3.38 \(\int \frac{a+b x^3+c x^6}{\sqrt{d+e x^3}} \, dx\)

Optimal. Leaf size=278 \[ \frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (55 a e^2-22 b d e+16 c d^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt{3}\right )}{55 \sqrt [4]{3} e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}-\frac{2 x \sqrt{d+e x^3} (8 c d-11 b e)}{55 e^2}+\frac{2 c x^4 \sqrt{d+e x^3}}{11 e} \]

[Out]

(-2*(8*c*d - 11*b*e)*x*Sqrt[d + e*x^3])/(55*e^2) + (2*c*x^4*Sqrt[d + e*x^3])/(11*e) + (2*Sqrt[2 + Sqrt[3]]*(16
*c*d^2 - 22*b*d*e + 55*a*e^2)*(d^(1/3) + e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x + e^(2/3)*x^2)/((1 + Sqr
t[3])*d^(1/3) + e^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*d^(1/3) + e^(1/3)*x)/((1 + Sqrt[3])*d^(1/3) + e^
(1/3)*x)], -7 - 4*Sqrt[3]])/(55*3^(1/4)*e^(7/3)*Sqrt[(d^(1/3)*(d^(1/3) + e^(1/3)*x))/((1 + Sqrt[3])*d^(1/3) +
e^(1/3)*x)^2]*Sqrt[d + e*x^3])

________________________________________________________________________________________

Rubi [A]  time = 0.182336, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1411, 388, 218} \[ \frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (55 a e^2-22 b d e+16 c d^2\right ) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{e} x+\left (1-\sqrt{3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt{3}\right ) \sqrt [3]{d}}\right )|-7-4 \sqrt{3}\right )}{55 \sqrt [4]{3} e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}-\frac{2 x \sqrt{d+e x^3} (8 c d-11 b e)}{55 e^2}+\frac{2 c x^4 \sqrt{d+e x^3}}{11 e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)/Sqrt[d + e*x^3],x]

[Out]

(-2*(8*c*d - 11*b*e)*x*Sqrt[d + e*x^3])/(55*e^2) + (2*c*x^4*Sqrt[d + e*x^3])/(11*e) + (2*Sqrt[2 + Sqrt[3]]*(16
*c*d^2 - 22*b*d*e + 55*a*e^2)*(d^(1/3) + e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x + e^(2/3)*x^2)/((1 + Sqr
t[3])*d^(1/3) + e^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*d^(1/3) + e^(1/3)*x)/((1 + Sqrt[3])*d^(1/3) + e^
(1/3)*x)], -7 - 4*Sqrt[3]])/(55*3^(1/4)*e^(7/3)*Sqrt[(d^(1/3)*(d^(1/3) + e^(1/3)*x))/((1 + Sqrt[3])*d^(1/3) +
e^(1/3)*x)^2]*Sqrt[d + e*x^3])

Rule 1411

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Simp[(c*x^(n + 1)*
(d + e*x^n)^(q + 1))/(e*(n*(q + 2) + 1)), x] + Dist[1/(e*(n*(q + 2) + 1)), Int[(d + e*x^n)^q*(a*e*(n*(q + 2) +
1) - (c*d*(n + 1) - b*e*(n*(q + 2) + 1))*x^n), x], x] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && N
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 388

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

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]

Rubi steps

\begin{align*} \int \frac{a+b x^3+c x^6}{\sqrt{d+e x^3}} \, dx &=\frac{2 c x^4 \sqrt{d+e x^3}}{11 e}+\frac{2 \int \frac{\frac{11 a e}{2}-\left (4 c d-\frac{11 b e}{2}\right ) x^3}{\sqrt{d+e x^3}} \, dx}{11 e}\\ &=-\frac{2 (8 c d-11 b e) x \sqrt{d+e x^3}}{55 e^2}+\frac{2 c x^4 \sqrt{d+e x^3}}{11 e}-\frac{1}{55} \left (-55 a-\frac{2 d (8 c d-11 b e)}{e^2}\right ) \int \frac{1}{\sqrt{d+e x^3}} \, dx\\ &=-\frac{2 (8 c d-11 b e) x \sqrt{d+e x^3}}{55 e^2}+\frac{2 c x^4 \sqrt{d+e x^3}}{11 e}+\frac{2 \sqrt{2+\sqrt{3}} \left (16 c d^2-22 b d e+55 a e^2\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right )|-7-4 \sqrt{3}\right )}{55 \sqrt [4]{3} e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0905538, size = 98, normalized size = 0.35 \[ \frac{x \left (\sqrt{\frac{e x^3}{d}+1} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{e x^3}{d}\right ) \left (11 e (5 a e-2 b d)+16 c d^2\right )-2 \left (d+e x^3\right ) \left (-11 b e+8 c d-5 c e x^3\right )\right )}{55 e^2 \sqrt{d+e x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3 + c*x^6)/Sqrt[d + e*x^3],x]

[Out]

(x*(-2*(d + e*x^3)*(8*c*d - 11*b*e - 5*c*e*x^3) + (16*c*d^2 + 11*e*(-2*b*d + 5*a*e))*Sqrt[1 + (e*x^3)/d]*Hyper
geometric2F1[1/3, 1/2, 4/3, -((e*x^3)/d)]))/(55*e^2*Sqrt[d + e*x^3])

________________________________________________________________________________________

Maple [B]  time = 0.028, size = 907, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^6+b*x^3+a)/(e*x^3+d)^(1/2),x)

[Out]

c*(2/11/e*x^4*(e*x^3+d)^(1/2)-16/55*d/e^2*x*(e*x^3+d)^(1/2)-32/165*I*d^2/e^3*3^(1/2)*(-d*e^2)^(1/3)*(I*(x+1/2/
e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3))/(-3/2
/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^
(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*
3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1
/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)))+b*(2/5/e*x*(e*x^3+d)^(1/2)+4/15*I*d/e^2*3^(1/2)*(-d*e^2)^(1/3)*(I*(x+1
/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3))/(-
3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^
2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2
*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3
)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)))-2/3*I*a*3^(1/2)/e*(-d*e^2)^(1/3)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3
^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3))/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*
3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2
)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)
)*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^
(1/3)))^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c x^{6} + b x^{3} + a}{\sqrt{e x^{3} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(1/2),x, algorithm="maxima")

[Out]

integrate((c*x^6 + b*x^3 + a)/sqrt(e*x^3 + d), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(1/2),x, algorithm="fricas")

[Out]

integral((c*x^6 + b*x^3 + a)/sqrt(e*x^3 + d), x)

________________________________________________________________________________________

Sympy [A]  time = 2.62862, size = 119, normalized size = 0.43 \begin{align*} \frac{a x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{1}{2} \\ \frac{4}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt{d} \Gamma \left (\frac{4}{3}\right )} + \frac{b x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt{d} \Gamma \left (\frac{7}{3}\right )} + \frac{c x^{7} \Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt{d} \Gamma \left (\frac{10}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**6+b*x**3+a)/(e*x**3+d)**(1/2),x)

[Out]

a*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), e*x**3*exp_polar(I*pi)/d)/(3*sqrt(d)*gamma(4/3)) + b*x**4*gamma(4/3)*
hyper((1/2, 4/3), (7/3,), e*x**3*exp_polar(I*pi)/d)/(3*sqrt(d)*gamma(7/3)) + c*x**7*gamma(7/3)*hyper((1/2, 7/3
), (10/3,), e*x**3*exp_polar(I*pi)/d)/(3*sqrt(d)*gamma(10/3))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c x^{6} + b x^{3} + a}{\sqrt{e x^{3} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(1/2),x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)/sqrt(e*x^3 + d), x)