\(\int (a+b x+c x^2)^{4/3} \, dx\) [682]

Optimal result

Integrand size = 14, antiderivative size = 88 \[ \int \left (a+b x+c x^2\right )^{4/3} \, dx=\frac {(b+2 c x) \left (a+b x+c x^2\right )^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8\ 2^{2/3} c \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{4/3}} \] Output:

1/16*(2*c*x+b)*(c*x^2+b*x+a)^(4/3)*hypergeom([-4/3, 1/2],[3/2],(2*c*x+b)^2 
/(-4*a*c+b^2))*2^(1/3)/c/(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(4/3)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08 \[ \int \left (a+b x+c x^2\right )^{4/3} \, dx=-\frac {\left (b^2-4 a c\right ) (b+2 c x) \sqrt [3]{a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8\ 2^{2/3} c^2 \sqrt [3]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(4/3),x]
 

Output:

-1/8*((b^2 - 4*a*c)*(b + 2*c*x)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[ 
-4/3, 1/2, 3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(2^(2/3)*c^2*((c*(a + x*(b + 
 c*x)))/(-b^2 + 4*a*c))^(1/3))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(533\) vs. \(2(88)=176\).

Time = 0.43 (sec) , antiderivative size = 533, normalized size of antiderivative = 6.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1087, 1087, 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.

\(\displaystyle \int \left (a+b x+c x^2\right )^{4/3} \, dx\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {2 \left (b^2-4 a c\right ) \int \sqrt [3]{c x^2+b x+a}dx}{11 c}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {2 \left (b^2-4 a c\right ) \left (\frac {3 (b+2 c x) \sqrt [3]{a+b x+c x^2}}{10 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\left (c x^2+b x+a\right )^{2/3}}dx}{10 c}\right )}{11 c}\)

\(\Big \downarrow \) 1095

\(\displaystyle \frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {2 \left (b^2-4 a c\right ) \left (\frac {3 (b+2 c x) \sqrt [3]{a+b x+c x^2}}{10 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [3]{c x^2+b x+a}}{10 c (b+2 c x)}\right )}{11 c}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {2 \left (b^2-4 a c\right ) \left (\frac {3 (b+2 c x) \sqrt [3]{a+b x+c x^2}}{10 c}-\frac {3^{3/4} \sqrt {2+\sqrt {3}} \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right ) \sqrt {\frac {-2^{2/3} \sqrt [3]{c} \sqrt [3]{b^2-4 a c} \sqrt [3]{a+b x+c x^2}+\left (b^2-4 a c\right )^{2/3}+2 \sqrt [3]{2} c^{2/3} \left (a+b x+c x^2\right )^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c x^2+b x+a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c x^2+b x+a}}\right ),-7-4 \sqrt {3}\right )}{5\ 2^{2/3} c^{4/3} (b+2 c x) \sqrt {\frac {\sqrt [3]{b^2-4 a c} \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}\right )}{11 c}\)

Input:

Int[(a + b*x + c*x^2)^(4/3),x]
 

Output:

(3*(b + 2*c*x)*(a + b*x + c*x^2)^(4/3))/(22*c) - (2*(b^2 - 4*a*c)*((3*(b + 
 2*c*x)*(a + b*x + c*x^2)^(1/3))/(10*c) - (3^(3/4)*Sqrt[2 + Sqrt[3]]*(b^2 
- 4*a*c)*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]])/(5*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)])))/(11*c)
 

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 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

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]
 

Maple [F]

\[\int \left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}d x\]

Input:

int((c*x^2+b*x+a)^(4/3),x)
 

Output:

int((c*x^2+b*x+a)^(4/3),x)
 

Fricas [F]

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

integrate((c*x^2+b*x+a)^(4/3),x, algorithm="fricas")
 

Output:

integral((c*x^2 + b*x + a)^(4/3), x)
 

Sympy [F]

\[ \int \left (a+b x+c x^2\right )^{4/3} \, dx=\int \left (a + b x + c x^{2}\right )^{\frac {4}{3}}\, dx \] Input:

integrate((c*x**2+b*x+a)**(4/3),x)
 

Output:

Integral((a + b*x + c*x**2)**(4/3), x)
 

Maxima [F]

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

integrate((c*x^2+b*x+a)^(4/3),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(4/3), x)
 

Giac [F]

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

integrate((c*x^2+b*x+a)^(4/3),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x+c x^2\right )^{4/3} \, dx=\int {\left (c\,x^2+b\,x+a\right )}^{4/3} \,d x \] Input:

int((a + b*x + c*x^2)^(4/3),x)
 

Output:

int((a + b*x + c*x^2)^(4/3), x)
 

Reduce [F]

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

int((c*x^2+b*x+a)^(4/3),x)
 

Output:

int((c*x^2+b*x+a)^(4/3),x)