3.2485 \(\int (a+b x+c x^2)^{4/3} \, dx\)

Optimal. Leaf size=490 \[ \frac {\sqrt [3]{2} 3^{3/4} \sqrt {2+\sqrt {3}} \left (b^2-4 a c\right )^2 \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right ) \sqrt {\frac {-2^{2/3} \sqrt [3]{c} \sqrt [3]{b^2-4 a c} \sqrt [3]{a+b x+c x^2}+\left (b^2-4 a c\right )^{2/3}+2 \sqrt [3]{2} c^{2/3} \left (a+b x+c x^2\right )^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c x^2+b x+a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c x^2+b x+a}}\right )|-7-4 \sqrt {3}\right )}{55 c^{7/3} (b+2 c x) \sqrt {\frac {\sqrt [3]{b^2-4 a c} \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}}}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [3]{a+b x+c x^2}}{55 c^2}+\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c} \]

[Out]

-3/55*(-4*a*c+b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1/3)/c^2+3/22*(2*c*x+b)*(c*x^2+b*x+a)^(4/3)/c+1/55*2^(1/3)*3^(3/4)
*(-4*a*c+b^2)^2*((-4*a*c+b^2)^(1/3)+2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3))*EllipticF((2^(2/3)*c^(1/3)*(c*x^2+b*x
+a)^(1/3)+(-4*a*c+b^2)^(1/3)*(1-3^(1/2)))/(2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3)+(-4*a*c+b^2)^(1/3)*(1+3^(1/2)))
,I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*(((-4*a*c+b^2)^(2/3)-2^(2/3)*c^(1/3)*(-4*a*c+b^2)^(1/3)*(c*x^2+b*x+a
)^(1/3)+2*2^(1/3)*c^(2/3)*(c*x^2+b*x+a)^(2/3))/(2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3)+(-4*a*c+b^2)^(1/3)*(1+3^(1
/2)))^2)^(1/2)/c^(7/3)/(2*c*x+b)/((-4*a*c+b^2)^(1/3)*((-4*a*c+b^2)^(1/3)+2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3))/
(2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3)+(-4*a*c+b^2)^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 490, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {623, 321, 218} \[ -\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [3]{a+b x+c x^2}}{55 c^2}+\frac {\sqrt [3]{2} 3^{3/4} \sqrt {2+\sqrt {3}} \left (b^2-4 a c\right )^2 \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right ) \sqrt {\frac {-2^{2/3} \sqrt [3]{c} \sqrt [3]{b^2-4 a c} \sqrt [3]{a+b x+c x^2}+\left (b^2-4 a c\right )^{2/3}+2 \sqrt [3]{2} c^{2/3} \left (a+b x+c x^2\right )^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c x^2+b x+a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c x^2+b x+a}}\right )|-7-4 \sqrt {3}\right )}{55 c^{7/3} (b+2 c x) \sqrt {\frac {\sqrt [3]{b^2-4 a c} \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}}}+\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(4/3),x]

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x)*(a + b*x + c*x^2)^(1/3))/(55*c^2) + (3*(b + 2*c*x)*(a + b*x + c*x^2)^(4/3))/(22*
c) + (2^(1/3)*3^(3/4)*Sqrt[2 + Sqrt[3]]*(b^2 - 4*a*c)^2*((b^2 - 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*(a + b*x + c*x^
2)^(1/3))*Sqrt[((b^2 - 4*a*c)^(2/3) - 2^(2/3)*c^(1/3)*(b^2 - 4*a*c)^(1/3)*(a + b*x + c*x^2)^(1/3) + 2*2^(1/3)*
c^(2/3)*(a + b*x + c*x^2)^(2/3))/((1 + Sqrt[3])*(b^2 - 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*(a + b*x + c*x^2)^(1/3))
^2]*EllipticF[ArcSin[((1 - Sqrt[3])*(b^2 - 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*(a + b*x + c*x^2)^(1/3))/((1 + Sqrt[
3])*(b^2 - 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*(a + b*x + c*x^2)^(1/3))], -7 - 4*Sqrt[3]])/(55*c^(7/3)*(b + 2*c*x)*
Sqrt[((b^2 - 4*a*c)^(1/3)*((b^2 - 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*(a + b*x + c*x^2)^(1/3)))/((1 + Sqrt[3])*(b^2
- 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*(a + b*x + c*x^2)^(1/3))^2])

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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]

Rubi steps

\begin {align*} \int \left (a+b x+c x^2\right )^{4/3} \, dx &=\frac {\left (3 \sqrt {(b+2 c x)^2}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt {b^2-4 a c+4 c x^3}} \, dx,x,\sqrt [3]{a+b x+c x^2}\right )}{b+2 c x}\\ &=\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {\left (6 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {b^2-4 a c+4 c x^3}} \, dx,x,\sqrt [3]{a+b x+c x^2}\right )}{11 c (b+2 c x)}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [3]{a+b x+c x^2}}{55 c^2}+\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}+\frac {\left (3 \left (b^2-4 a c\right )^2 \sqrt {(b+2 c x)^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c+4 c x^3}} \, dx,x,\sqrt [3]{a+b x+c x^2}\right )}{55 c^2 (b+2 c x)}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [3]{a+b x+c x^2}}{55 c^2}+\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}+\frac {\sqrt [3]{2} 3^{3/4} \sqrt {2+\sqrt {3}} \left (b^2-4 a c\right )^2 \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right ) \sqrt {\frac {\left (b^2-4 a c\right )^{2/3}-2^{2/3} \sqrt [3]{c} \sqrt [3]{b^2-4 a c} \sqrt [3]{a+b x+c x^2}+2 \sqrt [3]{2} c^{2/3} \left (a+b x+c x^2\right )^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}}\right )|-7-4 \sqrt {3}\right )}{55 c^{7/3} (b+2 c x) \sqrt {\frac {\sqrt [3]{b^2-4 a c} \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 95, normalized size = 0.19 \[ -\frac {\left (b^2-4 a c\right ) (b+2 c x) \sqrt [3]{a+x (b+c x)} \, _2F_1\left (-\frac {4}{3},\frac {1}{2};\frac {3}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8\ 2^{2/3} c^2 \sqrt [3]{\frac {c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(4/3),x]

[Out]

-1/8*((b^2 - 4*a*c)*(b + 2*c*x)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, 1/2, 3/2, (b + 2*c*x)^2/(b^2 -
4*a*c)])/(2^(2/3)*c^2*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x + a)^(4/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(4/3), x)

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maple [F]  time = 2.51, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(4/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,x^2+b\,x+a\right )}^{4/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(4/3),x)

[Out]

int((a + b*x + c*x^2)^(4/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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