\(\int \frac {1}{(d x)^{5/2} (a+b x^n+c x^{2 n})} \, dx\) [206]

Optimal result

Integrand size = 24, antiderivative size = 162 \[ \int \frac {1}{(d x)^{5/2} \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {4 c \operatorname {Hypergeometric2F1}\left (1,-\frac {3}{2 n},1-\frac {3}{2 n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d (d x)^{3/2}}+\frac {4 c \operatorname {Hypergeometric2F1}\left (1,-\frac {3}{2 n},1-\frac {3}{2 n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d (d x)^{3/2}} \] Output:

4/3*c*hypergeom([1, -3/2/n],[1-3/2/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^ 
2-4*a*c-b*(-4*a*c+b^2)^(1/2))/d/(d*x)^(3/2)+4/3*c*hypergeom([1, -3/2/n],[1 
-3/2/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)/ 
d/(d*x)^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.42 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(d x)^{5/2} \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {x \left (-2 \sqrt {b^2-4 a c}+2^{\left .\frac {3}{2}\right /n} \left (b+\sqrt {b^2-4 a c}\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{\left .\frac {3}{2}\right /n} \operatorname {Hypergeometric2F1}\left (\frac {3}{2 n},\frac {3}{2 n},1+\frac {3}{2 n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )+2^{\left .\frac {3}{2}\right /n} \left (-b+\sqrt {b^2-4 a c}\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{\left .\frac {3}{2}\right /n} \operatorname {Hypergeometric2F1}\left (\frac {3}{2 n},\frac {3}{2 n},1+\frac {3}{2 n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )\right )}{3 a \sqrt {b^2-4 a c} (d x)^{5/2}} \] Input:

Integrate[1/((d*x)^(5/2)*(a + b*x^n + c*x^(2*n))),x]
 

Output:

(x*(-2*Sqrt[b^2 - 4*a*c] + 2^(3/(2*n))*(b + Sqrt[b^2 - 4*a*c])*((c*x^n)/(b 
 - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(3/(2*n))*Hypergeometric2F1[3/(2*n), 3/(2 
*n), 1 + 3/(2*n), (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n 
)] + 2^(3/(2*n))*(-b + Sqrt[b^2 - 4*a*c])*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] 
+ 2*c*x^n))^(3/(2*n))*Hypergeometric2F1[3/(2*n), 3/(2*n), 1 + 3/(2*n), (b 
+ Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]))/(3*a*Sqrt[b^2 - 
4*a*c]*(d*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1719, 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.

\(\displaystyle \int \frac {1}{(d x)^{5/2} \left (a+b x^n+c x^{2 n}\right )} \, dx\)

\(\Big \downarrow \) 1719

\(\displaystyle \frac {2 c \int \frac {1}{(d x)^{5/2} \left (2 c x^n+b-\sqrt {b^2-4 a c}\right )}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{(d x)^{5/2} \left (2 c x^n+b+\sqrt {b^2-4 a c}\right )}dx}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {4 c \operatorname {Hypergeometric2F1}\left (1,-\frac {3}{2 n},1-\frac {3}{2 n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 d (d x)^{3/2} \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {4 c \operatorname {Hypergeometric2F1}\left (1,-\frac {3}{2 n},1-\frac {3}{2 n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 d (d x)^{3/2} \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}\)

Input:

Int[1/((d*x)^(5/2)*(a + b*x^n + c*x^(2*n))),x]
 

Output:

(-4*c*Hypergeometric2F1[1, -3/(2*n), 1 - 3/(2*n), (-2*c*x^n)/(b - Sqrt[b^2 
 - 4*a*c])])/(3*Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*(d*x)^(3/2)) + 
 (4*c*Hypergeometric2F1[1, -3/(2*n), 1 - 3/(2*n), (-2*c*x^n)/(b + Sqrt[b^2 
 - 4*a*c])])/(3*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*(d*x)^(3/2))
 

Defintions of rubi rules used

rule 888

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])
 

rule 1719

Int[((d_.)*(x_))^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symb 
ol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int[(d*x)^m/(b - q + 2 
*c*x^n), x], x] - Simp[2*(c/q)   Int[(d*x)^m/(b + q + 2*c*x^n), x], x]] /; 
FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
 

Maple [F]

\[\int \frac {1}{\left (d x \right )^{\frac {5}{2}} \left (a +b \,x^{n}+c \,x^{2 n}\right )}d x\]

Input:

int(1/(d*x)^(5/2)/(a+b*x^n+c*x^(2*n)),x)
 

Output:

int(1/(d*x)^(5/2)/(a+b*x^n+c*x^(2*n)),x)
 

Fricas [F]

\[ \int \frac {1}{(d x)^{5/2} \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} \left (d x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(d*x)^(5/2)/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")
 

Output:

integral(sqrt(d*x)/(c*d^3*x^3*x^(2*n) + b*d^3*x^3*x^n + a*d^3*x^3), x)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {1}{(d x)^{5/2} \left (a+b x^n+c x^{2 n}\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate(1/(d*x)**(5/2)/(a+b*x**n+c*x**(2*n)),x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int \frac {1}{(d x)^{5/2} \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} \left (d x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(d*x)^(5/2)/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")
 

Output:

-2/((c*d^(5/2)*(4*n + 3)*x^(2*n) + b*d^(5/2)*(4*n + 3)*x^n + a*d^(5/2)*(4* 
n + 3))*x^(3/2)) + integrate(2*(b*n*x^n + 2*a*n)/((c^2*d^(5/2)*(4*n + 3)*x 
^(4*n) + 2*b*c*d^(5/2)*(4*n + 3)*x^(3*n) + 2*a*b*d^(5/2)*(4*n + 3)*x^n + a 
^2*d^(5/2)*(4*n + 3) + (b^2*d^(5/2)*(4*n + 3) + 2*a*c*d^(5/2)*(4*n + 3))*x 
^(2*n))*x^(5/2)), x)
 

Giac [F]

\[ \int \frac {1}{(d x)^{5/2} \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} \left (d x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(d*x)^(5/2)/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")
 

Output:

integrate(1/((c*x^(2*n) + b*x^n + a)*(d*x)^(5/2)), x)
 

Mupad [F(-1)]

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

int(1/((d*x)^(5/2)*(a + b*x^n + c*x^(2*n))),x)
 

Output:

int(1/((d*x)^(5/2)*(a + b*x^n + c*x^(2*n))), x)
 

Reduce [F]

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

int(1/(d*x)^(5/2)/(a+b*x^n+c*x^(2*n)),x)
 

Output:

(sqrt(d)*int(1/(x**((4*n + 1)/2)*c*x**2 + x**((2*n + 1)/2)*b*x**2 + sqrt(x 
)*a*x**2),x))/d**3