\(\int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx\) [307]

Optimal result

Integrand size = 28, antiderivative size = 233 \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx=-\frac {4 c (e x)^{1+m} (a-b x)^{7/2} (a c+b c x)^{9/2}}{7 e}+\frac {a^7 c^5 (11+4 m) (e x)^{1+m} \sqrt {a-b x} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{7 e (1+m) \sqrt {1-\frac {b^2 x^2}{a^2}}}+\frac {a^6 b c^5 (29+4 m) (e x)^{2+m} \sqrt {a-b x} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )}{7 e^2 (2+m) \sqrt {1-\frac {b^2 x^2}{a^2}}} \] Output:

-4/7*c*(e*x)^(1+m)*(-b*x+a)^(7/2)*(b*c*x+a*c)^(9/2)/e+1/7*a^7*c^5*(11+4*m) 
*(e*x)^(1+m)*(-b*x+a)^(1/2)*(b*c*x+a*c)^(1/2)*hypergeom([-7/2, 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)+1/7*a^6*b*c^5*(29+4 
*m)*(e*x)^(2+m)*(-b*x+a)^(1/2)*(b*c*x+a*c)^(1/2)*hypergeom([-7/2, 1+1/2*m] 
,[2+1/2*m],b^2*x^2/a^2)/e^2/(2+m)/(1-b^2*x^2/a^2)^(1/2)
 

Mathematica [A] (verified)

Time = 10.16 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.03 \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx=\frac {a^4 c^5 x (e x)^m \sqrt {a-b x} \sqrt {c (a+b x)} \left (a^3 \left (24+26 m+9 m^2+m^3\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )+b (1+m) x \left (3 a^2 \left (12+7 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )+b (2+m) x \left (3 a (4+m) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {3+m}{2},\frac {5+m}{2},\frac {b^2 x^2}{a^2}\right )+b (3+m) x \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {4+m}{2},\frac {6+m}{2},\frac {b^2 x^2}{a^2}\right )\right )\right )\right )}{(1+m) (2+m) (3+m) (4+m) \sqrt {1-\frac {b^2 x^2}{a^2}}} \] Input:

Integrate[(e*x)^m*(a - b*x)^(5/2)*(a*c + b*c*x)^(11/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 366, 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.

\(\displaystyle \int (a-b x)^{5/2} (e x)^m (a c+b c x)^{11/2} \, dx\)

\(\Big \downarrow \) 147

\(\displaystyle \int \left (a^3 c^3 (a-b x)^{5/2} (e x)^m (a c+b c x)^{5/2}+\frac {3 a^2 b c^3 (a-b x)^{5/2} (e x)^{m+1} (a c+b c x)^{5/2}}{e}+\frac {b^3 c^3 (a-b x)^{5/2} (e x)^{m+3} (a c+b c x)^{5/2}}{e^3}+\frac {3 a b^2 c^3 (a-b x)^{5/2} (e x)^{m+2} (a c+b c x)^{5/2}}{e^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^7 c^5 \sqrt {a-b x} (e x)^{m+1} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {b^2 x^2}{a^2}\right )}{e (m+1) \sqrt {1-\frac {b^2 x^2}{a^2}}}+\frac {3 a^6 b c^5 \sqrt {a-b x} (e x)^{m+2} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {b^2 x^2}{a^2}\right )}{e^2 (m+2) \sqrt {1-\frac {b^2 x^2}{a^2}}}+\frac {3 a^5 b^2 c^5 \sqrt {a-b x} (e x)^{m+3} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {m+3}{2},\frac {m+5}{2},\frac {b^2 x^2}{a^2}\right )}{e^3 (m+3) \sqrt {1-\frac {b^2 x^2}{a^2}}}+\frac {a^4 b^3 c^5 \sqrt {a-b x} (e x)^{m+4} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {m+4}{2},\frac {m+6}{2},\frac {b^2 x^2}{a^2}\right )}{e^4 (m+4) \sqrt {1-\frac {b^2 x^2}{a^2}}}\)

Input:

Int[(e*x)^m*(a - b*x)^(5/2)*(a*c + b*c*x)^(11/2),x]
 

Output:

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

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]

\[\int \left (e x \right )^{m} \left (-b x +a \right )^{\frac {5}{2}} \left (b c x +a c \right )^{\frac {11}{2}}d x\]

Input:

int((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(11/2),x)
 

Output:

int((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(11/2),x)
 

Fricas [F]

\[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx=\int { {\left (b c x + a c\right )}^{\frac {11}{2}} {\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m} \,d x } \] Input:

integrate((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(11/2),x, algorithm="fricas")
 

Output:

integral((b^7*c^5*x^7 + 3*a*b^6*c^5*x^6 + a^2*b^5*c^5*x^5 - 5*a^3*b^4*c^5* 
x^4 - 5*a^4*b^3*c^5*x^3 + a^5*b^2*c^5*x^2 + 3*a^6*b*c^5*x + a^7*c^5)*sqrt( 
b*c*x + a*c)*sqrt(-b*x + a)*(e*x)^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx=\text {Timed out} \] Input:

integrate((e*x)**m*(-b*x+a)**(5/2)*(b*c*x+a*c)**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx=\int { {\left (b c x + a c\right )}^{\frac {11}{2}} {\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m} \,d x } \] Input:

integrate((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(11/2),x, algorithm="maxima")
 

Output:

integrate((b*c*x + a*c)^(11/2)*(-b*x + a)^(5/2)*(e*x)^m, x)
 

Giac [F]

\[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx=\int { {\left (b c x + a c\right )}^{\frac {11}{2}} {\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m} \,d x } \] Input:

integrate((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(11/2),x, algorithm="giac")
 

Output:

integrate((b*c*x + a*c)^(11/2)*(-b*x + a)^(5/2)*(e*x)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx=\int {\left (a\,c+b\,c\,x\right )}^{11/2}\,{\left (e\,x\right )}^m\,{\left (a-b\,x\right )}^{5/2} \,d x \] Input:

int((a*c + b*c*x)^(11/2)*(e*x)^m*(a - b*x)^(5/2),x)
 

Output:

int((a*c + b*c*x)^(11/2)*(e*x)^m*(a - b*x)^(5/2), x)
 

Reduce [F]

\[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{11/2} \, dx=\int \left (e x \right )^{m} \left (-b x +a \right )^{\frac {5}{2}} \left (b c x +a c \right )^{\frac {11}{2}}d x \] Input:

int((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(11/2),x)
 

Output:

int((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(11/2),x)