Integrand size = 29, antiderivative size = 208 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=-\frac {(b c-a d) (e x)^{1+m} \left (A+B x^2\right )}{4 c d e \left (c+d x^2\right )^2}+\frac {(b c (A d (1+m)-B c (3+m))+a d (A d (3-m)-B (c-c m))) (e x)^{1+m}}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {(a d (1-m) (A d (3-m)+B c (1+m))+b c (1+m) (A d (1-m)+B c (3+m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 d^2 e (1+m)} \]
-1/4*(-a*d+b*c)*(e*x)^(1+m)*(B*x^2+A)/c/d/e/(d*x^2+c)^2+1/8*(b*c*(A*d*(1+m )-B*c*(3+m))+a*d*(A*d*(3-m)-B*(-c*m+c)))*(e*x)^(1+m)/c^2/d^2/e/(d*x^2+c)+1 /8*(a*d*(1-m)*(A*d*(3-m)+B*c*(1+m))+b*c*(1+m)*(A*d*(1-m)+B*c*(3+m)))*(e*x) ^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c^3/d^2/e/(1+m)
Time = 0.43 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.64 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {x (e x)^m \left (b B c^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+c (-2 b B c+A b d+a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+(b c-a d) (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )}{c^3 d^2 (1+m)} \]
(x*(e*x)^m*(b*B*c^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c )] + c*(-2*b*B*c + A*b*d + a*B*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/ 2, -((d*x^2)/c)] + (b*c - a*d)*(B*c - A*d)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)]))/(c^3*d^2*(1 + m))
Time = 0.40 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {439, 25, 362, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) (e x)^m}{\left (c+d x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 439 |
| \(\displaystyle -\frac {\int -\frac {(e x)^m \left (B (a d (1-m)+b c (m+3)) x^2+A (a d (3-m)+b c (m+1))\right )}{\left (d x^2+c\right )^2}dx}{4 c d}-\frac {\left (A+B x^2\right ) (e x)^{m+1} (b c-a d)}{4 c d e \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(e x)^m \left (B (a d (1-m)+b c (m+3)) x^2+A (a d (3-m)+b c (m+1))\right )}{\left (d x^2+c\right )^2}dx}{4 c d}-\frac {\left (A+B x^2\right ) (e x)^{m+1} (b c-a d)}{4 c d e \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 362 |
| \(\displaystyle \frac {\frac {(a d (1-m) (A d (3-m)+B c (m+1))+b c (m+1) (A d (1-m)+B c (m+3))) \int \frac {(e x)^m}{d x^2+c}dx}{2 c d}+\frac {(e x)^{m+1} (a d (A d (3-m)-B (c-c m))+b c (A d (m+1)-B c (m+3)))}{2 c d e \left (c+d x^2\right )}}{4 c d}-\frac {\left (A+B x^2\right ) (e x)^{m+1} (b c-a d)}{4 c d e \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right ) (a d (1-m) (A d (3-m)+B c (m+1))+b c (m+1) (A d (1-m)+B c (m+3)))}{2 c^2 d e (m+1)}+\frac {(e x)^{m+1} (a d (A d (3-m)-B (c-c m))+b c (A d (m+1)-B c (m+3)))}{2 c d e \left (c+d x^2\right )}}{4 c d}-\frac {\left (A+B x^2\right ) (e x)^{m+1} (b c-a d)}{4 c d e \left (c+d x^2\right )^2}\) |
-1/4*((b*c - a*d)*(e*x)^(1 + m)*(A + B*x^2))/(c*d*e*(c + d*x^2)^2) + (((b* c*(A*d*(1 + m) - B*c*(3 + m)) + a*d*(A*d*(3 - m) - B*(c - c*m)))*(e*x)^(1 + m))/(2*c*d*e*(c + d*x^2)) + ((a*d*(1 - m)*(A*d*(3 - m) + B*c*(1 + m)) + b*c*(1 + m)*(A*d*(1 - m) + B*c*(3 + m)))*(e*x)^(1 + m)*Hypergeometric2F1[1 , (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(2*c^2*d*e*(1 + m)))/(4*c*d)
3.1.39.3.1 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[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e *(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1)) I nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
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/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 ))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G tQ[q, 0] && !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right ) \left (x^{2} B +A \right )}{\left (d \,x^{2}+c \right )^{3}}d x\]
\[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
integral((B*b*x^4 + (B*a + A*b)*x^2 + A*a)*(e*x)^m/(d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^3), x)
Result contains complex when optimal does not.
Time = 94.11 (sec) , antiderivative size = 6411, normalized size of antiderivative = 30.82 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\text {Too large to display} \]
A*a*(c**2*e**m*m**3*x**(m + 1)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*c**5*gamma(m/2 + 3/2) + 64*c**4*d*x**2*gamma(m/ 2 + 3/2) + 32*c**3*d**2*x**4*gamma(m/2 + 3/2)) - 3*c**2*e**m*m**2*x**(m + 1)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*c **5*gamma(m/2 + 3/2) + 64*c**4*d*x**2*gamma(m/2 + 3/2) + 32*c**3*d**2*x**4 *gamma(m/2 + 3/2)) - 2*c**2*e**m*m**2*x**(m + 1)*gamma(m/2 + 1/2)/(32*c**5 *gamma(m/2 + 3/2) + 64*c**4*d*x**2*gamma(m/2 + 3/2) + 32*c**3*d**2*x**4*ga mma(m/2 + 3/2)) - c**2*e**m*m*x**(m + 1)*lerchphi(d*x**2*exp_polar(I*pi)/c , 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*c**5*gamma(m/2 + 3/2) + 64*c**4*d*x** 2*gamma(m/2 + 3/2) + 32*c**3*d**2*x**4*gamma(m/2 + 3/2)) + 8*c**2*e**m*m*x **(m + 1)*gamma(m/2 + 1/2)/(32*c**5*gamma(m/2 + 3/2) + 64*c**4*d*x**2*gamm a(m/2 + 3/2) + 32*c**3*d**2*x**4*gamma(m/2 + 3/2)) + 3*c**2*e**m*x**(m + 1 )*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*c* *5*gamma(m/2 + 3/2) + 64*c**4*d*x**2*gamma(m/2 + 3/2) + 32*c**3*d**2*x**4* gamma(m/2 + 3/2)) + 10*c**2*e**m*x**(m + 1)*gamma(m/2 + 1/2)/(32*c**5*gamm a(m/2 + 3/2) + 64*c**4*d*x**2*gamma(m/2 + 3/2) + 32*c**3*d**2*x**4*gamma(m /2 + 3/2)) + 2*c*d*e**m*m**3*x**2*x**(m + 1)*lerchphi(d*x**2*exp_polar(I*p i)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*c**5*gamma(m/2 + 3/2) + 64*c**4*d *x**2*gamma(m/2 + 3/2) + 32*c**3*d**2*x**4*gamma(m/2 + 3/2)) - 6*c*d*e**m* m**2*x**2*x**(m + 1)*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*g...
\[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
\[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,\left (b\,x^2+a\right )}{{\left (d\,x^2+c\right )}^3} \,d x \]