\(\int \frac {d+e x}{(a+b x+c x^2)^{2/3}} \, dx\) [687]

Optimal result
Mathematica [A] (verified)
Rubi [B] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 117 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{2/3}} \, dx=\frac {3 e \sqrt [3]{a+b x+c x^2}}{2 c}+\frac {(2 c d-b e) (b+2 c x) \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2^{2/3} c^2 \left (a+b x+c x^2\right )^{2/3}} \] Output:

3/2*e*(c*x^2+b*x+a)^(1/3)/c+1/2*(-b*e+2*c*d)*(2*c*x+b)*(-c*(c*x^2+b*x+a)/( 
-4*a*c+b^2))^(2/3)*hypergeom([1/2, 2/3],[3/2],(2*c*x+b)^2/(-4*a*c+b^2))*2^ 
(1/3)/c^2/(c*x^2+b*x+a)^(2/3)
 

Mathematica [A] (verified)

Time = 10.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.95 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{2/3}} \, dx=\frac {3 c e (a+x (b+c x))+\sqrt [3]{2} (2 c d-b e) (b+2 c x) \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 c^2 (a+x (b+c x))^{2/3}} \] Input:

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

Output:

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

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(483\) vs. \(2(117)=234\).

Time = 0.37 (sec) , antiderivative size = 483, normalized size of antiderivative = 4.13, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1160, 1095, 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.

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1095

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

\(\Big \downarrow \) 759

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

Input:

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

Output:

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

Defintions of rubi rules used

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 1095
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[3*(Sqrt[(b 
+ 2*c*x)^2]/(b + 2*c*x))   Subst[Int[x^(3*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4 
*c*x^3], x], x, (a + b*x + c*x^2)^(1/3)], x] /; FreeQ[{a, b, c}, x] && Inte 
gerQ[3*p]
 

rule 1160
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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