Integrand size = 22, antiderivative size = 156 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right )}+\frac {b (b c (1-m)-a d (3-m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^2 (1+m)}+\frac {d^2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (b c-a d)^2 (1+m)} \]
1/2*b*x^(1+m)/a/(-a*d+b*c)/(b*x^2+a)+1/2*b*(b*c*(1-m)-a*d*(3-m))*x^(1+m)*h ypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a^2/(-a*d+b*c)^2/(1+m)+d^2*x ^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c/(-a*d+b*c)^2/(1+m)
Time = 0.06 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {x^{1+m} \left (-a b c d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^2 d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+b c (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{a^2 c (b c-a d)^2 (1+m)} \]
(x^(1 + m)*(-(a*b*c*d*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2) /a)]) + a^2*d^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] + 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 = 0.33 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {374, 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.
\(\displaystyle \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 374 |
\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)}-\frac {\int \frac {x^m \left (-b d (1-m) x^2+2 a d-b c (1-m)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 a (b c-a d)}\) |
\(\Big \downarrow \) 446 |
\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)}-\frac {\int \left (\frac {b (a d (3-m)-b c (1-m)) x^m}{(b c-a d) \left (b x^2+a\right )}+\frac {2 a d^2 x^m}{(a d-b c) \left (d x^2+c\right )}\right )dx}{2 a (b c-a d)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)}-\frac {\frac {b x^{m+1} (a d (3-m)-b (c-c m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a (m+1) (b c-a d)}-\frac {2 a d^2 x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)}}{2 a (b c-a d)}\) |
(b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)) - ((b*(a*d*(3 - m) - b*(c - c* m))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a *(b*c - a*d)*(1 + m)) - (2*a*d^2*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(b*c - a*d)*(1 + m)))/(2*a*(b*c - a*d))
3.4.40.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
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 ]
\[\int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )}d x\]
\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int \frac {x^m}{{\left (b\,x^2+a\right )}^2\,\left (d\,x^2+c\right )} \,d x \]