\(\int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx\) [278]

Optimal result

Integrand size = 21, antiderivative size = 492 \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx=\frac {\left (1+\sqrt {3}\right ) x^{3/2} \left (a+b x^3\right )}{b^{2/3} \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right ) \sqrt {a x^2+b x^5}}-\frac {\sqrt [4]{3} \sqrt [3]{a} x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{b^{2/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a} x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} b^{2/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}} \] Output:

(1+3^(1/2))*x^(3/2)*(b*x^3+a)/b^(2/3)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)/(b*x 
^5+a*x^2)^(1/2)-3^(1/4)*a^(1/3)*x^(3/2)*(a^(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1 
/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2)*Ellipt 
icE((1-(a^(1/3)+(1-3^(1/2))*b^(1/3)*x)^2/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2 
)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))/b^(2/3)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a 
^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2)/(b*x^5+a*x^2)^(1/2)-1/6*(1-3^(1/2)) 
*a^(1/3)*x^(3/2)*(a^(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x 
^2)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2)*InverseJacobiAM(arccos((a^(1/ 
3)+(1-3^(1/2))*b^(1/3)*x)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)),1/4*6^(1/2)+1/4 
*2^(1/2))*3^(3/4)/b^(2/3)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+(1+3^(1/ 
2))*b^(1/3)*x)^2)^(1/2)/(b*x^5+a*x^2)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.12 \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 x^{7/2} \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},-\frac {b x^3}{a}\right )}{5 \sqrt {x^2 \left (a+b x^3\right )}} \] Input:

Integrate[x^(5/2)/Sqrt[a*x^2 + b*x^5],x]
 

Output:

(2*x^(7/2)*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/2, 5/6, 11/6, -((b*x^3) 
/a)])/(5*Sqrt[x^2*(a + b*x^3)])
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1938, 851, 837, 25, 766, 2420}

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.

\(\displaystyle \int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {x \sqrt {a+b x^3} \int \frac {x^{3/2}}{\sqrt {b x^3+a}}dx}{\sqrt {a x^2+b x^5}}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {2 x \sqrt {a+b x^3} \int \frac {x^2}{\sqrt {b x^3+a}}d\sqrt {x}}{\sqrt {a x^2+b x^5}}\)

\(\Big \downarrow \) 837

\(\displaystyle \frac {2 x \sqrt {a+b x^3} \left (-\frac {\left (1-\sqrt {3}\right ) a^{2/3} \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {x}}{2 b^{2/3}}-\frac {\int -\frac {2 b^{2/3} x^2+\left (1-\sqrt {3}\right ) a^{2/3}}{\sqrt {b x^3+a}}d\sqrt {x}}{2 b^{2/3}}\right )}{\sqrt {a x^2+b x^5}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 x \sqrt {a+b x^3} \left (\frac {\int \frac {2 b^{2/3} x^2+\left (1-\sqrt {3}\right ) a^{2/3}}{\sqrt {b x^3+a}}d\sqrt {x}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) a^{2/3} \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {x}}{2 b^{2/3}}\right )}{\sqrt {a x^2+b x^5}}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {2 x \sqrt {a+b x^3} \left (\frac {\int \frac {2 b^{2/3} x^2+\left (1-\sqrt {3}\right ) a^{2/3}}{\sqrt {b x^3+a}}d\sqrt {x}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a} \sqrt {x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b^{2/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\right )}{\sqrt {a x^2+b x^5}}\)

\(\Big \downarrow \) 2420

\(\displaystyle \frac {2 x \sqrt {a+b x^3} \left (\frac {\frac {\left (1+\sqrt {3}\right ) \sqrt {x} \sqrt {a+b x^3}}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}-\frac {\sqrt [4]{3} \sqrt [3]{a} \sqrt {x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a} \sqrt {x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b^{2/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\right )}{\sqrt {a x^2+b x^5}}\)

Input:

Int[x^(5/2)/Sqrt[a*x^2 + b*x^5],x]
 

Output:

(2*x*Sqrt[a + b*x^3]*((((1 + Sqrt[3])*Sqrt[x]*Sqrt[a + b*x^3])/(a^(1/3) + 
(1 + Sqrt[3])*b^(1/3)*x) - (3^(1/4)*a^(1/3)*Sqrt[x]*(a^(1/3) + b^(1/3)*x)* 
Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])* 
b^(1/3)*x)^2]*EllipticE[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3 
) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(Sqrt[(b^(1/3)*x*(a^(1/3) 
 + b^(1/3)*x))/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3]))/(2 
*b^(2/3)) - ((1 - Sqrt[3])*a^(1/3)*Sqrt[x]*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^( 
2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x 
)^2]*EllipticF[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + 
Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(4*3^(1/4)*b^(2/3)*Sqrt[(b^(1/3)*x 
*(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b* 
x^3])))/Sqrt[a*x^2 + b*x^5]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 837

Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 
3]], s = Denom[Rt[b/a, 3]]}, Simp[(Sqrt[3] - 1)*(s^2/(2*r^2))   Int[1/Sqrt[ 
a + b*x^6], x], x] - Simp[1/(2*r^2)   Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4)/S 
qrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]
 

rule 851

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

rule 1938

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]))   Int[x^(m + j* 
p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 2420

Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = 
 Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqr 
t[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d*s*x* 
(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2 
*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]) 
)*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 
 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 
- Sqrt[3])*d, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.90 (sec) , antiderivative size = 2374, normalized size of antiderivative = 4.83

method	result	size
default	\(\text {Expression too large to display}\)	\(2374\)

Input:

int(x^(5/2)/(b*x^5+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/(b*x^5+a*x^2)^(1/2)*x^(3/2)*(b*x^3+a)/b^2*(-I*(-a*b^2)^(1/3)*3^(1/2)*(- 
(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*( 
-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^( 
1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+ 
(-a*b^2)^(1/3)))^(1/2)*EllipticE((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+( 
-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2 
)-3))^(1/2))*b*x^2+2*I*(-a*b^2)^(2/3)*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/ 
2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^ 
2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^ 
(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Ell 
ipticE((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I* 
3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x-2*(-a*b^2)^ 
(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3 
^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^( 
1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1 
)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1) 
/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/( 
I*3^(1/2)-3))^(1/2))*b*x^2+3*(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2) 
-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2) 
^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)...
 

Fricas [F]

\[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx=\int { \frac {x^{\frac {5}{2}}}{\sqrt {b x^{5} + a x^{2}}} \,d x } \] Input:

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

Output:

integral(sqrt(b*x^5 + a*x^2)*sqrt(x)/(b*x^3 + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx=\int { \frac {x^{\frac {5}{2}}}{\sqrt {b x^{5} + a x^{2}}} \,d x } \] Input:

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

Output:

integrate(x^(5/2)/sqrt(b*x^5 + a*x^2), x)
 

Giac [F]

\[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx=\int { \frac {x^{\frac {5}{2}}}{\sqrt {b x^{5} + a x^{2}}} \,d x } \] Input:

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

Output:

integrate(x^(5/2)/sqrt(b*x^5 + a*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx=\int \frac {x^{5/2}}{\sqrt {b\,x^5+a\,x^2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^{5/2}}{\sqrt {a x^2+b x^5}} \, dx=\int \frac {\sqrt {x}\, \sqrt {b \,x^{3}+a}\, x}{b \,x^{3}+a}d x \] Input:

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

Output:

int((sqrt(x)*sqrt(a + b*x**3)*x)/(a + b*x**3),x)