\(\int \sqrt [3]{a x+b x^2} \, dx\) [53]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \sqrt [3]{a x+b x^2} \, dx=\frac {3 \left (a x+b x^2\right )^{4/3} \operatorname {Hypergeometric2F1}\left (1,\frac {8}{3},\frac {7}{3},-\frac {b x}{a}\right )}{4 a} \] Output:

3/4*(b*x^2+a*x)^(4/3)*hypergeom([1, 8/3],[7/3],-b*x/a)/a
 

Mathematica [A] (verified)

Time = 10.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \sqrt [3]{a x+b x^2} \, dx=\frac {3 x \sqrt [3]{x (a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {b x}{a}\right )}{4 \sqrt [3]{1+\frac {b x}{a}}} \] Input:

Integrate[(a*x + b*x^2)^(1/3),x]
 

Output:

(3*x*(x*(a + b*x))^(1/3)*Hypergeometric2F1[-1/3, 4/3, 7/3, -((b*x)/a)])/(4 
*(1 + (b*x)/a)^(1/3))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(313\) vs. \(2(35)=70\).

Time = 0.53 (sec) , antiderivative size = 313, normalized size of antiderivative = 8.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1087, 1093, 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.

\(\displaystyle \int \sqrt [3]{a x+b x^2} \, dx\)

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1093

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 234

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

\(\Big \downarrow \) 760

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

Input:

Int[(a*x + b*x^2)^(1/3),x]
 

Output:

(3*(a + 2*b*x)*(a*x + b*x^2)^(1/3))/(10*b) + (3^(3/4)*Sqrt[2 - Sqrt[3]]*a^ 
2*(1 + b/a + (2*b^2*x)/a^2)*(-((b*(a*x + b*x^2))/a^2))^(2/3)*Sqrt[(1 + (-( 
b/a) - (2*b^2*x)/a^2)^2 + (1 - (a^2*(-(b/a) - (2*b^2*x)/a^2)^2)/b^2)^(1/3) 
)/(1 - Sqrt[3] + b/a + (2*b^2*x)/a^2)^2]*EllipticF[ArcSin[(1 + Sqrt[3] + b 
/a + (2*b^2*x)/a^2)/(1 - Sqrt[3] + b/a + (2*b^2*x)/a^2)], -7 + 4*Sqrt[3]]) 
/(5*2^(2/3)*b*(-(b/a) - (2*b^2*x)/a^2)*Sqrt[-((1 + b/a + (2*b^2*x)/a^2)/(1 
 - Sqrt[3] + b/a + (2*b^2*x)/a^2)^2)]*(a*x + b*x^2)^(2/3))
 

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]
 

rule 1093

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

Maple [F]

\[\int \left (b \,x^{2}+a x \right )^{\frac {1}{3}}d x\]

Input:

int((b*x^2+a*x)^(1/3),x)
 

Output:

int((b*x^2+a*x)^(1/3),x)
 

Fricas [F]

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

integrate((b*x^2+a*x)^(1/3),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a*x)^(1/3), x)
 

Sympy [F]

\[ \int \sqrt [3]{a x+b x^2} \, dx=\int \sqrt [3]{a x + b x^{2}}\, dx \] Input:

integrate((b*x**2+a*x)**(1/3),x)
 

Output:

Integral((a*x + b*x**2)**(1/3), x)
 

Maxima [F]

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

integrate((b*x^2+a*x)^(1/3),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a*x)^(1/3), x)
 

Giac [F]

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

integrate((b*x^2+a*x)^(1/3),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a*x)^(1/3), x)
 

Mupad [B] (verification not implemented)

Time = 9.67 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \sqrt [3]{a x+b x^2} \, dx=\frac {3\,x\,{\left (b\,x^2+a\,x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {4}{3};\ \frac {7}{3};\ -\frac {b\,x}{a}\right )}{4\,{\left (\frac {b\,x}{a}+1\right )}^{1/3}} \] Input:

int((a*x + b*x^2)^(1/3),x)
 

Output:

(3*x*(a*x + b*x^2)^(1/3)*hypergeom([-1/3, 4/3], 7/3, -(b*x)/a))/(4*((b*x)/ 
a + 1)^(1/3))
 

Reduce [F]

\[ \int \sqrt [3]{a x+b x^2} \, dx=\frac {3 \left (b x +a \right )^{\frac {1}{3}} a^{2}+3 \left (b x +a \right )^{\frac {1}{3}} a b x +6 \left (b x +a \right )^{\frac {1}{3}} b^{2} x^{2}+2 x^{\frac {2}{3}} \left (\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{x^{\frac {5}{3}} a +x^{\frac {8}{3}} b}d x \right ) a^{3}}{10 x^{\frac {2}{3}} b^{2}} \] Input:

int((b*x^2+a*x)^(1/3),x)
 

Output:

(3*(a + b*x)**(1/3)*a**2 + 3*(a + b*x)**(1/3)*a*b*x + 6*(a + b*x)**(1/3)*b 
**2*x**2 + 2*x**(2/3)*int((a + b*x)**(1/3)/(x**(2/3)*a*x + x**(2/3)*b*x**2 
),x)*a**3)/(10*x**(2/3)*b**2)