\(\int \frac {(e x)^m}{(c+d x)^3 (a+b x^2)} \, dx\) [122]

Optimal result

Integrand size = 22, antiderivative size = 315 \[ \int \frac {(e x)^m}{(c+d x)^3 \left (a+b x^2\right )} \, dx=\frac {b^2 c \left (b c^2-3 a d^2\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a \left (b c^2+a d^2\right )^3 e (1+m)}+\frac {b d^2 \left (3 b c^2-a d^2\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{c \left (b c^2+a d^2\right )^3 e (1+m)}-\frac {b^2 d \left (3 b c^2-a d^2\right ) (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a \left (b c^2+a d^2\right )^3 e^2 (2+m)}+\frac {2 b d^2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {d x}{c}\right )}{c \left (b c^2+a d^2\right )^2 e (1+m)}+\frac {d^2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {d x}{c}\right )}{c^3 \left (b c^2+a d^2\right ) e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.77 \[ \int \frac {(e x)^m}{(c+d x)^3 \left (a+b x^2\right )} \, dx=\frac {x (e x)^m \left (\frac {b^2 c \left (b c^2-3 a d^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a (1+m)}+\frac {b d^2 \left (3 b c^2-a d^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{c (1+m)}+\frac {b^2 d \left (-3 b c^2+a d^2\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a (2+m)}+\frac {2 b d^2 \left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {d x}{c}\right )}{c (1+m)}+\frac {\left (b c^2 d+a d^3\right )^2 \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {d x}{c}\right )}{c^3 (1+m)}\right )}{\left (b c^2+a d^2\right )^3} \] Input:

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

Output:

(x*(e*x)^m*((b^2*c*(b*c^2 - 3*a*d^2)*Hypergeometric2F1[1, (1 + m)/2, (3 + 
m)/2, -((b*x^2)/a)])/(a*(1 + m)) + (b*d^2*(3*b*c^2 - a*d^2)*Hypergeometric 
2F1[1, 1 + m, 2 + m, -((d*x)/c)])/(c*(1 + m)) + (b^2*d*(-3*b*c^2 + a*d^2)* 
x*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/(a*(2 + m)) + 
(2*b*d^2*(b*c^2 + a*d^2)*Hypergeometric2F1[2, 1 + m, 2 + m, -((d*x)/c)])/( 
c*(1 + m)) + ((b*c^2*d + a*d^3)^2*Hypergeometric2F1[3, 1 + m, 2 + m, -((d* 
x)/c)])/(c^3*(1 + m))))/(b*c^2 + a*d^2)^3
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 315, 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 = {615, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b^2*c*(b*c^2 - 3*a*d^2)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 
+ m)/2, -((b*x^2)/a)])/(a*(b*c^2 + a*d^2)^3*e*(1 + m)) + (b*d^2*(3*b*c^2 - 
 a*d^2)*(e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((d*x)/c)])/(c*( 
b*c^2 + a*d^2)^3*e*(1 + m)) - (b^2*d*(3*b*c^2 - a*d^2)*(e*x)^(2 + m)*Hyper 
geometric2F1[1, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/(a*(b*c^2 + a*d^2)^3* 
e^2*(2 + m)) + (2*b*d^2*(e*x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, - 
((d*x)/c)])/(c*(b*c^2 + a*d^2)^2*e*(1 + m)) + (d^2*(e*x)^(1 + m)*Hypergeom 
etric2F1[3, 1 + m, 2 + m, -((d*x)/c)])/(c^3*(b*c^2 + a*d^2)*e*(1 + m))
 

Defintions of rubi rules used

rule 615

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] && ILtQ[p, 0]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((e*x)^m/(b*d^3*x^5 + 3*b*c*d^2*x^4 + 3*a*c^2*d*x + a*c^3 + (3*b*c 
^2*d + a*d^3)*x^3 + (b*c^3 + 3*a*c*d^2)*x^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m}{(c+d x)^3 \left (a+b x^2\right )} \, dx=e^{m} \left (\int \frac {x^{m}}{b \,d^{3} x^{5}+3 b c \,d^{2} x^{4}+a \,d^{3} x^{3}+3 b \,c^{2} d \,x^{3}+3 a c \,d^{2} x^{2}+b \,c^{3} x^{2}+3 a \,c^{2} d x +a \,c^{3}}d x \right ) \] Input:

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

Output:

e**m*int(x**m/(a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b*c 
**3*x**2 + 3*b*c**2*d*x**3 + 3*b*c*d**2*x**4 + b*d**3*x**5),x)