\(\int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx\) [738]

Optimal result

Integrand size = 20, antiderivative size = 57 \[ \int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx=\frac {1}{4} (-3+4 x) \sqrt [4]{x^3+x^4}+\frac {7}{8} \arctan \left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\frac {7}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right ) \]

[Out]

1/4*(-3+4*x)*(x^4+x^3)^(1/4)+7/8*arctan(x/(x^4+x^3)^(1/4))-7/8*arctanh(x/(x^4+x^3)^(1/4))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2064, 2046, 2057, 65, 338, 304, 209, 212} \[ \int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx=\frac {7 (x+1)^{3/4} x^{9/4} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{8 \left (x^4+x^3\right )^{3/4}}-\frac {7 (x+1)^{3/4} x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{8 \left (x^4+x^3\right )^{3/4}}-\frac {7}{4} \sqrt [4]{x^4+x^3}+\frac {\left (x^4+x^3\right )^{5/4}}{x^3} \]

[In]

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

[Out]

(-7*(x^3 + x^4)^(1/4))/4 + (x^3 + x^4)^(5/4)/x^3 + (7*x^(9/4)*(1 + x)^(3/4)*ArcTan[x^(1/4)/(1 + x)^(1/4)])/(8*
(x^3 + x^4)^(3/4)) - (7*x^(9/4)*(1 + x)^(3/4)*ArcTanh[x^(1/4)/(1 + x)^(1/4)])/(8*(x^3 + x^4)^(3/4))

Rule 65

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 209

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

Rule 212

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

Rule 304

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 338

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

Rule 2046

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*(m + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2064

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^3+x^4\right )^{5/4}}{x^3}-\frac {7}{4} \int \frac {\sqrt [4]{x^3+x^4}}{x} \, dx \\ & = -\frac {7}{4} \sqrt [4]{x^3+x^4}+\frac {\left (x^3+x^4\right )^{5/4}}{x^3}-\frac {7}{16} \int \frac {x^2}{\left (x^3+x^4\right )^{3/4}} \, dx \\ & = -\frac {7}{4} \sqrt [4]{x^3+x^4}+\frac {\left (x^3+x^4\right )^{5/4}}{x^3}-\frac {\left (7 x^{9/4} (1+x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (1+x)^{3/4}} \, dx}{16 \left (x^3+x^4\right )^{3/4}} \\ & = -\frac {7}{4} \sqrt [4]{x^3+x^4}+\frac {\left (x^3+x^4\right )^{5/4}}{x^3}-\frac {\left (7 x^{9/4} (1+x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{4 \left (x^3+x^4\right )^{3/4}} \\ & = -\frac {7}{4} \sqrt [4]{x^3+x^4}+\frac {\left (x^3+x^4\right )^{5/4}}{x^3}-\frac {\left (7 x^{9/4} (1+x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{4 \left (x^3+x^4\right )^{3/4}} \\ & = -\frac {7}{4} \sqrt [4]{x^3+x^4}+\frac {\left (x^3+x^4\right )^{5/4}}{x^3}-\frac {\left (7 x^{9/4} (1+x)^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{8 \left (x^3+x^4\right )^{3/4}}+\frac {\left (7 x^{9/4} (1+x)^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{8 \left (x^3+x^4\right )^{3/4}} \\ & = -\frac {7}{4} \sqrt [4]{x^3+x^4}+\frac {\left (x^3+x^4\right )^{5/4}}{x^3}+\frac {7 x^{9/4} (1+x)^{3/4} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{8 \left (x^3+x^4\right )^{3/4}}-\frac {7 x^{9/4} (1+x)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{8 \left (x^3+x^4\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.32 \[ \int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx=\frac {x^{9/4} (1+x)^{3/4} \left (2 x^{3/4} \sqrt [4]{1+x} (-3+4 x)+7 \arctan \left (\sqrt [4]{\frac {x}{1+x}}\right )-7 \text {arctanh}\left (\sqrt [4]{\frac {x}{1+x}}\right )\right )}{8 \left (x^3 (1+x)\right )^{3/4}} \]

[In]

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

[Out]

(x^(9/4)*(1 + x)^(3/4)*(2*x^(3/4)*(1 + x)^(1/4)*(-3 + 4*x) + 7*ArcTan[(x/(1 + x))^(1/4)] - 7*ArcTanh[(x/(1 + x
))^(1/4)]))/(8*(x^3*(1 + x))^(3/4))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.42 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.53

method	result	size
meijerg	\(-\frac {4 x^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x \right )}{3}+\frac {8 x^{\frac {7}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], -x \right )}{7}\)	\(30\)
pseudoelliptic	\(\frac {x^{6} \left (16 x \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}+7 \ln \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )-14 \arctan \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )-7 \ln \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}+x}{x}\right )-12 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}\right )}{16 {\left (\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}-x \right )}^{2} \left (x^{2}+\sqrt {x^{3} \left (1+x \right )}\right )^{2} {\left (\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}+x \right )}^{2}}\)	\(127\)
trager	\(\left (-\frac {3}{4}+x \right ) \left (x^{4}+x^{3}\right )^{\frac {1}{4}}-\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}+x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}-2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )}{16}-\frac {7 \ln \left (\frac {2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}+x^{3}}\, x +2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}+2 x^{3}+x^{2}}{x^{2}}\right )}{16}\)	\(147\)
risch	\(\frac {\left (-3+4 x \right ) \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{4}+\frac {\left (\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {3}{4}}-2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -4 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}}}{\left (1+x \right )^{2}}\right )}{16}+\frac {7 \ln \left (\frac {2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, x +2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x^{2}-2 x^{3}-2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}+4 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x -5 x^{2}+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}}-4 x -1}{\left (1+x \right )^{2}}\right )}{16}\right ) \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}} \left (\left (1+x \right )^{3} x \right )^{\frac {1}{4}}}{x \left (1+x \right )}\)	\(375\)

[In]

int((-1+2*x)*(x^4+x^3)^(1/4)/x,x,method=_RETURNVERBOSE)

[Out]

-4/3*x^(3/4)*hypergeom([-1/4,3/4],[7/4],-x)+8/7*x^(7/4)*hypergeom([-1/4,7/4],[11/4],-x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx=\frac {1}{4} \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} {\left (4 \, x - 3\right )} - \frac {7}{8} \, \arctan \left (\frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {7}{16} \, \log \left (\frac {x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {7}{16} \, \log \left (-\frac {x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

integrate((-1+2*x)*(x^4+x^3)^(1/4)/x,x, algorithm="fricas")

[Out]

1/4*(x^4 + x^3)^(1/4)*(4*x - 3) - 7/8*arctan((x^4 + x^3)^(1/4)/x) - 7/16*log((x + (x^4 + x^3)^(1/4))/x) + 7/16
*log(-(x - (x^4 + x^3)^(1/4))/x)

Sympy [F]

\[ \int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x + 1\right )} \left (2 x - 1\right )}{x}\, dx \]

[In]

integrate((-1+2*x)*(x**4+x**3)**(1/4)/x,x)

[Out]

Integral((x**3*(x + 1))**(1/4)*(2*x - 1)/x, x)

Maxima [F]

\[ \int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx=\int { \frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} {\left (2 \, x - 1\right )}}{x} \,d x } \]

[In]

integrate((-1+2*x)*(x^4+x^3)^(1/4)/x,x, algorithm="maxima")

[Out]

integrate((x^4 + x^3)^(1/4)*(2*x - 1)/x, x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx=-\frac {1}{4} \, {\left (3 \, {\left (\frac {1}{x} + 1\right )}^{\frac {5}{4}} - 7 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )} x^{2} - \frac {7}{8} \, \arctan \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {7}{16} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {7}{16} \, \log \left ({\left | {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[In]

integrate((-1+2*x)*(x^4+x^3)^(1/4)/x,x, algorithm="giac")

[Out]

-1/4*(3*(1/x + 1)^(5/4) - 7*(1/x + 1)^(1/4))*x^2 - 7/8*arctan((1/x + 1)^(1/4)) - 7/16*log((1/x + 1)^(1/4) + 1)
+ 7/16*log(abs((1/x + 1)^(1/4) - 1))

Mupad [F(-1)]

Timed out. \[ \int \frac {(-1+2 x) \sqrt [4]{x^3+x^4}}{x} \, dx=\int \frac {{\left (x^4+x^3\right )}^{1/4}\,\left (2\,x-1\right )}{x} \,d x \]

[In]

int(((x^3 + x^4)^(1/4)*(2*x - 1))/x,x)

[Out]

int(((x^3 + x^4)^(1/4)*(2*x - 1))/x, x)