Integrand size = 22, antiderivative size = 325 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac {b^3 (a d (7-m)-b (c-c m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^4 (1+m)}+\frac {d^2 \left (b^2 c^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )\right ) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^4 (1+m)} \]
1/4*d*(a*d+2*b*c)*x^(1+m)/a/c/(-a*d+b*c)^2/(d*x^2+c)^2+1/2*b*x^(1+m)/a/(-a *d+b*c)/(b*x^2+a)/(d*x^2+c)^2+1/8*d*(4*b^2*c^2-a^2*d^2*(3-m)+a*b*c*d*(11-m ))*x^(1+m)/a/c^2/(-a*d+b*c)^3/(d*x^2+c)-1/2*b^3*(a*d*(7-m)-b*(-c*m+c))*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)^4/(1+m) +1/8*d^2*(b^2*c^2*(m^2-12*m+35)-2*a*b*c*d*(m^2-8*m+7)+a^2*d^2*(m^2-4*m+3)) *x^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c^3/(-a*d+b*c)^4/( 1+m)
Time = 0.75 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.66 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {x^{1+m} \left (-3 a b^3 c^3 d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+3 a^2 b^2 c^2 d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+(b c-a d) \left (b^3 c^3 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^2 d^2 \left (2 b c \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+(b c-a d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )\right )\right )}{a^2 c^3 (b c-a d)^4 (1+m)} \]
(x^(1 + m)*(-3*a*b^3*c^3*d*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b *x^2)/a)] + 3*a^2*b^2*c^2*d^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, - ((d*x^2)/c)] + (b*c - a*d)*(b^3*c^3*Hypergeometric2F1[2, (1 + m)/2, (3 + m )/2, -((b*x^2)/a)] + a^2*d^2*(2*b*c*Hypergeometric2F1[2, (1 + m)/2, (3 + m )/2, -((d*x^2)/c)] + (b*c - a*d)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2 , -((d*x^2)/c)]))))/(a^2*c^3*(b*c - a*d)^4*(1 + m))
Time = 0.75 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {374, 441, 27, 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.
| \(\displaystyle \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 374 |
| \(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\int \frac {x^m \left (-b d (5-m) x^2+2 a d-b c (1-m)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )^3}dx}{2 a (b c-a d)}\) |
| \(\Big \downarrow \) 441 |
| \(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\int \frac {2 x^m \left (-2 b^2 (1-m) c^2+8 a b d c-b d (2 b c+a d) (3-m) x^2-a^2 d^2 (3-m)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}-\frac {d x^{m+1} (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\int \frac {x^m \left (-2 b^2 (1-m) c^2+8 a b d c-b d (2 b c+a d) (3-m) x^2-a^2 d^2 (3-m)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{2 c (b c-a d)}-\frac {d x^{m+1} (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\) |
| \(\Big \downarrow \) 441 |
| \(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\frac {\int \frac {x^m \left (-4 b^3 (1-m) c^3+24 a b^2 d c^2-a^2 b d^2 \left (m^2-12 m+11\right ) c-b d \left (4 b^2 c^2+a b d (11-m) c-a^2 d^2 (3-m)\right ) (1-m) x^2+a^3 d^3 \left (m^2-4 m+3\right )\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {d x^{m+1} \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )}{2 c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d x^{m+1} (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\) |
| \(\Big \downarrow \) 446 |
| \(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\frac {\int \left (\frac {4 b^3 c^2 (a d (7-m)-b c (1-m)) x^m}{(b c-a d) \left (b x^2+a\right )}+\frac {a d^2 \left (-b^2 \left (m^2-12 m+35\right ) c^2+2 a b d \left (m^2-8 m+7\right ) c-a^2 d^2 \left (m^2-4 m+3\right )\right ) x^m}{(b c-a d) \left (d x^2+c\right )}\right )dx}{2 c (b c-a d)}-\frac {d x^{m+1} \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )}{2 c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d x^{m+1} (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\frac {\frac {4 b^3 c^2 x^{m+1} (a d (7-m)-b (c-c m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a (m+1) (b c-a d)}-\frac {a d^2 x^{m+1} \left (a^2 d^2 \left (m^2-4 m+3\right )-2 a b c d \left (m^2-8 m+7\right )+b^2 c^2 \left (m^2-12 m+35\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)}}{2 c (b c-a d)}-\frac {d x^{m+1} \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )}{2 c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d x^{m+1} (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\) |
(b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^2) - (-1/2*(d*(2*b* c + a*d)*x^(1 + m))/(c*(b*c - a*d)*(c + d*x^2)^2) + (-1/2*(d*(4*b^2*c^2 - a^2*d^2*(3 - m) + a*b*c*d*(11 - m))*x^(1 + m))/(c*(b*c - a*d)*(c + d*x^2)) + ((4*b^3*c^2*(a*d*(7 - 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)) - (a*d^2*(b^2 *c^2*(35 - 12*m + m^2) - 2*a*b*c*d*(7 - 8*m + m^2) + a^2*d^2*(3 - 4*m + m^ 2))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c *(b*c - a*d)*(1 + m)))/(2*c*(b*c - a*d)))/(2*c*(b*c - a*d)))/(2*a*(b*c - a *d))
3.4.42.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)*((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 ]
\[\int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )^{3}}d x\]
\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
integral(x^m/(b^2*d^3*x^10 + (3*b^2*c*d^2 + 2*a*b*d^3)*x^8 + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 + a^2*c^3 + (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c* d^2)*x^4 + (2*a*b*c^3 + 3*a^2*c^2*d)*x^2), x)
Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\int \frac {x^m}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^3} \,d x \]