Integrand size = 15, antiderivative size = 51 \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx=-\frac {\left (a+b x^n\right )^{7/2} \operatorname {Hypergeometric2F1}\left (1,\frac {7}{2}-\frac {2}{n},-\frac {2-n}{n},-\frac {b x^n}{a}\right )}{2 a x^2} \] Output:
-1/2*(a+b*x^n)^(7/2)*hypergeom([1, 7/2-2/n],[-(2-n)/n],-b*x^n/a)/a/x^2
Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx=-\frac {a^2 \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {2}{n},1-\frac {2}{n},-\frac {b x^n}{a}\right )}{2 x^2 \sqrt {1+\frac {b x^n}{a}}} \] Input:
Integrate[(a + b*x^n)^(5/2)/x^3,x]
Output:
-1/2*(a^2*Sqrt[a + b*x^n]*Hypergeometric2F1[-5/2, -2/n, 1 - 2/n, -((b*x^n) /a)])/(x^2*Sqrt[1 + (b*x^n)/a])
Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {889, 888}
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.
\(\displaystyle \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {a^2 \sqrt {a+b x^n} \int \frac {\left (\frac {b x^n}{a}+1\right )^{5/2}}{x^3}dx}{\sqrt {\frac {b x^n}{a}+1}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle -\frac {a^2 \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {2}{n},-\frac {2-n}{n},-\frac {b x^n}{a}\right )}{2 x^2 \sqrt {\frac {b x^n}{a}+1}}\) |
Input:
Int[(a + b*x^n)^(5/2)/x^3,x]
Output:
-1/2*(a^2*Sqrt[a + b*x^n]*Hypergeometric2F1[-5/2, -2/n, -((2 - n)/n), -((b *x^n)/a)])/(x^2*Sqrt[1 + (b*x^n)/a])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(c*x) ^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0 ] && !(ILtQ[p, 0] || GtQ[a, 0])
\[\int \frac {\left (a +b \,x^{n}\right )^{\frac {5}{2}}}{x^{3}}d x\]
Input:
int((a+b*x^n)^(5/2)/x^3,x)
Output:
int((a+b*x^n)^(5/2)/x^3,x)
Exception generated. \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*x^n)^(5/2)/x^3,x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
Result contains complex when optimal does not.
Time = 2.74 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx=\frac {a^{- \frac {2}{n}} a^{\frac {5}{2} + \frac {2}{n}} \Gamma \left (- \frac {2}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {2}{n} \\ 1 - \frac {2}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n x^{2} \Gamma \left (1 - \frac {2}{n}\right )} \] Input:
integrate((a+b*x**n)**(5/2)/x**3,x)
Output:
a**(5/2 + 2/n)*gamma(-2/n)*hyper((-5/2, -2/n), (1 - 2/n,), b*x**n*exp_pola r(I*pi)/a)/(a**(2/n)*n*x**2*gamma(1 - 2/n))
\[ \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{\frac {5}{2}}}{x^{3}} \,d x } \] Input:
integrate((a+b*x^n)^(5/2)/x^3,x, algorithm="maxima")
Output:
integrate((b*x^n + a)^(5/2)/x^3, x)
\[ \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{\frac {5}{2}}}{x^{3}} \,d x } \] Input:
integrate((a+b*x^n)^(5/2)/x^3,x, algorithm="giac")
Output:
integrate((b*x^n + a)^(5/2)/x^3, x)
Timed out. \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx=\int \frac {{\left (a+b\,x^n\right )}^{5/2}}{x^3} \,d x \] Input:
int((a + b*x^n)^(5/2)/x^3,x)
Output:
int((a + b*x^n)^(5/2)/x^3, x)
\[ \int \frac {\left (a+b x^n\right )^{5/2}}{x^3} \, dx=\frac {6 x^{2 n} \sqrt {x^{n} b +a}\, b^{2} n^{2}-32 x^{2 n} \sqrt {x^{n} b +a}\, b^{2} n +32 x^{2 n} \sqrt {x^{n} b +a}\, b^{2}+22 x^{n} \sqrt {x^{n} b +a}\, a b \,n^{2}-104 x^{n} \sqrt {x^{n} b +a}\, a b n +64 x^{n} \sqrt {x^{n} b +a}\, a b +46 \sqrt {x^{n} b +a}\, a^{2} n^{2}-72 \sqrt {x^{n} b +a}\, a^{2} n +32 \sqrt {x^{n} b +a}\, a^{2}+225 \left (\int \frac {\sqrt {x^{n} b +a}}{15 x^{n} b \,n^{3} x^{3}-92 x^{n} b \,n^{2} x^{3}+144 x^{n} b n \,x^{3}-64 x^{n} b \,x^{3}+15 a \,n^{3} x^{3}-92 a \,n^{2} x^{3}+144 a n \,x^{3}-64 a \,x^{3}}d x \right ) a^{3} n^{6} x^{2}-1380 \left (\int \frac {\sqrt {x^{n} b +a}}{15 x^{n} b \,n^{3} x^{3}-92 x^{n} b \,n^{2} x^{3}+144 x^{n} b n \,x^{3}-64 x^{n} b \,x^{3}+15 a \,n^{3} x^{3}-92 a \,n^{2} x^{3}+144 a n \,x^{3}-64 a \,x^{3}}d x \right ) a^{3} n^{5} x^{2}+2160 \left (\int \frac {\sqrt {x^{n} b +a}}{15 x^{n} b \,n^{3} x^{3}-92 x^{n} b \,n^{2} x^{3}+144 x^{n} b n \,x^{3}-64 x^{n} b \,x^{3}+15 a \,n^{3} x^{3}-92 a \,n^{2} x^{3}+144 a n \,x^{3}-64 a \,x^{3}}d x \right ) a^{3} n^{4} x^{2}-960 \left (\int \frac {\sqrt {x^{n} b +a}}{15 x^{n} b \,n^{3} x^{3}-92 x^{n} b \,n^{2} x^{3}+144 x^{n} b n \,x^{3}-64 x^{n} b \,x^{3}+15 a \,n^{3} x^{3}-92 a \,n^{2} x^{3}+144 a n \,x^{3}-64 a \,x^{3}}d x \right ) a^{3} n^{3} x^{2}}{x^{2} \left (15 n^{3}-92 n^{2}+144 n -64\right )} \] Input:
int((a+b*x^n)^(5/2)/x^3,x)
Output:
(6*x**(2*n)*sqrt(x**n*b + a)*b**2*n**2 - 32*x**(2*n)*sqrt(x**n*b + a)*b**2 *n + 32*x**(2*n)*sqrt(x**n*b + a)*b**2 + 22*x**n*sqrt(x**n*b + a)*a*b*n**2 - 104*x**n*sqrt(x**n*b + a)*a*b*n + 64*x**n*sqrt(x**n*b + a)*a*b + 46*sqr t(x**n*b + a)*a**2*n**2 - 72*sqrt(x**n*b + a)*a**2*n + 32*sqrt(x**n*b + a) *a**2 + 225*int(sqrt(x**n*b + a)/(15*x**n*b*n**3*x**3 - 92*x**n*b*n**2*x** 3 + 144*x**n*b*n*x**3 - 64*x**n*b*x**3 + 15*a*n**3*x**3 - 92*a*n**2*x**3 + 144*a*n*x**3 - 64*a*x**3),x)*a**3*n**6*x**2 - 1380*int(sqrt(x**n*b + a)/( 15*x**n*b*n**3*x**3 - 92*x**n*b*n**2*x**3 + 144*x**n*b*n*x**3 - 64*x**n*b* x**3 + 15*a*n**3*x**3 - 92*a*n**2*x**3 + 144*a*n*x**3 - 64*a*x**3),x)*a**3 *n**5*x**2 + 2160*int(sqrt(x**n*b + a)/(15*x**n*b*n**3*x**3 - 92*x**n*b*n* *2*x**3 + 144*x**n*b*n*x**3 - 64*x**n*b*x**3 + 15*a*n**3*x**3 - 92*a*n**2* x**3 + 144*a*n*x**3 - 64*a*x**3),x)*a**3*n**4*x**2 - 960*int(sqrt(x**n*b + a)/(15*x**n*b*n**3*x**3 - 92*x**n*b*n**2*x**3 + 144*x**n*b*n*x**3 - 64*x* *n*b*x**3 + 15*a*n**3*x**3 - 92*a*n**2*x**3 + 144*a*n*x**3 - 64*a*x**3),x) *a**3*n**3*x**2)/(x**2*(15*n**3 - 92*n**2 + 144*n - 64))