Integrand size = 31, antiderivative size = 206 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right )}+\frac {(A b (b c (1-m)-a d (3-m))+a B (a d (1-m)+b c (1+m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^2 e (1+m)}-\frac {d (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (b c-a d)^2 e (1+m)} \]
1/2*(A*b-B*a)*(e*x)^(1+m)/a/(-a*d+b*c)/e/(b*x^2+a)+1/2*(A*b*(b*c*(1-m)-a*d *(3-m))+a*B*(a*d*(1-m)+b*c*(1+m)))*(e*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3 /2+1/2*m],-b*x^2/a)/a^2/(-a*d+b*c)^2/e/(1+m)-d*(-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/(-a*d+b*c)^2/e/(1+m)
Time = 0.36 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.72 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {x (e x)^m \left (a b c (-B c+A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^2 d (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )-(A b-a B) c (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{a^2 c (b c-a d)^2 (1+m)} \]
-((x*(e*x)^m*(a*b*c*(-(B*c) + A*d)*Hypergeometric2F1[1, (1 + m)/2, (3 + m) /2, -((b*x^2)/a)] + a^2*d*(B*c - A*d)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] - (A*b - a*B)*c*(b*c - a*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]))/(a^2*c*(b*c - a*d)^2*(1 + m)))
Time = 0.48 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {441, 446, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) (e x)^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 441 |
\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) (b c-a d)}-\frac {\int \frac {(e x)^m \left (-\left ((A b-a B) d (1-m) x^2\right )+2 a A d-A b c (1-m)-a B c (m+1)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 a (b c-a d)}\) |
\(\Big \downarrow \) 446 |
\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) (b c-a d)}-\frac {\int \left (\frac {(-A b (b c (1-m)-a d (3-m))-a B (a d (1-m)+b c (m+1))) (e x)^m}{(b c-a d) \left (b x^2+a\right )}+\frac {2 a d (A d-B c) (e x)^m}{(a d-b c) \left (d x^2+c\right )}\right )dx}{2 a (b c-a d)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) (b c-a d)}-\frac {\frac {2 a d (e x)^{m+1} (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c e (m+1) (b c-a d)}-\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right ) (A b (b c (1-m)-a d (3-m))+a B (a d (1-m)+b c (m+1)))}{a e (m+1) (b c-a d)}}{2 a (b c-a d)}\) |
((A*b - a*B)*(e*x)^(1 + m))/(2*a*(b*c - a*d)*e*(a + b*x^2)) - (-(((A*b*(b* c*(1 - m) - a*d*(3 - m)) + a*B*(a*d*(1 - m) + b*c*(1 + m)))*(e*x)^(1 + m)* Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(b*c - a*d)*e *(1 + m))) + (2*a*d*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m) /2, (3 + m)/2, -((d*x^2)/c)])/(c*(b*c - a*d)*e*(1 + m)))/(2*a*(b*c - a*d))
3.1.28.3.1 Defintions of rubi rules used
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*g*2*(b*c - a*d)*(p + 1))), x] + Si mp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2 )^q*Simp[c*(b*e - a*f)*(m + 1) + e*2*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, x] && LtQ[p, -1]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((e_) + (f_.)*(x_)^2))/( (c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^2)^ p*((e + f*x^2)/(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ]
\[\int \frac {\left (e x \right )^{m} \left (x^{2} B +A \right )}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )}d x\]
\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
integral((B*x^2 + A)*(e*x)^m/(b^2*d*x^6 + (b^2*c + 2*a*b*d)*x^4 + a^2*c + (2*a*b*c + a^2*d)*x^2), x)
Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m}{{\left (b\,x^2+a\right )}^2\,\left (d\,x^2+c\right )} \,d x \]