### 3.2083 $$\int \frac{(d+e x)^{2/3}}{\sqrt{a d e+(c d^2+a e^2) x+c d e x^2}} \, dx$$

Optimal. Leaf size=566 $\frac{3^{3/4} (d+e x)^{2/3} \left (c d^2-a e^2\right )^{2/3} \sqrt{a d e+c d^2 x} \left (\sqrt [3]{c d^2-a e^2}-\sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}\right ) \sqrt{\frac{\sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1} \sqrt [3]{c d^2-a e^2}+\left (c d^2-a e^2\right )^{2/3}+c^{2/3} d^{4/3} \left (\frac{e x}{d}+1\right )^{2/3}}{\left (\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt{3}\right ) \sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{c d^2-a e^2}-\left (1-\sqrt{3}\right ) \sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}}{\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt{3}\right ) \sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{4 c d e \sqrt{d (a e+c d x)} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \sqrt{-\frac{\sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1} \left (\sqrt [3]{c d^2-a e^2}-\sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}\right )}{\left (\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt{3}\right ) \sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}\right )^2}}}+\frac{3 (d+e x)^{2/3} (a e+c d x)}{2 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

[Out]

(3*(a*e + c*d*x)*(d + e*x)^(2/3))/(2*c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (3^(3/4)*(c*d^2 - a*e^
2)^(2/3)*Sqrt[a*d*e + c*d^2*x]*(d + e*x)^(2/3)*((c*d^2 - a*e^2)^(1/3) - c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))*S
qrt[((c*d^2 - a*e^2)^(2/3) + c^(1/3)*d^(2/3)*(c*d^2 - a*e^2)^(1/3)*(1 + (e*x)/d)^(1/3) + c^(2/3)*d^(4/3)*(1 +
(e*x)/d)^(2/3))/((c*d^2 - a*e^2)^(1/3) - (1 + Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))^2]*EllipticF[ArcCo
s[((c*d^2 - a*e^2)^(1/3) - (1 - Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))/((c*d^2 - a*e^2)^(1/3) - (1 + Sq
rt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))], (2 + Sqrt[3])/4])/(4*c*d*e*Sqrt[d*(a*e + c*d*x)]*Sqrt[a*d*e + (c
*d^2 + a*e^2)*x + c*d*e*x^2]*Sqrt[-((c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3)*((c*d^2 - a*e^2)^(1/3) - c^(1/3)*d^(2
/3)*(1 + (e*x)/d)^(1/3)))/((c*d^2 - a*e^2)^(1/3) - (1 + Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))^2)])

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Rubi [A]  time = 0.765203, antiderivative size = 566, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.128, Rules used = {679, 677, 50, 63, 225} $\frac{3^{3/4} (d+e x)^{2/3} \left (c d^2-a e^2\right )^{2/3} \sqrt{a d e+c d^2 x} \left (\sqrt [3]{c d^2-a e^2}-\sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}\right ) \sqrt{\frac{\sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1} \sqrt [3]{c d^2-a e^2}+\left (c d^2-a e^2\right )^{2/3}+c^{2/3} d^{4/3} \left (\frac{e x}{d}+1\right )^{2/3}}{\left (\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt{3}\right ) \sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{c d^2-a e^2}-\left (1-\sqrt{3}\right ) \sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}}{\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt{3}\right ) \sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{4 c d e \sqrt{d (a e+c d x)} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \sqrt{-\frac{\sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1} \left (\sqrt [3]{c d^2-a e^2}-\sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}\right )}{\left (\sqrt [3]{c d^2-a e^2}-\left (1+\sqrt{3}\right ) \sqrt [3]{c} d^{2/3} \sqrt [3]{\frac{e x}{d}+1}\right )^2}}}+\frac{3 (d+e x)^{2/3} (a e+c d x)}{2 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(2/3)/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

(3*(a*e + c*d*x)*(d + e*x)^(2/3))/(2*c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (3^(3/4)*(c*d^2 - a*e^
2)^(2/3)*Sqrt[a*d*e + c*d^2*x]*(d + e*x)^(2/3)*((c*d^2 - a*e^2)^(1/3) - c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))*S
qrt[((c*d^2 - a*e^2)^(2/3) + c^(1/3)*d^(2/3)*(c*d^2 - a*e^2)^(1/3)*(1 + (e*x)/d)^(1/3) + c^(2/3)*d^(4/3)*(1 +
(e*x)/d)^(2/3))/((c*d^2 - a*e^2)^(1/3) - (1 + Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))^2]*EllipticF[ArcCo
s[((c*d^2 - a*e^2)^(1/3) - (1 - Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))/((c*d^2 - a*e^2)^(1/3) - (1 + Sq
rt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))], (2 + Sqrt[3])/4])/(4*c*d*e*Sqrt[d*(a*e + c*d*x)]*Sqrt[a*d*e + (c
*d^2 + a*e^2)*x + c*d*e*x^2]*Sqrt[-((c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3)*((c*d^2 - a*e^2)^(1/3) - c^(1/3)*d^(2
/3)*(1 + (e*x)/d)^(1/3)))/((c*d^2 - a*e^2)^(1/3) - (1 + Sqrt[3])*c^(1/3)*d^(2/3)*(1 + (e*x)/d)^(1/3))^2)])

Rule 679

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

Rule 677

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^m*(a + b*x + c*x^2
)^FracPart[p])/((1 + (e*x)/d)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)
/e)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !Int
egerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (IntegerQ[3*p] || IntegerQ[4*p]))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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

Mathematica [C]  time = 0.0612044, size = 95, normalized size = 0.17 $\frac{2 \sqrt{(d+e x) (a e+c d x)} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{c d \sqrt [3]{d+e x} \sqrt [6]{\frac{c d (d+e x)}{c d^2-a e^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(2/3)/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*Hypergeometric2F1[-1/6, 1/2, 3/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(c*d
*(d + e*x)^(1/3)*((c*d*(d + e*x))/(c*d^2 - a*e^2))^(1/6))

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Maple [F]  time = 0.326, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{{\frac{2}{3}}}{\frac{1}{\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(2/3)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x)

[Out]

int((e*x+d)^(2/3)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^(2/3)/sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^(2/3)/sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(2/3)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^(2/3)/sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x), x)