3.26.38 \(\int \frac {1}{(a+b x+c x^2)^{3/4}} \, dx\) [2538]

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

[Out]

(-4*a*c+b^2)^(1/4)*(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*arcta
n(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))*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^(1/4)/(2*c*x+b)

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Rubi [A]
time = 0.08, antiderivative size = 170, normalized size of antiderivative = 1.00, 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 = {637, 226} \begin {gather*} \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 \text {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 )}{\sqrt [4]{c} (b+2 c x)} \end {gather*}

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 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 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*} \int \frac {1}{\left (a+b x+c x^2\right )^{3/4}} \, dx &=\frac {\left (4 \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 )}{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*}

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Mathematica [A]
time = 10.04, size = 88, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {2} \sqrt {b^2-4 a c} \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )\right |2\right )}{c (a+x (b+c x))^{3/4}} \end {gather*}

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 = 0.65, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x + c x^{2}\right )^{\frac {3}{4}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (c\,x^2+b\,x+a\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x + c*x^2)^(3/4),x)

[Out]

int(1/(a + b*x + c*x^2)^(3/4), x)

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