\(\int (3+4 x+2 x^2)^{4/3} \, dx\) [139]

Optimal result

Integrand size = 14, antiderivative size = 21 \[ \int \left (3+4 x+2 x^2\right )^{4/3} \, dx=(1+x) \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},\frac {3}{2},-2 (1+x)^2\right ) \] Output:

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

Mathematica [A] (verified)

Time = 7.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \left (3+4 x+2 x^2\right )^{4/3} \, dx=(1+x) \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},\frac {3}{2},-2 (1+x)^2\right ) \] Input:

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

Output:

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

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(184\) vs. \(2(21)=42\).

Time = 0.48 (sec) , antiderivative size = 184, normalized size of antiderivative = 8.76, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1087, 1087, 1090, 234, 760}

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.

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {8}{11} \int \sqrt [3]{2 x^2+4 x+3}dx+\frac {3}{11} (x+1) \left (2 x^2+4 x+3\right )^{4/3}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {8}{11} \left (\frac {2}{5} \int \frac {1}{\left (2 x^2+4 x+3\right )^{2/3}}dx+\frac {3}{5} \sqrt [3]{2 x^2+4 x+3} (x+1)\right )+\frac {3}{11} (x+1) \left (2 x^2+4 x+3\right )^{4/3}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {8}{11} \left (\frac {1}{10} \int \frac {1}{\left (\frac {1}{8} (4 x+4)^2+1\right )^{2/3}}d(4 x+4)+\frac {3}{5} \sqrt [3]{2 x^2+4 x+3} (x+1)\right )+\frac {3}{11} (x+1) \left (2 x^2+4 x+3\right )^{4/3}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {8}{11} \left (\frac {3 \sqrt {(4 x+4)^2} \int \frac {2 \sqrt {2}}{\sqrt {(4 x+4)^2}}d\sqrt [3]{\frac {1}{8} (4 x+4)^2+1}}{5 \sqrt {2} (4 x+4)}+\frac {3}{5} \sqrt [3]{2 x^2+4 x+3} (x+1)\right )+\frac {3}{11} (x+1) \left (2 x^2+4 x+3\right )^{4/3}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {8}{11} \left (\frac {3}{5} (x+1) \sqrt [3]{2 x^2+4 x+3}-\frac {4\ 3^{3/4} \sqrt {2-\sqrt {3}} (-4 x-3) \sqrt {\frac {4 x+\left (\frac {1}{8} (4 x+4)^2+1\right )^{2/3}+5}{\left (-4 x-\sqrt {3}-3\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-4 x+\sqrt {3}-3}{-4 x-\sqrt {3}-3}\right ),-7+4 \sqrt {3}\right )}{5 \sqrt {-\frac {-4 x-3}{\left (-4 x-\sqrt {3}-3\right )^2}} (4 x+4)}\right )+\frac {3}{11} (x+1) \left (2 x^2+4 x+3\right )^{4/3}\)

Input:

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

Output:

(3*(1 + x)*(3 + 4*x + 2*x^2)^(4/3))/11 + (8*((3*(1 + x)*(3 + 4*x + 2*x^2)^ 
(1/3))/5 - (4*3^(3/4)*Sqrt[2 - Sqrt[3]]*(-3 - 4*x)*Sqrt[(5 + 4*x + (1 + (4 
 + 4*x)^2/8)^(2/3))/(-3 - Sqrt[3] - 4*x)^2]*EllipticF[ArcSin[(-3 + Sqrt[3] 
 - 4*x)/(-3 - Sqrt[3] - 4*x)], -7 + 4*Sqrt[3]])/(5*Sqrt[-((-3 - 4*x)/(-3 - 
 Sqrt[3] - 4*x)^2)]*(4 + 4*x))))/11
 

Defintions of rubi rules used

rule 234

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b 
}, x]
 

rule 760

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 
] && NegQ[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 1090

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((2*x**2+4*x+3)**(4/3),x)
 

Output:

Integral((2*x**2 + 4*x + 3)**(4/3), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (3+4 x+2 x^2\right )^{4/3} \, dx=\frac {6 \left (2 x^{2}+4 x +3\right )^{\frac {1}{3}} x^{3}}{11}+\frac {18 \left (2 x^{2}+4 x +3\right )^{\frac {1}{3}} x^{2}}{11}+\frac {129 \left (2 x^{2}+4 x +3\right )^{\frac {1}{3}} x}{55}+\frac {81 \left (2 x^{2}+4 x +3\right )^{\frac {1}{3}}}{55}-\frac {16 \left (\int \frac {x}{\left (2 x^{2}+4 x +3\right )^{\frac {2}{3}}}d x \right )}{55} \] Input:

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

Output:

(30*(2*x**2 + 4*x + 3)**(1/3)*x**3 + 90*(2*x**2 + 4*x + 3)**(1/3)*x**2 + 1 
29*(2*x**2 + 4*x + 3)**(1/3)*x + 81*(2*x**2 + 4*x + 3)**(1/3) - 16*int(((2 
*x**2 + 4*x + 3)**(1/3)*x)/(2*x**2 + 4*x + 3),x))/55