\(\int \frac {(c x)^m}{(a+b x)^{5/2}} \, dx\) [782]

Optimal result

Integrand size = 15, antiderivative size = 50 \[ \int \frac {(c x)^m}{(a+b x)^{5/2}} \, dx=-\frac {2 \left (-\frac {b x}{a}\right )^{-m} (c x)^m \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-m,-\frac {1}{2},1+\frac {b x}{a}\right )}{3 b (a+b x)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {(c x)^m}{(a+b x)^{5/2}} \, dx=-\frac {2 \left (-\frac {b x}{a}\right )^{-m} (c x)^m \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-m,-\frac {1}{2},1+\frac {b x}{a}\right )}{3 b (a+b x)^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {77, 75}

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.

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

\(\Big \downarrow \) 77

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 75

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

rule 77

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.78 \[ \int \frac {(c x)^m}{(a+b x)^{5/2}} \, dx=\frac {c^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{a^{\frac {5}{2}} \Gamma \left (m + 2\right )} \] Input:

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

Output:

c**m*x**(m + 1)*gamma(m + 1)*hyper((5/2, m + 1), (m + 2,), b*x*exp_polar(I 
*pi)/a)/(a**(5/2)*gamma(m + 2))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c x)^m}{(a+b x)^{5/2}} \, dx=\frac {2 c^{m} \left (x^{m} \sqrt {b x +a}-2 \left (\int \frac {x^{m} \sqrt {b x +a}}{2 b^{3} m \,x^{4}+6 a \,b^{2} m \,x^{3}-3 b^{3} x^{4}+6 a^{2} b m \,x^{2}-9 a \,b^{2} x^{3}+2 a^{3} m x -9 a^{2} b \,x^{2}-3 a^{3} x}d x \right ) a^{3} m^{2}+3 \left (\int \frac {x^{m} \sqrt {b x +a}}{2 b^{3} m \,x^{4}+6 a \,b^{2} m \,x^{3}-3 b^{3} x^{4}+6 a^{2} b m \,x^{2}-9 a \,b^{2} x^{3}+2 a^{3} m x -9 a^{2} b \,x^{2}-3 a^{3} x}d x \right ) a^{3} m -4 \left (\int \frac {x^{m} \sqrt {b x +a}}{2 b^{3} m \,x^{4}+6 a \,b^{2} m \,x^{3}-3 b^{3} x^{4}+6 a^{2} b m \,x^{2}-9 a \,b^{2} x^{3}+2 a^{3} m x -9 a^{2} b \,x^{2}-3 a^{3} x}d x \right ) a^{2} b \,m^{2} x +6 \left (\int \frac {x^{m} \sqrt {b x +a}}{2 b^{3} m \,x^{4}+6 a \,b^{2} m \,x^{3}-3 b^{3} x^{4}+6 a^{2} b m \,x^{2}-9 a \,b^{2} x^{3}+2 a^{3} m x -9 a^{2} b \,x^{2}-3 a^{3} x}d x \right ) a^{2} b m x -2 \left (\int \frac {x^{m} \sqrt {b x +a}}{2 b^{3} m \,x^{4}+6 a \,b^{2} m \,x^{3}-3 b^{3} x^{4}+6 a^{2} b m \,x^{2}-9 a \,b^{2} x^{3}+2 a^{3} m x -9 a^{2} b \,x^{2}-3 a^{3} x}d x \right ) a \,b^{2} m^{2} x^{2}+3 \left (\int \frac {x^{m} \sqrt {b x +a}}{2 b^{3} m \,x^{4}+6 a \,b^{2} m \,x^{3}-3 b^{3} x^{4}+6 a^{2} b m \,x^{2}-9 a \,b^{2} x^{3}+2 a^{3} m x -9 a^{2} b \,x^{2}-3 a^{3} x}d x \right ) a \,b^{2} m \,x^{2}\right )}{b \left (2 b^{2} m \,x^{2}+4 a b m x -3 b^{2} x^{2}+2 a^{2} m -6 a b x -3 a^{2}\right )} \] Input:

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

Output:

(2*c**m*(x**m*sqrt(a + b*x) - 2*int((x**m*sqrt(a + b*x))/(2*a**3*m*x - 3*a 
**3*x + 6*a**2*b*m*x**2 - 9*a**2*b*x**2 + 6*a*b**2*m*x**3 - 9*a*b**2*x**3 
+ 2*b**3*m*x**4 - 3*b**3*x**4),x)*a**3*m**2 + 3*int((x**m*sqrt(a + b*x))/( 
2*a**3*m*x - 3*a**3*x + 6*a**2*b*m*x**2 - 9*a**2*b*x**2 + 6*a*b**2*m*x**3 
- 9*a*b**2*x**3 + 2*b**3*m*x**4 - 3*b**3*x**4),x)*a**3*m - 4*int((x**m*sqr 
t(a + b*x))/(2*a**3*m*x - 3*a**3*x + 6*a**2*b*m*x**2 - 9*a**2*b*x**2 + 6*a 
*b**2*m*x**3 - 9*a*b**2*x**3 + 2*b**3*m*x**4 - 3*b**3*x**4),x)*a**2*b*m**2 
*x + 6*int((x**m*sqrt(a + b*x))/(2*a**3*m*x - 3*a**3*x + 6*a**2*b*m*x**2 - 
 9*a**2*b*x**2 + 6*a*b**2*m*x**3 - 9*a*b**2*x**3 + 2*b**3*m*x**4 - 3*b**3* 
x**4),x)*a**2*b*m*x - 2*int((x**m*sqrt(a + b*x))/(2*a**3*m*x - 3*a**3*x + 
6*a**2*b*m*x**2 - 9*a**2*b*x**2 + 6*a*b**2*m*x**3 - 9*a*b**2*x**3 + 2*b**3 
*m*x**4 - 3*b**3*x**4),x)*a*b**2*m**2*x**2 + 3*int((x**m*sqrt(a + b*x))/(2 
*a**3*m*x - 3*a**3*x + 6*a**2*b*m*x**2 - 9*a**2*b*x**2 + 6*a*b**2*m*x**3 - 
 9*a*b**2*x**3 + 2*b**3*m*x**4 - 3*b**3*x**4),x)*a*b**2*m*x**2))/(b*(2*a** 
2*m - 3*a**2 + 4*a*b*m*x - 6*a*b*x + 2*b**2*m*x**2 - 3*b**2*x**2))