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

Optimal result

Integrand size = 63, antiderivative size = 111 \[ \int \frac {(d x)^{-1+\frac {n}{2}} \left (-a h+c f x^{n/2}+c g x^{3 n/2}+c h x^{2 n}\right )}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=-\frac {2 h (d x)^{n/2}}{d n \sqrt {a+b x^n+c x^{2 n}}}-\frac {2 c x^{-n/2} (d x)^{n/2} \left (b f-2 a g+(2 c f-b g) x^n\right )}{\left (b^2-4 a c\right ) d n \sqrt {a+b x^n+c x^{2 n}}} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(242\) vs. \(2(111)=222\).

Time = 4.39 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.18 \[ \int \frac {(d x)^{-1+\frac {n}{2}} \left (-a h+c f x^{n/2}+c g x^{3 n/2}+c h x^{2 n}\right )}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=-\frac {x^{-n/2} (d x)^{n/2} \left (-2 a b^2 h x^{n/2}-2 a b c \left (f-g x^n\right )+4 a c \left (-c f x^n+a \left (g+2 h x^{n/2}\right )\right )+b \sqrt {c} (b f-2 a g) \sqrt {a+x^n \left (b+c x^n\right )} \text {arctanh}\left (\frac {b+2 c x^n}{2 \sqrt {c} \sqrt {a+x^n \left (b+c x^n\right )}}\right )+b \sqrt {c} (b f-2 a g) \sqrt {a+x^n \left (b+c x^n\right )} \log \left (b+2 c x^n-2 \sqrt {c} \sqrt {a+x^n \left (b+c x^n\right )}\right )\right )}{a \left (-b^2+4 a c\right ) d n \sqrt {a+x^n \left (b+c x^n\right )}} \] Input:

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

Output:

-(((d*x)^(n/2)*(-2*a*b^2*h*x^(n/2) - 2*a*b*c*(f - g*x^n) + 4*a*c*(-(c*f*x^ 
n) + a*(g + 2*h*x^(n/2))) + b*Sqrt[c]*(b*f - 2*a*g)*Sqrt[a + x^n*(b + c*x^ 
n)]*ArcTanh[(b + 2*c*x^n)/(2*Sqrt[c]*Sqrt[a + x^n*(b + c*x^n)])] + b*Sqrt[ 
c]*(b*f - 2*a*g)*Sqrt[a + x^n*(b + c*x^n)]*Log[b + 2*c*x^n - 2*Sqrt[c]*Sqr 
t[a + x^n*(b + c*x^n)]]))/(a*(-b^2 + 4*a*c)*d*n*x^(n/2)*Sqrt[a + x^n*(b + 
c*x^n)]))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2290, 25, 2289}

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.

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

\(\Big \downarrow \) 2290

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2289

\(\displaystyle -\frac {2 x^{1-\frac {n}{2}} (d x)^{\frac {n-2}{2}} \left (h x^{n/2} \left (b^2-4 a c\right )+c (b f-2 a g)+c x^n (2 c f-b g)\right )}{n \left (b^2-4 a c\right ) \sqrt {a+b x^n+c x^{2 n}}}\)

Input:

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

Output:

(-2*x^(1 - n/2)*(d*x)^((-2 + n)/2)*(c*(b*f - 2*a*g) + (b^2 - 4*a*c)*h*x^(n 
/2) + c*(2*c*f - b*g)*x^n))/((b^2 - 4*a*c)*n*Sqrt[a + b*x^n + c*x^(2*n)])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2289

Int[((x_)^(m_.)*((e_) + (f_.)*(x_)^(q_.) + (g_.)*(x_)^(r_.) + (h_.)*(x_)^(s 
_.)))/((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(3/2), x_Symbol] :> Sim 
p[-(2*c*(b*f - 2*a*g) + 2*h*(b^2 - 4*a*c)*x^(n/2) + 2*c*(2*c*f - b*g)*x^n)/ 
(c*n*(b^2 - 4*a*c)*Sqrt[a + b*x^n + c*x^(2*n)]), x] /; FreeQ[{a, b, c, e, f 
, g, h, m, n}, x] && EqQ[n2, 2*n] && EqQ[q, n/2] && EqQ[r, 3*(n/2)] && EqQ[ 
s, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*m - n + 2, 0] && EqQ[c*e + a*h, 0]
 

rule 2290

Int[(((d_)*(x_))^(m_.)*((e_) + (f_.)*(x_)^(q_.) + (g_.)*(x_)^(r_.) + (h_.)* 
(x_)^(s_.)))/((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(3/2), x_Symbol] 
 :> Simp[(d*x)^m/x^m   Int[x^m*((e + f*x^(n/2) + g*x^((3*n)/2) + h*x^(2*n)) 
/(a + b*x^n + c*x^(2*n))^(3/2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m 
, n}, x] && EqQ[n2, 2*n] && EqQ[q, n/2] && EqQ[r, 3*(n/2)] && EqQ[s, 2*n] & 
& NeQ[b^2 - 4*a*c, 0] && EqQ[2*m - n + 2, 0] && EqQ[c*e + a*h, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.19 \[ \int \frac {(d x)^{-1+\frac {n}{2}} \left (-a h+c f x^{n/2}+c g x^{3 n/2}+c h x^{2 n}\right )}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=-\frac {2 \, {\left ({\left (b^{2} - 4 \, a c\right )} d^{\frac {1}{2} \, n - 1} h x^{\frac {1}{2} \, n} + {\left (2 \, c^{2} f - b c g\right )} d^{\frac {1}{2} \, n - 1} x^{n} + {\left (b c f - 2 \, a c g\right )} d^{\frac {1}{2} \, n - 1}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{{\left (b^{2} c - 4 \, a c^{2}\right )} n x^{2 \, n} + {\left (b^{3} - 4 \, a b c\right )} n x^{n} + {\left (a b^{2} - 4 \, a^{2} c\right )} n} \] Input:

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

Output:

-2*((b^2 - 4*a*c)*d^(1/2*n - 1)*h*x^(1/2*n) + (2*c^2*f - b*c*g)*d^(1/2*n - 
 1)*x^n + (b*c*f - 2*a*c*g)*d^(1/2*n - 1))*sqrt(c*x^(2*n) + b*x^n + a)/((b 
^2*c - 4*a*c^2)*n*x^(2*n) + (b^3 - 4*a*b*c)*n*x^n + (a*b^2 - 4*a^2*c)*n)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.15 \[ \int \frac {(d x)^{-1+\frac {n}{2}} \left (-a h+c f x^{n/2}+c g x^{3 n/2}+c h x^{2 n}\right )}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {2 d^{\frac {n}{2}} \sqrt {x^{2 n} c +x^{n} b +a}\, \left (-4 x^{\frac {n}{2}} a c h +x^{\frac {n}{2}} b^{2} h -x^{n} b c g +2 x^{n} c^{2} f -2 a c g +b c f \right )}{d n \left (4 x^{2 n} a \,c^{2}-x^{2 n} b^{2} c +4 x^{n} a b c -x^{n} b^{3}+4 a^{2} c -a \,b^{2}\right )} \] Input:

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

Output:

(2*d**(n/2)*sqrt(x**(2*n)*c + x**n*b + a)*( - 4*x**(n/2)*a*c*h + x**(n/2)* 
b**2*h - x**n*b*c*g + 2*x**n*c**2*f - 2*a*c*g + b*c*f))/(d*n*(4*x**(2*n)*a 
*c**2 - x**(2*n)*b**2*c + 4*x**n*a*b*c - x**n*b**3 + 4*a**2*c - a*b**2))