3.279 \(\int \frac{-16+x^{3/4}}{16+x^{3/4}} \, dx\)

Optimal. Leaf size=104 \[ x-128 \sqrt [4]{x}+\frac{256}{3} \sqrt [3]{2} \log \left (\sqrt [4]{x}+2 \sqrt [3]{2}\right )-\frac{128}{3} \sqrt [3]{2} \log \left (\sqrt{x}-2 \sqrt [3]{2} \sqrt [4]{x}+4\ 2^{2/3}\right )-\frac{256 \sqrt [3]{2} \tan ^{-1}\left (\frac{\sqrt [3]{2}-\sqrt [4]{x}}{\sqrt [3]{2} \sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

-128*x^(1/4) + x - (256*2^(1/3)*ArcTan[(2^(1/3) - x^(1/4))/(2^(1/3)*Sqrt[3])])/Sqrt[3] + (256*2^(1/3)*Log[2*2^
(1/3) + x^(1/4)])/3 - (128*2^(1/3)*Log[4*2^(2/3) - 2*2^(1/3)*x^(1/4) + Sqrt[x]])/3

________________________________________________________________________________________

Rubi [A]  time = 0.0861164, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {376, 459, 321, 200, 31, 634, 617, 204, 628} \[ x-128 \sqrt [4]{x}+\frac{256}{3} \sqrt [3]{2} \log \left (\sqrt [4]{x}+2 \sqrt [3]{2}\right )-\frac{128}{3} \sqrt [3]{2} \log \left (\sqrt{x}-2 \sqrt [3]{2} \sqrt [4]{x}+4\ 2^{2/3}\right )-\frac{256 \sqrt [3]{2} \tan ^{-1}\left (\frac{\sqrt [3]{2}-\sqrt [4]{x}}{\sqrt [3]{2} \sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(-16 + x^(3/4))/(16 + x^(3/4)),x]

[Out]

-128*x^(1/4) + x - (256*2^(1/3)*ArcTan[(2^(1/3) - x^(1/4))/(2^(1/3)*Sqrt[3])])/Sqrt[3] + (256*2^(1/3)*Log[2*2^
(1/3) + x^(1/4)])/3 - (128*2^(1/3)*Log[4*2^(2/3) - 2*2^(1/3)*x^(1/4) + Sqrt[x]])/3

Rule 376

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, Dis
t[g, Subst[Int[x^(g - 1)*(a + b*x^(g*n))^p*(c + d*x^(g*n))^q, x], x, x^(1/g)], x]] /; FreeQ[{a, b, c, d, p, q}
, x] && NeQ[b*c - a*d, 0] && FractionQ[n]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 321

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 200

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{-16+x^{3/4}}{16+x^{3/4}} \, dx &=4 \operatorname{Subst}\left (\int \frac{x^3 \left (-16+x^3\right )}{16+x^3} \, dx,x,\sqrt [4]{x}\right )\\ &=x-128 \operatorname{Subst}\left (\int \frac{x^3}{16+x^3} \, dx,x,\sqrt [4]{x}\right )\\ &=-128 \sqrt [4]{x}+x+2048 \operatorname{Subst}\left (\int \frac{1}{16+x^3} \, dx,x,\sqrt [4]{x}\right )\\ &=-128 \sqrt [4]{x}+x+\frac{1}{3} \left (256 \sqrt [3]{2}\right ) \operatorname{Subst}\left (\int \frac{1}{2 \sqrt [3]{2}+x} \, dx,x,\sqrt [4]{x}\right )+\frac{1}{3} \left (256 \sqrt [3]{2}\right ) \operatorname{Subst}\left (\int \frac{4 \sqrt [3]{2}-x}{4\ 2^{2/3}-2 \sqrt [3]{2} x+x^2} \, dx,x,\sqrt [4]{x}\right )\\ &=-128 \sqrt [4]{x}+x+\frac{256}{3} \sqrt [3]{2} \log \left (2 \sqrt [3]{2}+\sqrt [4]{x}\right )-\frac{1}{3} \left (128 \sqrt [3]{2}\right ) \operatorname{Subst}\left (\int \frac{-2 \sqrt [3]{2}+2 x}{4\ 2^{2/3}-2 \sqrt [3]{2} x+x^2} \, dx,x,\sqrt [4]{x}\right )+\left (256\ 2^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{4\ 2^{2/3}-2 \sqrt [3]{2} x+x^2} \, dx,x,\sqrt [4]{x}\right )\\ &=-128 \sqrt [4]{x}+x+\frac{256}{3} \sqrt [3]{2} \log \left (2 \sqrt [3]{2}+\sqrt [4]{x}\right )-\frac{128}{3} \sqrt [3]{2} \log \left (4\ 2^{2/3}-2 \sqrt [3]{2} \sqrt [4]{x}+\sqrt{x}\right )+\left (256 \sqrt [3]{2}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{\sqrt [4]{x}}{\sqrt [3]{2}}\right )\\ &=-128 \sqrt [4]{x}+x-\frac{256 \sqrt [3]{2} \tan ^{-1}\left (\frac{2-2^{2/3} \sqrt [4]{x}}{2 \sqrt{3}}\right )}{\sqrt{3}}+\frac{256}{3} \sqrt [3]{2} \log \left (2 \sqrt [3]{2}+\sqrt [4]{x}\right )-\frac{128}{3} \sqrt [3]{2} \log \left (4\ 2^{2/3}-2 \sqrt [3]{2} \sqrt [4]{x}+\sqrt{x}\right )\\ \end{align*}

Mathematica [C]  time = 0.0031688, size = 22, normalized size = 0.21 \[ x-2 x \, _2F_1\left (1,\frac{4}{3};\frac{7}{3};-\frac{x^{3/4}}{16}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-16 + x^(3/4))/(16 + x^(3/4)),x]

[Out]

x - 2*x*Hypergeometric2F1[1, 4/3, 7/3, -x^(3/4)/16]

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 66, normalized size = 0.6 \begin{align*} x-128\,\sqrt [4]{x}+{\frac{128\,\sqrt [3]{16}}{3}\ln \left ( \sqrt [4]{x}+\sqrt [3]{16} \right ) }-{\frac{64\,\sqrt [3]{16}}{3}\ln \left ( \sqrt{x}-\sqrt [3]{16}\sqrt [4]{x}+{16}^{{\frac{2}{3}}} \right ) }+{\frac{128\,\sqrt [3]{16}\sqrt{3}}{3}\arctan \left ({\frac{\sqrt{3}}{3} \left ({\frac{{16}^{{\frac{2}{3}}}}{8}\sqrt [4]{x}}-1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16+x^(3/4))/(16+x^(3/4)),x)

[Out]

x-128*x^(1/4)+128/3*16^(1/3)*ln(x^(1/4)+16^(1/3))-64/3*16^(1/3)*ln(x^(1/2)-16^(1/3)*x^(1/4)+16^(2/3))+128/3*16
^(1/3)*3^(1/2)*arctan(1/3*3^(1/2)*(1/8*16^(2/3)*x^(1/4)-1))

________________________________________________________________________________________

Maxima [A]  time = 1.45643, size = 96, normalized size = 0.92 \begin{align*} \frac{256}{3} \, \sqrt{3} 2^{\frac{1}{3}} \arctan \left (-\frac{1}{6} \, \sqrt{3} 2^{\frac{2}{3}}{\left (2^{\frac{1}{3}} - x^{\frac{1}{4}}\right )}\right ) - \frac{128}{3} \cdot 2^{\frac{1}{3}} \log \left (4 \cdot 2^{\frac{2}{3}} - 2 \cdot 2^{\frac{1}{3}} x^{\frac{1}{4}} + \sqrt{x}\right ) + \frac{256}{3} \cdot 2^{\frac{1}{3}} \log \left (2 \cdot 2^{\frac{1}{3}} + x^{\frac{1}{4}}\right ) + x - 128 \, x^{\frac{1}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16+x^(3/4))/(16+x^(3/4)),x, algorithm="maxima")

[Out]

256/3*sqrt(3)*2^(1/3)*arctan(-1/6*sqrt(3)*2^(2/3)*(2^(1/3) - x^(1/4))) - 128/3*2^(1/3)*log(4*2^(2/3) - 2*2^(1/
3)*x^(1/4) + sqrt(x)) + 256/3*2^(1/3)*log(2*2^(1/3) + x^(1/4)) + x - 128*x^(1/4)

________________________________________________________________________________________

Fricas [A]  time = 1.57667, size = 261, normalized size = 2.51 \begin{align*} \frac{256}{3} \, \sqrt{3} 2^{\frac{1}{3}} \arctan \left (\frac{1}{6} \, \sqrt{3} 2^{\frac{2}{3}} x^{\frac{1}{4}} - \frac{1}{3} \, \sqrt{3}\right ) - \frac{128}{3} \cdot 2^{\frac{1}{3}} \log \left (4 \cdot 2^{\frac{2}{3}} - 2 \cdot 2^{\frac{1}{3}} x^{\frac{1}{4}} + \sqrt{x}\right ) + \frac{256}{3} \cdot 2^{\frac{1}{3}} \log \left (2 \cdot 2^{\frac{1}{3}} + x^{\frac{1}{4}}\right ) + x - 128 \, x^{\frac{1}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16+x^(3/4))/(16+x^(3/4)),x, algorithm="fricas")

[Out]

256/3*sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*x^(1/4) - 1/3*sqrt(3)) - 128/3*2^(1/3)*log(4*2^(2/3) - 2*2^(1
/3)*x^(1/4) + sqrt(x)) + 256/3*2^(1/3)*log(2*2^(1/3) + x^(1/4)) + x - 128*x^(1/4)

________________________________________________________________________________________

Sympy [A]  time = 8.5829, size = 102, normalized size = 0.98 \begin{align*} - 128 \sqrt [4]{x} + x + \frac{256 \sqrt [3]{2} \log{\left (\sqrt [4]{x} + 2 \sqrt [3]{2} \right )}}{3} - \frac{128 \sqrt [3]{2} \log{\left (- 2 \sqrt [3]{2} \sqrt [4]{x} + \sqrt{x} + 4 \cdot 2^{\frac{2}{3}} \right )}}{3} + \frac{256 \sqrt [3]{2} \sqrt{3} \operatorname{atan}{\left (\frac{2^{\frac{2}{3}} \sqrt{3} \sqrt [4]{x}}{6} - \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16+x**(3/4))/(16+x**(3/4)),x)

[Out]

-128*x**(1/4) + x + 256*2**(1/3)*log(x**(1/4) + 2*2**(1/3))/3 - 128*2**(1/3)*log(-2*2**(1/3)*x**(1/4) + sqrt(x
) + 4*2**(2/3))/3 + 256*2**(1/3)*sqrt(3)*atan(2**(2/3)*sqrt(3)*x**(1/4)/6 - sqrt(3)/3)/3

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16+x^(3/4))/(16+x^(3/4)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError