3.39 \(\int \frac{a+b x^3+c x^6}{(d+e x^3)^{3/2}} \, dx\)

Optimal. Leaf size=289 \[ -\frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (16 c d^2-5 e (a e+2 b d)\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt{3}\right )}{15 \sqrt [4]{3} d e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}+\frac{2 x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \sqrt{d+e x^3}}+\frac{2 c x \sqrt{d+e x^3}}{5 e^2} \]

[Out]

(2*(c*d^2 - b*d*e + a*e^2)*x)/(3*d*e^2*Sqrt[d + e*x^3]) + (2*c*x*Sqrt[d + e*x^3])/(5*e^2) - (2*Sqrt[2 + Sqrt[3
]]*(16*c*d^2 - 5*e*(2*b*d + a*e))*(d^(1/3) + e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x + e^(2/3)*x^2)/((1 +
Sqrt[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]])/(15*3^(1/4)*d*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])

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Rubi [A]  time = 0.189006, antiderivative size = 289, 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 = {1409, 388, 218} \[ \frac{2 x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \sqrt{d+e x^3}}-\frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (16 c d^2-5 e (a e+2 b d)\right ) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{e} x+\left (1-\sqrt{3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt{3}\right ) \sqrt [3]{d}}\right )|-7-4 \sqrt{3}\right )}{15 \sqrt [4]{3} d e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}+\frac{2 c x \sqrt{d+e x^3}}{5 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)/(d + e*x^3)^(3/2),x]

[Out]

(2*(c*d^2 - b*d*e + a*e^2)*x)/(3*d*e^2*Sqrt[d + e*x^3]) + (2*c*x*Sqrt[d + e*x^3])/(5*e^2) - (2*Sqrt[2 + Sqrt[3
]]*(16*c*d^2 - 5*e*(2*b*d + a*e))*(d^(1/3) + e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x + e^(2/3)*x^2)/((1 +
Sqrt[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]])/(15*3^(1/4)*d*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 1409

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

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}{\left (d+e x^3\right )^{3/2}} \, dx &=\frac{2 \left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \sqrt{d+e x^3}}-\frac{2 \int \frac{\frac{1}{2} \left (2 c d^2-e (2 b d+a e)\right )-\frac{3}{2} c d e x^3}{\sqrt{d+e x^3}} \, dx}{3 d e^2}\\ &=\frac{2 \left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \sqrt{d+e x^3}}+\frac{2 c x \sqrt{d+e x^3}}{5 e^2}-\frac{\left (16 c d^2-5 e (2 b d+a e)\right ) \int \frac{1}{\sqrt{d+e x^3}} \, dx}{15 d e^2}\\ &=\frac{2 \left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \sqrt{d+e x^3}}+\frac{2 c x \sqrt{d+e x^3}}{5 e^2}-\frac{2 \sqrt{2+\sqrt{3}} \left (16 c d^2-5 e (2 b d+a e)\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 )}{15 \sqrt [4]{3} d 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 = 0.105869, size = 102, normalized size = 0.35 \[ \frac{x \left (\sqrt{\frac{e x^3}{d}+1} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{e x^3}{d}\right ) \left (5 e (a e+2 b d)-16 c d^2\right )+2 \left (5 e (a e-b d)+c d \left (8 d+3 e x^3\right )\right )\right )}{15 d e^2 \sqrt{d+e x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3 + c*x^6)/(d + e*x^3)^(3/2),x]

[Out]

(x*(2*(5*e*(-(b*d) + a*e) + c*d*(8*d + 3*e*x^3)) + (-16*c*d^2 + 5*e*(2*b*d + a*e))*Sqrt[1 + (e*x^3)/d]*Hyperge
ometric2F1[1/3, 1/2, 4/3, -((e*x^3)/d)]))/(15*d*e^2*Sqrt[d + e*x^3])

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Maple [B]  time = 0.038, size = 934, normalized size = 3.2 \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)^(3/2),x)

[Out]

c*(2/3/e^2*d*x/((x^3+d/e)*e)^(1/2)+2/5/e^2*x*(e*x^3+d)^(1/2)+32/45*I*d/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*(-2/3/e*x/((x^3+d/e)*e)^(1/2)-4/9*I/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)))+a*(2/3/d*x/((x^3+d/e)*e)^(1/2)-2/9*I/d*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)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{6} + b x^{3} + a\right )} \sqrt{e x^{3} + d}}{e^{2} x^{6} + 2 \, d e x^{3} + d^{2}}, 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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 18.6978, size = 119, normalized size = 0.41 \begin{align*} \frac{a x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{3}{2} \\ \frac{4}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac{3}{2}} \Gamma \left (\frac{4}{3}\right )} + \frac{b x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{4}{3}, \frac{3}{2} \\ \frac{7}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac{3}{2}} \Gamma \left (\frac{7}{3}\right )} + \frac{c x^{7} \Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac{3}{2}} \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)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

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