3.40 \(\int \frac {a+b x^3+c x^6}{(d+e x^3)^{5/2}} \, dx\)

Optimal. Leaf size=309 \[ -\frac {2 x \left (-7 a e^2-2 b d e+11 c d^2\right )}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 x \left (a e^2-b d e+c d^2\right )}{9 d e^2 \left (d+e x^3\right )^{3/2}}+\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (e (7 a e+2 b d)+16 c d^2\right ) F\left (\sin ^{-1}\left (\frac {\sqrt [3]{e} x+\left (1-\sqrt {3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt {3}\right ) \sqrt [3]{d}}\right )|-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} d^2 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \]

[Out]

2/9*(a*e^2-b*d*e+c*d^2)*x/d/e^2/(e*x^3+d)^(3/2)-2/27*(-7*a*e^2-2*b*d*e+11*c*d^2)*x/d^2/e^2/(e*x^3+d)^(1/2)+2/8
1*(16*c*d^2+e*(7*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^2/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 = 0.21, antiderivative size = 309, 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, 385, 218} \[ -\frac {2 x \left (-7 a e^2-2 b d e+11 c d^2\right )}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 x \left (a e^2-b d e+c d^2\right )}{9 d e^2 \left (d+e x^3\right )^{3/2}}+\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (e (7 a e+2 b d)+16 c d^2\right ) F\left (\sin ^{-1}\left (\frac {\sqrt [3]{e} x+\left (1-\sqrt {3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt {3}\right ) \sqrt [3]{d}}\right )|-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} d^2 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)/(d + e*x^3)^(5/2),x]

[Out]

(2*(c*d^2 - b*d*e + a*e^2)*x)/(9*d*e^2*(d + e*x^3)^(3/2)) - (2*(11*c*d^2 - 2*b*d*e - 7*a*e^2)*x)/(27*d^2*e^2*S
qrt[d + e*x^3]) + (2*Sqrt[2 + Sqrt[3]]*(16*c*d^2 + e*(2*b*d + 7*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]])/(27*3^(1/4)*d^2*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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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 )^{5/2}} \, dx &=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (2 c d^2-e (2 b d+7 a e)\right )-\frac {9}{2} c d e x^3}{\left (d+e x^3\right )^{3/2}} \, dx}{9 d e^2}\\ &=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \left (11 c d^2-2 b d e-7 a e^2\right ) x}{27 d^2 e^2 \sqrt {d+e x^3}}-\frac {\left (4 \left (-\frac {9}{2} c d^2 e+\frac {1}{4} e \left (2 c d^2-e (2 b d+7 a e)\right )\right )\right ) \int \frac {1}{\sqrt {d+e x^3}} \, dx}{27 d^2 e^3}\\ &=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \left (11 c d^2-2 b d e-7 a e^2\right ) x}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2+e (2 b d+7 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 )}{27 \sqrt [4]{3} d^2 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 = 0.14, size = 129, normalized size = 0.42 \[ \frac {x \left (d+e x^3\right ) \sqrt {\frac {e x^3}{d}+1} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {e x^3}{d}\right ) \left (e (7 a e+2 b d)+16 c d^2\right )-2 x \left (e \left (b d \left (d-2 e x^3\right )-a e \left (10 d+7 e x^3\right )\right )+c d^2 \left (8 d+11 e x^3\right )\right )}{27 d^2 e^2 \left (d+e x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3 + c*x^6)/(d + e*x^3)^(5/2),x]

[Out]

(-2*x*(c*d^2*(8*d + 11*e*x^3) + e*(b*d*(d - 2*e*x^3) - a*e*(10*d + 7*e*x^3))) + (16*c*d^2 + e*(2*b*d + 7*a*e))
*x*(d + e*x^3)*Sqrt[1 + (e*x^3)/d]*Hypergeometric2F1[1/3, 1/2, 4/3, -((e*x^3)/d)])/(27*d^2*e^2*(d + e*x^3)^(3/
2))

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fricas [F]  time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{6} + b x^{3} + a\right )} \sqrt {e x^{3} + d}}{e^{3} x^{9} + 3 \, d e^{2} x^{6} + 3 \, d^{2} e x^{3} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(5/2),x, algorithm="fricas")

[Out]

integral((c*x^6 + b*x^3 + a)*sqrt(e*x^3 + d)/(e^3*x^9 + 3*d*e^2*x^6 + 3*d^2*e*x^3 + d^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)/(e*x^3 + d)^(5/2), x)

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maple [B]  time = 0.05, size = 1005, normalized size = 3.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^6+b*x^3+a)/(e*x^3+d)^(5/2),x)

[Out]

c*(2/9*d*x/e^4*(e*x^3+d)^(1/2)/(x^3+d/e)^2-22/27/e^2*x/((x^3+d/e)*e)^(1/2)-32/81*I/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*(-2/9*x/e^3*(e*x^3+d)^(1/2)/(x^3+d/e)^2+4/27/e/d*x/((x^3+d/e
)*e)^(1/2)-4/81*I/e^2/d*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*(2/9/d*x/e^2*
(e*x^3+d)^(1/2)/(x^3+d/e)^2+14/27/d^2*x/((x^3+d/e)*e)^(1/2)-14/81*I/d^2*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 \[ \int \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^6 + b*x^3 + a)/(e*x^3 + d)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {c\,x^6+b\,x^3+a}{{\left (e\,x^3+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3 + c*x^6)/(d + e*x^3)^(5/2),x)

[Out]

int((a + b*x^3 + c*x^6)/(d + e*x^3)^(5/2), x)

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sympy [A]  time = 86.88, size = 119, normalized size = 0.39 \[ \frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {5}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{3}, \frac {5}{2} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**6+b*x**3+a)/(e*x**3+d)**(5/2),x)

[Out]

a*x*gamma(1/3)*hyper((1/3, 5/2), (4/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(5/2)*gamma(4/3)) + b*x**4*gamma(4/3)
*hyper((4/3, 5/2), (7/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(5/2)*gamma(7/3)) + c*x**7*gamma(7/3)*hyper((7/3, 5
/2), (10/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(5/2)*gamma(10/3))

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