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

Optimal. Leaf size=61 \[ \frac {\tan ^{-1}\left (\frac {2^{3/4} \sqrt [4]{x^3+x}}{x+1}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {2^{3/4} \sqrt [4]{x^3+x}}{x+1}\right )}{2 \sqrt [4]{2}} \]

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Rubi [C]  time = 0.10, antiderivative size = 87, normalized size of antiderivative = 1.43, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2056, 959, 466, 510} \begin {gather*} -\frac {4 \sqrt [4]{x^2+1} x^2 F_1\left (\frac {7}{8};1,\frac {1}{4};\frac {15}{8};x^2,-x^2\right )}{7 \sqrt [4]{x^3+x}}-\frac {4 \sqrt [4]{x^2+1} x F_1\left (\frac {3}{8};1,\frac {1}{4};\frac {11}{8};x^2,-x^2\right )}{3 \sqrt [4]{x^3+x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[1/((-1 + x)*(x + x^3)^(1/4)),x]

[Out]

(-4*x*(1 + x^2)^(1/4)*AppellF1[3/8, 1, 1/4, 11/8, x^2, -x^2])/(3*(x + x^3)^(1/4)) - (4*x^2*(1 + x^2)^(1/4)*App
ellF1[7/8, 1, 1/4, 15/8, x^2, -x^2])/(7*(x + x^3)^(1/4))

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 959

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[(d*(g*x)^n)/x^n, In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[(e*(g*x)^n)/x^n, Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

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}{(-1+x) \sqrt [4]{x+x^3}} \, dx &=\frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{(-1+x) \sqrt [4]{x} \sqrt [4]{1+x^2}} \, dx}{\sqrt [4]{x+x^3}}\\ &=-\frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{\sqrt [4]{x} \left (1-x^2\right ) \sqrt [4]{1+x^2}} \, dx}{\sqrt [4]{x+x^3}}-\frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {x^{3/4}}{\left (1-x^2\right ) \sqrt [4]{1+x^2}} \, dx}{\sqrt [4]{x+x^3}}\\ &=-\frac {\left (4 \sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1-x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x+x^3}}-\frac {\left (4 \sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x+x^3}}\\ &=-\frac {4 x \sqrt [4]{1+x^2} F_1\left (\frac {3}{8};1,\frac {1}{4};\frac {11}{8};x^2,-x^2\right )}{3 \sqrt [4]{x+x^3}}-\frac {4 x^2 \sqrt [4]{1+x^2} F_1\left (\frac {7}{8};1,\frac {1}{4};\frac {15}{8};x^2,-x^2\right )}{7 \sqrt [4]{x+x^3}}\\ \end {align*}

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Mathematica [F]  time = 0.27, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(-1+x) \sqrt [4]{x+x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.26, size = 61, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {2^{3/4} \sqrt [4]{x+x^3}}{1+x}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {2^{3/4} \sqrt [4]{x+x^3}}{1+x}\right )}{2 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 6.23, size = 332, normalized size = 5.44 \begin {gather*} \frac {1}{4} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} {\left (x^{3} + x\right )}^{\frac {1}{4}} + 16 \cdot 2^{\frac {1}{4}} {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x + 1\right )} + 2^{\frac {3}{4}} {\left (4 \cdot 2^{\frac {3}{4}} \sqrt {x^{3} + x} {\left (x^{2} + 2 \, x + 1\right )} + 2^{\frac {1}{4}} {\left (x^{4} + 12 \, x^{3} + 6 \, x^{2} + 12 \, x + 1\right )}\right )}}{2 \, {\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1\right )}}\right ) - \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (x^{4} + 12 \, x^{3} + 6 \, x^{2} + 12 \, x + 1\right )} + 4 \, \sqrt {2} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} {\left (x^{3} + x\right )}^{\frac {1}{4}} + 8 \cdot 2^{\frac {1}{4}} \sqrt {x^{3} + x} {\left (x^{2} + 2 \, x + 1\right )} + 16 \, {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x + 1\right )}}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) + \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {3}{4}} {\left (x^{4} + 12 \, x^{3} + 6 \, x^{2} + 12 \, x + 1\right )} - 4 \, \sqrt {2} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} {\left (x^{3} + x\right )}^{\frac {1}{4}} + 8 \cdot 2^{\frac {1}{4}} \sqrt {x^{3} + x} {\left (x^{2} + 2 \, x + 1\right )} - 16 \, {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x + 1\right )}}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*2^(3/4)*arctan(1/2*(4*2^(3/4)*(x^3 + 3*x^2 + 3*x + 1)*(x^3 + x)^(1/4) + 16*2^(1/4)*(x^3 + x)^(3/4)*(x + 1)
+ 2^(3/4)*(4*2^(3/4)*sqrt(x^3 + x)*(x^2 + 2*x + 1) + 2^(1/4)*(x^4 + 12*x^3 + 6*x^2 + 12*x + 1)))/(x^4 - 4*x^3
+ 6*x^2 - 4*x + 1)) - 1/16*2^(3/4)*log((2^(3/4)*(x^4 + 12*x^3 + 6*x^2 + 12*x + 1) + 4*sqrt(2)*(x^3 + 3*x^2 +
3*x + 1)*(x^3 + x)^(1/4) + 8*2^(1/4)*sqrt(x^3 + x)*(x^2 + 2*x + 1) + 16*(x^3 + x)^(3/4)*(x + 1))/(x^4 - 4*x^3
+ 6*x^2 - 4*x + 1)) + 1/16*2^(3/4)*log(-(2^(3/4)*(x^4 + 12*x^3 + 6*x^2 + 12*x + 1) - 4*sqrt(2)*(x^3 + 3*x^2 +
3*x + 1)*(x^3 + x)^(1/4) + 8*2^(1/4)*sqrt(x^3 + x)*(x^2 + 2*x + 1) - 16*(x^3 + x)^(3/4)*(x + 1))/(x^4 - 4*x^3
+ 6*x^2 - 4*x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 9.64, size = 507, normalized size = 8.31

method	result	size
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {-2 \sqrt {x^{3}+x}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{2}-4 \sqrt {x^{3}+x}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x -2 \left (x^{3}+x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-2 \sqrt {x^{3}+x}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )-6 \left (x^{3}+x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+16 \left (x^{3}+x \right )^{\frac {3}{4}} x -6 \left (x^{3}+x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x +12 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}+16 \left (x^{3}+x \right )^{\frac {3}{4}}-2 \left (x^{3}+x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2}+6 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{2}+12 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (-1+x \right )^{4}}\right )}{8}-\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {2 \sqrt {x^{3}+x}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+4 \sqrt {x^{3}+x}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x +2 \left (x^{3}+x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+2 \sqrt {x^{3}+x}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3}+6 \left (x^{3}+x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+x^{4} \RootOf \left (\textit {\_Z}^{4}-8\right )+16 \left (x^{3}+x \right )^{\frac {3}{4}} x +6 \left (x^{3}+x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x +12 \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{3}+16 \left (x^{3}+x \right )^{\frac {3}{4}}+2 \left (x^{3}+x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2}+6 \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{2}+12 \RootOf \left (\textit {\_Z}^{4}-8\right ) x +\RootOf \left (\textit {\_Z}^{4}-8\right )}{\left (-1+x \right )^{4}}\right )}{8}\)	\(507\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*RootOf(_Z^2+RootOf(_Z^4-8)^2)*ln((-2*(x^3+x)^(1/2)*RootOf(_Z^4-8)^2*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^2-4*(
x^3+x)^(1/2)*RootOf(_Z^4-8)^2*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x-2*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x^3-2*(x^3+x)^(
1/2)*RootOf(_Z^4-8)^2*RootOf(_Z^2+RootOf(_Z^4-8)^2)-6*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x^2+RootOf(_Z^2+RootOf(_Z
^4-8)^2)*x^4+16*(x^3+x)^(3/4)*x-6*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x+12*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^3+16*(x^
3+x)^(3/4)-2*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2+6*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^2+12*RootOf(_Z^2+RootOf(_Z^4-8)^
2)*x+RootOf(_Z^2+RootOf(_Z^4-8)^2))/(-1+x)^4)-1/8*RootOf(_Z^4-8)*ln((2*(x^3+x)^(1/2)*RootOf(_Z^4-8)^3*x^2+4*(x
^3+x)^(1/2)*RootOf(_Z^4-8)^3*x+2*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x^3+2*(x^3+x)^(1/2)*RootOf(_Z^4-8)^3+6*(x^3+x)
^(1/4)*RootOf(_Z^4-8)^2*x^2+x^4*RootOf(_Z^4-8)+16*(x^3+x)^(3/4)*x+6*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x+12*RootOf
(_Z^4-8)*x^3+16*(x^3+x)^(3/4)+2*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2+6*RootOf(_Z^4-8)*x^2+12*RootOf(_Z^4-8)*x+RootOf
(_Z^4-8))/(-1+x)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x + x^3)^(1/4)*(x - 1)),x)

[Out]

int(1/((x + x^3)^(1/4)*(x - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{x \left (x^{2} + 1\right )} \left (x - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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