\(\int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx\) [724]

Optimal result

Integrand size = 20, antiderivative size = 163 \[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\frac {(2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{6 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}+\frac {\left (-b^2+4 a c\right )^{3/2} (2 c d-b e) \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right ),2\right )}{6 \sqrt {2} c^3 \left (a+b x+c x^2\right )^{3/4}} \] Output:

1/6*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c^2+2/5*e*(c*x^2+b*x+a)^(5/ 
4)/c+1/12*(4*a*c-b^2)^(3/2)*(-b*e+2*c*d)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^( 
3/4)*InverseJacobiAM(1/2*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)),2^(1/2))*2^(1 
/2)/c^3/(c*x^2+b*x+a)^(3/4)
 

Mathematica [A] (warning: unable to verify)

Time = 9.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86 \[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\frac {24 c^2 e (a+x (b+c x))^2+5 (2 c d-b e) \left (2 c (b+2 c x) (a+x (b+c x))-\sqrt {2} \left (b^2-4 a c\right )^{3/2} \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )\right )}{60 c^3 (a+x (b+c x))^{3/4}} \] Input:

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

Output:

(24*c^2*e*(a + x*(b + c*x))^2 + 5*(2*c*d - b*e)*(2*c*(b + 2*c*x)*(a + x*(b 
 + c*x)) - Sqrt[2]*(b^2 - 4*a*c)^(3/2)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a* 
c))^(3/4)*EllipticF[ArcSin[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]/2, 2]))/(60*c^3* 
(a + x*(b + c*x))^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.36 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.78, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1160, 1087, 1094, 761}

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.

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1094

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[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 1094

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((d + e*x)*(a + b*x + c*x**2)**(1/4), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\frac {-16 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} a b e +80 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} a c d +4 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} b^{2} e x +40 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} b c d x +24 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} b c e \,x^{2}+20 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) a b c e -40 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) a \,c^{2} d -5 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) b^{3} e +10 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) b^{2} c d}{60 b c} \] Input:

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

Output:

( - 16*(a + b*x + c*x**2)**(1/4)*a*b*e + 80*(a + b*x + c*x**2)**(1/4)*a*c* 
d + 4*(a + b*x + c*x**2)**(1/4)*b**2*e*x + 40*(a + b*x + c*x**2)**(1/4)*b* 
c*d*x + 24*(a + b*x + c*x**2)**(1/4)*b*c*e*x**2 + 20*int(((a + b*x + c*x** 
2)**(1/4)*x)/(a + b*x + c*x**2),x)*a*b*c*e - 40*int(((a + b*x + c*x**2)**( 
1/4)*x)/(a + b*x + c*x**2),x)*a*c**2*d - 5*int(((a + b*x + c*x**2)**(1/4)* 
x)/(a + b*x + c*x**2),x)*b**3*e + 10*int(((a + b*x + c*x**2)**(1/4)*x)/(a 
+ b*x + c*x**2),x)*b**2*c*d)/(60*b*c)