∫ 1____ (a+ b_ x3 )3∕2x8 dx [2049]

3.21.49 \(\int \frac {1}{(a+\frac {b}{x^3})^{3/2} x^8} \, dx\) [2049]

Optimal. Leaf size=267 \[ \frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x^4}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}} \]

[Out]

2/3/b/x^4/(a+b/x^3)^(1/2)-16/15*(a+b/x^3)^(1/2)/b^2/x+32/45*a*(a^(1/3)+b^(1/3)/x)*EllipticF((b^(1/3)/x+a^(1/3)

*(1-3^(1/2)))/(b^(1/3)/x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)+b^(2/3)/x^2-a

^(1/3)*b^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/b^(7/3)/(a+b/x^3)^(1/2)/(a^(1/3)*(a^(1/3)+b

^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {342, 294, 327, 224} \begin {gather*} \frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\text {ArcSin}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {2}{3 b x^4 \sqrt {a+\frac {b}{x^3}}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[1/((a + b/x^3)^(3/2)*x^8),x]

[Out]

2/(3*b*Sqrt[a + b/x^3]*x^4) - (16*Sqrt[a + b/x^3])/(15*b^2*x) + (32*Sqrt[2 + Sqrt[3]]*a*(a^(1/3) + b^(1/3)/x)*

Sqrt[(a^(2/3) + b^(2/3)/x^2 - (a^(1/3)*b^(1/3))/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)^2]*EllipticF[ArcSin[((1

- Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)], -7 - 4*Sqrt[3]])/(15*3^(1/4)*b^(7/3)*Sq

rt[a + b/x^3]*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)/x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)^2])

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt

[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq

rt[s*((s + r*x)/((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]], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^

n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x

], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && !ILtQ[(m + n*(p + 1) + 1)/n, 0]

&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n

)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p

, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,

c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;

FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx &=-\text {Subst}\left (\int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x^4}-\frac {8 \text {Subst}\left (\int \frac {x^3}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{3 b}\\ &=\frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x^4}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {(16 a) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{15 b^2}\\ &=\frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x^4}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 54, normalized size = 0.20 \begin {gather*} -\frac {2 \sqrt {1+\frac {a x^3}{b}} \, _2F_1\left (-\frac {5}{6},\frac {3}{2};\frac {1}{6};-\frac {a x^3}{b}\right )}{5 b \sqrt {a+\frac {b}{x^3}} x^4} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((a + b/x^3)^(3/2)*x^8),x]

[Out]

(-2*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[-5/6, 3/2, 1/6, -((a*x^3)/b)])/(5*b*Sqrt[a + b/x^3]*x^4)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2055 vs. \(2 (204 ) = 408\).
time = 0.07, size = 2056, normalized size = 7.70


method	result	size

risch	\(\text {Expression too large to display}\)	\(1431\)
default	\(\text {Expression too large to display}\)	\(2056\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^3)^(3/2)/x^8,x,method=_RETURNVERBOSE)

[Out]

-2/15/((a*x^3+b)/x^3)^(3/2)/x^8*(a*x^3+b)/(-a^2*b)^(1/3)/b^2*(-32*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(

-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1

/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(

I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^

(1/2)-3))^(1/2))*(x*(a*x^3+b))^(1/2)*3^(1/2)*a^2*x^5+64*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1

/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^

(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-

3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^

(1/2))*(x*(a*x^3+b))^(1/2)*(-a^2*b)^(1/3)*3^(1/2)*a*x^4-32*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)

^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I

*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/

2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3

))^(1/2))*(x*(a*x^3+b))^(1/2)*(-a^2*b)^(2/3)*3^(1/2)*x^3+32*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^

(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*

3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2

)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3)

)^(1/2))*(x*(a*x^3+b))^(1/2)*a^2*x^5-64*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^

(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/

3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/

2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(x*(a*x^3+b

))^(1/2)*(-a^2*b)^(1/3)*a*x^4+32*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(

-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*

x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a

*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(x*(a*x^3+b))^(1/2

)*(-a^2*b)^(2/3)*x^3+5*I*(1/a^2*x*(-a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))*(I*3^(

1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/2)*(-a^2*b)^(1/3)*3^(1/2)*a*x^4-15*(1/a^2*x*(-a*x+(-a^2*b)^(1/3)

)*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/2)*(-a^2

*b)^(1/3)*a*x^4+3*I*(a*x^4+b*x)^(1/2)*(1/a^2*x*(-a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^

(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/2)*(x*(a*x^3+b))^(1/2)*(-a^2*b)^(1/3)*3^(1/2)-9*(a*

x^4+b*x)^(1/2)*(1/a^2*x*(-a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2

*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/2)*(x*(a*x^3+b))^(1/2)*(-a^2*b)^(1/3))/(I*3^(1/2)-3)/(1/a^2*x*(-a*x+(-a^2*

b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^3)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate(1/((a + b/x^3)^(3/2)*x^8), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 75, normalized size = 0.28 \begin {gather*} \frac {2 \, {\left (16 \, {\left (a^{2} x^{4} + a b x\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, \frac {1}{x}\right ) - {\left (8 \, a b x^{3} + 3 \, b^{2}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right )}}{15 \, {\left (a b^{3} x^{4} + b^{4} x\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^3)^(3/2)/x^8,x, algorithm="fricas")

[Out]

2/15*(16*(a^2*x^4 + a*b*x)*sqrt(b)*weierstrassPInverse(0, -4*a/b, 1/x) - (8*a*b*x^3 + 3*b^2)*sqrt((a*x^3 + b)/

x^3))/(a*b^3*x^4 + b^4*x)

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Sympy [A]
time = 0.96, size = 39, normalized size = 0.15 \begin {gather*} - \frac {\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 {b e^{i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{2}} x^{7} \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**3)**(3/2)/x**8,x)

[Out]

-gamma(7/3)*hyper((3/2, 7/3), (10/3,), b*exp_polar(I*pi)/(a*x**3))/(3*a**(3/2)*x**7*gamma(10/3))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^3)^(3/2)/x^8,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^3)^(3/2)*x^8), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^8\,{\left (a+\frac {b}{x^3}\right )}^{3/2}} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8*(a + b/x^3)^(3/2)),x)

[Out]

int(1/(x^8*(a + b/x^3)^(3/2)), x)

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