3.13.87 \(\int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx\)

Optimal. Leaf size=93 \[ -2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^3}}\right )+2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-x^3}}\right )+2 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^3}}\right )-2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-x^3}}\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 185, normalized size of antiderivative = 1.99, number of steps used = 11, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {2042, 105, 63, 240, 212, 206, 203, 93, 298} \begin {gather*} \frac {2 \sqrt [4]{x^4-x^3} \tan ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}}+\frac {2 \sqrt [4]{2} \sqrt [4]{x^4-x^3} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{\sqrt [4]{x-1} x^{3/4}}+\frac {2 \sqrt [4]{x^4-x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}}-\frac {2 \sqrt [4]{2} \sqrt [4]{x^4-x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{\sqrt [4]{x-1} x^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^3 + x^4)^(1/4)/(x*(1 + x)),x]

[Out]

(2*(-x^3 + x^4)^(1/4)*ArcTan[(-1 + x)^(1/4)/x^(1/4)])/((-1 + x)^(1/4)*x^(3/4)) + (2*2^(1/4)*(-x^3 + x^4)^(1/4)
*ArcTan[(2^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/((-1 + x)^(1/4)*x^(3/4)) + (2*(-x^3 + x^4)^(1/4)*ArcTanh[(-1 + x)^(
1/4)/x^(1/4)])/((-1 + x)^(1/4)*x^(3/4)) - (2*2^(1/4)*(-x^3 + x^4)^(1/4)*ArcTanh[(2^(1/4)*x^(1/4))/(-1 + x)^(1/
4)])/((-1 + x)^(1/4)*x^(3/4))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 203

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

Rule 206

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 2042

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol]
:> Dist[(e^IntPart[m]*(e*x)^FracPart[m]*(a*x^j + b*x^(j + n))^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a
+ b*x^n)^FracPart[p]), Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n,
p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx &=\frac {\sqrt [4]{-x^3+x^4} \int \frac {\sqrt [4]{-1+x}}{\sqrt [4]{x} (1+x)} \, dx}{\sqrt [4]{-1+x} x^{3/4}}\\ &=\frac {\sqrt [4]{-x^3+x^4} \int \frac {1}{(-1+x)^{3/4} \sqrt [4]{x}} \, dx}{\sqrt [4]{-1+x} x^{3/4}}-\frac {\left (2 \sqrt [4]{-x^3+x^4}\right ) \int \frac {1}{(-1+x)^{3/4} \sqrt [4]{x} (1+x)} \, dx}{\sqrt [4]{-1+x} x^{3/4}}\\ &=\frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\sqrt [4]{-1+x} x^{3/4}}-\frac {\left (8 \sqrt [4]{-x^3+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-2 x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-1+x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}\\ &=\frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}-\frac {\left (2 \sqrt {2} \sqrt [4]{-x^3+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-1+x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}+\frac {\left (2 \sqrt {2} \sqrt [4]{-x^3+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-1+x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}\\ &=\frac {2 \sqrt [4]{2} \sqrt [4]{-x^3+x^4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{-1+x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}-\frac {2 \sqrt [4]{2} \sqrt [4]{-x^3+x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{-1+x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}+\frac {\left (2 \sqrt [4]{-x^3+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}+\frac {\left (2 \sqrt [4]{-x^3+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}\\ &=\frac {2 \sqrt [4]{-x^3+x^4} \tan ^{-1}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}+\frac {2 \sqrt [4]{2} \sqrt [4]{-x^3+x^4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{-1+x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}+\frac {2 \sqrt [4]{-x^3+x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}-\frac {2 \sqrt [4]{2} \sqrt [4]{-x^3+x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{-1+x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 58, normalized size = 0.62 \begin {gather*} \frac {4 \sqrt [4]{(x-1) x^3} \left (\sqrt [4]{x} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};1-x\right )-\, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {x-1}{2 x}\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^3 + x^4)^(1/4)/(x*(1 + x)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.32, size = 93, normalized size = 1.00 \begin {gather*} -2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^3+x^4}}\right )+2 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )-2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^3+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-x^3 + x^4)^(1/4)/(x*(1 + x)),x]

[Out]

-2*ArcTan[x/(-x^3 + x^4)^(1/4)] + 2*2^(1/4)*ArcTan[(2^(1/4)*x)/(-x^3 + x^4)^(1/4)] + 2*ArcTanh[x/(-x^3 + x^4)^
(1/4)] - 2*2^(1/4)*ArcTanh[(2^(1/4)*x)/(-x^3 + x^4)^(1/4)]

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fricas [B]  time = 0.46, size = 174, normalized size = 1.87 \begin {gather*} 4 \cdot 2^{\frac {1}{4}} \arctan \left (\frac {2^{\frac {3}{4}} x \sqrt {\frac {\sqrt {2} x^{2} + \sqrt {x^{4} - x^{3}}}{x^{2}}} - 2^{\frac {3}{4}} {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{2 \, x}\right ) - 2^{\frac {1}{4}} \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 2^{\frac {1}{4}} \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 2 \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x^3)^(1/4)/x/(1+x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 100, normalized size = 1.08 \begin {gather*} 2 \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) - 2 \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x^3)^(1/4)/x/(1+x),x, algorithm="giac")

[Out]

2*2^(1/4)*arctan(1/2*2^(3/4)*(-1/x + 1)^(1/4)) + 2^(1/4)*log(2^(1/4) + (-1/x + 1)^(1/4)) - 2^(1/4)*log(abs(-2^
(1/4) + (-1/x + 1)^(1/4))) - 2*arctan((-1/x + 1)^(1/4)) - log((-1/x + 1)^(1/4) + 1) + log(abs((-1/x + 1)^(1/4)
- 1))

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maple [C]  time = 1.51, size = 387, normalized size = 4.16

method	result	size
trager	\(\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{x^{2}}\right )+\ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+2 x^{3}-x^{2}}{x^{2}}\right )-\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {3 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{3}+4 \left (x^{4}-x^{3}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+4 \sqrt {x^{4}-x^{3}}\, \RootOf \left (\textit {\_Z}^{4}-2\right ) x +4 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}}{x^{2} \left (1+x \right )}\right )-\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \left (x^{4}-x^{3}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}+4 \sqrt {x^{4}-x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x +4 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}}{x^{2} \left (1+x \right )}\right )\)	\(387\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-x^3)^(1/4)/x/(1+x),x,method=_RETURNVERBOSE)

[Out]

RootOf(_Z^2+1)*ln((2*(x^4-x^3)^(1/2)*RootOf(_Z^2+1)*x-2*RootOf(_Z^2+1)*x^3+2*(x^4-x^3)^(3/4)-2*x^2*(x^4-x^3)^(
1/4)+RootOf(_Z^2+1)*x^2)/x^2)+ln((2*(x^4-x^3)^(3/4)+2*(x^4-x^3)^(1/2)*x+2*x^2*(x^4-x^3)^(1/4)+2*x^3-x^2)/x^2)-
RootOf(_Z^4-2)*ln((3*RootOf(_Z^4-2)^3*x^3+4*(x^4-x^3)^(1/4)*RootOf(_Z^4-2)^2*x^2-RootOf(_Z^4-2)^3*x^2+4*(x^4-x
^3)^(1/2)*RootOf(_Z^4-2)*x+4*(x^4-x^3)^(3/4))/x^2/(1+x))-RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln((-3*RootOf(_Z^2+Root
Of(_Z^4-2)^2)*RootOf(_Z^4-2)^2*x^3+RootOf(_Z^2+RootOf(_Z^4-2)^2)*RootOf(_Z^4-2)^2*x^2-4*(x^4-x^3)^(1/4)*RootOf
(_Z^4-2)^2*x^2+4*(x^4-x^3)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x+4*(x^4-x^3)^(3/4))/x^2/(1+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{{\left (x + 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x^3)^(1/4)/x/(1+x),x, algorithm="maxima")

[Out]

integrate((x^4 - x^3)^(1/4)/((x + 1)*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^4-x^3\right )}^{1/4}}{x\,\left (x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - x^3)^(1/4)/(x*(x + 1)),x)

[Out]

int((x^4 - x^3)^(1/4)/(x*(x + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x^{3} \left (x - 1\right )}}{x \left (x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-x**3)**(1/4)/x/(1+x),x)

[Out]

Integral((x**3*(x - 1))**(1/4)/(x*(x + 1)), x)

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