\(\int \frac {(c x)^m (A+B x+C x^2+D x^3)}{(a+b x^2)^3} \, dx\) [141]

Optimal result

Integrand size = 30, antiderivative size = 194 \[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {C (c x)^{1+m}}{b c (3-m) \left (a+b x^2\right )^2}-\frac {D (c x)^{2+m}}{b c^2 (2-m) \left (a+b x^2\right )^2}+\frac {\left (\frac {A}{1+m}+\frac {a C}{3 b-b m}\right ) (c x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^3 c}+\frac {(b B (2-m)+a D (2+m)) (c x)^{2+m} \operatorname {Hypergeometric2F1}\left (3,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a^3 b c^2 (2-m) (2+m)} \] Output:

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

Mathematica [A] (verified)

Time = 1.70 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.82 \[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x (c x)^m \left (\frac {a C \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+\frac {a D x \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{2+m}+\frac {(A b-a C) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+\frac {(b B-a D) x \operatorname {Hypergeometric2F1}\left (3,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{2+m}\right )}{a^3 b} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2337, 25, 27, 557, 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.

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

\(\Big \downarrow \) 2337

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((D*x^3 + C*x^2 + B*x + 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 = 127.58 (sec) , antiderivative size = 5258, normalized size of antiderivative = 27.10 \[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate((c*x)**m*(D*x**3+C*x**2+B*x+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]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(c**m*( - x**m*a*d*m**2 + x**m*a*d*m + 6*x**m*a*d + x**m*b**2*m**2 - 5*x** 
m*b**2*m + 6*x**m*b**2 + x**m*b*c*m**2*x - 6*x**m*b*c*m*x + 8*x**m*b*c*x + 
 x**m*b*d*m**2*x**2 - 7*x**m*b*d*m*x**2 + 12*x**m*b*d*x**2 + int(x**m/(a** 
3*m**3*x - 9*a**3*m**2*x + 26*a**3*m*x - 24*a**3*x + 3*a**2*b*m**3*x**3 - 
27*a**2*b*m**2*x**3 + 78*a**2*b*m*x**3 - 72*a**2*b*x**3 + 3*a*b**2*m**3*x* 
*5 - 27*a*b**2*m**2*x**5 + 78*a*b**2*m*x**5 - 72*a*b**2*x**5 + b**3*m**3*x 
**7 - 9*b**3*m**2*x**7 + 26*b**3*m*x**7 - 24*b**3*x**7),x)*a**4*d*m**6 - 1 
0*int(x**m/(a**3*m**3*x - 9*a**3*m**2*x + 26*a**3*m*x - 24*a**3*x + 3*a**2 
*b*m**3*x**3 - 27*a**2*b*m**2*x**3 + 78*a**2*b*m*x**3 - 72*a**2*b*x**3 + 3 
*a*b**2*m**3*x**5 - 27*a*b**2*m**2*x**5 + 78*a*b**2*m*x**5 - 72*a*b**2*x** 
5 + b**3*m**3*x**7 - 9*b**3*m**2*x**7 + 26*b**3*m*x**7 - 24*b**3*x**7),x)* 
a**4*d*m**5 + 29*int(x**m/(a**3*m**3*x - 9*a**3*m**2*x + 26*a**3*m*x - 24* 
a**3*x + 3*a**2*b*m**3*x**3 - 27*a**2*b*m**2*x**3 + 78*a**2*b*m*x**3 - 72* 
a**2*b*x**3 + 3*a*b**2*m**3*x**5 - 27*a*b**2*m**2*x**5 + 78*a*b**2*m*x**5 
- 72*a*b**2*x**5 + b**3*m**3*x**7 - 9*b**3*m**2*x**7 + 26*b**3*m*x**7 - 24 
*b**3*x**7),x)*a**4*d*m**4 + 4*int(x**m/(a**3*m**3*x - 9*a**3*m**2*x + 26* 
a**3*m*x - 24*a**3*x + 3*a**2*b*m**3*x**3 - 27*a**2*b*m**2*x**3 + 78*a**2* 
b*m*x**3 - 72*a**2*b*x**3 + 3*a*b**2*m**3*x**5 - 27*a*b**2*m**2*x**5 + 78* 
a*b**2*m*x**5 - 72*a*b**2*x**5 + b**3*m**3*x**7 - 9*b**3*m**2*x**7 + 26*b* 
*3*m*x**7 - 24*b**3*x**7),x)*a**4*d*m**3 - 132*int(x**m/(a**3*m**3*x - ...