\(\int \frac {(c x)^m (A+B x^2+C x^4+D x^6)}{(a+b x^2)^3} \, dx\) [258]

Optimal result

Integrand size = 32, antiderivative size = 242 \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {D (c x)^{1+m}}{b^3 c (1+m)}+\frac {\left (\frac {A}{a}-\frac {b^2 B-a b C+a^2 D}{b^3}\right ) (c x)^{1+m}}{4 c \left (a+b x^2\right )^2}+\frac {\left (\frac {A (3-m)}{a}+\frac {b^2 B (1+m)-a b C (5+m)+a^2 D (9+m)}{b^3}\right ) (c x)^{1+m}}{8 a c \left (a+b x^2\right )}+\frac {\left (A b^3 \left (3-4 m+m^2\right )+a \left (a b C \left (3+4 m+m^2\right )-a^2 D \left (15+8 m+m^2\right )+b^2 \left (B-B m^2\right )\right )\right ) (c x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{8 a^3 b^3 c (1+m)} \] Output:

D*(c*x)^(1+m)/b^3/c/(1+m)+1/4*(A/a-(B*b^2-C*a*b+D*a^2)/b^3)*(c*x)^(1+m)/c/ 
(b*x^2+a)^2+1/8*(A*(3-m)/a+(b^2*B*(1+m)-a*b*C*(5+m)+a^2*D*(9+m))/b^3)*(c*x 
)^(1+m)/a/c/(b*x^2+a)+1/8*(A*b^3*(m^2-4*m+3)+a*(a*b*C*(m^2+4*m+3)-a^2*D*(m 
^2+8*m+15)+b^2*(-B*m^2+B)))*(c*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2* 
m],-b*x^2/a)/a^3/b^3/c/(1+m)
 

Mathematica [A] (verified)

Time = 1.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.64 \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x (c x)^m \left (D+\frac {(b C-3 a D) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a}+\frac {\left (b^2 B-2 a b C+3 a^2 D\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2}+\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^3}\right )}{b^3 (1+m)} \] Input:

Integrate[((c*x)^m*(A + B*x^2 + C*x^4 + D*x^6))/(a + b*x^2)^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2337, 25, 1674, 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.

\(\displaystyle \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^3} \, dx\)

\(\Big \downarrow \) 2337

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1674

\(\displaystyle \frac {\int \left (\frac {4 a D (c x)^m}{b^3}-\frac {4 a (3 a D-b C) (c x)^m}{b^3 \left (b x^2+a\right )}+\frac {\left (A (3-m) b^3+a \left (D (m+9) a^2-b C (m+5) a+b^2 B (m+1)\right )\right ) (c x)^m}{b^3 \left (b x^2+a\right )^2}\right )dx}{4 a}+\frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{4 a c \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {(c x)^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right ) \left (a \left (a^2 D (m+9)-a b C (m+5)+b^2 B (m+1)\right )+A b^3 (3-m)\right )}{a^2 b^3 c (m+1)}+\frac {4 (c x)^{m+1} (b C-3 a D) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{b^3 c (m+1)}+\frac {4 a D (c x)^{m+1}}{b^3 c (m+1)}}{4 a}+\frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{4 a c \left (a+b x^2\right )^2}\)

Input:

Int[((c*x)^m*(A + B*x^2 + C*x^4 + D*x^6))/(a + b*x^2)^3,x]
 

Output:

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*(c*x)^(1 + m))/(4*a*c*(a + b*x^2)^2 
) + ((4*a*D*(c*x)^(1 + m))/(b^3*c*(1 + m)) + (4*(b*C - 3*a*D)*(c*x)^(1 + m 
)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(b^3*c*(1 + m) 
) + ((A*b^3*(3 - m) + a*(b^2*B*(1 + m) - a*b*C*(5 + m) + a^2*D*(9 + m)))*( 
c*x)^(1 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a^ 
2*b^3*c*(1 + m)))/(4*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1674

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && N 
eQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])
 

rule 2009

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

rule 2337

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq 
, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 1]}, Simp[(-(c*x)^(m + 1))*(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))) 
, x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2 
*a*(p + 1)*Q + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a 
, b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] &&  !GtQ[m, 0]
 

Maple [F]

\[\int \frac {\left (c x \right )^{m} \left (D x^{6}+C \,x^{4}+x^{2} B +A \right )}{\left (b \,x^{2}+a \right )^{3}}d x\]

Input:

int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x)
 

Output:

int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x)
 

Fricas [F]

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

integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")
 

Output:

integral((D*x^6 + C*x^4 + B*x^2 + A)*(c*x)^m/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^ 
2*b*x^2 + a^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 107.43 (sec) , antiderivative size = 6484, normalized size of antiderivative = 26.79 \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate((c*x)**m*(D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**3,x)
 

Output:

A*(a**2*c**m*m**3*x**(m + 1)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1 
/2)*gamma(m/2 + 1/2)/(32*a**5*gamma(m/2 + 3/2) + 64*a**4*b*x**2*gamma(m/2 
+ 3/2) + 32*a**3*b**2*x**4*gamma(m/2 + 3/2)) - 3*a**2*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)/(32*a** 
5*gamma(m/2 + 3/2) + 64*a**4*b*x**2*gamma(m/2 + 3/2) + 32*a**3*b**2*x**4*g 
amma(m/2 + 3/2)) - 2*a**2*c**m*m**2*x**(m + 1)*gamma(m/2 + 1/2)/(32*a**5*g 
amma(m/2 + 3/2) + 64*a**4*b*x**2*gamma(m/2 + 3/2) + 32*a**3*b**2*x**4*gamm 
a(m/2 + 3/2)) - a**2*c**m*m*x**(m + 1)*lerchphi(b*x**2*exp_polar(I*pi)/a, 
1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*a**5*gamma(m/2 + 3/2) + 64*a**4*b*x**2* 
gamma(m/2 + 3/2) + 32*a**3*b**2*x**4*gamma(m/2 + 3/2)) + 8*a**2*c**m*m*x** 
(m + 1)*gamma(m/2 + 1/2)/(32*a**5*gamma(m/2 + 3/2) + 64*a**4*b*x**2*gamma( 
m/2 + 3/2) + 32*a**3*b**2*x**4*gamma(m/2 + 3/2)) + 3*a**2*c**m*x**(m + 1)* 
lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*a**5 
*gamma(m/2 + 3/2) + 64*a**4*b*x**2*gamma(m/2 + 3/2) + 32*a**3*b**2*x**4*ga 
mma(m/2 + 3/2)) + 10*a**2*c**m*x**(m + 1)*gamma(m/2 + 1/2)/(32*a**5*gamma( 
m/2 + 3/2) + 64*a**4*b*x**2*gamma(m/2 + 3/2) + 32*a**3*b**2*x**4*gamma(m/2 
 + 3/2)) + 2*a*b*c**m*m**3*x**2*x**(m + 1)*lerchphi(b*x**2*exp_polar(I*pi) 
/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*a**5*gamma(m/2 + 3/2) + 64*a**4*b*x 
**2*gamma(m/2 + 3/2) + 32*a**3*b**2*x**4*gamma(m/2 + 3/2)) - 6*a*b*c**m*m* 
*2*x**2*x**(m + 1)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gam...
 

Maxima [F]

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

integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)*(c*x)^m/(b*x^2 + a)^3, x)
 

Giac [F]

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

integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)*(c*x)^m/(b*x^2 + a)^3, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^3} \, dx=\int \frac {{\left (c\,x\right )}^m\,\left (A+B\,x^2+C\,x^4+x^6\,D\right )}{{\left (b\,x^2+a\right )}^3} \,d x \] Input:

int(((c*x)^m*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^3,x)
 

Output:

int(((c*x)^m*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^3, x)
 

Reduce [F]

\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^3} \, dx=\text {too large to display} \] Input:

int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x)
 

Output:

(c**m*(x**m*a**2*d*m**2*x + 8*x**m*a**2*d*m*x + 15*x**m*a**2*d*x - x**m*a* 
b*c*m**2*x - 4*x**m*a*b*c*m*x - 3*x**m*a*b*c*x - x**m*a*b*d*m**2*x**3 - 2* 
x**m*a*b*d*m*x**3 + 15*x**m*a*b*d*x**3 + x**m*b**3*m**2*x - x**m*b**3*x + 
x**m*b**2*c*m**2*x**3 - 2*x**m*b**2*c*m*x**3 - 3*x**m*b**2*c*x**3 + x**m*b 
**2*d*m**2*x**5 - 4*x**m*b**2*d*m*x**5 + 3*x**m*b**2*d*x**5 - int(x**m/(a* 
*3*m**2 - 4*a**3*m + 3*a**3 + 3*a**2*b*m**2*x**2 - 12*a**2*b*m*x**2 + 9*a* 
*2*b*x**2 + 3*a*b**2*m**2*x**4 - 12*a*b**2*m*x**4 + 9*a*b**2*x**4 + b**3*m 
**2*x**6 - 4*b**3*m*x**6 + 3*b**3*x**6),x)*a**5*d*m**5 - 5*int(x**m/(a**3* 
m**2 - 4*a**3*m + 3*a**3 + 3*a**2*b*m**2*x**2 - 12*a**2*b*m*x**2 + 9*a**2* 
b*x**2 + 3*a*b**2*m**2*x**4 - 12*a*b**2*m*x**4 + 9*a*b**2*x**4 + b**3*m**2 
*x**6 - 4*b**3*m*x**6 + 3*b**3*x**6),x)*a**5*d*m**4 + 10*int(x**m/(a**3*m* 
*2 - 4*a**3*m + 3*a**3 + 3*a**2*b*m**2*x**2 - 12*a**2*b*m*x**2 + 9*a**2*b* 
x**2 + 3*a*b**2*m**2*x**4 - 12*a*b**2*m*x**4 + 9*a*b**2*x**4 + b**3*m**2*x 
**6 - 4*b**3*m*x**6 + 3*b**3*x**6),x)*a**5*d*m**3 + 50*int(x**m/(a**3*m**2 
 - 4*a**3*m + 3*a**3 + 3*a**2*b*m**2*x**2 - 12*a**2*b*m*x**2 + 9*a**2*b*x* 
*2 + 3*a*b**2*m**2*x**4 - 12*a*b**2*m*x**4 + 9*a*b**2*x**4 + b**3*m**2*x** 
6 - 4*b**3*m*x**6 + 3*b**3*x**6),x)*a**5*d*m**2 - 9*int(x**m/(a**3*m**2 - 
4*a**3*m + 3*a**3 + 3*a**2*b*m**2*x**2 - 12*a**2*b*m*x**2 + 9*a**2*b*x**2 
+ 3*a*b**2*m**2*x**4 - 12*a*b**2*m*x**4 + 9*a*b**2*x**4 + b**3*m**2*x**6 - 
 4*b**3*m*x**6 + 3*b**3*x**6),x)*a**5*d*m - 45*int(x**m/(a**3*m**2 - 4*...