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

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

[Out]

2/3*(a*e^2-b*d*e+c*d^2)*x/d/e^2/(e*x^3+d)^(1/2)+2/5*c*x*(e*x^3+d)^(1/2)/e^2-2/45*(16*c*d^2-5*e*(a*e+2*b*d))*(d
^(1/3)+e^(1/3)*x)*EllipticF((e^(1/3)*x+d^(1/3)*(1-3^(1/2)))/(e^(1/3)*x+d^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/
2*6^(1/2)+1/2*2^(1/2))*((d^(2/3)-d^(1/3)*e^(1/3)*x+e^(2/3)*x^2)/(e^(1/3)*x+d^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/
4)/d/e^(7/3)/(e*x^3+d)^(1/2)/(d^(1/3)*(d^(1/3)+e^(1/3)*x)/(e^(1/3)*x+d^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 289, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [C]  time = 0.10, 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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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{6} + b x^{3} + a\right )} \sqrt {e x^{3} + d}}{e^{2} x^{6} + 2 \, d e x^{3} + d^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.05, size = 934, normalized size = 3.23 \[ \text {result too large to display} \]

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*(-
d*e^2)^(1/3)/e-1/2*I*3^(1/2)*(-d*e^2)^(1/3)/e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2)*((x-(-d*e^2)^(1/3)/e)/(-3/2*(-d
*e^2)^(1/3)/e+1/2*I*3^(1/2)*(-d*e^2)^(1/3)/e))^(1/2)*(-I*(x+1/2*(-d*e^2)^(1/3)/e+1/2*I*3^(1/2)*(-d*e^2)^(1/3)/
e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-d*e^2)^(1/3)/e-1/2*I*3^(1
/2)*(-d*e^2)^(1/3)/e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2),(I*3^(1/2)*(-d*e^2)^(1/3)/(-3/2*(-d*e^2)^(1/3)/e+1/2*I*3
^(1/2)*(-d*e^2)^(1/3)/e)/e)^(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
*(-d*e^2)^(1/3)/e-1/2*I*3^(1/2)*(-d*e^2)^(1/3)/e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2)*((x-(-d*e^2)^(1/3)/e)/(-3/2*
(-d*e^2)^(1/3)/e+1/2*I*3^(1/2)*(-d*e^2)^(1/3)/e))^(1/2)*(-I*(x+1/2*(-d*e^2)^(1/3)/e+1/2*I*3^(1/2)*(-d*e^2)^(1/
3)/e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-d*e^2)^(1/3)/e-1/2*I*3
^(1/2)*(-d*e^2)^(1/3)/e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2),(I*3^(1/2)*(-d*e^2)^(1/3)/(-3/2*(-d*e^2)^(1/3)/e+1/2*
I*3^(1/2)*(-d*e^2)^(1/3)/e)/e)^(1/2)))+a*(2/3/d*x/((x^3+d/e)*e)^(1/2)-2/9*I/d*3^(1/2)*(-d*e^2)^(1/3)/e*(I*(x+1
/2*(-d*e^2)^(1/3)/e-1/2*I*3^(1/2)*(-d*e^2)^(1/3)/e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2)*((x-(-d*e^2)^(1/3)/e)/(-3/
2*(-d*e^2)^(1/3)/e+1/2*I*3^(1/2)*(-d*e^2)^(1/3)/e))^(1/2)*(-I*(x+1/2*(-d*e^2)^(1/3)/e+1/2*I*3^(1/2)*(-d*e^2)^(
1/3)/e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-d*e^2)^(1/3)/e-1/2*I
*3^(1/2)*(-d*e^2)^(1/3)/e)*3^(1/2)/(-d*e^2)^(1/3)*e)^(1/2),(I*3^(1/2)*(-d*e^2)^(1/3)/(-3/2*(-d*e^2)^(1/3)/e+1/
2*I*3^(1/2)*(-d*e^2)^(1/3)/e)/e)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {c\,x^6+b\,x^3+a}{{\left (e\,x^3+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 15.07, size = 119, normalized size = 0.41 \[ \frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {3}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {10}{3}\right )} \]

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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