\(\int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx\) [202]

Optimal result

Integrand size = 17, antiderivative size = 77 \[ \int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx=-\frac {4}{\sqrt [4]{2 x+3 x^2}}+\frac {2 \sqrt {2} \sqrt [4]{3} \sqrt [4]{x} \sqrt [4]{2+3 x} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {2}}{\sqrt {2+3 x}}\right )\right |2\right )}{\sqrt [4]{2 x+3 x^2}} \] Output:

-4/(3*x^2+2*x)^(1/4)+2*2^(1/2)*3^(1/4)*x^(1/4)*(2+3*x)^(1/4)*EllipticE(sin 
(1/2*arcsin(2^(1/2)/(2+3*x)^(1/2))),2^(1/2))/(3*x^2+2*x)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx=-\frac {2\ 2^{3/4} \sqrt [4]{2+3 x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},-\frac {3 x}{2}\right )}{\sqrt [4]{x (2+3 x)}} \] Input:

Integrate[1/(x*(2*x + 3*x^2)^(1/4)),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.46 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.58, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {1137, 61, 73, 839, 813, 27, 858, 807, 212}

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.

\(\displaystyle \int \frac {1}{x \sqrt [4]{3 x^2+2 x}} \, dx\)

\(\Big \downarrow \) 1137

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \int \frac {1}{x^{5/4} \sqrt [4]{3 x+2}}dx}{\sqrt [4]{3 x^2+2 x}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \left (3 \int \frac {1}{\sqrt [4]{x} \sqrt [4]{3 x+2}}dx-\frac {2 (3 x+2)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{3 x^2+2 x}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \left (12 \int \frac {\sqrt {x}}{\sqrt [4]{3 x+2}}d\sqrt [4]{x}-\frac {2 (3 x+2)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{3 x^2+2 x}}\)

\(\Big \downarrow \) 839

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \left (12 \left (\frac {x^{3/4}}{2 \sqrt [4]{3 x+2}}-\int \frac {\sqrt {x}}{(3 x+2)^{5/4}}d\sqrt [4]{x}\right )-\frac {2 (3 x+2)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{3 x^2+2 x}}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \left (12 \left (\frac {x^{3/4}}{2 \sqrt [4]{3 x+2}}-\frac {\sqrt [4]{\frac {2}{x}+3} \sqrt [4]{x} \int \frac {3 \sqrt [4]{3}}{\left (3+\frac {2}{x}\right )^{5/4} x^{3/4}}d\sqrt [4]{x}}{3 \sqrt [4]{3} \sqrt [4]{3 x+2}}\right )-\frac {2 (3 x+2)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{3 x^2+2 x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \left (12 \left (\frac {x^{3/4}}{2 \sqrt [4]{3 x+2}}-\frac {\sqrt [4]{\frac {2}{x}+3} \sqrt [4]{x} \int \frac {1}{\left (3+\frac {2}{x}\right )^{5/4} x^{3/4}}d\sqrt [4]{x}}{\sqrt [4]{3 x+2}}\right )-\frac {2 (3 x+2)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{3 x^2+2 x}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \left (12 \left (\frac {\sqrt [4]{\frac {2}{x}+3} \sqrt [4]{x} \int \frac {1}{\sqrt [4]{x} (2 x+3)^{5/4}}d\frac {1}{\sqrt [4]{x}}}{\sqrt [4]{3 x+2}}+\frac {x^{3/4}}{2 \sqrt [4]{3 x+2}}\right )-\frac {2 (3 x+2)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{3 x^2+2 x}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \left (12 \left (\frac {\sqrt [4]{\frac {2}{x}+3} \sqrt [4]{x} \int \frac {1}{\left (2 \sqrt {x}+3\right )^{5/4}}d\sqrt {x}}{2 \sqrt [4]{3 x+2}}+\frac {x^{3/4}}{2 \sqrt [4]{3 x+2}}\right )-\frac {2 (3 x+2)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{3 x^2+2 x}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{3 x+2} \left (12 \left (\frac {\sqrt [4]{\frac {2}{x}+3} \sqrt [4]{x} E\left (\left .\frac {1}{2} \arctan \left (\sqrt {\frac {2}{3}} \sqrt {x}\right )\right |2\right )}{\sqrt {2} 3^{3/4} \sqrt [4]{3 x+2}}+\frac {x^{3/4}}{2 \sqrt [4]{3 x+2}}\right )-\frac {2 (3 x+2)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{3 x^2+2 x}}\)

Input:

Int[1/(x*(2*x + 3*x^2)^(1/4)),x]
 

Output:

(x^(1/4)*(2 + 3*x)^(1/4)*((-2*(2 + 3*x)^(3/4))/x^(1/4) + 12*(x^(3/4)/(2*(2 
 + 3*x)^(1/4)) + ((3 + 2/x)^(1/4)*x^(1/4)*EllipticE[ArcTan[Sqrt[2/3]*Sqrt[ 
x]]/2, 2])/(Sqrt[2]*3^(3/4)*(2 + 3*x)^(1/4)))))/(2*x + 3*x^2)^(1/4)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/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 813

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 839

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 
)^(1/4)), x] - Simp[a/2   Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} 
, x] && PosQ[b/a]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 1137

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( 
e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p))   Int[x^(m + p)*(b + c*x)^ 
p, x], x] /; FreeQ[{b, c, e, m}, x]
 

Maple [C] (verified)

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

Time = 0.38 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.23

method	result	size
meijerg	\(-\frac {2 \,2^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {1}{4}\right ], \left [\frac {3}{4}\right ], -\frac {3 x}{2}\right )}{x^{\frac {1}{4}}}\)	\(18\)
risch	\(-\frac {2 \left (3 x +2\right )}{\left (x \left (3 x +2\right )\right )^{\frac {1}{4}}}+2 \,2^{\frac {3}{4}} x^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -\frac {3 x}{2}\right )\)	\(35\)

Input:

int(1/x/(3*x^2+2*x)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

-2*2^(3/4)/x^(1/4)*hypergeom([-1/4,1/4],[3/4],-3/2*x)
 

Fricas [F]

\[ \int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} + 2 \, x\right )}^{\frac {1}{4}} x} \,d x } \] Input:

integrate(1/x/(3*x^2+2*x)^(1/4),x, algorithm="fricas")
 

Output:

integral((3*x^2 + 2*x)^(3/4)/(3*x^3 + 2*x^2), x)
 

Sympy [F]

\[ \int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx=\int \frac {1}{x \sqrt [4]{x \left (3 x + 2\right )}}\, dx \] Input:

integrate(1/x/(3*x**2+2*x)**(1/4),x)
 

Output:

Integral(1/(x*(x*(3*x + 2))**(1/4)), x)
 

Maxima [F]

\[ \int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} + 2 \, x\right )}^{\frac {1}{4}} x} \,d x } \] Input:

integrate(1/x/(3*x^2+2*x)^(1/4),x, algorithm="maxima")
 

Output:

integrate(1/((3*x^2 + 2*x)^(1/4)*x), x)
 

Giac [F]

\[ \int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} + 2 \, x\right )}^{\frac {1}{4}} x} \,d x } \] Input:

integrate(1/x/(3*x^2+2*x)^(1/4),x, algorithm="giac")
 

Output:

integrate(1/((3*x^2 + 2*x)^(1/4)*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx=\int \frac {1}{x\,{\left (3\,x^2+2\,x\right )}^{1/4}} \,d x \] Input:

int(1/(x*(2*x + 3*x^2)^(1/4)),x)
 

Output:

int(1/(x*(2*x + 3*x^2)^(1/4)), x)
 

Reduce [F]

\[ \int \frac {1}{x \sqrt [4]{2 x+3 x^2}} \, dx=\int \frac {1}{x^{\frac {5}{4}} \left (3 x +2\right )^{\frac {1}{4}}}d x \] Input:

int(1/x/(3*x^2+2*x)^(1/4),x)
 

Output:

int(1/(x**(1/4)*(3*x + 2)**(1/4)*x),x)