\(\int \frac {(e x)^m (a+b x^2)^2}{(c+d x)^2} \, dx\) [1705]

Optimal result

Integrand size = 22, antiderivative size = 181 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^2} \, dx=\frac {b \left (3 b c^2+2 a d^2\right ) (e x)^{1+m}}{d^4 e (1+m)}-\frac {2 b^2 c (e x)^{2+m}}{d^3 e^2 (2+m)}+\frac {b^2 (e x)^{3+m}}{d^2 e^3 (3+m)}+\frac {\left (b c^2+a d^2\right )^2 (e x)^{1+m}}{c d^4 e (c+d x)}-\frac {\left (b c^2+a d^2\right ) \left (a d^2 m+b c^2 (4+m)\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{c^2 d^4 e (1+m)} \] Output:

b*(2*a*d^2+3*b*c^2)*(e*x)^(1+m)/d^4/e/(1+m)-2*b^2*c*(e*x)^(2+m)/d^3/e^2/(2 
+m)+b^2*(e*x)^(3+m)/d^2/e^3/(3+m)+(a*d^2+b*c^2)^2*(e*x)^(1+m)/c/d^4/e/(d*x 
+c)-(a*d^2+b*c^2)*(a*d^2*m+b*c^2*(4+m))*(e*x)^(1+m)*hypergeom([1, 1+m],[2+ 
m],-d*x/c)/c^2/d^4/e/(1+m)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.72 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^2} \, dx=\frac {x (e x)^m \left (\frac {b \left (3 b c^2+2 a d^2\right )}{1+m}-\frac {2 b^2 c d x}{2+m}+\frac {b^2 d^2 x^2}{3+m}-\frac {4 b \left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{1+m}+\frac {\left (b c^2+a d^2\right )^2 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {d x}{c}\right )}{c^2 (1+m)}\right )}{d^4} \] Input:

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

Output:

(x*(e*x)^m*((b*(3*b*c^2 + 2*a*d^2))/(1 + m) - (2*b^2*c*d*x)/(2 + m) + (b^2 
*d^2*x^2)/(3 + m) - (4*b*(b*c^2 + a*d^2)*Hypergeometric2F1[1, 1 + m, 2 + m 
, -((d*x)/c)])/(1 + m) + ((b*c^2 + a*d^2)^2*Hypergeometric2F1[2, 1 + m, 2 
+ m, -((d*x)/c)])/(c^2*(1 + m))))/d^4
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {519, 2123, 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.

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

\(\Big \downarrow \) 519

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

\(\Big \downarrow \) 2123

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((b*c^2 + a*d^2)^2*(e*x)^(1 + m))/(c*d^4*e*(c + d*x)) - (-((b*c*(3*b*c^2 + 
 2*a*d^2)*(e*x)^(1 + m))/(d^4*e*(1 + m))) + (2*b^2*c^2*(e*x)^(2 + m))/(d^3 
*e^2*(2 + m)) - (b^2*c*(e*x)^(3 + m))/(d^2*e^3*(3 + m)) + ((b*c^2 + a*d^2) 
*(a*d^2*m + b*c^2*(4 + m))*(e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m 
, -((d*x)/c)])/(c*d^4*e*(1 + m)))/c
 

Defintions of rubi rules used

rule 519

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, c + d*x, x], R = 
 PolynomialRemainder[(a + b*x^2)^p, c + d*x, x]}, Simp[(-R)*(e*x)^(m + 1)*( 
(c + d*x)^(n + 1)/(c*e*(n + 1))), x] + Simp[1/(c*(n + 1))   Int[(e*x)^m*(c 
+ d*x)^(n + 1)*ExpandToSum[c*(n + 1)*Qx + R*(m + n + 2), x], x], x]] /; Fre 
eQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0] && LtQ[n, -1] &&  !IntegerQ[m]
 

rule 2009

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

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*(x**m*a**2*d**4*m**4 + 6*x**m*a**2*d**4*m**3 + 11*x**m*a**2*d**4*m** 
2 + 6*x**m*a**2*d**4*m + 2*x**m*a*b*c**2*d**2*m**4 + 16*x**m*a*b*c**2*d**2 
*m**3 + 46*x**m*a*b*c**2*d**2*m**2 + 56*x**m*a*b*c**2*d**2*m + 24*x**m*a*b 
*c**2*d**2 - 2*x**m*a*b*c*d**3*m**4*x - 12*x**m*a*b*c*d**3*m**3*x - 18*x** 
m*a*b*c*d**3*m**2*x + 8*x**m*a*b*c*d**3*m*x + 24*x**m*a*b*c*d**3*x + 2*x** 
m*a*b*d**4*m**4*x**2 + 8*x**m*a*b*d**4*m**3*x**2 + 2*x**m*a*b*d**4*m**2*x* 
*2 - 12*x**m*a*b*d**4*m*x**2 + x**m*b**2*c**4*m**4 + 10*x**m*b**2*c**4*m** 
3 + 35*x**m*b**2*c**4*m**2 + 50*x**m*b**2*c**4*m + 24*x**m*b**2*c**4 - x** 
m*b**2*c**3*d*m**4*x - 8*x**m*b**2*c**3*d*m**3*x - 17*x**m*b**2*c**3*d*m** 
2*x + 2*x**m*b**2*c**3*d*m*x + 24*x**m*b**2*c**3*d*x + x**m*b**2*c**2*d**2 
*m**4*x**2 + 6*x**m*b**2*c**2*d**2*m**3*x**2 + 5*x**m*b**2*c**2*d**2*m**2* 
x**2 - 12*x**m*b**2*c**2*d**2*m*x**2 - x**m*b**2*c*d**3*m**4*x**3 - 4*x**m 
*b**2*c*d**3*m**3*x**3 + x**m*b**2*c*d**3*m**2*x**3 + 4*x**m*b**2*c*d**3*m 
*x**3 + x**m*b**2*d**4*m**4*x**4 + 2*x**m*b**2*d**4*m**3*x**4 - x**m*b**2* 
d**4*m**2*x**4 - 2*x**m*b**2*d**4*m*x**4 - int(x**m/(c**2*m*x - c**2*x + 2 
*c*d*m*x**2 - 2*c*d*x**2 + d**2*m*x**3 - d**2*x**3),x)*a**2*c**2*d**4*m**6 
 - 5*int(x**m/(c**2*m*x - c**2*x + 2*c*d*m*x**2 - 2*c*d*x**2 + d**2*m*x**3 
 - d**2*x**3),x)*a**2*c**2*d**4*m**5 - 5*int(x**m/(c**2*m*x - c**2*x + 2*c 
*d*m*x**2 - 2*c*d*x**2 + d**2*m*x**3 - d**2*x**3),x)*a**2*c**2*d**4*m**4 + 
 5*int(x**m/(c**2*m*x - c**2*x + 2*c*d*m*x**2 - 2*c*d*x**2 + d**2*m*x**...