\(\int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx\) [1701]

Optimal result

Integrand size = 28, antiderivative size = 59 \[ \int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx=\frac {(1+a x)^{7/2} (c+a c x)^m \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {7}{2}+m,\frac {9}{2}+m,\frac {1}{2} (1+a x)\right )}{2 \sqrt {2} a (7+2 m)} \] Output:

1/4*(a*x+1)^(7/2)*(a*c*x+c)^m*hypergeom([5/2, 7/2+m],[9/2+m],1/2*a*x+1/2)* 
2^(1/2)/a/(7+2*m)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.10 \[ \int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx=\frac {2^{\frac {7}{2}+m} (1+a x)^{-m} (c+a c x)^m \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{2}-m,-\frac {1}{2},\frac {1}{2} (1-a x)\right )}{3 a (1-a x)^{3/2}} \] Input:

Integrate[((1 + a*x)^(5/2)*(c + a*c*x)^m)/(1 - a*x)^(5/2),x]
 

Output:

(2^(7/2 + m)*(c + a*c*x)^m*Hypergeometric2F1[-3/2, -5/2 - m, -1/2, (1 - a* 
x)/2])/(3*a*(1 - a*x)^(3/2)*(1 + a*x)^m)
 

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {37, 79}

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.

\(\displaystyle \int \frac {(a x+1)^{5/2} (a c x+c)^m}{(1-a x)^{5/2}} \, dx\)

\(\Big \downarrow \) 37

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

\(\Big \downarrow \) 79

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

Input:

Int[((1 + a*x)^(5/2)*(c + a*c*x)^m)/(1 - a*x)^(5/2),x]
 

Output:

(2^(7/2 + m)*(c + a*c*x)^m*Hypergeometric2F1[-3/2, -5/2 - m, -1/2, (1 - a* 
x)/2])/(3*a*(1 - a*x)^(3/2)*(1 + a*x)^m)
 

Defintions of rubi rules used

rule 37

Int[(u_.)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> S 
imp[(a + b*x)^m/(c + d*x)^m   Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, 
b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !SimplerQ[a + b*x, c + d*x]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

Maple [F]

\[\int \frac {\left (a x +1\right )^{\frac {5}{2}} \left (a c x +c \right )^{m}}{\left (-a x +1\right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )}^{\frac {5}{2}} {\left (a c x + c\right )}^{m}}{{\left (-a x + 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

integral(-(a^2*x^2 + 2*a*x + 1)*sqrt(a*x + 1)*sqrt(-a*x + 1)*(a*c*x + c)^m 
/(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )}^{\frac {5}{2}} {\left (a c x + c\right )}^{m}}{{\left (-a x + 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )}^{\frac {5}{2}} {\left (a c x + c\right )}^{m}}{{\left (-a x + 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx=\int \frac {{\left (c+a\,c\,x\right )}^m\,{\left (a\,x+1\right )}^{5/2}}{{\left (1-a\,x\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(1+a x)^{5/2} (c+a c x)^m}{(1-a x)^{5/2}} \, dx=\left (\int \frac {\sqrt {a x +1}\, \left (a c x +c \right )^{m} x^{2}}{\sqrt {-a x +1}\, a^{2} x^{2}-2 \sqrt {-a x +1}\, a x +\sqrt {-a x +1}}d x \right ) a^{2}+2 \left (\int \frac {\sqrt {a x +1}\, \left (a c x +c \right )^{m} x}{\sqrt {-a x +1}\, a^{2} x^{2}-2 \sqrt {-a x +1}\, a x +\sqrt {-a x +1}}d x \right ) a +\int \frac {\sqrt {a x +1}\, \left (a c x +c \right )^{m}}{\sqrt {-a x +1}\, a^{2} x^{2}-2 \sqrt {-a x +1}\, a x +\sqrt {-a x +1}}d x \] Input:

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

Output:

int((sqrt(a*x + 1)*(a*c*x + c)**m*x**2)/(sqrt( - a*x + 1)*a**2*x**2 - 2*sq 
rt( - a*x + 1)*a*x + sqrt( - a*x + 1)),x)*a**2 + 2*int((sqrt(a*x + 1)*(a*c 
*x + c)**m*x)/(sqrt( - a*x + 1)*a**2*x**2 - 2*sqrt( - a*x + 1)*a*x + sqrt( 
 - a*x + 1)),x)*a + int((sqrt(a*x + 1)*(a*c*x + c)**m)/(sqrt( - a*x + 1)*a 
**2*x**2 - 2*sqrt( - a*x + 1)*a*x + sqrt( - a*x + 1)),x)