\(\int \frac {x^m}{(a+b x^2)^2 (c+d x^2)^2} \, dx\) [341]

Optimal result

Integrand size = 22, antiderivative size = 230 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {b^2 (a d (5-m)-b (c-c m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^3 (1+m)}-\frac {d^2 (a d (1-m)-b c (5-m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^3 (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {483, 593, 598, 371} \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=-\frac {b^2 x^{m+1} (a d (5-m)-b (c-c m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^3}-\frac {d^2 x^{m+1} (a d (1-m)-b c (5-m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{2 c^2 (m+1) (b c-a d)^3}+\frac {d x^{m+1} (a d+b c)}{2 a c \left (c+d x^2\right ) (b c-a d)^2}+\frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)} \]

[In]

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

[Out]

(d*(b*c + a*d)*x^(1 + m))/(2*a*c*(b*c - a*d)^2*(c + d*x^2)) + (b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)*(c +
d*x^2)) - (b^2*(a*d*(5 - m) - b*(c - c*m))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])
/(2*a^2*(b*c - a*d)^3*(1 + m)) - (d^2*(a*d*(1 - m) - b*c*(5 - m))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3
+ m)/2, -((d*x^2)/c)])/(2*c^2*(b*c - a*d)^3*(1 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 483

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*(e*
x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a*d)
*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n
*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ
[p, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 593

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p +
1))), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)
*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \frac {x^m \left (2 a d-b c (1-m)-b d (3-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{2 a (b c-a d)} \\ & = \frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \frac {x^m \left (2 \left (4 a b c d-b^2 c^2 (1-m)-a^2 d^2 (1-m)\right )-2 b d (b c+a d) (1-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{4 a c (b c-a d)^2} \\ & = \frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \left (\frac {2 b^2 c (-b c (1-m)+a d (5-m)) x^m}{(b c-a d) \left (a+b x^2\right )}+\frac {2 a d^2 (a d (1-m)-b c (5-m)) x^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{4 a c (b c-a d)^2} \\ & = \frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\left (d^2 (a d (1-m)-b c (5-m))\right ) \int \frac {x^m}{c+d x^2} \, dx}{2 c (b c-a d)^3}+\frac {\left (b^2 (b c (1-m)-a d (5-m))\right ) \int \frac {x^m}{a+b x^2} \, dx}{2 a (b c-a d)^3} \\ & = \frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b^2 (b c (1-m)-a d (5-m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^3 (1+m)}-\frac {d^2 (a d (1-m)-b c (5-m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^3 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.75 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {x^{1+m} \left (2 a b^2 c^2 d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )-2 a^2 b c d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )-(b c-a d) \left (b^2 c^2 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^2 d^2 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )\right )}{a^2 c^2 (-b c+a d)^3 (1+m)} \]

[In]

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

[Out]

(x^(1 + m)*(2*a*b^2*c^2*d*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] - 2*a^2*b*c*d^2*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] - (b*c - a*d)*(b^2*c^2*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2
, -((b*x^2)/a)] + a^2*d^2*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])))/(a^2*c^2*(-(b*c) + a*d)^
3*(1 + m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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