\(\int \frac {(d+e x)^{2/3}}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [330]

Optimal result

Integrand size = 39, antiderivative size = 502 \[ \int \frac {(d+e x)^{2/3}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c d \sqrt [3]{d+e x}}+\frac {3^{3/4} \left (c d^2-a e^2\right )^{2/3} (d+e x)^{2/3} \left (\sqrt [3]{c d^2-a e^2}-\sqrt [3]{c} \sqrt [3]{d} \sqrt [3]{d+e x}\right ) \sqrt {\frac {\left (c d^2-a e^2\right )^{2/3}+\sqrt [3]{c} \sqrt [3]{d} \sqrt [3]{c d^2-a e^2} \sqrt [3]{d+e x}+c^{2/3} d^{2/3} (d+e x)^{2/3}}{\left (\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{d} \sqrt [3]{d+e x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{c d^2-a e^2}-\left (1-\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{d} \sqrt [3]{d+e x}}{\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{d} \sqrt [3]{d+e x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 c d e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \sqrt {-\frac {\sqrt [3]{c} \sqrt [3]{d} \sqrt [3]{d+e x} \left (\sqrt [3]{c d^2-a e^2}-\sqrt [3]{c} \sqrt [3]{d} \sqrt [3]{d+e x}\right )}{\left (\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{d} \sqrt [3]{d+e x}\right )^2}}} \] Output:

3/2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/(e*x+d)^(1/3)+1/4*3^(3/4)* 
(-a*e^2+c*d^2)^(2/3)*(e*x+d)^(2/3)*((-a*e^2+c*d^2)^(1/3)-c^(1/3)*d^(1/3)*( 
e*x+d)^(1/3))*(((-a*e^2+c*d^2)^(2/3)+c^(1/3)*d^(1/3)*(-a*e^2+c*d^2)^(1/3)* 
(e*x+d)^(1/3)+c^(2/3)*d^(2/3)*(e*x+d)^(2/3))/((-a*e^2+c*d^2)^(1/3)-(1+3^(1 
/2))*c^(1/3)*d^(1/3)*(e*x+d)^(1/3))^2)^(1/2)*InverseJacobiAM(arccos(((-a*e 
^2+c*d^2)^(1/3)-(1-3^(1/2))*c^(1/3)*d^(1/3)*(e*x+d)^(1/3))/((-a*e^2+c*d^2) 
^(1/3)-(1+3^(1/2))*c^(1/3)*d^(1/3)*(e*x+d)^(1/3))),1/4*6^(1/2)+1/4*2^(1/2) 
)/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(-c^(1/3)*d^(1/3)*(e*x+d)^ 
(1/3)*((-a*e^2+c*d^2)^(1/3)-c^(1/3)*d^(1/3)*(e*x+d)^(1/3))/((-a*e^2+c*d^2) 
^(1/3)-(1+3^(1/2))*c^(1/3)*d^(1/3)*(e*x+d)^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.19 \[ \int \frac {(d+e x)^{2/3}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {e (a e+c d x)}{-c d^2+a e^2}\right )}{c d \sqrt [3]{d+e x} \sqrt [6]{\frac {c d (d+e x)}{c d^2-a e^2}}} \] Input:

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

Output:

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*Hypergeometric2F1[-1/6, 1/2, 3/2, (e*(a*e 
 + c*d*x))/(-(c*d^2) + a*e^2)])/(c*d*(d + e*x)^(1/3)*((c*d*(d + e*x))/(c*d 
^2 - a*e^2))^(1/6))
 

Rubi [A] (warning: unable to verify)

Time = 0.97 (sec) , antiderivative size = 604, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {1139, 1138, 60, 73, 766}

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.

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

\(\Big \downarrow \) 1139

\(\displaystyle \frac {(d+e x)^{2/3} \int \frac {\left (\frac {e x}{d}+1\right )^{2/3}}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{\left (\frac {e x}{d}+1\right )^{2/3}}\)

\(\Big \downarrow \) 1138

\(\displaystyle \frac {(d+e x)^{2/3} \sqrt {a d e+c d^2 x} \int \frac {\sqrt [6]{\frac {e x}{d}+1}}{\sqrt {c x d^2+a e d}}dx}{\sqrt [6]{\frac {e x}{d}+1} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(d+e x)^{2/3} \sqrt {a d e+c d^2 x} \left (\frac {1}{4} \left (1-\frac {a e^2}{c d^2}\right ) \int \frac {1}{\sqrt {c x d^2+a e d} \left (\frac {e x}{d}+1\right )^{5/6}}dx+\frac {3 \sqrt [6]{\frac {e x}{d}+1} \sqrt {a d e+c d^2 x}}{2 c d^2}\right )}{\sqrt [6]{\frac {e x}{d}+1} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(d+e x)^{2/3} \sqrt {a d e+c d^2 x} \left (\frac {3 d \left (1-\frac {a e^2}{c d^2}\right ) \int \frac {1}{\sqrt {\frac {c d^3 \left (\frac {e x}{d}+1\right )}{e}-d \left (\frac {c d^2}{e}-a e\right )}}d\sqrt [6]{\frac {e x}{d}+1}}{2 e}+\frac {3 \sqrt [6]{\frac {e x}{d}+1} \sqrt {a d e+c d^2 x}}{2 c d^2}\right )}{\sqrt [6]{\frac {e x}{d}+1} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 766

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

Input:

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

Output:

(Sqrt[a*d*e + c*d^2*x]*(d + e*x)^(2/3)*((3*Sqrt[a*d*e + c*d^2*x]*(1 + (e*x 
)/d)^(1/6))/(2*c*d^2) + (3^(3/4)*d*(1 - (a*e^2)/(c*d^2))*(1 + (e*x)/d)^(1/ 
6)*((c*d^2 - a*e^2)^(1/3) - c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))*Sqrt[((c* 
d^2 - a*e^2)^(2/3) + c^(1/3)*d^(2/3)*(c*d^2 - a*e^2)^(1/3)*(1 + (e*x)/d)^( 
1/3) + c^(2/3)*d^(4/3)*(1 + (e*x)/d)^(2/3))/((c*d^2 - a*e^2)^(1/3) - (1 + 
Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))^2]*EllipticF[ArcCos[((c*d^2 
- a*e^2)^(1/3) - (1 - Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))/((c*d^ 
2 - a*e^2)^(1/3) - (1 + Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))], (2 
 + Sqrt[3])/4])/(4*e*(c*d^2 - a*e^2)^(1/3)*Sqrt[-((c^(1/3)*d^(2/3)*(1 + (e 
*x)/d)^(1/3)*((c*d^2 - a*e^2)^(1/3) - c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3)) 
)/((c*d^2 - a*e^2)^(1/3) - (1 + Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/ 
3))^2)]*Sqrt[-(d*((c*d^2)/e - a*e)) + (c*d^3*(1 + (e*x)/d))/e])))/((1 + (e 
*x)/d)^(1/6)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 1138

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d^m*((a + b*x + c*x^2)^FracPart[p]/((1 + e*(x/d))^FracPart[p] 
*(a/d + (c*x)/e)^FracPart[p]))   Int[(1 + e*(x/d))^(m + p)*(a/d + (c/e)*x)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
&& (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (IntegerQ[3*p] || Integer 
Q[4*p]))
 

rule 1139

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d^IntPart[m]*((d + e*x)^FracPart[m]/(1 + e*(x/d))^FracPart[m] 
)   Int[(1 + e*(x/d))^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e 
, m}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !(IntegerQ[m] || GtQ[d, 0])
 

Maple [F]

\[\int \frac {\left (e x +d \right )^{\frac {2}{3}}}{\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((e*x+d)^(2/3)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorit 
hm="fricas")
 

Output:

integral((e*x + d)^(2/3)/sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x), x)
 

Sympy [F]

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

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

Output:

Integral((d + e*x)**(2/3)/sqrt((d + e*x)*(a*e + c*d*x)), x)
 

Maxima [F]

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

integrate((e*x+d)^(2/3)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorit 
hm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x+d)^(2/3)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorit 
hm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d+e x)^{2/3}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {6 \left (e x +d \right )^{\frac {1}{6}} \sqrt {c d x +a e}-\left (\int \frac {\left (e x +d \right )^{\frac {1}{6}} \sqrt {c d x +a e}}{c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}d x \right ) a \,e^{2}+\left (\int \frac {\left (e x +d \right )^{\frac {1}{6}} \sqrt {c d x +a e}}{c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}d x \right ) c \,d^{2}}{4 c d} \] Input:

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

Output:

(6*(d + e*x)**(1/6)*sqrt(a*e + c*d*x) - int(((d + e*x)**(1/6)*sqrt(a*e + c 
*d*x))/(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2),x)*a*e**2 + int(((d + e* 
x)**(1/6)*sqrt(a*e + c*d*x))/(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2),x) 
*c*d**2)/(4*c*d)