\(\int \frac {1}{x (a x+b x^3)^{3/2}} \, dx\) [80]

Optimal result

Integrand size = 17, antiderivative size = 139 \[ \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx=\frac {1}{a x \sqrt {a x+b x^3}}-\frac {5 \sqrt {a x+b x^3}}{3 a^2 x^2}-\frac {5 b^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 a^{9/4} \sqrt {a x+b x^3}} \] Output:

1/a/x/(b*x^3+a*x)^(1/2)-5/3*(b*x^3+a*x)^(1/2)/a^2/x^2-5/6*b^(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)*InverseJacobiA 
M(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)),1/2*2^(1/2))/a^(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 = 56, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx=-\frac {2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},-\frac {b x^2}{a}\right )}{3 a x \sqrt {x \left (a+b x^2\right )}} \] Input:

Integrate[1/(x*(a*x + b*x^3)^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 147, 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 = {1929, 1931, 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.

\(\displaystyle \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1929

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[1/(x*(a*x + b*x^3)^(3/2)),x]
 

Output:

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

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 1929

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

rule 1931

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p 
+ 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*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]) && LtQ[ 
m + j*p + 1, 0]
 

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {b x}{a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}-\frac {2 \sqrt {b \,x^{3}+a x}}{3 a^{2} x^{2}}-\frac {5 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{6 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(150\)
elliptic	\(-\frac {b x}{a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}-\frac {2 \sqrt {b \,x^{3}+a x}}{3 a^{2} x^{2}}-\frac {5 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{6 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(150\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )}{3 a^{2} x \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {b \left (\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b \,x^{3}+a x}}+3 a \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b \,x^{3}+a x}}\right )\right )}{3 a^{2}}\)	\(276\)

Input:

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

Output:

-b*x/a^2/((x^2+a/b)*b*x)^(1/2)-2/3*(b*x^3+a*x)^(1/2)/a^2/x^2-5/6/a^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)*Ellipti 
cF(((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 = 0.49 \[ \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx=-\frac {5 \, {\left (b x^{4} + a x^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + \sqrt {b x^{3} + a x} {\left (5 \, b x^{2} + 2 \, a\right )}}{3 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx=\int \frac {1}{x \left (x \left (a + b x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x\right )}^{\frac {3}{2}} x} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x\right )}^{\frac {3}{2}} x} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (b\,x^3+a\,x\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{2} x^{7}+2 a b \,x^{5}+a^{2} x^{3}}d x \] Input:

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

Output:

int((sqrt(x)*sqrt(a + b*x**2))/(a**2*x**3 + 2*a*b*x**5 + b**2*x**7),x)