\(\int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{(a+b x^n)^{3/2}} \, dx\) [30]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 58, antiderivative size = 60 \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=-\frac {2 g}{n \sqrt {a+b x^n}}-\frac {2 x^{n/4} \left (2 a h-b f x^{n/4}\right )}{a n \sqrt {a+b x^n}} \] Output:

-2*g/n/(a+b*x^n)^(1/2)-2*x^(1/4*n)*(2*a*h-b*f*x^(1/4*n))/a/n/(a+b*x^n)^(1/ 
2)
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75 \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\frac {2 b f x^{n/2}-2 a \left (g+2 h x^{n/4}\right )}{a n \sqrt {a+b x^n}} \] Input:

Integrate[(-(a*h*x^(-1 + n/4)) + b*f*x^(-1 + n/2) + b*g*x^(-1 + n) + b*h*x 
^(-1 + (5*n)/4))/(a + b*x^n)^(3/2),x]
 

Output:

(2*b*f*x^(n/2) - 2*a*(g + 2*h*x^(n/4)))/(a*n*Sqrt[a + b*x^n])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2029, 2356}

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.

\(\displaystyle \int \frac {-a h x^{\frac {n}{4}-1}+b f x^{\frac {n}{2}-1}+b g x^{n-1}+b h x^{\frac {5 n}{4}-1}}{\left (a+b x^n\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2029

\(\displaystyle \int \frac {x^{\frac {n}{4}-1} \left (-a h+b f x^{n/4}+b g x^{3 n/4}+b h x^n\right )}{\left (a+b x^n\right )^{3/2}}dx\)

\(\Big \downarrow \) 2356

\(\displaystyle -\frac {2 \left (a g+2 a h x^{n/4}-b f x^{n/2}\right )}{a n \sqrt {a+b x^n}}\)

Input:

Int[(-(a*h*x^(-1 + n/4)) + b*f*x^(-1 + n/2) + b*g*x^(-1 + n) + b*h*x^(-1 + 
 (5*n)/4))/(a + b*x^n)^(3/2),x]
 

Output:

(-2*(a*g + 2*a*h*x^(n/4) - b*f*x^(n/2)))/(a*n*Sqrt[a + b*x^n])
 

Defintions of rubi rules used

rule 2029
Int[(Fx_.)*((d_.)*(x_)^(q_.) + (a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)* 
(x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r) 
+ d*x^(q - r))^p*Fx, x] /; FreeQ[{a, b, c, d, r, s, t, q}, x] && IntegerQ[p 
] && PosQ[s - r] && PosQ[t - r] && PosQ[q - r] &&  !(EqQ[p, 1] && EqQ[u, 1] 
)
 

rule 2356
Int[((x_)^(m_.)*((e_) + (h_.)*(x_)^(n_.) + (f_.)*(x_)^(q_.) + (g_.)*(x_)^(r 
_.)))/((a_) + (c_.)*(x_)^(n_.))^(3/2), x_Symbol] :> Simp[-(2*a*g + 4*a*h*x^ 
(n/4) - 2*c*f*x^(n/2))/(a*c*n*Sqrt[a + c*x^n]), x] /; FreeQ[{a, c, e, f, g, 
 h, m, n}, x] && EqQ[q, n/4] && EqQ[r, 3*(n/4)] && EqQ[4*m - n + 4, 0] && E 
qQ[c*e + a*h, 0]
 
Maple [F]

\[\int \frac {-a h \,x^{-1+\frac {n}{4}}+b f \,x^{-1+\frac {n}{2}}+b g \,x^{-1+n}+b h \,x^{-1+\frac {5 n}{4}}}{\left (a +b \,x^{n}\right )^{\frac {3}{2}}}d x\]

Input:

int((-a*h*x^(-1+1/4*n)+b*f*x^(-1+1/2*n)+b*g*x^(-1+n)+b*h*x^(-1+5/4*n))/(a+ 
b*x^n)^(3/2),x)
 

Output:

int((-a*h*x^(-1+1/4*n)+b*f*x^(-1+1/2*n)+b*g*x^(-1+n)+b*h*x^(-1+5/4*n))/(a+ 
b*x^n)^(3/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\frac {2 \, \sqrt {b x^{4} x^{n - 4} + a} {\left (b f x^{2} x^{\frac {1}{2} \, n - 2} - 2 \, a h x x^{\frac {1}{4} \, n - 1} - a g\right )}}{a b n x^{4} x^{n - 4} + a^{2} n} \] Input:

integrate((-a*h*x^(-1+1/4*n)+b*f*x^(-1+1/2*n)+b*g*x^(-1+n)+b*h*x^(-1+5/4*n 
))/(a+b*x^n)^(3/2),x, algorithm="fricas")
 

Output:

2*sqrt(b*x^4*x^(n - 4) + a)*(b*f*x^2*x^(1/2*n - 2) - 2*a*h*x*x^(1/4*n - 1) 
 - a*g)/(a*b*n*x^4*x^(n - 4) + a^2*n)
 

Sympy [A] (verification not implemented)

Time = 136.23 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.70 \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=b g \left (\begin {cases} \frac {\log {\left (x \right )}}{a^{\frac {3}{2}}} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x x^{n - 1}}{a^{\frac {3}{2}} n} & \text {for}\: b = 0 \\\frac {\log {\left (x \right )}}{\left (a + b\right )^{\frac {3}{2}}} & \text {for}\: n = 0 \\- \frac {2}{b n \sqrt {a + b x^{n}}} & \text {otherwise} \end {cases}\right ) + \frac {2 \sqrt {b} f}{a n \sqrt {\frac {a x^{- n}}{b} + 1}} - \frac {h x^{\frac {n}{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{\sqrt {a} n \Gamma \left (\frac {5}{4}\right )} + \frac {b h x^{\frac {5 n}{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{a^{\frac {3}{2}} n \Gamma \left (\frac {9}{4}\right )} \] Input:

integrate((-a*h*x**(-1+1/4*n)+b*f*x**(-1+1/2*n)+b*g*x**(-1+n)+b*h*x**(-1+5 
/4*n))/(a+b*x**n)**(3/2),x)
 

Output:

b*g*Piecewise((log(x)/a**(3/2), Eq(b, 0) & Eq(n, 0)), (x*x**(n - 1)/(a**(3 
/2)*n), Eq(b, 0)), (log(x)/(a + b)**(3/2), Eq(n, 0)), (-2/(b*n*sqrt(a + b* 
x**n)), True)) + 2*sqrt(b)*f/(a*n*sqrt(a/(b*x**n) + 1)) - h*x**(n/4)*gamma 
(1/4)*hyper((1/4, 3/2), (5/4,), b*x**n*exp_polar(I*pi)/a)/(sqrt(a)*n*gamma 
(5/4)) + b*h*x**(5*n/4)*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**n*exp_po 
lar(I*pi)/a)/(a**(3/2)*n*gamma(9/4))
 

Maxima [F]

\[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\int { \frac {b h x^{\frac {5}{4} \, n - 1} + b g x^{n - 1} + b f x^{\frac {1}{2} \, n - 1} - a h x^{\frac {1}{4} \, n - 1}}{{\left (b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-a*h*x^(-1+1/4*n)+b*f*x^(-1+1/2*n)+b*g*x^(-1+n)+b*h*x^(-1+5/4*n 
))/(a+b*x^n)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*h*x^(5/4*n - 1) + b*g*x^(n - 1) + b*f*x^(1/2*n - 1) - a*h*x^( 
1/4*n - 1))/(b*x^n + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\int { \frac {b h x^{\frac {5}{4} \, n - 1} + b g x^{n - 1} + b f x^{\frac {1}{2} \, n - 1} - a h x^{\frac {1}{4} \, n - 1}}{{\left (b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-a*h*x^(-1+1/4*n)+b*f*x^(-1+1/2*n)+b*g*x^(-1+n)+b*h*x^(-1+5/4*n 
))/(a+b*x^n)^(3/2),x, algorithm="giac")
 

Output:

integrate((b*h*x^(5/4*n - 1) + b*g*x^(n - 1) + b*f*x^(1/2*n - 1) - a*h*x^( 
1/4*n - 1))/(b*x^n + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\int \frac {b\,f\,x^{\frac {n}{2}-1}-a\,h\,x^{\frac {n}{4}-1}+b\,h\,x^{\frac {5\,n}{4}-1}+b\,g\,x^{n-1}}{{\left (a+b\,x^n\right )}^{3/2}} \,d x \] Input:

int((b*f*x^(n/2 - 1) - a*h*x^(n/4 - 1) + b*h*x^((5*n)/4 - 1) + b*g*x^(n - 
1))/(a + b*x^n)^(3/2),x)
 

Output:

int((b*f*x^(n/2 - 1) - a*h*x^(n/4 - 1) + b*h*x^((5*n)/4 - 1) + b*g*x^(n - 
1))/(a + b*x^n)^(3/2), x)
 

Reduce [F]

\[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\frac {2 x^{\frac {n}{2}} \sqrt {x^{n} b +a}\, b f -2 \sqrt {x^{n} b +a}\, a g +x^{n} \left (\int \frac {x^{\frac {5 n}{4}} \sqrt {x^{n} b +a}}{x^{2 n} b^{2} x +2 x^{n} a b x +a^{2} x}d x \right ) a \,b^{2} h n -x^{n} \left (\int \frac {x^{\frac {n}{4}} \sqrt {x^{n} b +a}}{x^{2 n} b^{2} x +2 x^{n} a b x +a^{2} x}d x \right ) a^{2} b h n +\left (\int \frac {x^{\frac {5 n}{4}} \sqrt {x^{n} b +a}}{x^{2 n} b^{2} x +2 x^{n} a b x +a^{2} x}d x \right ) a^{2} b h n -\left (\int \frac {x^{\frac {n}{4}} \sqrt {x^{n} b +a}}{x^{2 n} b^{2} x +2 x^{n} a b x +a^{2} x}d x \right ) a^{3} h n}{a n \left (x^{n} b +a \right )} \] Input:

int((-a*h*x^(-1+1/4*n)+b*f*x^(-1+1/2*n)+b*g*x^(-1+n)+b*h*x^(-1+5/4*n))/(a+ 
b*x^n)^(3/2),x)
 

Output:

(2*x**(n/2)*sqrt(x**n*b + a)*b*f - 2*sqrt(x**n*b + a)*a*g + x**n*int((x**( 
(5*n)/4)*sqrt(x**n*b + a))/(x**(2*n)*b**2*x + 2*x**n*a*b*x + a**2*x),x)*a* 
b**2*h*n - x**n*int((x**(n/4)*sqrt(x**n*b + a))/(x**(2*n)*b**2*x + 2*x**n* 
a*b*x + a**2*x),x)*a**2*b*h*n + int((x**((5*n)/4)*sqrt(x**n*b + a))/(x**(2 
*n)*b**2*x + 2*x**n*a*b*x + a**2*x),x)*a**2*b*h*n - int((x**(n/4)*sqrt(x** 
n*b + a))/(x**(2*n)*b**2*x + 2*x**n*a*b*x + a**2*x),x)*a**3*h*n)/(a*n*(x** 
n*b + a))