\(\int \frac {(e x)^m}{(a+b x) (a c-b c x)^2} \, dx\) [324]

Optimal result

Integrand size = 24, antiderivative size = 98 \[ \int \frac {(e x)^m}{(a+b x) (a c-b c x)^2} \, dx=\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{a^3 c^2 e (1+m)}+\frac {b (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )}{a^4 c^2 e^2 (2+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.89 \[ \int \frac {(e x)^m}{(a+b x) (a c-b c x)^2} \, dx=\frac {x (e x)^m \left (b (1+m) x \operatorname {Hypergeometric2F1}\left (2,1+\frac {m}{2},2+\frac {m}{2},\frac {b^2 x^2}{a^2}\right )+a (2+m) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )\right )}{a^4 c^2 (1+m) (2+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {92, 27, 82, 278}

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.

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

\(\Big \downarrow \) 92

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 82

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 82

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

rule 92

Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), 
x_] :> Simp[a   Int[(a + b*x)^n*(c + d*x)^n*(f*x)^p, x], x] + Simp[b/f   In 
t[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, f, m, 
 n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] &&  !RationalQ[p] && 
!IGtQ[m, 0] && NeQ[m + n + p + 2, 0]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.45 (sec) , antiderivative size = 440, normalized size of antiderivative = 4.49 \[ \int \frac {(e x)^m}{(a+b x) (a c-b c x)^2} \, dx=- \frac {2 a e^{m} m^{2} x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} + \frac {a e^{m} m x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} - \frac {a e^{m} m x^{m} \Phi \left (\frac {a e^{i \pi }}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} + \frac {2 b e^{m} m^{2} x x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} - \frac {b e^{m} m x x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} + \frac {b e^{m} m x x^{m} \Phi \left (\frac {a e^{i \pi }}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} + \frac {2 b e^{m} m x x^{m} \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} \] Input:

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

Output:

-2*a*e**m*m**2*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4* 
a**3*b*c**2*gamma(1 - m) + 4*a**2*b**2*c**2*x*gamma(1 - m)) + a*e**m*m*x** 
m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma( 
1 - m) + 4*a**2*b**2*c**2*x*gamma(1 - m)) - a*e**m*m*x**m*lerchphi(a*exp_p 
olar(I*pi)/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma(1 
- m) + 4*a**2*b**2*c**2*x*gamma(1 - m)) + 2*b*e**m*m**2*x*x**m*lerchphi(a/ 
(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma(1 - m) + 4*a* 
*2*b**2*c**2*x*gamma(1 - m)) - b*e**m*m*x*x**m*lerchphi(a/(b*x), 1, m*exp_ 
polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma(1 - m) + 4*a**2*b**2*c**2*x*g 
amma(1 - m)) + b*e**m*m*x*x**m*lerchphi(a*exp_polar(I*pi)/(b*x), 1, m*exp_ 
polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma(1 - m) + 4*a**2*b**2*c**2*x*g 
amma(1 - m)) + 2*b*e**m*m*x*x**m*gamma(-m)/(-4*a**3*b*c**2*gamma(1 - m) + 
4*a**2*b**2*c**2*x*gamma(1 - m))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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