\(\int (e x)^m (a-b x)^{5/2} (a c+b c x)^{5/2} \, dx\) [310]

Optimal result

Integrand size = 28, antiderivative size = 89 \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{5/2} \, dx=\frac {a^4 c^2 (e x)^{1+m} \sqrt {a-b x} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{e (1+m) \sqrt {1-\frac {b^2 x^2}{a^2}}} \] Output:

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

Mathematica [A] (verified)

Time = 10.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{5/2} \, dx=\frac {a^4 c^2 x (e x)^m \sqrt {a-b x} \sqrt {c (a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{(1+m) \sqrt {1-\frac {b^2 x^2}{a^2}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {136, 279, 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.

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

\(\Big \downarrow \) 136

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

\(\Big \downarrow \) 279

\(\displaystyle \frac {a^4 c^2 \sqrt {a-b x} \sqrt {a c+b c x} \int (e x)^m \left (1-\frac {b^2 x^2}{a^2}\right )^{5/2}dx}{\sqrt {1-\frac {b^2 x^2}{a^2}}}\)

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 136

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

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])
 

rule 279

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b^4*c^2*x^4 - 2*a^2*b^2*c^2*x^2 + a^4*c^2)*sqrt(b*c*x + a*c)*sqr 
t(-b*x + a)*(e*x)^m, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

e**m*sqrt(c)*c**2*(int(x**m*sqrt(a + b*x)*sqrt(a - b*x)*x**4,x)*b**4 - 2*i 
nt(x**m*sqrt(a + b*x)*sqrt(a - b*x)*x**2,x)*a**2*b**2 + int(x**m*sqrt(a + 
b*x)*sqrt(a - b*x),x)*a**4)