\(\int \frac {(e x)^m (a c-b c x)^3}{a+b x} \, dx\) [320]

Optimal result

Integrand size = 24, antiderivative size = 107 \[ \int \frac {(e x)^m (a c-b c x)^3}{a+b x} \, dx=-\frac {7 a^2 c^3 (e x)^{1+m}}{e (1+m)}+\frac {4 a b c^3 (e x)^{2+m}}{e^2 (2+m)}-\frac {b^2 c^3 (e x)^{3+m}}{e^3 (3+m)}+\frac {8 a^2 c^3 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )}{e (1+m)} \] Output:

-7*a^2*c^3*(e*x)^(1+m)/e/(1+m)+4*a*b*c^3*(e*x)^(2+m)/e^2/(2+m)-b^2*c^3*(e* 
x)^(3+m)/e^3/(3+m)+8*a^2*c^3*(e*x)^(1+m)*hypergeom([1, 1+m],[2+m],-b*x/a)/ 
e/(1+m)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64 \[ \int \frac {(e x)^m (a c-b c x)^3}{a+b x} \, dx=c^3 x (e x)^m \left (-\frac {7 a^2}{1+m}+\frac {4 a b x}{2+m}-\frac {b^2 x^2}{3+m}+\frac {8 a^2 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )}{1+m}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {99, 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.

\(\displaystyle \int \frac {(e x)^m (a c-b c x)^3}{a+b x} \, dx\)

\(\Big \downarrow \) 99

\(\displaystyle \int \left (\frac {8 a^3 c^3 (e x)^m}{a+b x}-4 a^2 c^3 (e x)^m-2 a c^2 (e x)^m (a c-b c x)-c (e x)^m (a c-b c x)^2\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {8 a^2 c^3 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {b x}{a}\right )}{e (m+1)}-\frac {7 a^2 c^3 (e x)^{m+1}}{e (m+1)}+\frac {4 a b c^3 (e x)^{m+2}}{e^2 (m+2)}-\frac {b^2 c^3 (e x)^{m+3}}{e^3 (m+3)}\)

Input:

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

Output:

(-7*a^2*c^3*(e*x)^(1 + m))/(e*(1 + m)) + (4*a*b*c^3*(e*x)^(2 + m))/(e^2*(2 
 + m)) - (b^2*c^3*(e*x)^(3 + m))/(e^3*(3 + m)) + (8*a^2*c^3*(e*x)^(1 + m)* 
Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(e*(1 + m))
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.93 (sec) , antiderivative size = 330, normalized size of antiderivative = 3.08 \[ \int \frac {(e x)^m (a c-b c x)^3}{a+b x} \, dx=\frac {a^{2} c^{3} e^{m} m x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} + \frac {a^{2} c^{3} e^{m} x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} - \frac {3 a b c^{3} e^{m} m x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} - \frac {6 a b c^{3} e^{m} x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} + \frac {3 b^{2} c^{3} e^{m} m x^{m + 3} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{\Gamma \left (m + 4\right )} + \frac {9 b^{2} c^{3} e^{m} x^{m + 3} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{\Gamma \left (m + 4\right )} - \frac {b^{3} c^{3} e^{m} m x^{m + 4} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 4\right ) \Gamma \left (m + 4\right )}{a \Gamma \left (m + 5\right )} - \frac {4 b^{3} c^{3} e^{m} x^{m + 4} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 4\right ) \Gamma \left (m + 4\right )}{a \Gamma \left (m + 5\right )} \] Input:

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

Output:

a**2*c**3*e**m*m*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamm 
a(m + 1)/gamma(m + 2) + a**2*c**3*e**m*x**(m + 1)*lerchphi(b*x*exp_polar(I 
*pi)/a, 1, m + 1)*gamma(m + 1)/gamma(m + 2) - 3*a*b*c**3*e**m*m*x**(m + 2) 
*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m + 2)/gamma(m + 3) - 6*a 
*b*c**3*e**m*x**(m + 2)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m 
+ 2)/gamma(m + 3) + 3*b**2*c**3*e**m*m*x**(m + 3)*lerchphi(b*x*exp_polar(I 
*pi)/a, 1, m + 3)*gamma(m + 3)/gamma(m + 4) + 9*b**2*c**3*e**m*x**(m + 3)* 
lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 3)*gamma(m + 3)/gamma(m + 4) - b**3 
*c**3*e**m*m*x**(m + 4)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 4)*gamma(m 
+ 4)/(a*gamma(m + 5)) - 4*b**3*c**3*e**m*x**(m + 4)*lerchphi(b*x*exp_polar 
(I*pi)/a, 1, m + 4)*gamma(m + 4)/(a*gamma(m + 5))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m (a c-b c x)^3}{a+b x} \, dx=\frac {e^{m} c^{3} \left (8 x^{m} a^{3} m^{3}+48 x^{m} a^{3} m^{2}+88 x^{m} a^{3} m +48 x^{m} a^{3}-7 x^{m} a^{2} b \,m^{3} x -35 x^{m} a^{2} b \,m^{2} x -42 x^{m} a^{2} b m x +4 x^{m} a \,b^{2} m^{3} x^{2}+16 x^{m} a \,b^{2} m^{2} x^{2}+12 x^{m} a \,b^{2} m \,x^{2}-x^{m} b^{3} m^{3} x^{3}-3 x^{m} b^{3} m^{2} x^{3}-2 x^{m} b^{3} m \,x^{3}-8 \left (\int \frac {x^{m}}{b \,x^{2}+a x}d x \right ) a^{4} m^{4}-48 \left (\int \frac {x^{m}}{b \,x^{2}+a x}d x \right ) a^{4} m^{3}-88 \left (\int \frac {x^{m}}{b \,x^{2}+a x}d x \right ) a^{4} m^{2}-48 \left (\int \frac {x^{m}}{b \,x^{2}+a x}d x \right ) a^{4} m \right )}{b m \left (m^{3}+6 m^{2}+11 m +6\right )} \] Input:

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

Output:

(e**m*c**3*(8*x**m*a**3*m**3 + 48*x**m*a**3*m**2 + 88*x**m*a**3*m + 48*x** 
m*a**3 - 7*x**m*a**2*b*m**3*x - 35*x**m*a**2*b*m**2*x - 42*x**m*a**2*b*m*x 
 + 4*x**m*a*b**2*m**3*x**2 + 16*x**m*a*b**2*m**2*x**2 + 12*x**m*a*b**2*m*x 
**2 - x**m*b**3*m**3*x**3 - 3*x**m*b**3*m**2*x**3 - 2*x**m*b**3*m*x**3 - 8 
*int(x**m/(a*x + b*x**2),x)*a**4*m**4 - 48*int(x**m/(a*x + b*x**2),x)*a**4 
*m**3 - 88*int(x**m/(a*x + b*x**2),x)*a**4*m**2 - 48*int(x**m/(a*x + b*x** 
2),x)*a**4*m))/(b*m*(m**3 + 6*m**2 + 11*m + 6))