\(\int \frac {x^2}{\sqrt {a x+b x^3}} \, dx\) [66]

Optimal result

Integrand size = 17, antiderivative size = 116 \[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b x^3}} \] Output:

2/3*(b*x^3+a*x)^(1/2)/b-1/3*a^(3/4)*x^(1/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a) 
/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x^(1/2)/a^( 
1/4)),1/2*2^(1/2))/b^(5/4)/(b*x^3+a*x)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.55 \[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\frac {2 x \left (a+b x^2-a \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 b \sqrt {x \left (a+b x^2\right )}} \] Input:

Integrate[x^2/Sqrt[a*x + b*x^3],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1930, 1917, 266, 761}

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.

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[x^2/Sqrt[a*x + b*x^3],x]
 

Output:

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

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 1930

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {2 \sqrt {b \,x^{3}+a x}}{3 b}-\frac {a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(127\)
elliptic	\(\frac {2 \sqrt {b \,x^{3}+a x}}{3 b}-\frac {a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(127\)
risch	\(\frac {2 x \left (b \,x^{2}+a \right )}{3 b \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(135\)

Input:

int(x^2/(b*x^3+a*x)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/3*(b*x^3+a*x)^(1/2)/b-1/3*a/b^2*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*b)^ 
(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-1/(-a*b)^(1/ 
2)*b*x)^(1/2)/(b*x^3+a*x)^(1/2)*EllipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2) 
*b)^(1/2),1/2*2^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.29 \[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=-\frac {2 \, {\left (a \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - \sqrt {b x^{3} + a x} b\right )}}{3 \, b^{2}} \] Input:

integrate(x^2/(b*x^3+a*x)^(1/2),x, algorithm="fricas")
 

Output:

-2/3*(a*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) - sqrt(b*x^3 + a*x)*b)/b 
^2
 

Sympy [F]

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

integrate(x**2/(b*x**3+a*x)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(x^2/(b*x^3+a*x)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^2/(b*x^3+a*x)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(x^2/(a*x + b*x^3)^(1/2),x)
 

Output:

int(x^2/(a*x + b*x^3)^(1/2), x)
 

Reduce [F]

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

int(x^2/(b*x^3+a*x)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(2*sqrt(x)*sqrt(a + b*x**2) - int((sqrt(x)*sqrt(a + b*x**2))/(a*x + b*x**3 
),x)*a)/(3*b)