\(\int \frac {x^m}{(a+b \sqrt {x})^2} \, dx\) [152]

Optimal result

Integrand size = 15, antiderivative size = 37 \[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,2 (1+m),3+2 m,-\frac {b \sqrt {x}}{a}\right )}{a^2 (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,2 (1+m),1+2 (1+m),-\frac {b \sqrt {x}}{a}\right )}{a^2 (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {864, 74}

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.

\(\displaystyle \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx\)

\(\Big \downarrow \) 864

\(\displaystyle 2 \int \frac {x^{\frac {1}{2} (2 m+1)}}{\left (a+b \sqrt {x}\right )^2}d\sqrt {x}\)

\(\Big \downarrow \) 74

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 74

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

rule 864

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi 
nator[n]}, Simp[k   Subst[Int[x^(k*(m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x 
^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]
 

Maple [F]

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

Input:

int(x^m/(a+b*x^(1/2))^2,x)
 

Output:

int(x^m/(a+b*x^(1/2))^2,x)
 

Fricas [F]

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

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

Output:

integral(-(2*a*b*sqrt(x)*x^m - (b^2*x + a^2)*x^m)/(b^4*x^2 - 2*a^2*b^2*x + 
 a^4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.20 (sec) , antiderivative size = 478, normalized size of antiderivative = 12.92 \[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=- \frac {8 a m^{2} x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {12 a m x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} + \frac {4 a m x^{m + 1} \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {4 a x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} + \frac {4 a x^{m + 1} \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {8 b m^{2} \sqrt {x} x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {12 b m \sqrt {x} x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {4 b \sqrt {x} x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} \] Input:

integrate(x**m/(a+b*x**(1/2))**2,x)
 

Output:

-8*a*m**2*x**(m + 1)*lerchphi(b*sqrt(x)*exp_polar(I*pi)/a, 1, 2*m + 2)*gam 
ma(2*m + 2)/(a**3*gamma(2*m + 3) + a**2*b*sqrt(x)*gamma(2*m + 3)) - 12*a*m 
*x**(m + 1)*lerchphi(b*sqrt(x)*exp_polar(I*pi)/a, 1, 2*m + 2)*gamma(2*m + 
2)/(a**3*gamma(2*m + 3) + a**2*b*sqrt(x)*gamma(2*m + 3)) + 4*a*m*x**(m + 1 
)*gamma(2*m + 2)/(a**3*gamma(2*m + 3) + a**2*b*sqrt(x)*gamma(2*m + 3)) - 4 
*a*x**(m + 1)*lerchphi(b*sqrt(x)*exp_polar(I*pi)/a, 1, 2*m + 2)*gamma(2*m 
+ 2)/(a**3*gamma(2*m + 3) + a**2*b*sqrt(x)*gamma(2*m + 3)) + 4*a*x**(m + 1 
)*gamma(2*m + 2)/(a**3*gamma(2*m + 3) + a**2*b*sqrt(x)*gamma(2*m + 3)) - 8 
*b*m**2*sqrt(x)*x**(m + 1)*lerchphi(b*sqrt(x)*exp_polar(I*pi)/a, 1, 2*m + 
2)*gamma(2*m + 2)/(a**3*gamma(2*m + 3) + a**2*b*sqrt(x)*gamma(2*m + 3)) - 
12*b*m*sqrt(x)*x**(m + 1)*lerchphi(b*sqrt(x)*exp_polar(I*pi)/a, 1, 2*m + 2 
)*gamma(2*m + 2)/(a**3*gamma(2*m + 3) + a**2*b*sqrt(x)*gamma(2*m + 3)) - 4 
*b*sqrt(x)*x**(m + 1)*lerchphi(b*sqrt(x)*exp_polar(I*pi)/a, 1, 2*m + 2)*ga 
mma(2*m + 2)/(a**3*gamma(2*m + 3) + a**2*b*sqrt(x)*gamma(2*m + 3))
 

Maxima [F]

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

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

Output:

-(2*m + 1)*integrate(x^m/(a*b*sqrt(x) + a^2), x) + 2*x*x^m/(a*b*sqrt(x) + 
a^2)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

int(x^m/(a + b*x^(1/2))^2,x)
 

Output:

int(x^m/(a + b*x^(1/2))^2, x)
 

Reduce [F]

\[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\text {too large to display} \] Input:

int(x^m/(a+b*x^(1/2))^2,x)
 

Output:

( - 4*x**((2*m + 1)/2)*a**2*b*m + x**((2*m + 1)/2)*a**2*b - 2*x**((2*m + 1 
)/2)*b**3*m**2*x + 3*x**((2*m + 1)/2)*b**3*m*x - x**((2*m + 1)/2)*b**3*x - 
 8*sqrt(x)*int(x**((2*m + 1)/2)/(2*a**4*m*x - a**4*x - 4*a**2*b**2*m*x**2 
+ 2*a**2*b**2*x**2 + 2*b**4*m*x**3 - b**4*x**3),x)*a**5*b**2*m**4 + 8*sqrt 
(x)*int(x**((2*m + 1)/2)/(2*a**4*m*x - a**4*x - 4*a**2*b**2*m*x**2 + 2*a** 
2*b**2*x**2 + 2*b**4*m*x**3 - b**4*x**3),x)*a**5*b**2*m**3 + 2*sqrt(x)*int 
(x**((2*m + 1)/2)/(2*a**4*m*x - a**4*x - 4*a**2*b**2*m*x**2 + 2*a**2*b**2* 
x**2 + 2*b**4*m*x**3 - b**4*x**3),x)*a**5*b**2*m**2 - 2*sqrt(x)*int(x**((2 
*m + 1)/2)/(2*a**4*m*x - a**4*x - 4*a**2*b**2*m*x**2 + 2*a**2*b**2*x**2 + 
2*b**4*m*x**3 - b**4*x**3),x)*a**5*b**2*m + 8*sqrt(x)*int(x**((2*m + 1)/2) 
/(2*a**4*m*x - a**4*x - 4*a**2*b**2*m*x**2 + 2*a**2*b**2*x**2 + 2*b**4*m*x 
**3 - b**4*x**3),x)*a**3*b**4*m**4*x - 8*sqrt(x)*int(x**((2*m + 1)/2)/(2*a 
**4*m*x - a**4*x - 4*a**2*b**2*m*x**2 + 2*a**2*b**2*x**2 + 2*b**4*m*x**3 - 
 b**4*x**3),x)*a**3*b**4*m**3*x - 2*sqrt(x)*int(x**((2*m + 1)/2)/(2*a**4*m 
*x - a**4*x - 4*a**2*b**2*m*x**2 + 2*a**2*b**2*x**2 + 2*b**4*m*x**3 - b**4 
*x**3),x)*a**3*b**4*m**2*x + 2*sqrt(x)*int(x**((2*m + 1)/2)/(2*a**4*m*x - 
a**4*x - 4*a**2*b**2*m*x**2 + 2*a**2*b**2*x**2 + 2*b**4*m*x**3 - b**4*x**3 
),x)*a**3*b**4*m*x + 4*sqrt(x)*int(x**m/(2*a**4*m**2*x - 3*a**4*m*x + a**4 
*x - 4*a**2*b**2*m**2*x**2 + 6*a**2*b**2*m*x**2 - 2*a**2*b**2*x**2 + 2*b** 
4*m**2*x**3 - 3*b**4*m*x**3 + b**4*x**3),x)*a**6*b*m**5 - 4*sqrt(x)*int...