\(\int \frac {(e x)^m (A+B x^2) (c+d x^2)^3}{(a+b x^2)^2} \, dx\) [20]

Optimal result

Integrand size = 31, antiderivative size = 347 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^3}{\left (a+b x^2\right )^2} \, dx=-\frac {d \left (A b \left (3 b^2 c^2 (1+m)-3 a b c d (3+m)+a^2 d^2 (5+m)\right )-a B \left (3 b^2 c^2 (3+m)-3 a b c d (5+m)+a^2 d^2 (7+m)\right )\right ) (e x)^{1+m}}{2 a b^4 e (1+m)}-\frac {d^2 (A b (3 b c (3+m)-a d (5+m))-a B (3 b c (5+m)-a d (7+m))) (e x)^{3+m}}{2 a b^3 e^3 (3+m)}-\frac {d^3 (A b (5+m)-a B (7+m)) (e x)^{5+m}}{2 a b^2 e^5 (5+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{2 a b e \left (a+b x^2\right )}+\frac {(b c-a d)^2 (a B (b c (1+m)-a d (7+m))+A b (a d (5+m)+b (c-c m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 b^4 e (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 347, 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, 584, 371} \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^3}{\left (a+b x^2\right )^2} \, dx=\frac {(e x)^{m+1} (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right ) (A b (a d (m+5)+b (c-c m))+a B (b c (m+1)-a d (m+7)))}{2 a^2 b^4 e (m+1)}-\frac {d (e x)^{m+1} \left (A b \left (a^2 d^2 (m+5)-3 a b c d (m+3)+3 b^2 c^2 (m+1)\right )-a B \left (a^2 d^2 (m+7)-3 a b c d (m+5)+3 b^2 c^2 (m+3)\right )\right )}{2 a b^4 e (m+1)}-\frac {d^2 (e x)^{m+3} (A b (3 b c (m+3)-a d (m+5))-a B (3 b c (m+5)-a d (m+7)))}{2 a b^3 e^3 (m+3)}-\frac {d^3 (e x)^{m+5} (A b (m+5)-a B (m+7))}{2 a b^2 e^5 (m+5)}+\frac {\left (c+d x^2\right )^3 (e x)^{m+1} (A b-a B)}{2 a b e \left (a+b x^2\right )} \]

[In]

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

[Out]

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

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 {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{2 a b e \left (a+b x^2\right )}-\frac {\int \frac {(e x)^m \left (c+d x^2\right )^2 \left (-c (A b (1-m)+a B (1+m))+d (A b (5+m)-a B (7+m)) x^2\right )}{a+b x^2} \, dx}{2 a b} \\ & = \frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{2 a b e \left (a+b x^2\right )}-\frac {\int \left (\frac {d \left (A b \left (3 b^2 c^2 (1+m)-3 a b c d (3+m)+a^2 d^2 (5+m)\right )-a B \left (3 b^2 c^2 (3+m)-3 a b c d (5+m)+a^2 d^2 (7+m)\right )\right ) (e x)^m}{b^3}+\frac {d^2 (A b (3 b c (3+m)-a d (5+m))-a B (3 b c (5+m)-a d (7+m))) (e x)^{2+m}}{b^2 e^2}+\frac {d^3 (A b (5+m)-a B (7+m)) (e x)^{4+m}}{b e^4}+\frac {\left (-A b^4 c^3-a b^3 B c^3-3 a A b^3 c^2 d+9 a^2 b^2 B c^2 d+9 a^2 A b^2 c d^2-15 a^3 b B c d^2-5 a^3 A b d^3+7 a^4 B d^3+A b^4 c^3 m-a b^3 B c^3 m-3 a A b^3 c^2 d m+3 a^2 b^2 B c^2 d m+3 a^2 A b^2 c d^2 m-3 a^3 b B c d^2 m-a^3 A b d^3 m+a^4 B d^3 m\right ) (e x)^m}{b^3 \left (a+b x^2\right )}\right ) \, dx}{2 a b} \\ & = -\frac {d \left (A b \left (3 b^2 c^2 (1+m)-3 a b c d (3+m)+a^2 d^2 (5+m)\right )-a B \left (3 b^2 c^2 (3+m)-3 a b c d (5+m)+a^2 d^2 (7+m)\right )\right ) (e x)^{1+m}}{2 a b^4 e (1+m)}-\frac {d^2 (A b (3 b c (3+m)-a d (5+m))-a B (3 b c (5+m)-a d (7+m))) (e x)^{3+m}}{2 a b^3 e^3 (3+m)}-\frac {d^3 (A b (5+m)-a B (7+m)) (e x)^{5+m}}{2 a b^2 e^5 (5+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{2 a b e \left (a+b x^2\right )}-\frac {\left (-A b^4 c^3-a b^3 B c^3-3 a A b^3 c^2 d+9 a^2 b^2 B c^2 d+9 a^2 A b^2 c d^2-15 a^3 b B c d^2-5 a^3 A b d^3+7 a^4 B d^3+A b^4 c^3 m-a b^3 B c^3 m-3 a A b^3 c^2 d m+3 a^2 b^2 B c^2 d m+3 a^2 A b^2 c d^2 m-3 a^3 b B c d^2 m-a^3 A b d^3 m+a^4 B d^3 m\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{2 a b^4} \\ & = -\frac {d \left (A b \left (3 b^2 c^2 (1+m)-3 a b c d (3+m)+a^2 d^2 (5+m)\right )-a B \left (3 b^2 c^2 (3+m)-3 a b c d (5+m)+a^2 d^2 (7+m)\right )\right ) (e x)^{1+m}}{2 a b^4 e (1+m)}-\frac {d^2 (A b (3 b c (3+m)-a d (5+m))-a B (3 b c (5+m)-a d (7+m))) (e x)^{3+m}}{2 a b^3 e^3 (3+m)}-\frac {d^3 (A b (5+m)-a B (7+m)) (e x)^{5+m}}{2 a b^2 e^5 (5+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{2 a b e \left (a+b x^2\right )}+\frac {(b c-a d)^2 (A b (b c (1-m)+a d (5+m))+a B (b c (1+m)-a d (7+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 b^4 e (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.60 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^3}{\left (a+b x^2\right )^2} \, dx=\frac {x (e x)^m \left (\frac {d \left (3 a^2 B d^2+3 b^2 c (B c+A d)-2 a b d (3 B c+A d)\right )}{1+m}+\frac {b d^2 (3 b B c+A b d-2 a B d) x^2}{3+m}+\frac {b^2 B d^3 x^4}{5+m}+\frac {(b c-a d)^2 (b B c+3 A b d-4 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a (1+m)}+\frac {(-A b+a B) (-b c+a d)^3 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2 (1+m)}\right )}{b^4} \]

[In]

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

[Out]

(x*(e*x)^m*((d*(3*a^2*B*d^2 + 3*b^2*c*(B*c + A*d) - 2*a*b*d*(3*B*c + A*d)))/(1 + m) + (b*d^2*(3*b*B*c + A*b*d
- 2*a*B*d)*x^2)/(3 + m) + (b^2*B*d^3*x^4)/(5 + m) + ((b*c - a*d)^2*(b*B*c + 3*A*b*d - 4*a*B*d)*Hypergeometric2
F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(1 + m)) + ((-(A*b) + a*B)*(-(b*c) + a*d)^3*Hypergeometric2F1[2,
(1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a^2*(1 + m))))/b^4

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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