\(\int \frac {(e x)^m (a-b x)^{5/2}}{\sqrt {a c+b c x}} \, dx\) [312]

Optimal result

Integrand size = 28, antiderivative size = 223 \[ \int \frac {(e x)^m (a-b x)^{5/2}}{\sqrt {a c+b c x}} \, dx=-\frac {4 (e x)^{1+m} (a-b x)^{3/2} \sqrt {a c+b c x}}{c e}+\frac {a (5+4 m) (e x)^{1+m} \sqrt {a-b x} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{c e (1+m) \sqrt {1-\frac {b^2 x^2}{a^2}}}-\frac {b (11+4 m) (e x)^{2+m} \sqrt {a-b x} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )}{c e^2 (2+m) \sqrt {1-\frac {b^2 x^2}{a^2}}} \] Output:

-4*(e*x)^(1+m)*(-b*x+a)^(3/2)*(b*c*x+a*c)^(1/2)/c/e+a*(5+4*m)*(e*x)^(1+m)* 
(-b*x+a)^(1/2)*(b*c*x+a*c)^(1/2)*hypergeom([-1/2, 1/2+1/2*m],[3/2+1/2*m],b 
^2*x^2/a^2)/c/e/(1+m)/(1-b^2*x^2/a^2)^(1/2)-b*(11+4*m)*(e*x)^(2+m)*(-b*x+a 
)^(1/2)*(b*c*x+a*c)^(1/2)*hypergeom([-1/2, 1+1/2*m],[2+1/2*m],b^2*x^2/a^2) 
/c/e^2/(2+m)/(1-b^2*x^2/a^2)^(1/2)
 

Mathematica [A] (verified)

Time = 8.58 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.05 \[ \int \frac {(e x)^m (a-b x)^{5/2}}{\sqrt {a c+b c x}} \, dx=\frac {x (e x)^m \sqrt {1-\frac {b^2 x^2}{a^2}} \left (a^3 \left (24+26 m+9 m^2+m^3\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )-b (1+m) x \left (3 a^2 \left (12+7 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )+b (2+m) x \left (-3 a (4+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\frac {b^2 x^2}{a^2}\right )+b (3+m) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+m}{2},\frac {6+m}{2},\frac {b^2 x^2}{a^2}\right )\right )\right )\right )}{(1+m) (2+m) (3+m) (4+m) \sqrt {a-b x} \sqrt {c (a+b x)}} \] Input:

Integrate[((e*x)^m*(a - b*x)^(5/2))/Sqrt[a*c + b*c*x],x]
 

Output:

(x*(e*x)^m*Sqrt[1 - (b^2*x^2)/a^2]*(a^3*(24 + 26*m + 9*m^2 + m^3)*Hypergeo 
metric2F1[1/2, (1 + m)/2, (3 + m)/2, (b^2*x^2)/a^2] - b*(1 + m)*x*(3*a^2*( 
12 + 7*m + m^2)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, (b^2*x^2)/a^2 
] + b*(2 + m)*x*(-3*a*(4 + m)*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, 
 (b^2*x^2)/a^2] + b*(3 + m)*x*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, 
 (b^2*x^2)/a^2]))))/((1 + m)*(2 + m)*(3 + m)*(4 + m)*Sqrt[a - b*x]*Sqrt[c* 
(a + b*x)])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.57, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {147, 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.

\(\displaystyle \int \frac {(a-b x)^{5/2} (e x)^m}{\sqrt {a c+b c x}} \, dx\)

\(\Big \downarrow \) 147

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 a b^2 \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},\frac {b^2 x^2}{a^2}\right )}{e^3 (m+3) \sqrt {a-b x} \sqrt {a c+b c x}}-\frac {3 a^2 b \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {b^2 x^2}{a^2}\right )}{e^2 (m+2) \sqrt {a-b x} \sqrt {a c+b c x}}-\frac {b^3 \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+4}{2},\frac {m+6}{2},\frac {b^2 x^2}{a^2}\right )}{e^4 (m+4) \sqrt {a-b x} \sqrt {a c+b c x}}+\frac {a^3 \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {b^2 x^2}{a^2}\right )}{e (m+1) \sqrt {a-b x} \sqrt {a c+b c x}}\)

Input:

Int[((e*x)^m*(a - b*x)^(5/2))/Sqrt[a*c + b*c*x],x]
 

Output:

(a^3*(e*x)^(1 + m)*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeometric2F1[1/2, (1 + m)/ 
2, (3 + m)/2, (b^2*x^2)/a^2])/(e*(1 + m)*Sqrt[a - b*x]*Sqrt[a*c + b*c*x]) 
- (3*a^2*b*(e*x)^(2 + m)*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeometric2F1[1/2, (2 
 + m)/2, (4 + m)/2, (b^2*x^2)/a^2])/(e^2*(2 + m)*Sqrt[a - b*x]*Sqrt[a*c + 
b*c*x]) + (3*a*b^2*(e*x)^(3 + m)*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeometric2F1 
[1/2, (3 + m)/2, (5 + m)/2, (b^2*x^2)/a^2])/(e^3*(3 + m)*Sqrt[a - b*x]*Sqr 
t[a*c + b*c*x]) - (b^3*(e*x)^(4 + m)*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeometri 
c2F1[1/2, (4 + m)/2, (6 + m)/2, (b^2*x^2)/a^2])/(e^4*(4 + m)*Sqrt[a - b*x] 
*Sqrt[a*c + b*c*x])
 

Defintions of rubi rules used

rule 147

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

rule 2009

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

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (-b x +a \right )^{\frac {5}{2}}}{\sqrt {b c x +a c}}d x\]

Input:

int((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(1/2),x)
 

Output:

int((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(1/2),x)
 

Fricas [F]

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

integrate((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((e*x)**m*(-b*x+a)**(5/2)/(b*c*x+a*c)**(1/2),x)
 

Output:

Integral((e*x)**m*(a - b*x)**(5/2)/sqrt(c*(a + b*x)), x)
 

Maxima [F]

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

integrate((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(1/2),x, algorithm="maxima")
 

Output:

integrate((-b*x + a)^(5/2)*(e*x)^m/sqrt(b*c*x + a*c), x)
 

Giac [F]

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

integrate((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(1/2),x, algorithm="giac")
 

Output:

integrate((-b*x + a)^(5/2)*(e*x)^m/sqrt(b*c*x + a*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m (a-b x)^{5/2}}{\sqrt {a c+b c x}} \, dx=\int \frac {{\left (e\,x\right )}^m\,{\left (a-b\,x\right )}^{5/2}}{\sqrt {a\,c+b\,c\,x}} \,d x \] Input:

int(((e*x)^m*(a - b*x)^(5/2))/(a*c + b*c*x)^(1/2),x)
 

Output:

int(((e*x)^m*(a - b*x)^(5/2))/(a*c + b*c*x)^(1/2), x)
 

Reduce [F]

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

int((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(1/2),x)
 

Output:

(e**m*(int((x**m*sqrt(a - b*x)*x**2)/sqrt(a + b*x),x)*b**2 - 2*int((x**m*s 
qrt(a - b*x)*x)/sqrt(a + b*x),x)*a*b + int((x**m*sqrt(a - b*x))/sqrt(a + b 
*x),x)*a**2))/sqrt(c)