\(\int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx\) [743]

Optimal result

Integrand size = 14, antiderivative size = 88 \[ \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx=\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt [4]{-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )\right |2\right )}{c \sqrt [4]{a+b x+c x^2}} \] Output:

2^(1/2)*(-4*a*c+b^2)^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/4)*EllipticE 
(sin(1/2*arcsin((2*c*x+b)/(-4*a*c+b^2)^(1/2))),2^(1/2))/c/(c*x^2+b*x+a)^(1 
/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 10.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx=\frac {(b+2 c x) \sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\sqrt {2} c \sqrt [4]{a+x (b+c x)}} \] Input:

Integrate[(a + b*x + c*x^2)^(-1/4),x]
 

Output:

((b + 2*c*x)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/4)*Hypergeometric2F 
1[1/4, 1/2, 3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(Sqrt[2]*c*(a + x*(b + c*x) 
)^(1/4))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(536\) vs. \(2(88)=176\).

Time = 0.46 (sec) , antiderivative size = 536, normalized size of antiderivative = 6.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1094, 834, 761, 1510}

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.

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

\(\Big \downarrow \) 1094

\(\displaystyle \frac {4 \sqrt {(b+2 c x)^2} \int \frac {\sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{b+2 c x}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {4 \sqrt {(b+2 c x)^2} \left (\frac {\sqrt {b^2-4 a c} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{b+2 c x}\)

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

Int[(a + b*x + c*x^2)^(-1/4),x]
 

Output:

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

Defintions of rubi rules used

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 834

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

rule 1094

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[4*(Sqrt[(b 
+ 2*c*x)^2]/(b + 2*c*x))   Subst[Int[x^(4*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4 
*c*x^4], x], x, (a + b*x + c*x^2)^(1/4)], x] /; FreeQ[{a, b, c}, x] && Inte 
gerQ[4*p]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [F]

\[\int \frac {1}{\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x\]

Input:

int(1/(c*x^2+b*x+a)^(1/4),x)
 

Output:

int(1/(c*x^2+b*x+a)^(1/4),x)
 

Fricas [F]

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

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

Output:

integral((c*x^2 + b*x + a)^(-1/4), x)
 

Sympy [F]

\[ \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {1}{\sqrt [4]{a + b x + c x^{2}}}\, dx \] Input:

integrate(1/(c*x**2+b*x+a)**(1/4),x)
 

Output:

Integral((a + b*x + c*x**2)**(-1/4), x)
 

Maxima [F]

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

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

Output:

integrate((c*x^2 + b*x + a)^(-1/4), x)
 

Giac [F]

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

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

Output:

integrate((c*x^2 + b*x + a)^(-1/4), x)
 

Mupad [F(-1)]

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

int(1/(a + b*x + c*x^2)^(1/4),x)
 

Output:

int(1/(a + b*x + c*x^2)^(1/4), x)
 

Reduce [F]

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

int(1/(c*x^2+b*x+a)^(1/4),x)
 

Output:

int(1/(a + b*x + c*x**2)**(1/4),x)