### 3.2513 $$\int \sqrt [4]{a+b x+c x^2} \, dx$$

Optimal. Leaf size=201 $\frac{(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac{\left (b^2-4 a c\right )^{5/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{6 \sqrt{2} c^{5/4} (b+2 c x)}$

[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))

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Rubi [A]  time = 0.131318, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {612, 623, 220} $\frac{(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac{\left (b^2-4 a c\right )^{5/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{6 \sqrt{2} c^{5/4} (b+2 c x)}$

Antiderivative was successfully veriﬁed.

[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 612

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 623

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]

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \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 ) \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 ) \operatorname{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]  time = 0.242979, size = 100, normalized size = 0.5 $\frac{\sqrt [4]{a+x (b+c x)} \left (\frac{\sqrt{2} \sqrt{b^2-4 a c} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ),2\right )}{\sqrt [4]{\frac{c (a+x (b+c x))}{4 a c-b^2}}}+2 (b+2 c x)\right )}{6 c}$

Antiderivative was successfully veriﬁed.

[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)

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Maple [F]  time = 2.113, size = 0, normalized size = 0. \begin{align*} \int \sqrt [4]{c{x}^{2}+bx+a}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(1/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x + a)^(1/4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [4]{a + b x + c x^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(1/4), x)