\(\int \frac {a+b x^3+c x^6}{(d+e x^3)^{3/2}} \, dx\) [13]

Optimal result

Integrand size = 24, antiderivative size = 289 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \sqrt {d+e x^3}}+\frac {2 c x \sqrt {d+e x^3}}{5 e^2}-\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2-5 e (2 b d+a e)\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} d e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \] Output:

2/3*(a*e^2-b*d*e+c*d^2)*x/d/e^2/(e*x^3+d)^(1/2)+2/5*c*x*(e*x^3+d)^(1/2)/e^ 
2-2/45*(1/2*6^(1/2)+1/2*2^(1/2))*(16*c*d^2-5*e*(a*e+2*b*d))*(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)/d/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 = 10.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.35 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\frac {x \left (2 \left (5 e (-b d+a e)+c d \left (8 d+3 e x^3\right )\right )+\left (-16 c d^2+5 e (2 b d+a e)\right ) \sqrt {1+\frac {e x^3}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {e x^3}{d}\right )\right )}{15 d e^2 \sqrt {d+e x^3}} \] Input:

Integrate[(a + b*x^3 + c*x^6)/(d + e*x^3)^(3/2),x]
 

Output:

(x*(2*(5*e*(-(b*d) + a*e) + c*d*(8*d + 3*e*x^3)) + (-16*c*d^2 + 5*e*(2*b*d 
 + a*e))*Sqrt[1 + (e*x^3)/d]*Hypergeometric2F1[1/3, 1/2, 4/3, -((e*x^3)/d) 
]))/(15*d*e^2*Sqrt[d + e*x^3])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1739, 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.

\(\displaystyle \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1739

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 913

\(\displaystyle \frac {2 x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \sqrt {d+e x^3}}-\frac {\frac {1}{5} \left (16 c d^2-5 e (a e+2 b d)\right ) \int \frac {1}{\sqrt {e x^3+d}}dx-\frac {6}{5} c d x \sqrt {d+e x^3}}{3 d e^2}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {2 x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \sqrt {d+e x^3}}-\frac {\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (16 c d^2-5 e (a e+2 b d)\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{e} x+\left (1-\sqrt {3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt {3}\right ) \sqrt [3]{d}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt [3]{e} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}}-\frac {6}{5} c d x \sqrt {d+e x^3}}{3 d e^2}\)

Input:

Int[(a + b*x^3 + c*x^6)/(d + e*x^3)^(3/2),x]
 

Output:

(2*(c*d^2 - b*d*e + a*e^2)*x)/(3*d*e^2*Sqrt[d + e*x^3]) - ((-6*c*d*x*Sqrt[ 
d + e*x^3])/5 + (2*Sqrt[2 + Sqrt[3]]*(16*c*d^2 - 5*e*(2*b*d + 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 + Sqr 
t[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^(1/3)*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]))/(3*d*e^2)
 

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 1739

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.32

method	result	size
elliptic	\(\frac {2 x \left (a \,e^{2}-b d e +c \,d^{2}\right )}{3 e^{2} d \sqrt {\left (x^{3}+\frac {d}{e}\right ) e}}+\frac {2 c x \sqrt {e \,x^{3}+d}}{5 e^{2}}-\frac {2 i \left (\frac {e b -c d}{e^{2}}+\frac {a \,e^{2}-b d e +c \,d^{2}}{3 d \,e^{2}}-\frac {2 c d}{5 e^{2}}\right ) \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{e}}{-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{e \left (-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right )}}\right )}{3 e \sqrt {e \,x^{3}+d}}\)	\(382\)
default	\(\text {Expression too large to display}\)	\(934\)
risch	\(\text {Expression too large to display}\)	\(959\)

Input:

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

Output:

2/3/e^2*x/d*(a*e^2-b*d*e+c*d^2)/((x^3+d/e)*e)^(1/2)+2/5*c*x*(e*x^3+d)^(1/2 
)/e^2-2/3*I*((b*e-c*d)/e^2+1/3*(a*e^2-b*d*e+c*d^2)/d/e^2-2/5*c*d/e^2)*3^(1 
/2)/e*(-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 = 0.07 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.43 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left ({\left (16 \, c d^{3} - 10 \, b d^{2} e - 5 \, a d e^{2} + {\left (16 \, c d^{2} e - 10 \, b d e^{2} - 5 \, a e^{3}\right )} x^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (0, -\frac {4 \, d}{e}, x\right ) - {\left (3 \, c d e^{2} x^{4} + {\left (8 \, c d^{2} e - 5 \, b d e^{2} + 5 \, a e^{3}\right )} x\right )} \sqrt {e x^{3} + d}\right )}}{15 \, {\left (d e^{4} x^{3} + d^{2} e^{3}\right )}} \] Input:

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

Output:

-2/15*((16*c*d^3 - 10*b*d^2*e - 5*a*d*e^2 + (16*c*d^2*e - 10*b*d*e^2 - 5*a 
*e^3)*x^3)*sqrt(e)*weierstrassPInverse(0, -4*d/e, x) - (3*c*d*e^2*x^4 + (8 
*c*d^2*e - 5*b*d*e^2 + 5*a*e^3)*x)*sqrt(e*x^3 + d))/(d*e^4*x^3 + d^2*e^3)
 

Sympy [A] (verification not implemented)

Time = 4.72 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.41 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {3}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {10}{3}\right )} \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\int { \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\int { \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\int \frac {c\,x^6+b\,x^3+a}{{\left (e\,x^3+d\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\frac {-10 \sqrt {e \,x^{3}+d}\, b e x +16 \sqrt {e \,x^{3}+d}\, c d x +2 \sqrt {e \,x^{3}+d}\, c e \,x^{4}+5 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e^{2} x^{6}+2 d e \,x^{3}+d^{2}}d x \right ) a d \,e^{2}+5 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e^{2} x^{6}+2 d e \,x^{3}+d^{2}}d x \right ) a \,e^{3} x^{3}+10 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e^{2} x^{6}+2 d e \,x^{3}+d^{2}}d x \right ) b \,d^{2} e +10 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e^{2} x^{6}+2 d e \,x^{3}+d^{2}}d x \right ) b d \,e^{2} x^{3}-16 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e^{2} x^{6}+2 d e \,x^{3}+d^{2}}d x \right ) c \,d^{3}-16 \left (\int \frac {\sqrt {e \,x^{3}+d}}{e^{2} x^{6}+2 d e \,x^{3}+d^{2}}d x \right ) c \,d^{2} e \,x^{3}}{5 e^{2} \left (e \,x^{3}+d \right )} \] Input:

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

Output:

( - 10*sqrt(d + e*x**3)*b*e*x + 16*sqrt(d + e*x**3)*c*d*x + 2*sqrt(d + e*x 
**3)*c*e*x**4 + 5*int(sqrt(d + e*x**3)/(d**2 + 2*d*e*x**3 + e**2*x**6),x)* 
a*d*e**2 + 5*int(sqrt(d + e*x**3)/(d**2 + 2*d*e*x**3 + e**2*x**6),x)*a*e** 
3*x**3 + 10*int(sqrt(d + e*x**3)/(d**2 + 2*d*e*x**3 + e**2*x**6),x)*b*d**2 
*e + 10*int(sqrt(d + e*x**3)/(d**2 + 2*d*e*x**3 + e**2*x**6),x)*b*d*e**2*x 
**3 - 16*int(sqrt(d + e*x**3)/(d**2 + 2*d*e*x**3 + e**2*x**6),x)*c*d**3 - 
16*int(sqrt(d + e*x**3)/(d**2 + 2*d*e*x**3 + e**2*x**6),x)*c*d**2*e*x**3)/ 
(5*e**2*(d + e*x**3))