\(\int \frac {x^5}{(a x+b x^3)^{3/2}} \, dx\) [74]

Optimal result

Integrand size = 17, antiderivative size = 137 \[ \int \frac {x^5}{\left (a x+b x^3\right )^{3/2}} \, dx=-\frac {x^3}{b \sqrt {a x+b x^3}}+\frac {5 \sqrt {a x+b x^3}}{3 b^2}-\frac {5 a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 b^{9/4} \sqrt {a x+b x^3}} \] Output:

-x^3/b/(b*x^3+a*x)^(1/2)+5/3*(b*x^3+a*x)^(1/2)/b^2-5/6*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^(9/4)/(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 = 10.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.49 \[ \int \frac {x^5}{\left (a x+b x^3\right )^{3/2}} \, dx=\frac {x \left (5 a+2 b x^2-5 a \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 b^2 \sqrt {x \left (a+b x^2\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1928, 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.

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

\(\Big \downarrow \) 1928

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

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

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 1928

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*(n - j)*( 
p + 1))), x] - Simp[c^n*((m + j*p - n + j + 1)/(b*(n - j)*(p + 1)))   Int[( 
c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] &&  !In 
tegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] & 
& GtQ[m + j*p + 1, 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 = 0.83 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07

method	result	size
default	\(\frac {x a}{b^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {2 \sqrt {b \,x^{3}+a x}}{3 b^{2}}-\frac {5 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{6 b^{3} \sqrt {b \,x^{3}+a x}}\)	\(147\)
elliptic	\(\frac {x a}{b^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {2 \sqrt {b \,x^{3}+a x}}{3 b^{2}}-\frac {5 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{6 b^{3} \sqrt {b \,x^{3}+a x}}\)	\(147\)
risch	\(\frac {2 \left (b \,x^{2}+a \right ) x}{3 b^{2} \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {a \left (\frac {4 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b \,x^{3}+a x}}-3 a \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b \,x^{3}+a x}}\right )\right )}{3 b^{2}}\)	\(275\)

Input:

int(x^5/(b*x^3+a*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/b^2*x*a/((x^2+a/b)*b*x)^(1/2)+2/3*(b*x^3+a*x)^(1/2)/b^2-5/6*a/b^3*(-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)*Elliptic 
F(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.50 \[ \int \frac {x^5}{\left (a x+b x^3\right )^{3/2}} \, dx=-\frac {5 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (2 \, b^{2} x^{2} + 5 \, a b\right )} \sqrt {b x^{3} + a x}}{3 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \] Input:

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

Output:

-1/3*(5*(a*b*x^2 + a^2)*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) - (2*b^2 
*x^2 + 5*a*b)*sqrt(b*x^3 + a*x))/(b^4*x^2 + a*b^3)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^5}{\left (a x+b x^3\right )^{3/2}} \, dx=\frac {10 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a +2 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, b \,x^{2}-5 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{2} x^{5}+2 a b \,x^{3}+a^{2} x}d x \right ) a^{3}-5 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{2} x^{5}+2 a b \,x^{3}+a^{2} x}d x \right ) a^{2} b \,x^{2}}{3 b^{2} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

(10*sqrt(x)*sqrt(a + b*x**2)*a + 2*sqrt(x)*sqrt(a + b*x**2)*b*x**2 - 5*int 
((sqrt(x)*sqrt(a + b*x**2))/(a**2*x + 2*a*b*x**3 + b**2*x**5),x)*a**3 - 5* 
int((sqrt(x)*sqrt(a + b*x**2))/(a**2*x + 2*a*b*x**3 + b**2*x**5),x)*a**2*b 
*x**2)/(3*b**2*(a + b*x**2))