\(\int x \sqrt [4]{a x+b x^2} \, dx\) [147]

Optimal result

Integrand size = 15, antiderivative size = 139 \[ \int x \sqrt [4]{a x+b x^2} \, dx=-\frac {a^2 \sqrt [4]{a x+b x^2}}{6 b^2}+\frac {a x \sqrt [4]{a x+b x^2}}{15 b}+\frac {2}{5} x^2 \sqrt [4]{a x+b x^2}-\frac {a^{5/2} \left (\frac {b x}{a+b x}\right )^{3/4} \sqrt {a+b x} \sqrt [4]{a x+b x^2} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {a+b x}}\right ),2\right )}{6 b^3 x} \] Output:

-1/6*a^2*(b*x^2+a*x)^(1/4)/b^2+1/15*a*x*(b*x^2+a*x)^(1/4)/b+2/5*x^2*(b*x^2 
+a*x)^(1/4)-1/6*a^(5/2)*(b*x/(b*x+a))^(3/4)*(b*x+a)^(1/2)*(b*x^2+a*x)^(1/4 
)*InverseJacobiAM(1/2*arcsin(a^(1/2)/(b*x+a)^(1/2)),2^(1/2))/b^3/x
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.34 \[ \int x \sqrt [4]{a x+b x^2} \, dx=\frac {4 x^2 \sqrt [4]{x (a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {9}{4},\frac {13}{4},-\frac {b x}{a}\right )}{9 \sqrt [4]{1+\frac {b x}{a}}} \] Input:

Integrate[x*(a*x + b*x^2)^(1/4),x]
 

Output:

(4*x^2*(x*(a + b*x))^(1/4)*Hypergeometric2F1[-1/4, 9/4, 13/4, -((b*x)/a)]) 
/(9*(1 + (b*x)/a)^(1/4))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1160, 1087, 1093, 1090, 230}

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.

\(\displaystyle \int x \sqrt [4]{a x+b x^2} \, dx\)

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {2 \left (a x+b x^2\right )^{5/4}}{5 b}-\frac {a \left (\frac {(a+2 b x) \sqrt [4]{a x+b x^2}}{3 b}-\frac {a^2 \int \frac {1}{\left (b x^2+a x\right )^{3/4}}dx}{12 b}\right )}{2 b}\)

\(\Big \downarrow \) 1093

\(\displaystyle \frac {2 \left (a x+b x^2\right )^{5/4}}{5 b}-\frac {a \left (\frac {(a+2 b x) \sqrt [4]{a x+b x^2}}{3 b}-\frac {a^2 \left (-\frac {b \left (a x+b x^2\right )}{a^2}\right )^{3/4} \int \frac {1}{\left (-\frac {b^2 x^2}{a^2}-\frac {b x}{a}\right )^{3/4}}dx}{12 b \left (a x+b x^2\right )^{3/4}}\right )}{2 b}\)

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 230

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

Input:

Int[x*(a*x + b*x^2)^(1/4),x]
 

Output:

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

Defintions of rubi rules used

rule 230

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2] 
))*EllipticF[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[b/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])
 

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]

\[\int x \left (b \,x^{2}+a x \right )^{\frac {1}{4}}d x\]

Input:

int(x*(b*x^2+a*x)^(1/4),x)
 

Output:

int(x*(b*x^2+a*x)^(1/4),x)
 

Fricas [F]

\[ \int x \sqrt [4]{a x+b x^2} \, dx=\int { {\left (b x^{2} + a x\right )}^{\frac {1}{4}} x \,d x } \] Input:

integrate(x*(b*x^2+a*x)^(1/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a*x)^(1/4)*x, x)
 

Sympy [F]

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

integrate(x*(b*x**2+a*x)**(1/4),x)
 

Output:

Integral(x*(x*(a + b*x))**(1/4), x)
 

Maxima [F]

\[ \int x \sqrt [4]{a x+b x^2} \, dx=\int { {\left (b x^{2} + a x\right )}^{\frac {1}{4}} x \,d x } \] Input:

integrate(x*(b*x^2+a*x)^(1/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a*x)^(1/4)*x, x)
 

Giac [F]

\[ \int x \sqrt [4]{a x+b x^2} \, dx=\int { {\left (b x^{2} + a x\right )}^{\frac {1}{4}} x \,d x } \] Input:

integrate(x*(b*x^2+a*x)^(1/4),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a*x)^(1/4)*x, x)
 

Mupad [F(-1)]

Timed out. \[ \int x \sqrt [4]{a x+b x^2} \, dx=\int x\,{\left (b\,x^2+a\,x\right )}^{1/4} \,d x \] Input:

int(x*(a*x + b*x^2)^(1/4),x)
 

Output:

int(x*(a*x + b*x^2)^(1/4), x)
 

Reduce [F]

\[ \int x \sqrt [4]{a x+b x^2} \, dx=\frac {-20 x^{\frac {1}{4}} \left (b x +a \right )^{\frac {1}{4}} a^{2}+8 x^{\frac {5}{4}} \left (b x +a \right )^{\frac {1}{4}} a b +48 x^{\frac {9}{4}} \left (b x +a \right )^{\frac {1}{4}} b^{2}+5 \left (\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{x^{\frac {3}{4}} a +x^{\frac {7}{4}} b}d x \right ) a^{3}}{120 b^{2}} \] Input:

int(x*(b*x^2+a*x)^(1/4),x)
 

Output:

( - 20*x**(1/4)*(a + b*x)**(1/4)*a**2 + 8*x**(1/4)*(a + b*x)**(1/4)*a*b*x 
+ 48*x**(1/4)*(a + b*x)**(1/4)*b**2*x**2 + 5*int((a + b*x)**(1/4)/(x**(3/4 
)*a + x**(3/4)*b*x),x)*a**3)/(120*b**2)