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

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

[Out]

-2/55*(-11*b*e+8*c*d)*x*(e*x^3+d)^(1/2)/e^2+2/11*c*x^4*(e*x^3+d)^(1/2)/e+2/165*(55*a*e^2-22*b*d*e+16*c*d^2)*(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)/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 = 0.18, antiderivative size = 278, 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 = {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 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 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]

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*}

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Mathematica [C]  time = 0.09, 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])

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {c x^{6} + b x^{3} + a}{\sqrt {e x^{3} + d}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 907, normalized size = 3.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^6+b*x^3+a)/(e*x^3+d)^(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*
(-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*(2/5/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
*(-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)))-2/3*I*a*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 \[ \int \frac {c x^{6} + b x^{3} + a}{\sqrt {e x^{3} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {c\,x^6+b\,x^3+a}{\sqrt {e\,x^3+d}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.90, size = 119, normalized size = 0.43 \[ \frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt {d} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt {d} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 \sqrt {d} \Gamma \left (\frac {10}{3}\right )} \]

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))

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