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

Optimal. Leaf size=57 \[ \frac {1}{4} \sqrt [4]{x^4+x^3} (4 x-3)+\frac {7}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^3}}\right )-\frac {7}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^3}}\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 113, normalized size of antiderivative = 1.98, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2039, 2021, 2032, 63, 331, 298, 203, 206} \begin {gather*} -\frac {7}{4} \sqrt [4]{x^4+x^3}+\frac {\left (x^4+x^3\right )^{5/4}}{x^3}+\frac {7 (x+1)^{3/4} x^{9/4} \tan ^{-1}\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} \tanh ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{8 \left (x^4+x^3\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[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 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 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 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 331

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 2021

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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 2039

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

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Mathematica [C]  time = 0.02, size = 47, normalized size = 0.82 \begin {gather*} \frac {\sqrt [4]{x^3 (x+1)} \left (3 (x+1)^{5/4}-7 \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-x\right )\right )}{3 \sqrt [4]{x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x^3*(1 + x))^(1/4)*(3*(1 + x)^(5/4) - 7*Hypergeometric2F1[-1/4, 3/4, 7/4, -x]))/(3*(1 + x)^(1/4))

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IntegrateAlgebraic [A]  time = 0.29, size = 57, normalized size = 1.00 \begin {gather*} \frac {1}{4} (-3+4 x) \sqrt [4]{x^3+x^4}+\frac {7}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\frac {7}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 72, normalized size = 1.26 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.58, size = 60, normalized size = 1.05 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 0.77, size = 30, normalized size = 0.53

method	result	size
meijerg	\(-\frac {4 x^{\frac {3}{4}} \hypergeom \left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x \right )}{3}+\frac {8 x^{\frac {7}{4}} \hypergeom \left (\left [-\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], -x \right )}{7}\)	\(30\)
trager	\(\left (-\frac {3}{4}+x \right ) \left (x^{4}+x^{3}\right )^{\frac {1}{4}}-\frac {7 \ln \left (\frac {2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}+x^{3}}\, x +2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}+2 x^{3}+x^{2}}{x^{2}}\right )}{16}+\frac {7 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \sqrt {x^{4}+x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x +2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}-2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )}{16}\)	\(146\)
risch	\(\frac {\left (-3+4 x \right ) \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{4}+\frac {\left (\frac {7 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, \RootOf \left (\textit {\_Z}^{2}+1\right )-5 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {3}{4}}-2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x^{2}-4 \RootOf \left (\textit {\_Z}^{2}+1\right ) x -4 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x -\RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}}}{\left (1+x \right )^{2}}\right )}{16}-\frac {7 \ln \left (\frac {2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {3}{4}}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, x +2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x^{2}+2 x^{3}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}+4 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x +5 x^{2}+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}}+4 x +1}{\left (1+x \right )^{2}}\right )}{16}\right ) \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}} \left (\left (1+x \right )^{3} x \right )^{\frac {1}{4}}}{x \left (1+x \right )}\)	\(375\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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 (2 \, x - 1\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (x^4+x^3\right )}^{1/4}\,\left (2\,x-1\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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 )} \left (2 x - 1\right )}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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