Integrand size = 58, antiderivative size = 45 \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=-\frac {2 \left (a g+2 a h x^{n/4}-b f x^{n/2}\right )}{a n \sqrt {a+b x^n}} \]
Time = 0.95 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\frac {2 b f x^{n/2}-2 a \left (g+2 h x^{n/4}\right )}{a n \sqrt {a+b x^n}} \]
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]
Time = 0.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, 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 |
\(\displaystyle -\frac {2 \left (a g+2 a h x^{n/4}-b f x^{n/2}\right )}{a n \sqrt {a+b x^n}}\) |
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]
3.6.85.3.1 Defintions of rubi rules used
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] )
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]
\[\int \frac {-a h \,x^{-1+\frac {n}{4}}+b f \,x^{-1+\frac {n}{2}}+b g \,x^{-1+n}+b h \,x^{-1+\frac {5 n}{4}}}{\left (a +b \,x^{n}\right )^{\frac {3}{2}}}d x\]
Time = 0.40 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\frac {2 \, \sqrt {b x^{4} x^{n - 4} + a} {\left (b f x^{2} x^{\frac {1}{2} \, n - 2} - 2 \, a h x x^{\frac {1}{4} \, n - 1} - a g\right )}}{a b n x^{4} x^{n - 4} + a^{2} n} \]
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")
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)
Time = 137.42 (sec) , antiderivative size = 162, normalized size of antiderivative = 3.60 \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=b g \left (\begin {cases} \frac {\log {\left (x \right )}}{a^{\frac {3}{2}}} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x x^{n - 1}}{a^{\frac {3}{2}} n} & \text {for}\: b = 0 \\\frac {\log {\left (x \right )}}{\left (a + b\right )^{\frac {3}{2}}} & \text {for}\: n = 0 \\- \frac {2}{b n \sqrt {a + b x^{n}}} & \text {otherwise} \end {cases}\right ) + \frac {2 \sqrt {b} f}{a n \sqrt {\frac {a x^{- n}}{b} + 1}} - \frac {h x^{\frac {n}{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{\sqrt {a} n \Gamma \left (\frac {5}{4}\right )} + \frac {b h x^{\frac {5 n}{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{a^{\frac {3}{2}} n \Gamma \left (\frac {9}{4}\right )} \]
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)
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))
\[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\int { \frac {b h x^{\frac {5}{4} \, n - 1} + b g x^{n - 1} + b f x^{\frac {1}{2} \, n - 1} - a h x^{\frac {1}{4} \, n - 1}}{{\left (b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \]
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")
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)
\[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\int { \frac {b h x^{\frac {5}{4} \, n - 1} + b g x^{n - 1} + b f x^{\frac {1}{2} \, n - 1} - a h x^{\frac {1}{4} \, n - 1}}{{\left (b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \]
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")
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)
Timed out. \[ \int \frac {-a h x^{-1+\frac {n}{4}}+b f x^{-1+\frac {n}{2}}+b g x^{-1+n}+b h x^{-1+\frac {5 n}{4}}}{\left (a+b x^n\right )^{3/2}} \, dx=\int \frac {b\,f\,x^{\frac {n}{2}-1}-a\,h\,x^{\frac {n}{4}-1}+b\,h\,x^{\frac {5\,n}{4}-1}+b\,g\,x^{n-1}}{{\left (a+b\,x^n\right )}^{3/2}} \,d x \]
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)