\(\int \frac {(e x)^m (a+b x^2)^2}{(c+d x)^3} \, dx\) [113]

Optimal result

Integrand size = 22, antiderivative size = 216 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^3} \, dx=-\frac {3 b^2 c (e x)^{1+m}}{d^4 e (1+m)}+\frac {b^2 (e x)^{2+m}}{d^3 e^2 (2+m)}+\frac {\left (b c^2+a d^2\right )^2 (e x)^{1+m}}{2 c d^4 e (c+d x)^2}+\frac {2 b \left (3 b c^2+a d^2\right ) (e x)^{1+m}}{d^4 e m (c+d x)}+\frac {\left (a^2 d^4 (1-m) m-2 a b c^2 d^2 \left (2+3 m+m^2\right )-b^2 c^4 \left (12+7 m+m^2\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {d x}{c}\right )}{2 c^3 d^4 e m (1+m)} \] Output:

-3*b^2*c*(e*x)^(1+m)/d^4/e/(1+m)+b^2*(e*x)^(2+m)/d^3/e^2/(2+m)+1/2*(a*d^2+ 
b*c^2)^2*(e*x)^(1+m)/c/d^4/e/(d*x+c)^2+2*b*(a*d^2+3*b*c^2)*(e*x)^(1+m)/d^4 
/e/m/(d*x+c)+1/2*(a^2*d^4*(1-m)*m-2*a*b*c^2*d^2*(m^2+3*m+2)-b^2*c^4*(m^2+7 
*m+12))*(e*x)^(1+m)*hypergeom([2, 1+m],[2+m],-d*x/c)/c^3/d^4/e/m/(1+m)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.67 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^3} \, dx=\frac {x (e x)^m \left (-\frac {3 b^2 c}{1+m}+\frac {b^2 d x}{2+m}+\frac {2 b \left (3 b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{c (1+m)}-\frac {4 b \left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {d x}{c}\right )}{c (1+m)}+\frac {\left (b c^2+a d^2\right )^2 \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {d x}{c}\right )}{c^3 (1+m)}\right )}{d^4} \] Input:

Integrate[((e*x)^m*(a + b*x^2)^2)/(c + d*x)^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {519, 25, 2124, 25, 1195, 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.

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

\(\Big \downarrow \) 519

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2124

\(\displaystyle \frac {\frac {\int -\frac {(e x)^m \left (\frac {4 b^2 e x c^3}{d^3}-\frac {2 b^2 e x^2 c^2}{d^2}+e \left (-\frac {b^2 \left (m^2+7 m+6\right ) c^4}{d^4}-\frac {2 a b \left (m^2+3 m+2\right ) c^2}{d^2}+a^2 (1-m) m\right )\right )}{c+d x}dx}{c e}+\frac {(e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-m)-b c^2 (m+7)\right )}{c d^4 e (c+d x)}}{2 c}+\frac {(e x)^{m+1} \left (a d^2+b c^2\right )^2}{2 c d^4 e (c+d x)^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {(e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-m)-b c^2 (m+7)\right )}{c d^4 e (c+d x)}-\frac {\int \frac {(e x)^m \left (\frac {4 b^2 e x c^3}{d^3}-\frac {2 b^2 e x^2 c^2}{d^2}+\frac {e \left (-b^2 \left (m^2+7 m+6\right ) c^4-2 a b d^2 \left (m^2+3 m+2\right ) c^2+a^2 d^4 (1-m) m\right )}{d^4}\right )}{c+d x}dx}{c e}}{2 c}+\frac {(e x)^{m+1} \left (a d^2+b c^2\right )^2}{2 c d^4 e (c+d x)^2}\)

\(\Big \downarrow \) 1195

\(\displaystyle \frac {\frac {(e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-m)-b c^2 (m+7)\right )}{c d^4 e (c+d x)}-\frac {\int \left (\frac {6 b^2 c^3 e (e x)^m}{d^4}+\frac {e \left (-b^2 \left (m^2+7 m+12\right ) c^4-2 a b d^2 \left (m^2+3 m+2\right ) c^2+a^2 d^4 (1-m) m\right ) (e x)^m}{d^4 (c+d x)}-\frac {2 b^2 c^2 (e x)^{m+1}}{d^3}\right )dx}{c e}}{2 c}+\frac {(e x)^{m+1} \left (a d^2+b c^2\right )^2}{2 c d^4 e (c+d x)^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {(e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-m)-b c^2 (m+7)\right )}{c d^4 e (c+d x)}-\frac {\frac {(e x)^{m+1} \left (a^2 d^4 (1-m) m-2 a b c^2 d^2 \left (m^2+3 m+2\right )-b^2 c^4 \left (m^2+7 m+12\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {d x}{c}\right )}{c d^4 (m+1)}+\frac {6 b^2 c^3 (e x)^{m+1}}{d^4 (m+1)}-\frac {2 b^2 c^2 (e x)^{m+2}}{d^3 e (m+2)}}{c e}}{2 c}+\frac {(e x)^{m+1} \left (a d^2+b c^2\right )^2}{2 c d^4 e (c+d x)^2}\)

Input:

Int[((e*x)^m*(a + b*x^2)^2)/(c + d*x)^3,x]
 

Output:

((b*c^2 + a*d^2)^2*(e*x)^(1 + m))/(2*c*d^4*e*(c + d*x)^2) + (((b*c^2 + a*d 
^2)*(a*d^2*(1 - m) - b*c^2*(7 + m))*(e*x)^(1 + m))/(c*d^4*e*(c + d*x)) - ( 
(6*b^2*c^3*(e*x)^(1 + m))/(d^4*(1 + m)) - (2*b^2*c^2*(e*x)^(2 + m))/(d^3*e 
*(2 + m)) + ((a^2*d^4*(1 - m)*m - 2*a*b*c^2*d^2*(2 + 3*m + m^2) - b^2*c^4* 
(12 + 7*m + m^2))*(e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((d*x) 
/c)])/(c*d^4*(1 + m)))/(c*e))/(2*c)
 

Defintions of rubi rules used

rule 25

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

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 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 2009

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

rule 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

Maple [F]

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

Input:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^3,x)
 

Output:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^3,x)
 

Fricas [F]

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

integrate((e*x)^m*(b*x^2+a)^2/(d*x+c)^3,x, algorithm="fricas")
 

Output:

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*(e*x)^m/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^ 
2*d*x + c^3), x)
 

Sympy [F]

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

integrate((e*x)**m*(b*x**2+a)**2/(d*x+c)**3,x)
 

Output:

Integral((e*x)**m*(a + b*x**2)**2/(c + d*x)**3, x)
 

Maxima [F]

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

integrate((e*x)^m*(b*x^2+a)^2/(d*x+c)^3,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^2*(e*x)^m/(d*x + c)^3, x)
 

Giac [F]

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

integrate((e*x)^m*(b*x^2+a)^2/(d*x+c)^3,x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^2*(e*x)^m/(d*x + c)^3, x)
 

Mupad [F(-1)]

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

int(((e*x)^m*(a + b*x^2)^2)/(c + d*x)^3,x)
 

Output:

int(((e*x)^m*(a + b*x^2)^2)/(c + d*x)^3, x)
 

Reduce [F]

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

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^3,x)
 

Output:

(e**m*(x**m*a**2*d**4*m**4 + 2*x**m*a**2*d**4*m**3 - x**m*a**2*d**4*m**2 - 
 2*x**m*a**2*d**4*m + 2*x**m*a*b*c**2*d**2*m**4 + 12*x**m*a*b*c**2*d**2*m* 
*3 + 26*x**m*a*b*c**2*d**2*m**2 + 24*x**m*a*b*c**2*d**2*m + 8*x**m*a*b*c** 
2*d**2 - 2*x**m*a*b*c*d**3*m**4*x - 6*x**m*a*b*c*d**3*m**3*x + 4*x**m*a*b* 
c*d**3*m**2*x + 24*x**m*a*b*c*d**3*m*x + 16*x**m*a*b*c*d**3*x + 2*x**m*a*b 
*d**4*m**4*x**2 - 10*x**m*a*b*d**4*m**2*x**2 + 8*x**m*a*b*d**4*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 - 7*x**m*b**2* 
c**3*d*m**3*x - 8*x**m*b**2*c**3*d*m**2*x + 28*x**m*b**2*c**3*d*m*x + 48*x 
**m*b**2*c**3*d*x + x**m*b**2*c**2*d**2*m**4*x**2 + 4*x**m*b**2*c**2*d**2* 
m**3*x**2 - 7*x**m*b**2*c**2*d**2*m**2*x**2 - 22*x**m*b**2*c**2*d**2*m*x** 
2 + 24*x**m*b**2*c**2*d**2*x**2 - x**m*b**2*c*d**3*m**4*x**3 - x**m*b**2*c 
*d**3*m**3*x**3 + 10*x**m*b**2*c*d**3*m**2*x**3 - 8*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**3*m**2*x - 3*c**3*m*x 
+ 2*c**3*x + 3*c**2*d*m**2*x**2 - 9*c**2*d*m*x**2 + 6*c**2*d*x**2 + 3*c*d* 
*2*m**2*x**3 - 9*c*d**2*m*x**3 + 6*c*d**2*x**3 + d**3*m**2*x**4 - 3*d**3*m 
*x**4 + 2*d**3*x**4),x)*a**2*c**3*d**4*m**7 + int(x**m/(c**3*m**2*x - 3*c* 
*3*m*x + 2*c**3*x + 3*c**2*d*m**2*x**2 - 9*c**2*d*m*x**2 + 6*c**2*d*x**2 + 
 3*c*d**2*m**2*x**3 - 9*c*d**2*m*x**3 + 6*c*d**2*x**3 + d**3*m**2*x**4 ...