\(\int \frac {(e x)^m}{(c+d x) (a+b x^2)} \, dx\) [120]

Optimal result

Integrand size = 22, antiderivative size = 170 \[ \int \frac {(e x)^m}{(c+d x) \left (a+b x^2\right )} \, dx=\frac {b c (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a \left (b c^2+a d^2\right ) e (1+m)}+\frac {d^2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{c \left (b c^2+a d^2\right ) e (1+m)}-\frac {b d (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a \left (b c^2+a d^2\right ) e^2 (2+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.75 \[ \int \frac {(e x)^m}{(c+d x) \left (a+b x^2\right )} \, dx=\frac {x (e x)^m \left (b c^2 (2+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+d \left (a d (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )-b c (1+m) x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{a c \left (b c^2+a d^2\right ) (1+m) (2+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 170, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.35 (sec) , antiderivative size = 983, normalized size of antiderivative = 5.78 \[ \int \frac {(e x)^m}{(c+d x) \left (a+b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

c*e**m*m*x**(m - 1)*lerchphi(a*exp_polar(I*pi)/(b*x**2), 1, 1/2 - m/2)*gam 
ma(1/2 - m/2)*gamma(1 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2)/(4*a*d**2*gam 
ma(1 - m/2)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2) + 4*b*c**2*ga 
mma(1 - m/2)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2)) - c*e**m*x* 
*(m - 1)*lerchphi(a*exp_polar(I*pi)/(b*x**2), 1, 1/2 - m/2)*gamma(1/2 - m/ 
2)*gamma(1 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2)/(4*a*d**2*gamma(1 - m/2) 
*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2) + 4*b*c**2*gamma(1 - m/2 
)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2)) + c**m*c**(2 - m)*d**( 
1 - m)*d**(m - 2)*e**m*m*x**(m - 2)*lerchphi(c**2/(d**2*x**2), 1, 1 - m/2) 
*gamma(1 - m/2)**2*gamma(3/2 - m/2)*gamma(5/2 - m/2)/(4*a*d**2*gamma(1 - m 
/2)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2) + 4*b*c**2*gamma(1 - 
m/2)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2)) - 2*c**m*c**(2 - m) 
*d**(1 - m)*d**(m - 2)*e**m*x**(m - 2)*lerchphi(c**2/(d**2*x**2), 1, 1 - m 
/2)*gamma(1 - m/2)**2*gamma(3/2 - m/2)*gamma(5/2 - m/2)/(4*a*d**2*gamma(1 
- m/2)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2) + 4*b*c**2*gamma(1 
 - m/2)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2)) - c**m*c**(3 - m 
)*d**(1 - m)*d**(m - 3)*e**m*m*x**(m - 3)*lerchphi(c**2/(d**2*x**2), 1, 3/ 
2 - m/2)*gamma(1 - m/2)*gamma(3/2 - m/2)**2*gamma(2 - m/2)/(4*a*d**2*gamma 
(1 - m/2)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2) + 4*b*c**2*gamm 
a(1 - m/2)*gamma(3/2 - m/2)*gamma(2 - m/2)*gamma(5/2 - m/2)) + 3*c**m*c...
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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