\(\int \frac {x^m}{(a+b x^2)^3 (c+d x^2)} \, dx\) [1600]

Optimal result

Integrand size = 22, antiderivative size = 234 \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right ) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{8 a^3 (b c-a d)^3 (1+m)}-\frac {d^3 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (b c-a d)^3 (1+m)} \] Output:

1/4*b*x^(1+m)/a/(-a*d+b*c)/(b*x^2+a)^2+1/8*b*(b*c*(3-m)-a*d*(7-m))*x^(1+m) 
/a^2/(-a*d+b*c)^2/(b*x^2+a)+1/8*b*(a^2*d^2*(m^2-8*m+15)-2*a*b*c*d*(m^2-6*m 
+5)+b^2*c^2*(m^2-4*m+3))*x^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-b*x 
^2/a)/a^3/(-a*d+b*c)^3/(1+m)-d^3*x^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2 
*m],-d*x^2/c)/c/(-a*d+b*c)^3/(1+m)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.73 \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\frac {x^{1+m} \left (-a^2 b c d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^3 d^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )-b c (-b c+a d) \left (a d \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+(-b c+a d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{a^3 c (-b c+a d)^3 (1+m)} \] Input:

Integrate[x^m/((a + b*x^2)^3*(c + d*x^2)),x]
 

Output:

(x^(1 + m)*(-(a^2*b*c*d^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b* 
x^2)/a)]) + a^3*d^3*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c 
)] - b*c*(-(b*c) + a*d)*(a*d*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -( 
(b*x^2)/a)] + (-(b*c) + a*d)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -( 
(b*x^2)/a)])))/(a^3*c*(-(b*c) + a*d)^3*(1 + m))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {374, 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.

\(\displaystyle \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx\)

\(\Big \downarrow \) 374

\(\displaystyle \frac {b x^{m+1}}{4 a \left (a+b x^2\right )^2 (b c-a d)}-\frac {\int \frac {x^m \left (-b d (3-m) x^2+4 a d-b c (3-m)\right )}{\left (b x^2+a\right )^2 \left (d x^2+c\right )}dx}{4 a (b c-a d)}\)

\(\Big \downarrow \) 441

\(\displaystyle \frac {b x^{m+1}}{4 a \left (a+b x^2\right )^2 (b c-a d)}-\frac {-\frac {\int \frac {x^m \left (b^2 \left (m^2-4 m+3\right ) c^2-a b d \left (m^2-8 m+7\right ) c+8 a^2 d^2+b d (b c (3-m)-a d (7-m)) (1-m) x^2\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 a (b c-a d)}-\frac {b x^{m+1} (b c (3-m)-a d (7-m))}{2 a \left (a+b x^2\right ) (b c-a d)}}{4 a (b c-a d)}\)

\(\Big \downarrow \) 446

\(\displaystyle \frac {b x^{m+1}}{4 a \left (a+b x^2\right )^2 (b c-a d)}-\frac {-\frac {\int \left (\frac {b \left (b^2 \left (m^2-4 m+3\right ) c^2-2 a b d \left (m^2-6 m+5\right ) c+a^2 d^2 \left (m^2-8 m+15\right )\right ) x^m}{(b c-a d) \left (b x^2+a\right )}+\frac {8 a^2 d^3 x^m}{(a d-b c) \left (d x^2+c\right )}\right )dx}{2 a (b c-a d)}-\frac {b x^{m+1} (b c (3-m)-a d (7-m))}{2 a \left (a+b x^2\right ) (b c-a d)}}{4 a (b c-a d)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b x^{m+1}}{4 a \left (a+b x^2\right )^2 (b c-a d)}-\frac {-\frac {\frac {b x^{m+1} \left (a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (m^2-6 m+5\right )+b^2 c^2 \left (m^2-4 m+3\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a (m+1) (b c-a d)}-\frac {8 a^2 d^3 x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)}}{2 a (b c-a d)}-\frac {b x^{m+1} (b c (3-m)-a d (7-m))}{2 a \left (a+b x^2\right ) (b c-a d)}}{4 a (b c-a d)}\)

Input:

Int[x^m/((a + b*x^2)^3*(c + d*x^2)),x]
 

Output:

(b*x^(1 + m))/(4*a*(b*c - a*d)*(a + b*x^2)^2) - (-1/2*(b*(b*c*(3 - m) - a* 
d*(7 - m))*x^(1 + m))/(a*(b*c - a*d)*(a + b*x^2)) - ((b*(a^2*d^2*(15 - 8*m 
 + m^2) - 2*a*b*c*d*(5 - 6*m + m^2) + b^2*c^2*(3 - 4*m + m^2))*x^(1 + m)*H 
ypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(b*c - a*d)*(1 
 + m)) - (8*a^2*d^3*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)))/(4*a*(b*c - a*d) 
)
 

Defintions of rubi rules used

rule 374

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]
 

rule 441

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]
 

rule 446

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 
]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{3} \left (x^{2} d +c \right )}d x\]

Input:

int(x^m/(b*x^2+a)^3/(d*x^2+c),x)
 

Output:

int(x^m/(b*x^2+a)^3/(d*x^2+c),x)
 

Fricas [F]

\[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}} \,d x } \] Input:

integrate(x^m/(b*x^2+a)^3/(d*x^2+c),x, algorithm="fricas")
 

Output:

integral(x^m/(b^3*d*x^8 + (b^3*c + 3*a*b^2*d)*x^6 + 3*(a*b^2*c + a^2*b*d)* 
x^4 + a^3*c + (3*a^2*b*c + a^3*d)*x^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(x**m/(b*x**2+a)**3/(d*x**2+c),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}} \,d x } \] Input:

integrate(x^m/(b*x^2+a)^3/(d*x^2+c),x, algorithm="maxima")
 

Output:

integrate(x^m/((b*x^2 + a)^3*(d*x^2 + c)), x)
 

Giac [F]

\[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}} \,d x } \] Input:

integrate(x^m/(b*x^2+a)^3/(d*x^2+c),x, algorithm="giac")
 

Output:

integrate(x^m/((b*x^2 + a)^3*(d*x^2 + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int \frac {x^m}{{\left (b\,x^2+a\right )}^3\,\left (d\,x^2+c\right )} \,d x \] Input:

int(x^m/((a + b*x^2)^3*(c + d*x^2)),x)
 

Output:

int(x^m/((a + b*x^2)^3*(c + d*x^2)), x)
 

Reduce [F]

\[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int \frac {x^{m}}{b^{3} d \,x^{8}+3 a \,b^{2} d \,x^{6}+b^{3} c \,x^{6}+3 a^{2} b d \,x^{4}+3 a \,b^{2} c \,x^{4}+a^{3} d \,x^{2}+3 a^{2} b c \,x^{2}+a^{3} c}d x \] Input:

int(x^m/(b*x^2+a)^3/(d*x^2+c),x)
 

Output:

int(x**m/(a**3*c + a**3*d*x**2 + 3*a**2*b*c*x**2 + 3*a**2*b*d*x**4 + 3*a*b 
**2*c*x**4 + 3*a*b**2*d*x**6 + b**3*c*x**6 + b**3*d*x**8),x)