3.7.55 \(\int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx\) [655]

Optimal. Leaf size=52 \[ \frac {\left (-4-x^3\right ) \sqrt [4]{1+x^3}}{24 x^6}+\frac {1}{16} \text {ArcTan}\left (\sqrt [4]{1+x^3}\right )+\frac {1}{16} \tanh ^{-1}\left (\sqrt [4]{1+x^3}\right ) \]

[Out]

1/24*(-x^3-4)*(x^3+1)^(1/4)/x^6+1/16*arctan((x^3+1)^(1/4))+1/16*arctanh((x^3+1)^(1/4))

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Rubi [A]
time = 0.01, antiderivative size = 61, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {272, 43, 44, 65, 218, 212, 209} \begin {gather*} \frac {1}{16} \text {ArcTan}\left (\sqrt [4]{x^3+1}\right )-\frac {\sqrt [4]{x^3+1}}{24 x^3}+\frac {1}{16} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {\sqrt [4]{x^3+1}}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)^(1/4)/x^7,x]

[Out]

-1/6*(1 + x^3)^(1/4)/x^6 - (1 + x^3)^(1/4)/(24*x^3) + ArcTan[(1 + x^3)^(1/4)]/16 + ArcTanh[(1 + x^3)^(1/4)]/16

Rule 43

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 272

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]]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {\sqrt [4]{1+x}}{x^3} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{6 x^6}+\frac {1}{24} \text {Subst}\left (\int \frac {1}{x^2 (1+x)^{3/4}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{6 x^6}-\frac {\sqrt [4]{1+x^3}}{24 x^3}-\frac {1}{32} \text {Subst}\left (\int \frac {1}{x (1+x)^{3/4}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{6 x^6}-\frac {\sqrt [4]{1+x^3}}{24 x^3}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{1+x^3}\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{6 x^6}-\frac {\sqrt [4]{1+x^3}}{24 x^3}+\frac {1}{16} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^3}\right )+\frac {1}{16} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^3}\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{6 x^6}-\frac {\sqrt [4]{1+x^3}}{24 x^3}+\frac {1}{16} \tan ^{-1}\left (\sqrt [4]{1+x^3}\right )+\frac {1}{16} \tanh ^{-1}\left (\sqrt [4]{1+x^3}\right )\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 48, normalized size = 0.92 \begin {gather*} \frac {1}{48} \left (-\frac {2 \sqrt [4]{1+x^3} \left (4+x^3\right )}{x^6}+3 \text {ArcTan}\left (\sqrt [4]{1+x^3}\right )+3 \tanh ^{-1}\left (\sqrt [4]{1+x^3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)^(1/4)/x^7,x]

[Out]

((-2*(1 + x^3)^(1/4)*(4 + x^3))/x^6 + 3*ArcTan[(1 + x^3)^(1/4)] + 3*ArcTanh[(1 + x^3)^(1/4)])/48

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Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 3.
time = 1.64, size = 58, normalized size = 1.12

method	result	size
meijerg	\(-\frac {-\frac {7 \Gamma \left (\frac {3}{4}\right ) x^{3} \hypergeom \left (\left [1, 1, \frac {11}{4}\right ], \left [2, 4\right ], -x^{3}\right )}{32}+\frac {3 \left (-\frac {1}{6}-3 \ln \left (2\right )+\frac {\pi }{2}+3 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{8}+\frac {2 \Gamma \left (\frac {3}{4}\right )}{x^{6}}+\frac {\Gamma \left (\frac {3}{4}\right )}{x^{3}}}{12 \Gamma \left (\frac {3}{4}\right )}\)	\(58\)
risch	\(-\frac {x^{6}+5 x^{3}+4}{24 x^{6} \left (x^{3}+1\right )^{\frac {3}{4}}}-\frac {-\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{3} \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{3}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+3 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{32 \Gamma \left (\frac {3}{4}\right )}\)	\(66\)
trager	\(-\frac {\left (x^{3}+4\right ) \left (x^{3}+1\right )^{\frac {1}{4}}}{24 x^{6}}-\frac {\ln \left (\frac {2 \left (x^{3}+1\right )^{\frac {3}{4}}-x^{3}-2 \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {1}{4}}-2}{x^{3}}\right )}{32}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{3}+1\right )^{\frac {3}{4}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}+1}-2 \left (x^{3}+1\right )^{\frac {1}{4}}+2 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x^{3}}\right )}{32}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/GAMMA(3/4)*(-7/32*GAMMA(3/4)*x^3*hypergeom([1,1,11/4],[2,4],-x^3)+3/8*(-1/6-3*ln(2)+1/2*Pi+3*ln(x))*GAMM
A(3/4)+2*GAMMA(3/4)/x^6+GAMMA(3/4)/x^3)

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Maxima [A]
time = 0.47, size = 72, normalized size = 1.38 \begin {gather*} \frac {{\left (x^{3} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{24 \, {\left (2 \, x^{3} - {\left (x^{3} + 1\right )}^{2} + 1\right )}} + \frac {1}{16} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{32} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{32} \, \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((x^3+1)^(1/4)/x^7,x, algorithm="maxima")

[Out]

1/24*((x^3 + 1)^(5/4) + 3*(x^3 + 1)^(1/4))/(2*x^3 - (x^3 + 1)^2 + 1) + 1/16*arctan((x^3 + 1)^(1/4)) + 1/32*log
((x^3 + 1)^(1/4) + 1) - 1/32*log((x^3 + 1)^(1/4) - 1)

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Fricas [A]
time = 0.38, size = 63, normalized size = 1.21 \begin {gather*} \frac {6 \, x^{6} \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + 3 \, x^{6} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - 3 \, x^{6} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) - 4 \, {\left (x^{3} + 4\right )} {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{96 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.11, size = 34, normalized size = 0.65 \begin {gather*} - \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {21}{4}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(7/4)*hyper((-1/4, 7/4), (11/4,), exp_polar(I*pi)/x**3)/(3*x**(21/4)*gamma(11/4))

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Giac [A]
time = 0.38, size = 58, normalized size = 1.12 \begin {gather*} -\frac {{\left (x^{3} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{24 \, x^{6}} + \frac {1}{16} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{32} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{32} \, \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((x^3+1)^(1/4)/x^7,x, algorithm="giac")

[Out]

-1/24*((x^3 + 1)^(5/4) + 3*(x^3 + 1)^(1/4))/x^6 + 1/16*arctan((x^3 + 1)^(1/4)) + 1/32*log((x^3 + 1)^(1/4) + 1)
- 1/32*log(abs((x^3 + 1)^(1/4) - 1))

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Mupad [B]
time = 0.69, size = 45, normalized size = 0.87 \begin {gather*} \frac {\mathrm {atan}\left ({\left (x^3+1\right )}^{1/4}\right )}{16}+\frac {\mathrm {atanh}\left ({\left (x^3+1\right )}^{1/4}\right )}{16}-\frac {{\left (x^3+1\right )}^{1/4}}{8\,x^6}-\frac {{\left (x^3+1\right )}^{5/4}}{24\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 + 1)^(1/4)/x^7,x)

[Out]

atan((x^3 + 1)^(1/4))/16 + atanh((x^3 + 1)^(1/4))/16 - (x^3 + 1)^(1/4)/(8*x^6) - (x^3 + 1)^(5/4)/(24*x^6)

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