### 3.1417 $$\int \frac{(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{8/3}} \, dx$$

Optimal. Leaf size=581 $-\frac{\left (-4 a c+b^2-(b+2 c x)^2\right ) \sqrt [3]{d (b+2 c x)} \left (2 \sqrt [3]{c} d^{2/3}-\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt{\frac{\frac{2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\frac{(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+2 \sqrt [3]{2} c^{2/3} d^{4/3}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1-\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 \sqrt [4]{3} c^{10/3} d^{11/3} \left (a+b x+c x^2\right )^{2/3} \sqrt{-\frac{(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}}+\frac{\sqrt [3]{a+b x+c x^2} \sqrt [3]{d (b+2 c x)}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}$

[Out]

((d*(b + 2*c*x))^(1/3)*(a + b*x + c*x^2)^(1/3))/(5*c^2*d^3) - (3*(a + b*x + c*x^2)^(4/3))/(10*c*d*(d*(b + 2*c*
x))^(5/3)) - ((d*(b + 2*c*x))^(1/3)*(b^2 - 4*a*c - (b + 2*c*x)^2)*(2*c^(1/3)*d^(2/3) - (2^(1/3)*(d*(b + 2*c*x)
)^(2/3))/(a + b*x + c*x^2)^(1/3))*Sqrt[(2*2^(1/3)*c^(2/3)*d^(4/3) + (d*(b + 2*c*x))^(4/3)/(a + b*x + c*x^2)^(2
/3) + (2^(2/3)*c^(1/3)*d^(2/3)*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3) - ((1
+ Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))^2]*EllipticF[ArcCos[(2^(2/3)*c^(1/3)*d^(2/3) - ((1
- Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3) - ((1 + Sqrt[3])*(d*(b + 2
*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))], (2 + Sqrt[3])/4])/(40*3^(1/4)*c^(10/3)*d^(11/3)*(a + b*x + c*x^2)^(2/
3)*Sqrt[-(((d*(b + 2*c*x))^(2/3)*(2^(2/3)*c^(1/3)*d^(2/3) - (d*(b + 2*c*x))^(2/3)/(a + b*x + c*x^2)^(1/3)))/((
a + b*x + c*x^2)^(1/3)*(2^(2/3)*c^(1/3)*d^(2/3) - ((1 + Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3
))^2))])

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Rubi [A]  time = 1.94999, antiderivative size = 581, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {694, 277, 279, 329, 241, 225} $-\frac{\left (-4 a c+b^2-(b+2 c x)^2\right ) \sqrt [3]{d (b+2 c x)} \left (2 \sqrt [3]{c} d^{2/3}-\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt{\frac{\frac{2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\frac{(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+2 \sqrt [3]{2} c^{2/3} d^{4/3}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1-\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 \sqrt [4]{3} c^{10/3} d^{11/3} \left (a+b x+c x^2\right )^{2/3} \sqrt{-\frac{(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}}+\frac{\sqrt [3]{a+b x+c x^2} \sqrt [3]{d (b+2 c x)}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(8/3),x]

[Out]

((d*(b + 2*c*x))^(1/3)*(a + b*x + c*x^2)^(1/3))/(5*c^2*d^3) - (3*(a + b*x + c*x^2)^(4/3))/(10*c*d*(d*(b + 2*c*
x))^(5/3)) - ((d*(b + 2*c*x))^(1/3)*(b^2 - 4*a*c - (b + 2*c*x)^2)*(2*c^(1/3)*d^(2/3) - (2^(1/3)*(d*(b + 2*c*x)
)^(2/3))/(a + b*x + c*x^2)^(1/3))*Sqrt[(2*2^(1/3)*c^(2/3)*d^(4/3) + (d*(b + 2*c*x))^(4/3)/(a + b*x + c*x^2)^(2
/3) + (2^(2/3)*c^(1/3)*d^(2/3)*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3) - ((1
+ Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))^2]*EllipticF[ArcCos[(2^(2/3)*c^(1/3)*d^(2/3) - ((1
- Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3) - ((1 + Sqrt[3])*(d*(b + 2
*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))], (2 + Sqrt[3])/4])/(40*3^(1/4)*c^(10/3)*d^(11/3)*(a + b*x + c*x^2)^(2/
3)*Sqrt[-(((d*(b + 2*c*x))^(2/3)*(2^(2/3)*c^(1/3)*d^(2/3) - (d*(b + 2*c*x))^(2/3)/(a + b*x + c*x^2)^(1/3)))/((
a + b*x + c*x^2)^(1/3)*(2^(2/3)*c^(1/3)*d^(2/3) - ((1 + Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3
))^2))])

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rule 277

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

Rule 279

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 241

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a/(a + b*x^n))^(p + 1/n)*(a + b*x^n)^(p + 1/n), Subst[In
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p,
0] && NeQ[p, -2^(-1)] && LtQ[Denominator[p + 1/n], Denominator[p]]

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^{4/3}}{x^{8/3}} \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt [3]{a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}}}{x^{2/3}} \, dx,x,b d+2 c d x\right )}{5 c^2 d^3}\\ &=\frac{\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac{\left (b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^{2/3}} \, dx,x,b d+2 c d x\right )}{30 c^3 d^3}\\ &=\frac{\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac{\left (b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a-\frac{b^2}{4 c}+\frac{x^6}{4 c d^2}\right )^{2/3}} \, dx,x,\sqrt [3]{d (b+2 c x)}\right )}{10 c^3 d^3}\\ &=\frac{\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac{\left (b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^6}{4 c d^2}}} \, dx,x,\frac{\sqrt [3]{d (b+2 c x)}}{\sqrt [6]{a+x (b+c x)}}\right )}{10 c^3 d^3 \sqrt{\frac{a-\frac{b^2}{4 c}}{a+x (b+c x)}} \sqrt{a+x (b+c x)}}\\ &=\frac{\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac{\left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right ) \sqrt{\frac{2 \sqrt [3]{2} c^{2/3} d^{4/3}+\frac{(d (b+2 c x))^{4/3}}{(a+x (b+c x))^{2/3}}+\frac{2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )^2}} F\left (\cos ^{-1}\left (\frac{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1-\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5\ 2^{2/3} \sqrt [4]{3} c^{10/3} d^{11/3} \sqrt{\frac{4 a-\frac{b^2}{c}}{a+x (b+c x)}} (a+x (b+c x))^{2/3} \sqrt{4-\frac{(b+2 c x)^2}{c (a+x (b+c x))}} \sqrt{-\frac{(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{\sqrt [3]{a+x (b+c x)} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0645814, size = 104, normalized size = 0.18 $\frac{3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \, _2F_1\left (-\frac{4}{3},-\frac{5}{6};\frac{1}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{40\ 2^{2/3} c^2 d \sqrt [3]{\frac{c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^{5/3}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(8/3),x]

[Out]

(3*(b^2 - 4*a*c)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, -5/6, 1/6, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(40*
2^(2/3)*c^2*d*(d*(b + 2*c*x))^(5/3)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))

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Maple [F]  time = 0.776, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c{x}^{2}+bx+a \right ) ^{{\frac{4}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{8}{3}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(8/3),x)

[Out]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(8/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{8}{3}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(8/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(8/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (2 \, c d x + b d\right )}^{\frac{1}{3}}{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{8 \, c^{3} d^{3} x^{3} + 12 \, b c^{2} d^{3} x^{2} + 6 \, b^{2} c d^{3} x + b^{3} d^{3}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(8/3),x, algorithm="fricas")

[Out]

integral((2*c*d*x + b*d)^(1/3)*(c*x^2 + b*x + a)^(4/3)/(8*c^3*d^3*x^3 + 12*b*c^2*d^3*x^2 + 6*b^2*c*d^3*x + b^3
*d^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x + c x^{2}\right )^{\frac{4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac{8}{3}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(8/3),x)

[Out]

Integral((a + b*x + c*x**2)**(4/3)/(d*(b + 2*c*x))**(8/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{8}{3}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(8/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(8/3), x)