\(\int \frac {a+b x^3+c x^6}{\sqrt {d+e x^3}} \, dx\) [68]

Optimal result

Integrand size = 24, antiderivative size = 278 \[ \int \frac {a+b x^3+c x^6}{\sqrt {d+e x^3}} \, dx=-\frac {2 (8 c d-11 b e) x \sqrt {d+e x^3}}{55 e^2}+\frac {2 c x^4 \sqrt {d+e x^3}}{11 e}+\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2-22 b d e+55 a e^2\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt {3}\right )}{55 \sqrt [4]{3} e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \] Output:

-2/55*(-11*b*e+8*c*d)*x*(e*x^3+d)^(1/2)/e^2+2/11*c*x^4*(e*x^3+d)^(1/2)/e+2 
/165*(1/2*6^(1/2)+1/2*2^(1/2))*(55*a*e^2-22*b*d*e+16*c*d^2)*(d^(1/3)+e^(1/ 
3)*x)*((d^(2/3)-d^(1/3)*e^(1/3)*x+e^(2/3)*x^2)/((1+3^(1/2))*d^(1/3)+e^(1/3 
)*x)^2)^(1/2)*EllipticF(((1-3^(1/2))*d^(1/3)+e^(1/3)*x)/((1+3^(1/2))*d^(1/ 
3)+e^(1/3)*x),I*3^(1/2)+2*I)*3^(3/4)/e^(7/3)/(d^(1/3)*(d^(1/3)+e^(1/3)*x)/ 
((1+3^(1/2))*d^(1/3)+e^(1/3)*x)^2)^(1/2)/(e*x^3+d)^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.35 \[ \int \frac {a+b x^3+c x^6}{\sqrt {d+e x^3}} \, dx=\frac {x \left (-2 \left (d+e x^3\right ) \left (8 c d-11 b e-5 c e x^3\right )+\left (16 c d^2+11 e (-2 b d+5 a e)\right ) \sqrt {1+\frac {e x^3}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {e x^3}{d}\right )\right )}{55 e^2 \sqrt {d+e x^3}} \] Input:

Integrate[(a + b*x^3 + c*x^6)/Sqrt[d + e*x^3],x]
 

Output:

(x*(-2*(d + e*x^3)*(8*c*d - 11*b*e - 5*c*e*x^3) + (16*c*d^2 + 11*e*(-2*b*d 
 + 5*a*e))*Sqrt[1 + (e*x^3)/d]*Hypergeometric2F1[1/3, 1/2, 4/3, -((e*x^3)/ 
d)]))/(55*e^2*Sqrt[d + e*x^3])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1741, 27, 913, 759}

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.

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

\(\Big \downarrow \) 1741

\(\displaystyle \frac {2 \int \frac {11 a e-(8 c d-11 b e) x^3}{2 \sqrt {e x^3+d}}dx}{11 e}+\frac {2 c x^4 \sqrt {d+e x^3}}{11 e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {11 a e-(8 c d-11 b e) x^3}{\sqrt {e x^3+d}}dx}{11 e}+\frac {2 c x^4 \sqrt {d+e x^3}}{11 e}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {\left (16 c d^2-11 e (2 b d-5 a e)\right ) \int \frac {1}{\sqrt {e x^3+d}}dx}{5 e}-\frac {2 x \sqrt {d+e x^3} (8 c d-11 b e)}{5 e}}{11 e}+\frac {2 c x^4 \sqrt {d+e x^3}}{11 e}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (16 c d^2-11 e (2 b d-5 a e)\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{e} x+\left (1-\sqrt {3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt {3}\right ) \sqrt [3]{d}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} e^{4/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}}-\frac {2 x \sqrt {d+e x^3} (8 c d-11 b e)}{5 e}}{11 e}+\frac {2 c x^4 \sqrt {d+e x^3}}{11 e}\)

Input:

Int[(a + b*x^3 + c*x^6)/Sqrt[d + e*x^3],x]
 

Output:

(2*c*x^4*Sqrt[d + e*x^3])/(11*e) + ((-2*(8*c*d - 11*b*e)*x*Sqrt[d + e*x^3] 
)/(5*e) + (2*Sqrt[2 + Sqrt[3]]*(16*c*d^2 - 11*e*(2*b*d - 5*a*e))*(d^(1/3) 
+ e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x + e^(2/3)*x^2)/((1 + Sqrt[3 
])*d^(1/3) + e^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*d^(1/3) + e^(1/ 
3)*x)/((1 + Sqrt[3])*d^(1/3) + e^(1/3)*x)], -7 - 4*Sqrt[3]])/(5*3^(1/4)*e^ 
(4/3)*Sqrt[(d^(1/3)*(d^(1/3) + e^(1/3)*x))/((1 + Sqrt[3])*d^(1/3) + e^(1/3 
)*x)^2]*Sqrt[d + e*x^3]))/(11*e)
 

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 759

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* 
((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s 
+ r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & 
& PosQ[a]
 

rule 913

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

rule 1741

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ 
)), x_Symbol] :> Simp[c*x^(n + 1)*((d + e*x^n)^(q + 1)/(e*(n*(q + 2) + 1))) 
, x] + Simp[1/(e*(n*(q + 2) + 1))   Int[(d + e*x^n)^q*(a*e*(n*(q + 2) + 1) 
- (c*d*(n + 1) - b*e*(n*(q + 2) + 1))*x^n), x], x] /; FreeQ[{a, b, c, d, e, 
 n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a* 
e^2, 0]
 

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.20

method	result	size
risch	\(\frac {2 x \left (5 c e \,x^{3}+11 e b -8 c d \right ) \sqrt {e \,x^{3}+d}}{55 e^{2}}-\frac {2 i \left (55 a \,e^{2}-22 b d e +16 c \,d^{2}\right ) \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{e}}{-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{e \left (-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right )}}\right )}{165 e^{3} \sqrt {e \,x^{3}+d}}\)	\(333\)
elliptic	\(\frac {2 c \,x^{4} \sqrt {e \,x^{3}+d}}{11 e}+\frac {2 \left (b -\frac {8 d c}{11 e}\right ) x \sqrt {e \,x^{3}+d}}{5 e}-\frac {2 i \left (a -\frac {2 d \left (b -\frac {8 d c}{11 e}\right )}{5 e}\right ) \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{e}}{-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{e \left (-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right )}}\right )}{3 e \sqrt {e \,x^{3}+d}}\)	\(343\)
default	\(\text {Expression too large to display}\)	\(907\)

Input:

int((c*x^6+b*x^3+a)/(e*x^3+d)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/55*x*(5*c*e*x^3+11*b*e-8*c*d)/e^2*(e*x^3+d)^(1/2)-2/165*I*(55*a*e^2-22*b 
*d*e+16*c*d^2)/e^3*3^(1/2)*(-d*e^2)^(1/3)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I 
*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2 
)^(1/3))/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I 
*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2 
)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^ 
(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3 
^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1 
/3)))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.26 \[ \int \frac {a+b x^3+c x^6}{\sqrt {d+e x^3}} \, dx=\frac {2 \, {\left ({\left (16 \, c d^{2} - 22 \, b d e + 55 \, a e^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (0, -\frac {4 \, d}{e}, x\right ) + {\left (5 \, c e^{2} x^{4} - {\left (8 \, c d e - 11 \, b e^{2}\right )} x\right )} \sqrt {e x^{3} + d}\right )}}{55 \, e^{3}} \] Input:

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

Output:

2/55*((16*c*d^2 - 22*b*d*e + 55*a*e^2)*sqrt(e)*weierstrassPInverse(0, -4*d 
/e, x) + (5*c*e^2*x^4 - (8*c*d*e - 11*b*e^2)*x)*sqrt(e*x^3 + d))/e^3
 

Sympy [A] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.43 \[ \int \frac {a+b x^3+c x^6}{\sqrt {d+e x^3}} \, dx=\frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt {d} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt {d} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt {d} \Gamma \left (\frac {10}{3}\right )} \] Input:

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

Output:

a*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), e*x**3*exp_polar(I*pi)/d)/(3*sqrt 
(d)*gamma(4/3)) + b*x**4*gamma(4/3)*hyper((1/2, 4/3), (7/3,), e*x**3*exp_p 
olar(I*pi)/d)/(3*sqrt(d)*gamma(7/3)) + c*x**7*gamma(7/3)*hyper((1/2, 7/3), 
 (10/3,), e*x**3*exp_polar(I*pi)/d)/(3*sqrt(d)*gamma(10/3))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b x^3+c x^6}{\sqrt {d+e x^3}} \, dx=\frac {22 \sqrt {e \,x^{3}+d}\, b e x -16 \sqrt {e \,x^{3}+d}\, c d x +10 \sqrt {e \,x^{3}+d}\, c e \,x^{4}+55 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e \,x^{3}+d}d x \right ) a \,e^{2}-22 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e \,x^{3}+d}d x \right ) b d e +16 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e \,x^{3}+d}d x \right ) c \,d^{2}}{55 e^{2}} \] Input:

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

Output:

(22*sqrt(d + e*x**3)*b*e*x - 16*sqrt(d + e*x**3)*c*d*x + 10*sqrt(d + e*x** 
3)*c*e*x**4 + 55*int(sqrt(d + e*x**3)/(d + e*x**3),x)*a*e**2 - 22*int(sqrt 
(d + e*x**3)/(d + e*x**3),x)*b*d*e + 16*int(sqrt(d + e*x**3)/(d + e*x**3), 
x)*c*d**2)/(55*e**2)