3.7.54 \(\int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx\)

Optimal. Leaf size=52 \[ \frac {5}{48} \tan ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {5}{48} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right )+\frac {\left (x^3+1\right )^{3/4} \left (5 x^3-4\right )}{24 x^6} \]

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {266, 51, 63, 298, 203, 206} \begin {gather*} \frac {5 \left (x^3+1\right )^{3/4}}{24 x^3}+\frac {5}{48} \tan ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {5}{48} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {\left (x^3+1\right )^{3/4}}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(1 + x^3)^(1/4)),x]

[Out]

-1/6*(1 + x^3)^(3/4)/x^6 + (5*(1 + x^3)^(3/4))/(24*x^3) + (5*ArcTan[(1 + x^3)^(1/4)])/48 - (5*ArcTanh[(1 + x^3
)^(1/4)])/48

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 266

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

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]

Rubi steps

\begin {align*} \int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt [4]{1+x}} \, dx,x,x^3\right )\\ &=-\frac {\left (1+x^3\right )^{3/4}}{6 x^6}-\frac {5}{24} \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{1+x}} \, dx,x,x^3\right )\\ &=-\frac {\left (1+x^3\right )^{3/4}}{6 x^6}+\frac {5 \left (1+x^3\right )^{3/4}}{24 x^3}+\frac {5}{96} \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{1+x}} \, dx,x,x^3\right )\\ &=-\frac {\left (1+x^3\right )^{3/4}}{6 x^6}+\frac {5 \left (1+x^3\right )^{3/4}}{24 x^3}+\frac {5}{24} \operatorname {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt [4]{1+x^3}\right )\\ &=-\frac {\left (1+x^3\right )^{3/4}}{6 x^6}+\frac {5 \left (1+x^3\right )^{3/4}}{24 x^3}-\frac {5}{48} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^3}\right )+\frac {5}{48} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^3}\right )\\ &=-\frac {\left (1+x^3\right )^{3/4}}{6 x^6}+\frac {5 \left (1+x^3\right )^{3/4}}{24 x^3}+\frac {5}{48} \tan ^{-1}\left (\sqrt [4]{1+x^3}\right )-\frac {5}{48} \tanh ^{-1}\left (\sqrt [4]{1+x^3}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 26, normalized size = 0.50 \begin {gather*} -\frac {4}{9} \left (x^3+1\right )^{3/4} \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};x^3+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(1 + x^3)^(1/4)),x]

[Out]

(-4*(1 + x^3)^(3/4)*Hypergeometric2F1[3/4, 3, 7/4, 1 + x^3])/9

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IntegrateAlgebraic [A]  time = 0.06, size = 52, normalized size = 1.00 \begin {gather*} \frac {\left (1+x^3\right )^{3/4} \left (-4+5 x^3\right )}{24 x^6}+\frac {5}{48} \tan ^{-1}\left (\sqrt [4]{1+x^3}\right )-\frac {5}{48} \tanh ^{-1}\left (\sqrt [4]{1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + x^3)^(3/4)*(-4 + 5*x^3))/(24*x^6) + (5*ArcTan[(1 + x^3)^(1/4)])/48 - (5*ArcTanh[(1 + x^3)^(1/4)])/48

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fricas [A]  time = 0.46, size = 65, normalized size = 1.25 \begin {gather*} \frac {10 \, x^{6} \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - 5 \, x^{6} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + 5 \, x^{6} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) + 4 \, {\left (5 \, x^{3} - 4\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{96 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^3+1)^(1/4),x, algorithm="fricas")

[Out]

1/96*(10*x^6*arctan((x^3 + 1)^(1/4)) - 5*x^6*log((x^3 + 1)^(1/4) + 1) + 5*x^6*log((x^3 + 1)^(1/4) - 1) + 4*(5*
x^3 - 4)*(x^3 + 1)^(3/4))/x^6

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giac [A]  time = 0.22, size = 60, normalized size = 1.15 \begin {gather*} \frac {5 \, {\left (x^{3} + 1\right )}^{\frac {7}{4}} - 9 \, {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{24 \, x^{6}} + \frac {5}{48} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {5}{96} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {5}{96} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^3+1)^(1/4),x, algorithm="giac")

[Out]

1/24*(5*(x^3 + 1)^(7/4) - 9*(x^3 + 1)^(3/4))/x^6 + 5/48*arctan((x^3 + 1)^(1/4)) - 5/96*log((x^3 + 1)^(1/4) + 1
) + 5/96*log(abs((x^3 + 1)^(1/4) - 1))

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maple [C]  time = 1.90, size = 82, normalized size = 1.58

method	result	size
risch	\(\frac {5 x^{6}+x^{3}-4}{24 x^{6} \left (x^{3}+1\right )^{\frac {1}{4}}}+\frac {5 \sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {\pi \sqrt {2}\, x^{3} \hypergeom \left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], -x^{3}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \relax (2)-\frac {\pi }{2}+3 \ln \relax (x )\right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{192 \pi }\)	\(82\)
meijerg	\(\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {15 \pi \sqrt {2}\, x^{3} \hypergeom \left (\left [1, 1, \frac {13}{4}\right ], \left [2, 4\right ], -x^{3}\right )}{128 \Gamma \left (\frac {3}{4}\right )}+\frac {5 \left (\frac {33}{10}-3 \ln \relax (2)-\frac {\pi }{2}+3 \ln \relax (x )\right ) \pi \sqrt {2}}{32 \Gamma \left (\frac {3}{4}\right )}-\frac {\pi \sqrt {2}}{2 \Gamma \left (\frac {3}{4}\right ) x^{6}}+\frac {\pi \sqrt {2}}{4 \Gamma \left (\frac {3}{4}\right ) x^{3}}\right )}{6 \pi }\)	\(87\)
trager	\(\frac {\left (x^{3}+1\right )^{\frac {3}{4}} \left (5 x^{3}-4\right )}{24 x^{6}}-\frac {5 \ln \left (\frac {2 \left (x^{3}+1\right )^{\frac {3}{4}}+x^{3}+2 \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {1}{4}}+2}{x^{3}}\right )}{96}-\frac {5 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{3}+1\right )^{\frac {3}{4}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}+1}-2 \left (x^{3}+1\right )^{\frac {1}{4}}+2 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x^{3}}\right )}{96}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^3+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/24*(5*x^6+x^3-4)/x^6/(x^3+1)^(1/4)+5/192/Pi*2^(1/2)*GAMMA(3/4)*(-1/4*Pi*2^(1/2)/GAMMA(3/4)*x^3*hypergeom([1,
1,5/4],[2,2],-x^3)+(-3*ln(2)-1/2*Pi+3*ln(x))*Pi*2^(1/2)/GAMMA(3/4))

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maxima [A]  time = 0.42, size = 74, normalized size = 1.42 \begin {gather*} -\frac {5 \, {\left (x^{3} + 1\right )}^{\frac {7}{4}} - 9 \, {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{24 \, {\left (2 \, x^{3} - {\left (x^{3} + 1\right )}^{2} + 1\right )}} + \frac {5}{48} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {5}{96} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {5}{96} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^3+1)^(1/4),x, algorithm="maxima")

[Out]

-1/24*(5*(x^3 + 1)^(7/4) - 9*(x^3 + 1)^(3/4))/(2*x^3 - (x^3 + 1)^2 + 1) + 5/48*arctan((x^3 + 1)^(1/4)) - 5/96*
log((x^3 + 1)^(1/4) + 1) + 5/96*log((x^3 + 1)^(1/4) - 1)

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mupad [B]  time = 0.79, size = 45, normalized size = 0.87 \begin {gather*} \frac {5\,\mathrm {atan}\left ({\left (x^3+1\right )}^{1/4}\right )}{48}-\frac {5\,\mathrm {atanh}\left ({\left (x^3+1\right )}^{1/4}\right )}{48}-\frac {3\,{\left (x^3+1\right )}^{3/4}}{8\,x^6}+\frac {5\,{\left (x^3+1\right )}^{7/4}}{24\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(x^3 + 1)^(1/4)),x)

[Out]

(5*atan((x^3 + 1)^(1/4)))/48 - (5*atanh((x^3 + 1)^(1/4)))/48 - (3*(x^3 + 1)^(3/4))/(8*x^6) + (5*(x^3 + 1)^(7/4
))/(24*x^6)

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sympy [C]  time = 1.16, size = 32, normalized size = 0.62 \begin {gather*} - \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {27}{4}} \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(x**3+1)**(1/4),x)

[Out]

-gamma(9/4)*hyper((1/4, 9/4), (13/4,), exp_polar(I*pi)/x**3)/(3*x**(27/4)*gamma(13/4))

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