\(\int \frac {(c x)^m (A+B x^2)}{(a+b x^2)^2} \, dx\) [503]

Optimal result

Integrand size = 22, antiderivative size = 103 \[ \int \frac {(c x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(A b-a B) (c x)^{1+m}}{2 a b c \left (a+b x^2\right )}+\frac {(a B (1+m)+A (b-b m)) (c x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 b c (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {(c x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {x (c x)^m \left (a B \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+(A b-a B) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{a^2 b (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {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.

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 278

\(\displaystyle \frac {(c x)^{m+1} (a B (m+1)+A (b-b m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{2 a^2 b c (m+1)}+\frac {(c x)^{m+1} (A b-a B)}{2 a b c \left (a+b x^2\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 278

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])
 

rule 362

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)]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((B*x^2 + A)*(c*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 13.85 (sec) , antiderivative size = 954, normalized size of antiderivative = 9.26 \[ \int \frac {(c x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

A*(-a*c**m*m**2*x**(m + 1)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2 
)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/ 
2)) + 2*a*c**m*m*x**(m + 1)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8* 
a**2*b*x**2*gamma(m/2 + 3/2)) + a*c**m*x**(m + 1)*lerchphi(b*x**2*exp_pola 
r(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a** 
2*b*x**2*gamma(m/2 + 3/2)) + 2*a*c**m*x**(m + 1)*gamma(m/2 + 1/2)/(8*a**3* 
gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) - b*c**m*m**2*x**2*x**( 
m + 1)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/( 
8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) + b*c**m*x**2*x* 
*(m + 1)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2) 
/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2))) + B*(-a*c**m* 
m**2*x**(m + 3)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 
 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 4*a*c 
**m*m*x**(m + 3)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/ 
2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) + 2*a* 
c**m*m*x**(m + 3)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x** 
2*gamma(m/2 + 5/2)) - 3*a*c**m*x**(m + 3)*lerchphi(b*x**2*exp_polar(I*pi)/ 
a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2 
*gamma(m/2 + 5/2)) + 6*a*c**m*x**(m + 3)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/ 
2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - b*c**m*m**2*x**2*x**(m + 3...
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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