### 3.2531 $$\int \frac{1}{\sqrt [4]{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=418 $\frac{\left (b^2-4 a c\right )^{3/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{\sqrt{2} c^{3/4} (b+2 c x)}-\frac{\sqrt{2} \left (b^2-4 a c\right )^{3/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{c^{3/4} (b+2 c x)}+\frac{2 (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\sqrt{c} \sqrt{b^2-4 a c} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )}$

[Out]

(2*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(Sqrt[c]*Sqrt[b^2 - 4*a*c]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt
[b^2 - 4*a*c])) - (Sqrt[2]*(b^2 - 4*a*c)^(3/4)*Sqrt[(b + 2*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)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*EllipticE[2*ArcTan[
(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(c^(3/4)*(b + 2*c*x)) + ((b^2 - 4*a*c)^(
3/4)*Sqrt[(b + 2*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)]*(1 + (2*S
qrt[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])/(Sqrt[2]*c^(3/4)*(b + 2*c*x))

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Rubi [A]  time = 0.305418, antiderivative size = 418, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {623, 305, 220, 1196} $\frac{\left (b^2-4 a c\right )^{3/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt{2} c^{3/4} (b+2 c x)}-\frac{\sqrt{2} \left (b^2-4 a c\right )^{3/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{c^{3/4} (b+2 c x)}+\frac{2 (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\sqrt{c} \sqrt{b^2-4 a c} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(-1/4),x]

[Out]

(2*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(Sqrt[c]*Sqrt[b^2 - 4*a*c]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt
[b^2 - 4*a*c])) - (Sqrt[2]*(b^2 - 4*a*c)^(3/4)*Sqrt[(b + 2*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)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*EllipticE[2*ArcTan[
(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(c^(3/4)*(b + 2*c*x)) + ((b^2 - 4*a*c)^(
3/4)*Sqrt[(b + 2*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)]*(1 + (2*S
qrt[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])/(Sqrt[2]*c^(3/4)*(b + 2*c*x))

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 305

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

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [4]{a+b x+c x^2}} \, dx &=\frac{\left (4 \sqrt{(b+2 c x)^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{b+2 c x}\\ &=\frac{\left (2 \sqrt{b^2-4 a c} \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 )}{\sqrt{c} (b+2 c x)}-\frac{\left (2 \sqrt{b^2-4 a c} \sqrt{(b+2 c x)^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{2 \sqrt{c} x^2}{\sqrt{b^2-4 a c}}}{\sqrt{b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{\sqrt{c} (b+2 c x)}\\ &=\frac{2 (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\sqrt{c} \sqrt{b^2-4 a c} \left (1+\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}-\frac{\sqrt{2} \left (b^2-4 a c\right )^{3/4} \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 ) E\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 )}{c^{3/4} (b+2 c x)}+\frac{\left (b^2-4 a c\right )^{3/4} \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 )}{\sqrt{2} c^{3/4} (b+2 c x)}\\ \end{align*}

Mathematica [C]  time = 0.0333465, size = 84, normalized size = 0.2 $\frac{(b+2 c x) \sqrt [4]{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{\sqrt{2} c \sqrt [4]{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(-1/4),x]

[Out]

((b + 2*c*x)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (b + 2*c*x)^2/(b^2
- 4*a*c)])/(Sqrt[2]*c*(a + x*(b + c*x))^(1/4))

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Maple [F]  time = 2.052, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt [4]{c{x}^{2}+bx+a}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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 \frac{1}{{\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(1/(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 (\frac{1}{{\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(1/(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 \frac{1}{\sqrt [4]{a + b x + c x^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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