\(\int \frac {(d+e x)^2}{(a+b x+c x^2)^{2/3}} \, dx\) [686]

Optimal result

Integrand size = 22, antiderivative size = 151 \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{2/3}} \, dx=\frac {3 e (5 c d-2 b e+c e x) \sqrt [3]{a+b x+c x^2}}{5 c^2}+\frac {\sqrt [3]{2} \left (\left (2 b^2-3 a c\right ) e^2+5 c d (c d-b e)\right ) (b+2 c x) \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 c^3 \left (a+b x+c x^2\right )^{2/3}} \] Output:

3/5*e*(c*e*x-2*b*e+5*c*d)*(c*x^2+b*x+a)^(1/3)/c^2+1/5*2^(1/3)*((-3*a*c+2*b 
^2)*e^2+5*c*d*(-b*e+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))/c^3/(c*x^2+b*x+a)^(2/ 
3)
 

Mathematica [A] (verified)

Time = 10.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{2/3}} \, dx=\frac {3 \left (-\frac {4 e (-2 c d+b e) (a+x (b+c x))}{c}+2 e (d+e x) (a+x (b+c x))+\frac {2 \sqrt [3]{2} \left (5 c^2 d^2+2 b^2 e^2-c e (5 b d+3 a e)\right ) (b+2 c x) \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 c^2}\right )}{10 c (a+x (b+c x))^{2/3}} \] Input:

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

Output:

(3*((-4*e*(-2*c*d + b*e)*(a + x*(b + c*x)))/c + 2*e*(d + e*x)*(a + x*(b + 
c*x)) + (2*2^(1/3)*(5*c^2*d^2 + 2*b^2*e^2 - c*e*(5*b*d + 3*a*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)])/(3*c^2)))/(10*c*(a + x*(b + c*x))^(2/ 
3))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(546\) vs. \(2(151)=302\).

Time = 0.51 (sec) , antiderivative size = 546, normalized size of antiderivative = 3.62, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1166, 27, 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.

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

\(\Big \downarrow \) 1166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1095

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

\(\Big \downarrow \) 759

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

Input:

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

Output:

(3*e*(d + e*x)*(a + b*x + c*x^2)^(1/3))/(5*c) + ((6*e*(2*c*d - b*e)*(a + b 
*x + c*x^2)^(1/3))/c + (2^(1/3)*3^(3/4)*Sqrt[2 + Sqrt[3]]*(5*c^2*d^2 + 2*b 
^2*e^2 - c*e*(5*b*d + 3*a*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]] 
)/(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)]))/(5*c)
 

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 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]
 

rule 1166

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[1/(c*(m + 2*p + 1))   Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m 
+ 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* 
(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration 
alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat 
icQ[a, b, c, d, e, m, p, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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