\(\int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx\) [175]

Optimal result

Integrand size = 17, antiderivative size = 415 \[ \int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx=\frac {4 \sqrt {1+b x^{9/2}}}{3 b^{2/3} \left (1+\sqrt {3}+\sqrt [3]{b} x^{3/2}\right )}-\frac {2 \sqrt {2-\sqrt {3}} \left (1+\sqrt [3]{b} x^{3/2}\right ) \sqrt {\frac {1-\sqrt [3]{b} x^{3/2}+b^{2/3} x^3}{\left (1+\sqrt {3}+\sqrt [3]{b} x^{3/2}\right )^2}} E\left (\arcsin \left (\frac {1-\sqrt {3}+\sqrt [3]{b} x^{3/2}}{1+\sqrt {3}+\sqrt [3]{b} x^{3/2}}\right )|-7-4 \sqrt {3}\right )}{3^{3/4} b^{2/3} \sqrt {\frac {1+\sqrt [3]{b} x^{3/2}}{\left (1+\sqrt {3}+\sqrt [3]{b} x^{3/2}\right )^2}} \sqrt {1+b x^{9/2}}}+\frac {4 \sqrt {2} \left (1+\sqrt [3]{b} x^{3/2}\right ) \sqrt {\frac {1-\sqrt [3]{b} x^{3/2}+b^{2/3} x^3}{\left (1+\sqrt {3}+\sqrt [3]{b} x^{3/2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+\sqrt [3]{b} x^{3/2}}{1+\sqrt {3}+\sqrt [3]{b} x^{3/2}}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} b^{2/3} \sqrt {\frac {1+\sqrt [3]{b} x^{3/2}}{\left (1+\sqrt {3}+\sqrt [3]{b} x^{3/2}\right )^2}} \sqrt {1+b x^{9/2}}} \] Output:

4/3*(1+b*x^(9/2))^(1/2)/b^(2/3)/(1+3^(1/2)+b^(1/3)*x^(3/2))-2/3*(1/2*6^(1/ 
2)-1/2*2^(1/2))*(1+b^(1/3)*x^(3/2))*((1-b^(1/3)*x^(3/2)+b^(2/3)*x^3)/(1+3^ 
(1/2)+b^(1/3)*x^(3/2))^2)^(1/2)*EllipticE((1-3^(1/2)+b^(1/3)*x^(3/2))/(1+3 
^(1/2)+b^(1/3)*x^(3/2)),I*3^(1/2)+2*I)*3^(1/4)/b^(2/3)/((1+b^(1/3)*x^(3/2) 
)/(1+3^(1/2)+b^(1/3)*x^(3/2))^2)^(1/2)/(1+b*x^(9/2))^(1/2)+4/9*2^(1/2)*(1+ 
b^(1/3)*x^(3/2))*((1-b^(1/3)*x^(3/2)+b^(2/3)*x^3)/(1+3^(1/2)+b^(1/3)*x^(3/ 
2))^2)^(1/2)*EllipticF((1-3^(1/2)+b^(1/3)*x^(3/2))/(1+3^(1/2)+b^(1/3)*x^(3 
/2)),I*3^(1/2)+2*I)*3^(3/4)/b^(2/3)/((1+b^(1/3)*x^(3/2))/(1+3^(1/2)+b^(1/3 
)*x^(3/2))^2)^(1/2)/(1+b*x^(9/2))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.06 \[ \int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx=\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-b x^{9/2}\right ) \] Input:

Integrate[x^2/Sqrt[1 + b*x^(9/2)],x]
 

Output:

(x^3*Hypergeometric2F1[1/2, 2/3, 5/3, -(b*x^(9/2))])/3
 

Rubi [A] (warning: unable to verify)

Time = 0.75 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {864, 807, 832, 759, 2416}

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.

\(\displaystyle \int \frac {x^2}{\sqrt {b x^{9/2}+1}} \, dx\)

\(\Big \downarrow \) 864

\(\displaystyle 2 \int \frac {x^{5/2}}{\sqrt {b x^{9/2}+1}}d\sqrt {x}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {2}{3} \int \frac {x^{3/2}}{\sqrt {b x^{3/2}+1}}dx^{3/2}\)

\(\Big \downarrow \) 832

\(\displaystyle \frac {2}{3} \left (\frac {\int \frac {\sqrt [3]{b} x^{3/2}-\sqrt {3}+1}{\sqrt {b x^{3/2}+1}}dx^{3/2}}{\sqrt [3]{b}}-\frac {\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {b x^{3/2}+1}}dx^{3/2}}{\sqrt [3]{b}}\right )\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {2}{3} \left (\frac {\int \frac {\sqrt [3]{b} x^{3/2}-\sqrt {3}+1}{\sqrt {b x^{3/2}+1}}dx^{3/2}}{\sqrt [3]{b}}-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} \left (\sqrt [3]{b} x^{3/2}+1\right ) \sqrt {\frac {b^{2/3} x-\sqrt [3]{b} x^{3/2}+1}{\left (\sqrt [3]{b} x^{3/2}+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x^{3/2}-\sqrt {3}+1}{\sqrt [3]{b} x^{3/2}+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} b^{2/3} \sqrt {\frac {\sqrt [3]{b} x^{3/2}+1}{\left (\sqrt [3]{b} x^{3/2}+\sqrt {3}+1\right )^2}} \sqrt {b x^{3/2}+1}}\right )\)

\(\Big \downarrow \) 2416

\(\displaystyle \frac {2}{3} \left (\frac {\frac {2 \sqrt {b x^{3/2}+1}}{\sqrt [3]{b} \left (\sqrt [3]{b} x^{3/2}+\sqrt {3}+1\right )}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \left (\sqrt [3]{b} x^{3/2}+1\right ) \sqrt {\frac {b^{2/3} x-\sqrt [3]{b} x^{3/2}+1}{\left (\sqrt [3]{b} x^{3/2}+\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{b} x^{3/2}-\sqrt {3}+1}{\sqrt [3]{b} x^{3/2}+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt [3]{b} \sqrt {\frac {\sqrt [3]{b} x^{3/2}+1}{\left (\sqrt [3]{b} x^{3/2}+\sqrt {3}+1\right )^2}} \sqrt {b x^{3/2}+1}}}{\sqrt [3]{b}}-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} \left (\sqrt [3]{b} x^{3/2}+1\right ) \sqrt {\frac {b^{2/3} x-\sqrt [3]{b} x^{3/2}+1}{\left (\sqrt [3]{b} x^{3/2}+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x^{3/2}-\sqrt {3}+1}{\sqrt [3]{b} x^{3/2}+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} b^{2/3} \sqrt {\frac {\sqrt [3]{b} x^{3/2}+1}{\left (\sqrt [3]{b} x^{3/2}+\sqrt {3}+1\right )^2}} \sqrt {b x^{3/2}+1}}\right )\)

Input:

Int[x^2/Sqrt[1 + b*x^(9/2)],x]
 

Output:

(2*(((2*Sqrt[1 + b*x^(3/2)])/(b^(1/3)*(1 + Sqrt[3] + b^(1/3)*x^(3/2))) - ( 
3^(1/4)*Sqrt[2 - Sqrt[3]]*(1 + b^(1/3)*x^(3/2))*Sqrt[(1 + b^(2/3)*x - b^(1 
/3)*x^(3/2))/(1 + Sqrt[3] + b^(1/3)*x^(3/2))^2]*EllipticE[ArcSin[(1 - Sqrt 
[3] + b^(1/3)*x^(3/2))/(1 + Sqrt[3] + b^(1/3)*x^(3/2))], -7 - 4*Sqrt[3]])/ 
(b^(1/3)*Sqrt[(1 + b^(1/3)*x^(3/2))/(1 + Sqrt[3] + b^(1/3)*x^(3/2))^2]*Sqr 
t[1 + b*x^(3/2)]))/b^(1/3) - (2*(1 - Sqrt[3])*Sqrt[2 + Sqrt[3]]*(1 + b^(1/ 
3)*x^(3/2))*Sqrt[(1 + b^(2/3)*x - b^(1/3)*x^(3/2))/(1 + Sqrt[3] + b^(1/3)* 
x^(3/2))^2]*EllipticF[ArcSin[(1 - Sqrt[3] + b^(1/3)*x^(3/2))/(1 + Sqrt[3] 
+ b^(1/3)*x^(3/2))], -7 - 4*Sqrt[3]])/(3^(1/4)*b^(2/3)*Sqrt[(1 + b^(1/3)*x 
^(3/2))/(1 + Sqrt[3] + b^(1/3)*x^(3/2))^2]*Sqrt[1 + b*x^(3/2)])))/3
 

Defintions of rubi rules used

rule 759

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* 
((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s 
+ r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & 
& PosQ[a]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 832

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] 
], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 - Sqrt[3]))*(s/r)   Int[1/Sqrt[a + b*x 
^3], x], x] + Simp[1/r   Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x 
]] /; FreeQ[{a, b}, x] && PosQ[a]
 

rule 864

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi 
nator[n]}, Simp[k   Subst[Int[x^(k*(m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x 
^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]
 

rule 2416

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S 
imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt 
[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) 
*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq 
Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.47 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.04

method	result	size
meijerg	\(\frac {x^{3} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -b \,x^{\frac {9}{2}}\right )}{3}\)	\(18\)

Input:

int(x^2/(1+b*x^(9/2))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*x^3*hypergeom([1/2,2/3],[5/3],-b*x^(9/2))
 

Fricas [F]

\[ \int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx=\int { \frac {x^{2}}{\sqrt {b x^{\frac {9}{2}} + 1}} \,d x } \] Input:

integrate(x^2/(1+b*x^(9/2))^(1/2),x, algorithm="fricas")
 

Output:

integral((b*x^(13/2) - x^2)*sqrt(b*x^(9/2) + 1)/(b^2*x^9 - 1), x)
 

Sympy [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.08 \[ \int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx=\frac {2 x^{3} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {b x^{\frac {9}{2}} e^{i \pi }} \right )}}{9 \Gamma \left (\frac {5}{3}\right )} \] Input:

integrate(x**2/(1+b*x**(9/2))**(1/2),x)
 

Output:

2*x**3*gamma(2/3)*hyper((1/2, 2/3), (5/3,), b*x**(9/2)*exp_polar(I*pi))/(9 
*gamma(5/3))
 

Maxima [F]

\[ \int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx=\int { \frac {x^{2}}{\sqrt {b x^{\frac {9}{2}} + 1}} \,d x } \] Input:

integrate(x^2/(1+b*x^(9/2))^(1/2),x, algorithm="maxima")
 

Output:

integrate(x^2/sqrt(b*x^(9/2) + 1), x)
 

Giac [F]

\[ \int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx=\int { \frac {x^{2}}{\sqrt {b x^{\frac {9}{2}} + 1}} \,d x } \] Input:

integrate(x^2/(1+b*x^(9/2))^(1/2),x, algorithm="giac")
 

Output:

integrate(x^2/sqrt(b*x^(9/2) + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx=\int \frac {x^2}{\sqrt {b\,x^{9/2}+1}} \,d x \] Input:

int(x^2/(b*x^(9/2) + 1)^(1/2),x)
 

Output:

int(x^2/(b*x^(9/2) + 1)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^2}{\sqrt {1+b x^{9/2}}} \, dx=-\left (\int \frac {\sqrt {\sqrt {x}\, b \,x^{4}+1}\, x^{2}}{b^{2} x^{9}-1}d x \right )+\left (\int \frac {\sqrt {x}\, \sqrt {\sqrt {x}\, b \,x^{4}+1}\, x^{6}}{b^{2} x^{9}-1}d x \right ) b \] Input:

int(x^2/(1+b*x^(9/2))^(1/2),x)
 

Output:

 - int((sqrt(sqrt(x)*b*x**4 + 1)*x**2)/(b**2*x**9 - 1),x) + int((sqrt(x)*s 
qrt(sqrt(x)*b*x**4 + 1)*x**6)/(b**2*x**9 - 1),x)*b