\(\int \frac {(d x)^m}{(a^2+2 a b x^3+b^2 x^6)^{5/2}} \, dx\) [108]

Optimal result

Integrand size = 28, antiderivative size = 73 \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {(d x)^{1+m} \left (a+b x^3\right ) \operatorname {Hypergeometric2F1}\left (5,\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{a^5 d (1+m) \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:

(d*x)^(1+m)*(b*x^3+a)*hypergeom([5, 1/3+1/3*m],[4/3+1/3*m],-b*x^3/a)/a^5/d 
/(1+m)/((b*x^3+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {x (d x)^m \left (a+b x^3\right ) \operatorname {Hypergeometric2F1}\left (5,\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{a^5 (1+m) \sqrt {\left (a+b x^3\right )^2}} \] Input:

Integrate[(d*x)^m/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1384, 27, 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.

\(\displaystyle \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1384

\(\displaystyle \frac {b^5 \left (a+b x^3\right ) \int \frac {(d x)^m}{b^5 \left (b x^3+a\right )^5}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {(d x)^m}{\left (b x^3+a\right )^5}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {\left (a+b x^3\right ) (d x)^{m+1} \operatorname {Hypergeometric2F1}\left (5,\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{a^5 d (m+1) \sqrt {a^2+2 a b x^3+b^2 x^6}}\)

Input:

Int[(d*x)^m/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
 

Output:

((d*x)^(1 + m)*(a + b*x^3)*Hypergeometric2F1[5, (1 + m)/3, (4 + m)/3, -((b 
*x^3)/a)])/(a^5*d*(1 + m)*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 1384

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac 
Part[p]))   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n 
- 1)] && NeQ[u, x^(2*n - 1)] &&  !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
 

Maple [F]

\[\int \frac {\left (d x \right )^{m}}{\left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {5}{2}}}d x\]

Input:

int((d*x)^m/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
 

Output:

int((d*x)^m/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
 

Fricas [F]

\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x)^m/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*(d*x)^m/(b^6*x^18 + 6*a*b^5*x^15 
+ 15*a^2*b^4*x^12 + 20*a^3*b^3*x^9 + 15*a^4*b^2*x^6 + 6*a^5*b*x^3 + a^6), 
x)
 

Sympy [F]

\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {\left (d x\right )^{m}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((d*x)**m/(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)
 

Output:

Integral((d*x)**m/((a + b*x**3)**2)**(5/2), x)
 

Maxima [F]

\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x)^m/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate((d*x)^m/(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2), x)
 

Giac [F]

\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x)^m/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")
 

Output:

integrate((d*x)^m/(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {{\left (d\,x\right )}^m}{{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}} \,d x \] Input:

int((d*x)^m/(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)
 

Output:

int((d*x)^m/(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=d^{m} \left (\int \frac {x^{m}}{b^{5} x^{15}+5 a \,b^{4} x^{12}+10 a^{2} b^{3} x^{9}+10 a^{3} b^{2} x^{6}+5 a^{4} b \,x^{3}+a^{5}}d x \right ) \] Input:

int((d*x)^m/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
 

Output:

d**m*int(x**m/(a**5 + 5*a**4*b*x**3 + 10*a**3*b**2*x**6 + 10*a**2*b**3*x** 
9 + 5*a*b**4*x**12 + b**5*x**15),x)