Integrand size = 22, antiderivative size = 171 \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {(b c-a d)^2 x^{1+m}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (a d (3-m)+b c (5+m)) x^{1+m}}{8 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (2 a b c d \left (1-m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )+b^2 c^2 \left (3+4 m+m^2\right )\right ) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 d^2 (1+m)} \] Output:
1/4*(-a*d+b*c)^2*x^(1+m)/c/d^2/(d*x^2+c)^2-1/8*(-a*d+b*c)*(a*d*(3-m)+b*c*( 5+m))*x^(1+m)/c^2/d^2/(d*x^2+c)+1/8*(2*a*b*c*d*(-m^2+1)+a^2*d^2*(m^2-4*m+3 )+b^2*c^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/d^2/(1+m)
Time = 0.61 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {x^{1+m} \left (b^2 c^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )-(b c-a d) \left (2 b c \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+(-b c+a d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )\right )}{c^3 d^2 (1+m)} \] Input:
Integrate[(x^m*(a + b*x^2)^2)/(c + d*x^2)^3,x]
Output:
(x^(1 + m)*(b^2*c^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c )] - (b*c - a*d)*(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)])))/(c^3*d^2*(1 + m))
Time = 0.31 (sec) , antiderivative size = 176, 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, 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.
| \(\displaystyle \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 366 |
| \(\displaystyle \frac {x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}-\frac {\int -\frac {x^m \left (4 a^2 d^2+4 b^2 c x^2 d-(b c-a d)^2 (m+1)\right )}{\left (d x^2+c\right )^2}dx}{4 c d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x^m \left (4 a^2 d^2+4 b^2 c x^2 d-(b c-a d)^2 (m+1)\right )}{\left (d x^2+c\right )^2}dx}{4 c d^2}+\frac {x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 362 |
| \(\displaystyle \frac {\frac {\left (a^2 d^2 \left (m^2-4 m+3\right )+2 a b c d \left (1-m^2\right )+b^2 c^2 \left (m^2+4 m+3\right )\right ) \int \frac {x^m}{d x^2+c}dx}{2 c}-\frac {x^{m+1} (b c-a d) (a d (3-m)+b c (m+5))}{2 c \left (c+d x^2\right )}}{4 c d^2}+\frac {x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {\frac {x^{m+1} \left (a^2 d^2 \left (m^2-4 m+3\right )+2 a b c d \left (1-m^2\right )+b^2 c^2 \left (m^2+4 m+3\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{2 c^2 (m+1)}-\frac {x^{m+1} (b c-a d) (a d (3-m)+b c (m+5))}{2 c \left (c+d x^2\right )}}{4 c d^2}+\frac {x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\) |
Input:
Int[(x^m*(a + b*x^2)^2)/(c + d*x^2)^3,x]
Output:
((b*c - a*d)^2*x^(1 + m))/(4*c*d^2*(c + d*x^2)^2) + (-1/2*((b*c - a*d)*(a* d*(3 - m) + b*c*(5 + m))*x^(1 + m))/(c*(c + d*x^2)) + ((2*a*b*c*d*(1 - m^2 ) + a^2*d^2*(3 - 4*m + m^2) + b^2*c^2*(3 + 4*m + m^2))*x^(1 + m)*Hypergeom etric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(2*c^2*(1 + m)))/(4*c*d^2 )
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[((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]
\[\int \frac {x^{m} \left (b \,x^{2}+a \right )^{2}}{\left (x^{2} d +c \right )^{3}}d x\]
Input:
int(x^m*(b*x^2+a)^2/(d*x^2+c)^3,x)
Output:
int(x^m*(b*x^2+a)^2/(d*x^2+c)^3,x)
\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \] Input:
integrate(x^m*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="fricas")
Output:
integral((b^2*x^4 + 2*a*b*x^2 + a^2)*x^m/(d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d* x^2 + c^3), x)
\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int \frac {x^{m} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{3}}\, dx \] Input:
integrate(x**m*(b*x**2+a)**2/(d*x**2+c)**3,x)
Output:
Integral(x**m*(a + b*x**2)**2/(c + d*x**2)**3, x)
\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \] Input:
integrate(x^m*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^2*x^m/(d*x^2 + c)^3, x)
\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \] Input:
integrate(x^m*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^2*x^m/(d*x^2 + c)^3, x)
Timed out. \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int \frac {x^m\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^3} \,d x \] Input:
int((x^m*(a + b*x^2)^2)/(c + d*x^2)^3,x)
Output:
int((x^m*(a + b*x^2)^2)/(c + d*x^2)^3, x)
\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\text {too large to display} \] Input:
int(x^m*(b*x^2+a)^2/(d*x^2+c)^3,x)
Output:
(2*x**m*a*b*d*m*x - 2*x**m*a*b*d*x - x**m*b**2*c*m*x - 3*x**m*b**2*c*x + x **m*b**2*d*m*x**3 - 3*x**m*b**2*d*x**3 + int(x**m/(c**3*m**2 - 4*c**3*m + 3*c**3 + 3*c**2*d*m**2*x**2 - 12*c**2*d*m*x**2 + 9*c**2*d*x**2 + 3*c*d**2* m**2*x**4 - 12*c*d**2*m*x**4 + 9*c*d**2*x**4 + d**3*m**2*x**6 - 4*d**3*m*x **6 + 3*d**3*x**6),x)*a**2*c**2*d**2*m**4 - 8*int(x**m/(c**3*m**2 - 4*c**3 *m + 3*c**3 + 3*c**2*d*m**2*x**2 - 12*c**2*d*m*x**2 + 9*c**2*d*x**2 + 3*c* d**2*m**2*x**4 - 12*c*d**2*m*x**4 + 9*c*d**2*x**4 + d**3*m**2*x**6 - 4*d** 3*m*x**6 + 3*d**3*x**6),x)*a**2*c**2*d**2*m**3 + 22*int(x**m/(c**3*m**2 - 4*c**3*m + 3*c**3 + 3*c**2*d*m**2*x**2 - 12*c**2*d*m*x**2 + 9*c**2*d*x**2 + 3*c*d**2*m**2*x**4 - 12*c*d**2*m*x**4 + 9*c*d**2*x**4 + d**3*m**2*x**6 - 4*d**3*m*x**6 + 3*d**3*x**6),x)*a**2*c**2*d**2*m**2 - 24*int(x**m/(c**3*m **2 - 4*c**3*m + 3*c**3 + 3*c**2*d*m**2*x**2 - 12*c**2*d*m*x**2 + 9*c**2*d *x**2 + 3*c*d**2*m**2*x**4 - 12*c*d**2*m*x**4 + 9*c*d**2*x**4 + d**3*m**2* x**6 - 4*d**3*m*x**6 + 3*d**3*x**6),x)*a**2*c**2*d**2*m + 9*int(x**m/(c**3 *m**2 - 4*c**3*m + 3*c**3 + 3*c**2*d*m**2*x**2 - 12*c**2*d*m*x**2 + 9*c**2 *d*x**2 + 3*c*d**2*m**2*x**4 - 12*c*d**2*m*x**4 + 9*c*d**2*x**4 + d**3*m** 2*x**6 - 4*d**3*m*x**6 + 3*d**3*x**6),x)*a**2*c**2*d**2 + 2*int(x**m/(c**3 *m**2 - 4*c**3*m + 3*c**3 + 3*c**2*d*m**2*x**2 - 12*c**2*d*m*x**2 + 9*c**2 *d*x**2 + 3*c*d**2*m**2*x**4 - 12*c*d**2*m*x**4 + 9*c*d**2*x**4 + d**3*m** 2*x**6 - 4*d**3*m*x**6 + 3*d**3*x**6),x)*a**2*c*d**3*m**4*x**2 - 16*int...