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

Optimal. Leaf size=48 \[ \sqrt [4]{x^4+x^3}+\frac {7}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^3}}\right )-\frac {7}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^3}}\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 94, normalized size of antiderivative = 1.96, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2056, 80, 63, 331, 298, 203, 206} \begin {gather*} \sqrt [4]{x^4+x^3}+\frac {7 \sqrt [4]{x^4+x^3} \tan ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{2 x^{3/4} \sqrt [4]{x+1}}-\frac {7 \sqrt [4]{x^4+x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{2 x^{3/4} \sqrt [4]{x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 45, normalized size = 0.94 \begin {gather*} \frac {x^3 \left (-7 (x+1)^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-x\right )+3 x+3\right )}{3 \left (x^3 (x+1)\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.24, size = 48, normalized size = 1.00 \begin {gather*} \sqrt [4]{x^3+x^4}+\frac {7}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\frac {7}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 65, normalized size = 1.35 \begin {gather*} {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} - \frac {7}{2} \, \arctan \left (\frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {7}{4} \, \log \left (\frac {x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {7}{4} \, \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+x)*(x^4+x^3)^(1/4)/x/(1+x),x, algorithm="fricas")

[Out]

(x^4 + x^3)^(1/4) - 7/2*arctan((x^4 + x^3)^(1/4)/x) - 7/4*log((x + (x^4 + x^3)^(1/4))/x) + 7/4*log(-(x - (x^4
+ x^3)^(1/4))/x)

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giac [A]  time = 0.39, size = 45, normalized size = 0.94 \begin {gather*} x {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} - \frac {7}{2} \, \arctan \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {7}{4} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {7}{4} \, \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+x)*(x^4+x^3)^(1/4)/x/(1+x),x, algorithm="giac")

[Out]

x*(1/x + 1)^(1/4) - 7/2*arctan((1/x + 1)^(1/4)) - 7/4*log((1/x + 1)^(1/4) + 1) + 7/4*log(abs((1/x + 1)^(1/4) -
1))

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maple [C]  time = 0.76, size = 30, normalized size = 0.62

method	result	size
meijerg	\(-\frac {4 \hypergeom \left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x \right ) x^{\frac {3}{4}}}{3}+\frac {4 \hypergeom \left (\left [\frac {3}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], -x \right ) x^{\frac {7}{4}}}{7}\)	\(30\)
trager	\(\left (x^{4}+x^{3}\right )^{\frac {1}{4}}+\frac {7 \ln \left (\frac {2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}+x^{3}}\, x +2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}-2 x^{3}-x^{2}}{x^{2}}\right )}{4}-\frac {7 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}+x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}-2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )}{4}\)	\(145\)
risch	\(\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}+\frac {\left (-\frac {7 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, \RootOf \left (\textit {\_Z}^{2}+1\right )-5 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {3}{4}}+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x^{2}-4 \RootOf \left (\textit {\_Z}^{2}+1\right ) x +4 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x -\RootOf \left (\textit {\_Z}^{2}+1\right )+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}}}{\left (1+x \right )^{2}}\right )}{4}-\frac {7 \ln \left (\frac {2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {3}{4}}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, x +2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x^{2}+2 x^{3}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}+4 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x +5 x^{2}+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}}+4 x +1}{\left (1+x \right )^{2}}\right )}{4}\right ) \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}} \left (\left (1+x \right )^{3} x \right )^{\frac {1}{4}}}{x \left (1+x \right )}\)	\(369\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}}{{\left (x + 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + x^3)^(1/4)*(x - 1)/((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 (x-1\right )}{x\,\left (x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((x^3 + x^4)^(1/4)*(x - 1))/(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 )} \left (x - 1\right )}{x \left (x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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