3.2538 \(\int \frac{1}{(a+b x+c x^2)^{3/4}} \, dx\)

Optimal. Leaf size=170 \[ \frac{\sqrt{2} \sqrt [4]{b^2-4 a c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{\sqrt [4]{c} (b+2 c x)} \]

[Out]

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

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Rubi [A]  time = 0.111645, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {623, 220} \[ \frac{\sqrt{2} \sqrt [4]{b^2-4 a c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt [4]{c} (b+2 c x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(-3/4),x]

[Out]

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

Mathematica [A]  time = 0.0456623, size = 88, normalized size = 0.52 \[ \frac{2 \sqrt{2} \sqrt{b^2-4 a c} \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ),2\right )}{c (a+x (b+c x))^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(-3/4),x]

[Out]

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

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Maple [F]  time = 2.028, size = 0, normalized size = 0. \begin{align*} \int \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+b*x+a)^(3/4),x)

[Out]

int(1/(c*x^2+b*x+a)^(3/4),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{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(-3/4), 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{3}{4}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x+a)^(3/4),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^(-3/4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x + c x^{2}\right )^{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x+a)**(3/4),x)

[Out]

Integral((a + b*x + c*x**2)**(-3/4), 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{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x+a)^(3/4),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(-3/4), x)