\(\int \sqrt [4]{a+b x+c x^2} \, dx\) [2513]

Optimal result

Integrand size = 14, antiderivative size = 201 \[ \int \sqrt [4]{a+b x+c x^2} \, dx=\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} (b+2 c x)} \]

[Out]

1/3*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c-1/12*(-4*a*c+b^2)^(5/4)*(cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/
(-4*a*c+b^2)^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4)))*EllipticF(
sin(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4))),1/2*2^(1/2))*(1+2*c^(1/2)*(c*x^2+b*x+a)^
(1/2)/(-4*a*c+b^2)^(1/2))*((2*c*x+b)^2/(-4*a*c+b^2)/(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))^2)^(1
/2)/c^(5/4)/(2*c*x+b)*2^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {626, 637, 226} \[ \int \sqrt [4]{a+b x+c x^2} \, dx=\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} (b+2 c x)} \]

[In]

Int[(a + b*x + c*x^2)^(1/4),x]

[Out]

((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/((b^2 - 4*a*c)*(1 + (2*S
qrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])
*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))

Rule 226

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 626

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] - Dist[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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 637

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[d*(Sqrt[(b + 2*c*x)
^2]/(b + 2*c*x)), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rubi steps \begin{align*} \text {integral}& = \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 (a+b x+c x^2\right )^{3/4}} \, dx}{12 c} \\ & = \frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (\left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{3 c (b+2 c x)} \\ & = \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 {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} (b+2 c x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.50 \[ \int \sqrt [4]{a+b x+c x^2} \, dx=\frac {\sqrt [4]{a+x (b+c x)} \left (2 (b+2 c x)+\frac {\sqrt {2} \sqrt {b^2-4 a c} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )}{\sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}}\right )}{6 c} \]

[In]

Integrate[(a + b*x + c*x^2)^(1/4),x]

[Out]

((a + x*(b + c*x))^(1/4)*(2*(b + 2*c*x) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*EllipticF[ArcSin[(b + 2*c*x)/Sqrt[b^2 - 4
*a*c]]/2, 2])/((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/4)))/(6*c)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((c*x^2 + b*x + a)^(1/4), x)

Sympy [F]

\[ \int \sqrt [4]{a+b x+c x^2} \, dx=\int \sqrt [4]{a + b x + c x^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(1/4), x)

Giac [F]

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

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(1/4), x)

Mupad [F(-1)]

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

[In]

int((a + b*x + c*x^2)^(1/4),x)

[Out]

int((a + b*x + c*x^2)^(1/4), x)