\(\int \frac {(e x)^m (a+b x^2)^2}{c+d x} \, dx\) [111]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{c+d x} \, dx=-\frac {b c \left (b c^2+2 a d^2\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac {b \left (b c^2+2 a d^2\right ) (e x)^{2+m}}{d^3 e^2 (2+m)}-\frac {b^2 c (e x)^{3+m}}{d^2 e^3 (3+m)}+\frac {b^2 (e x)^{4+m}}{d e^4 (4+m)}+\frac {\left (b c^2+a d^2\right )^2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{c d^4 e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.75 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{c+d x} \, dx=\frac {x (e x)^m \left (-\frac {b c \left (b c^2+2 a d^2\right )}{1+m}+\frac {b d \left (b c^2+2 a d^2\right ) x}{2+m}-\frac {b^2 c d^2 x^2}{3+m}+\frac {b^2 d^3 x^3}{4+m}+\frac {\left (b c^2+a d^2\right )^2 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{c (1+m)}\right )}{d^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 163, 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 = {522, 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.

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

\(\Big \downarrow \) 522

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 522

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

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.86 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.43 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{c+d x} \, dx=\frac {a^{2} e^{m} m x^{m + 1} \Phi \left (\frac {d x e^{i \pi }}{c}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{c \Gamma \left (m + 2\right )} + \frac {a^{2} e^{m} x^{m + 1} \Phi \left (\frac {d x e^{i \pi }}{c}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{c \Gamma \left (m + 2\right )} + \frac {2 a b e^{m} m x^{m + 3} \Phi \left (\frac {d x e^{i \pi }}{c}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{c \Gamma \left (m + 4\right )} + \frac {6 a b e^{m} x^{m + 3} \Phi \left (\frac {d x e^{i \pi }}{c}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{c \Gamma \left (m + 4\right )} + \frac {b^{2} e^{m} m x^{m + 5} \Phi \left (\frac {d x e^{i \pi }}{c}, 1, m + 5\right ) \Gamma \left (m + 5\right )}{c \Gamma \left (m + 6\right )} + \frac {5 b^{2} e^{m} x^{m + 5} \Phi \left (\frac {d x e^{i \pi }}{c}, 1, m + 5\right ) \Gamma \left (m + 5\right )}{c \Gamma \left (m + 6\right )} \] Input:

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

Output:

a**2*e**m*m*x**(m + 1)*lerchphi(d*x*exp_polar(I*pi)/c, 1, m + 1)*gamma(m + 
 1)/(c*gamma(m + 2)) + a**2*e**m*x**(m + 1)*lerchphi(d*x*exp_polar(I*pi)/c 
, 1, m + 1)*gamma(m + 1)/(c*gamma(m + 2)) + 2*a*b*e**m*m*x**(m + 3)*lerchp 
hi(d*x*exp_polar(I*pi)/c, 1, m + 3)*gamma(m + 3)/(c*gamma(m + 4)) + 6*a*b* 
e**m*x**(m + 3)*lerchphi(d*x*exp_polar(I*pi)/c, 1, m + 3)*gamma(m + 3)/(c* 
gamma(m + 4)) + b**2*e**m*m*x**(m + 5)*lerchphi(d*x*exp_polar(I*pi)/c, 1, 
m + 5)*gamma(m + 5)/(c*gamma(m + 6)) + 5*b**2*e**m*x**(m + 5)*lerchphi(d*x 
*exp_polar(I*pi)/c, 1, m + 5)*gamma(m + 5)/(c*gamma(m + 6))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*(x**m*a**2*d**4*m**4 + 10*x**m*a**2*d**4*m**3 + 35*x**m*a**2*d**4*m* 
*2 + 50*x**m*a**2*d**4*m + 24*x**m*a**2*d**4 + 2*x**m*a*b*c**2*d**2*m**4 + 
 20*x**m*a*b*c**2*d**2*m**3 + 70*x**m*a*b*c**2*d**2*m**2 + 100*x**m*a*b*c* 
*2*d**2*m + 48*x**m*a*b*c**2*d**2 - 2*x**m*a*b*c*d**3*m**4*x - 18*x**m*a*b 
*c*d**3*m**3*x - 52*x**m*a*b*c*d**3*m**2*x - 48*x**m*a*b*c*d**3*m*x + 2*x* 
*m*a*b*d**4*m**4*x**2 + 16*x**m*a*b*d**4*m**3*x**2 + 38*x**m*a*b*d**4*m**2 
*x**2 + 24*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 - 9*x**m*b**2*c**3*d*m**3*x - 26*x**m*b**2*c**3*d* 
m**2*x - 24*x**m*b**2*c**3*d*m*x + x**m*b**2*c**2*d**2*m**4*x**2 + 8*x**m* 
b**2*c**2*d**2*m**3*x**2 + 19*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 - 7*x**m*b**2*c*d**3*m**3*x 
**3 - 14*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 + 6*x**m*b**2*d**4*m**3*x**4 + 11*x**m*b**2*d**4*m**2*x** 
4 + 6*x**m*b**2*d**4*m*x**4 - int(x**m/(c*x + d*x**2),x)*a**2*c*d**4*m**5 
- 10*int(x**m/(c*x + d*x**2),x)*a**2*c*d**4*m**4 - 35*int(x**m/(c*x + d*x* 
*2),x)*a**2*c*d**4*m**3 - 50*int(x**m/(c*x + d*x**2),x)*a**2*c*d**4*m**2 - 
 24*int(x**m/(c*x + d*x**2),x)*a**2*c*d**4*m - 2*int(x**m/(c*x + d*x**2),x 
)*a*b*c**3*d**2*m**5 - 20*int(x**m/(c*x + d*x**2),x)*a*b*c**3*d**2*m**4 - 
70*int(x**m/(c*x + d*x**2),x)*a*b*c**3*d**2*m**3 - 100*int(x**m/(c*x + ...