3.7.100 \(\int \frac {(a+b x^2)^{4/3}}{x^4} \, dx\) [700]

Optimal. Leaf size=284 \[ -\frac {8 b \sqrt [3]{a+b x^2}}{9 x}-\frac {\left (a+b x^2\right )^{4/3}}{3 x^3}-\frac {16 \sqrt {2-\sqrt {3}} b \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right )|-7+4 \sqrt {3}\right )}{9 \sqrt [4]{3} x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \]

[Out]

-8/9*b*(b*x^2+a)^(1/3)/x-1/3*(b*x^2+a)^(4/3)/x^3-16/27*b*(a^(1/3)-(b*x^2+a)^(1/3))*EllipticF((-(b*x^2+a)^(1/3)
+a^(1/3)*(1+3^(1/2)))/(-(b*x^2+a)^(1/3)+a^(1/3)*(1-3^(1/2))),2*I-I*3^(1/2))*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+
(b*x^2+a)^(2/3))/(-(b*x^2+a)^(1/3)+a^(1/3)*(1-3^(1/2)))^2)^(1/2)*(1/2*6^(1/2)-1/2*2^(1/2))*3^(3/4)/x/(-a^(1/3)
*(a^(1/3)-(b*x^2+a)^(1/3))/(-(b*x^2+a)^(1/3)+a^(1/3)*(1-3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {283, 242, 225} \begin {gather*} -\frac {16 \sqrt {2-\sqrt {3}} b \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\text {ArcSin}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right )|-7+4 \sqrt {3}\right )}{9 \sqrt [4]{3} x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac {8 b \sqrt [3]{a+b x^2}}{9 x}-\frac {\left (a+b x^2\right )^{4/3}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/x^4,x]

[Out]

(-8*b*(a + b*x^2)^(1/3))/(9*x) - (a + b*x^2)^(4/3)/(3*x^3) - (16*Sqrt[2 - Sqrt[3]]*b*(a^(1/3) - (a + b*x^2)^(1
/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3)
)^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))]
, -7 + 4*Sqrt[3]])/(9*3^(1/4)*x*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - (a + b
*x^2)^(1/3))^2)])

Rule 225

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] && NegQ[a]

Rule 242

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[3*(Sqrt[b*x^2]/(2*b*x)), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{4/3}}{x^4} \, dx &=-\frac {\left (a+b x^2\right )^{4/3}}{3 x^3}+\frac {1}{9} (8 b) \int \frac {\sqrt [3]{a+b x^2}}{x^2} \, dx\\ &=-\frac {8 b \sqrt [3]{a+b x^2}}{9 x}-\frac {\left (a+b x^2\right )^{4/3}}{3 x^3}+\frac {1}{27} \left (16 b^2\right ) \int \frac {1}{\left (a+b x^2\right )^{2/3}} \, dx\\ &=-\frac {8 b \sqrt [3]{a+b x^2}}{9 x}-\frac {\left (a+b x^2\right )^{4/3}}{3 x^3}+\frac {\left (8 b \sqrt {b x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a+x^3}} \, dx,x,\sqrt [3]{a+b x^2}\right )}{9 x}\\ &=-\frac {8 b \sqrt [3]{a+b x^2}}{9 x}-\frac {\left (a+b x^2\right )^{4/3}}{3 x^3}-\frac {16 \sqrt {2-\sqrt {3}} b \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right )|-7+4 \sqrt {3}\right )}{9 \sqrt [4]{3} x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\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 = 52, normalized size = 0.18 \begin {gather*} -\frac {a \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {3}{2},-\frac {4}{3};-\frac {1}{2};-\frac {b x^2}{a}\right )}{3 x^3 \sqrt [3]{1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/x^4,x]

[Out]

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

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(4/3)/x^4,x)

[Out]

int((b*x^2+a)^(4/3)/x^4,x)

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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((b*x^2+a)^(4/3)/x^4,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(4/3)/x^4, x)

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Fricas [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((b*x^2+a)^(4/3)/x^4,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(4/3)/x^4, x)

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Sympy [A]
time = 0.53, size = 34, normalized size = 0.12 \begin {gather*} - \frac {a^{\frac {4}{3}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {4}{3} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(4/3)/x**4,x)

[Out]

-a**(4/3)*hyper((-3/2, -4/3), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*x**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((b*x^2+a)^(4/3)/x^4,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(4/3)/x^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{4/3}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(4/3)/x^4,x)

[Out]

int((a + b*x^2)^(4/3)/x^4, x)

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