Integrand size = 22, antiderivative size = 120 \[ \int \frac {x^m \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {d^2 x^{1+m}}{b^2 (1+m)}+\frac {(b c-a d)^2 x^{1+m}}{2 a b^2 \left (a+b x^2\right )}+\frac {(b c-a d) (a d (3+m)+b (c-c m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 b^2 (1+m)} \]
d^2*x^(1+m)/b^2/(1+m)+1/2*(-a*d+b*c)^2*x^(1+m)/a/b^2/(b*x^2+a)+1/2*(-a*d+b *c)*(a*d*(3+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/b^2/(1+m)
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 1.62 (sec) , antiderivative size = 895, normalized size of antiderivative = 7.46 \[ \int \frac {x^m \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {x^{1+m} \left (a \left (105+71 m+15 m^2+m^3\right ) \left (c^2 \left (9-5 m+3 m^2+m^3\right )+2 c d (1+m)^3 x^2+d^2 (1+m)^3 x^4\right ) \Phi \left (-\frac {b x^2}{a},1,\frac {1+m}{2}\right )-2 a \left (105+71 m+15 m^2+m^3\right ) \left (c^2 (3+m)^3+2 c d \left (31+31 m+9 m^2+m^3\right ) x^2+d^2 (3+m)^3 x^4\right ) \Phi \left (-\frac {b x^2}{a},1,\frac {3+m}{2}\right )+13125 a c^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+16750 a c^2 m \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+8775 a c^2 m^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+2420 a c^2 m^3 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+371 a c^2 m^4 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+30 a c^2 m^5 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+a c^2 m^6 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+26250 a c d x^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+33500 a c d m x^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+17550 a c d m^2 x^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+4840 a c d m^3 x^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+742 a c d m^4 x^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+60 a c d m^5 x^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+2 a c d m^6 x^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+10605 a d^2 x^4 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+14206 a d^2 m x^4 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+7847 a d^2 m^2 x^4 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+2276 a d^2 m^3 x^4 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+363 a d^2 m^4 x^4 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+30 a d^2 m^5 x^4 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )+a d^2 m^6 x^4 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )-128 b c^2 x^2 \, _4F_3\left (2,2,2,\frac {3}{2}+\frac {m}{2};1,1,\frac {9}{2}+\frac {m}{2};-\frac {b x^2}{a}\right )-256 b c d x^4 \, _4F_3\left (2,2,2,\frac {3}{2}+\frac {m}{2};1,1,\frac {9}{2}+\frac {m}{2};-\frac {b x^2}{a}\right )-128 b d^2 x^6 \, _4F_3\left (2,2,2,\frac {3}{2}+\frac {m}{2};1,1,\frac {9}{2}+\frac {m}{2};-\frac {b x^2}{a}\right )\right )}{32 a^3 (3+m) (5+m) (7+m)} \]
(x^(1 + m)*(a*(105 + 71*m + 15*m^2 + m^3)*(c^2*(9 - 5*m + 3*m^2 + m^3) + 2 *c*d*(1 + m)^3*x^2 + d^2*(1 + m)^3*x^4)*HurwitzLerchPhi[-((b*x^2)/a), 1, ( 1 + m)/2] - 2*a*(105 + 71*m + 15*m^2 + m^3)*(c^2*(3 + m)^3 + 2*c*d*(31 + 3 1*m + 9*m^2 + m^3)*x^2 + d^2*(3 + m)^3*x^4)*HurwitzLerchPhi[-((b*x^2)/a), 1, (3 + m)/2] + 13125*a*c^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 16750*a*c^2*m*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 8775*a*c^2*m^2 *HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 2420*a*c^2*m^3*HurwitzLerch Phi[-((b*x^2)/a), 1, (5 + m)/2] + 371*a*c^2*m^4*HurwitzLerchPhi[-((b*x^2)/ a), 1, (5 + m)/2] + 30*a*c^2*m^5*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/ 2] + a*c^2*m^6*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 26250*a*c*d*x ^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 33500*a*c*d*m*x^2*Hurwitz LerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 17550*a*c*d*m^2*x^2*HurwitzLerchPhi [-((b*x^2)/a), 1, (5 + m)/2] + 4840*a*c*d*m^3*x^2*HurwitzLerchPhi[-((b*x^2 )/a), 1, (5 + m)/2] + 742*a*c*d*m^4*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, ( 5 + m)/2] + 60*a*c*d*m^5*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 2*a*c*d*m^6*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 10605*a*d^2 *x^4*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 14206*a*d^2*m*x^4*Hurwi tzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 7847*a*d^2*m^2*x^4*HurwitzLerchPh i[-((b*x^2)/a), 1, (5 + m)/2] + 2276*a*d^2*m^3*x^4*HurwitzLerchPhi[-((b*x^ 2)/a), 1, (5 + m)/2] + 363*a*d^2*m^4*x^4*HurwitzLerchPhi[-((b*x^2)/a), ...
Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {366, 25, 363, 278}
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.
| \(\displaystyle \int \frac {x^m \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 366 |
| \(\displaystyle \frac {x^{m+1} (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}-\frac {\int -\frac {x^m \left (2 b^2 c^2+2 a b d^2 x^2-(b c-a d)^2 (m+1)\right )}{b x^2+a}dx}{2 a b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x^m \left (2 b^2 c^2+2 a b d^2 x^2-(b c-a d)^2 (m+1)\right )}{b x^2+a}dx}{2 a b^2}+\frac {x^{m+1} (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(b c-a d) (a d (m+3)+b c (1-m)) \int \frac {x^m}{b x^2+a}dx+\frac {2 a d^2 x^{m+1}}{m+1}}{2 a b^2}+\frac {x^{m+1} (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {\frac {x^{m+1} (b c-a d) (a d (m+3)+b c (1-m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a (m+1)}+\frac {2 a d^2 x^{m+1}}{m+1}}{2 a b^2}+\frac {x^{m+1} (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}\) |
((b*c - a*d)^2*x^(1 + m))/(2*a*b^2*(a + b*x^2)) + ((2*a*d^2*x^(1 + m))/(1 + m) + ((b*c - a*d)*(b*c*(1 - m) + a*d*(3 + m))*x^(1 + m)*Hypergeometric2F 1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(1 + m)))/(2*a*b^2)
3.4.38.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
\[\int \frac {x^{m} \left (d \,x^{2}+c \right )^{2}}{\left (b \,x^{2}+a \right )^{2}}d x\]
\[ \int \frac {x^m \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{2} x^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \]
\[ \int \frac {x^m \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^{m} \left (c + d x^{2}\right )^{2}}{\left (a + b x^{2}\right )^{2}}\, dx \]
\[ \int \frac {x^m \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{2} x^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \]
\[ \int \frac {x^m \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{2} x^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^m \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^m\,{\left (d\,x^2+c\right )}^2}{{\left (b\,x^2+a\right )}^2} \,d x \]