\(\int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx\) [1287]

Optimal result

Integrand size = 22, antiderivative size = 93 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx=-2 \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+2 \sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^3+x^4}}\right )+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )-2 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^3+x^4}}\right ) \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.44, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2067, 129, 525, 524} \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx=\frac {4 \sqrt [4]{x^4-x^3} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},x,-x\right )}{3 \sqrt [4]{1-x}} \]

[In]

Int[(-x^3 + x^4)^(1/4)/(x*(1 + x)),x]

[Out]

(4*(-x^3 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x, -x])/(3*(1 - x)^(1/4))

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

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

Rule 2067

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*} \text {integral}& = \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 {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{1+x^4} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-1+x} x^{3/4}} \\ & = \frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{1+x^4} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{1-x} x^{3/4}} \\ & = \frac {4 \sqrt [4]{-x^3+x^4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},x,-x\right )}{3 \sqrt [4]{1-x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx=-\frac {2 (-1+x)^{3/4} x^{9/4} \left (\arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )-\sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2}}{\sqrt [4]{\frac {-1+x}{x}}}\right )-\text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )+\sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2}}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )}{\left ((-1+x) x^3\right )^{3/4}} \]

[In]

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

[Out]

(-2*(-1 + x)^(3/4)*x^(9/4)*(ArcTan[((-1 + x)/x)^(-1/4)] - 2^(1/4)*ArcTan[2^(1/4)/((-1 + x)/x)^(1/4)] - ArcTanh
[((-1 + x)/x)^(-1/4)] + 2^(1/4)*ArcTanh[2^(1/4)/((-1 + x)/x)^(1/4)]))/((-1 + x)*x^3)^(3/4)

Maple [A] (verified)

Time = 4.86 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.30

method	result	size
pseudoelliptic	\(-\ln \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}-x}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{x}\right )+\ln \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}+x}{x}\right )-\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}}-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {1}{4}}\)	\(121\)
trager	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}} x^{2}-4 \sqrt {x^{4}-x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {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 )+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}} x^{2}+4 \sqrt {x^{4}-x^{3}}\, \operatorname {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 )+\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 )-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {2 \sqrt {x^{4}-x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{3} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{2}-4 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}+4 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{x^{2}}\right )}{2}\)	\(449\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx=-2^{\frac {1}{4}} \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 2^{\frac {1}{4}} \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - i \cdot 2^{\frac {1}{4}} \log \left (\frac {i \cdot 2^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + i \cdot 2^{\frac {1}{4}} \log \left (\frac {-i \cdot 2^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 2 \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{{\left (x + 1\right )} x} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx=-2 \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) + 2 \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x (1+x)} \, dx=\int \frac {{\left (x^4-x^3\right )}^{1/4}}{x\,\left (x+1\right )} \,d x \]

[In]

int((x^4 - x^3)^(1/4)/(x*(x + 1)),x)

[Out]

int((x^4 - x^3)^(1/4)/(x*(x + 1)), x)