\(\int \sqrt {a+b x^3+c x^6} \, dx\) [183]

Optimal result

Integrand size = 16, antiderivative size = 135 \[ \int \sqrt {a+b x^3+c x^6} \, dx=\frac {x \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {1}{2},-\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}} \] Output:

x*(c*x^6+b*x^3+a)^(1/2)*AppellF1(1/3,-1/2,-1/2,4/3,-2*c*x^3/(b-(-4*a*c+b^2 
)^(1/2)),-2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))/(1+2*c*x^3/(b-(-4*a*c+b^2)^(1/2) 
))^(1/2)/(1+2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(335\) vs. \(2(135)=270\).

Time = 10.42 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.48 \[ \int \sqrt {a+b x^3+c x^6} \, dx=\frac {x \left (8 \left (a+b x^3+c x^6\right )+24 a \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+3 b x^3 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )\right )}{32 \sqrt {a+b x^3+c x^6}} \] Input:

Integrate[Sqrt[a + b*x^3 + c*x^6],x]
 

Output:

(x*(8*(a + b*x^3 + c*x^6) + 24*a*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b 
 - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^ 
2 - 4*a*c])]*AppellF1[1/3, 1/2, 1/2, 4/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c 
]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])] + 3*b*x^3*Sqrt[(b - Sqrt[b^2 - 4*a 
*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c* 
x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[4/3, 1/2, 1/2, 7/3, (-2*c*x^3)/(b + 
 Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])]))/(32*Sqrt[a + b* 
x^3 + c*x^6])
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1686, 936}

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.

\(\displaystyle \int \sqrt {a+b x^3+c x^6} \, dx\)

\(\Big \downarrow \) 1686

\(\displaystyle \frac {\sqrt {a+b x^3+c x^6} \int \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}+1}dx}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}}\)

\(\Big \downarrow \) 936

\(\displaystyle \frac {x \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {1}{2},-\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}}\)

Input:

Int[Sqrt[a + b*x^3 + c*x^6],x]
 

Output:

(x*Sqrt[a + b*x^3 + c*x^6]*AppellF1[1/3, -1/2, -1/2, 4/3, (-2*c*x^3)/(b - 
Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[1 + (2*c*x^ 
3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])
 

Defintions of rubi rules used

rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1686

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^ 
IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 
- 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^2 - 4*a*c, 2])))^FracPar 
t[p]))   Int[(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - S 
qrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] 
&& NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]
 

Maple [F]

\[\int \sqrt {c \,x^{6}+b \,x^{3}+a}d x\]

Input:

int((c*x^6+b*x^3+a)^(1/2),x)
 

Output:

int((c*x^6+b*x^3+a)^(1/2),x)
 

Fricas [F]

\[ \int \sqrt {a+b x^3+c x^6} \, dx=\int { \sqrt {c x^{6} + b x^{3} + a} \,d x } \] Input:

integrate((c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(c*x^6 + b*x^3 + a), x)
 

Sympy [F]

\[ \int \sqrt {a+b x^3+c x^6} \, dx=\int \sqrt {a + b x^{3} + c x^{6}}\, dx \] Input:

integrate((c*x**6+b*x**3+a)**(1/2),x)
 

Output:

Integral(sqrt(a + b*x**3 + c*x**6), x)
 

Maxima [F]

\[ \int \sqrt {a+b x^3+c x^6} \, dx=\int { \sqrt {c x^{6} + b x^{3} + a} \,d x } \] Input:

integrate((c*x^6+b*x^3+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^6 + b*x^3 + a), x)
 

Giac [F]

\[ \int \sqrt {a+b x^3+c x^6} \, dx=\int { \sqrt {c x^{6} + b x^{3} + a} \,d x } \] Input:

integrate((c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*x^6 + b*x^3 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b x^3+c x^6} \, dx=\int \sqrt {c\,x^6+b\,x^3+a} \,d x \] Input:

int((a + b*x^3 + c*x^6)^(1/2),x)
 

Output:

int((a + b*x^3 + c*x^6)^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {a+b x^3+c x^6} \, dx=\frac {\sqrt {c \,x^{6}+b \,x^{3}+a}\, x}{4}+\frac {3 \left (\int \frac {\sqrt {c \,x^{6}+b \,x^{3}+a}}{c \,x^{6}+b \,x^{3}+a}d x \right ) a}{4}+\frac {3 \left (\int \frac {\sqrt {c \,x^{6}+b \,x^{3}+a}\, x^{3}}{c \,x^{6}+b \,x^{3}+a}d x \right ) b}{8} \] Input:

int((c*x^6+b*x^3+a)^(1/2),x)
 

Output:

(2*sqrt(a + b*x**3 + c*x**6)*x + 6*int(sqrt(a + b*x**3 + c*x**6)/(a + b*x* 
*3 + c*x**6),x)*a + 3*int((sqrt(a + b*x**3 + c*x**6)*x**3)/(a + b*x**3 + c 
*x**6),x)*b)/8