^m/(d*x^2 + c)^3, x)

Giac [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]

[In]

integrate((e*x)^m*(b*x^2+a)^2*(B*x^2+A)/(d*x^2+c)^3,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^2*(e*x)^m/(d*x^2 + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^3} \,d x \]

[In]

int(((A + B*x^2)*(e*x)^m*(a + b*x^2)^2)/(c + d*x^2)^3,x)

[Out]

int(((A + B*x^2)*(e*x)^m*(a + b*x^2)^2)/(c + d*x^2)^3, x)