3.8.4 \(\int x^6 \sqrt [4]{1+x^4} \, dx\)

Optimal. Leaf size=55 \[ \frac {3}{64} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3}{64} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{32} \sqrt [4]{x^4+1} \left (4 x^7+x^3\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {279, 321, 331, 298, 203, 206} \begin {gather*} \frac {3}{64} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3}{64} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{8} \sqrt [4]{x^4+1} x^7+\frac {1}{32} \sqrt [4]{x^4+1} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6*(1 + x^4)^(1/4),x]

[Out]

(x^3*(1 + x^4)^(1/4))/32 + (x^7*(1 + x^4)^(1/4))/8 + (3*ArcTan[x/(1 + x^4)^(1/4)])/64 - (3*ArcTanh[x/(1 + x^4)
^(1/4)])/64

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 279

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

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 321

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

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]

Rubi steps

\begin {align*} \int x^6 \sqrt [4]{1+x^4} \, dx &=\frac {1}{8} x^7 \sqrt [4]{1+x^4}+\frac {1}{8} \int \frac {x^6}{\left (1+x^4\right )^{3/4}} \, dx\\ &=\frac {1}{32} x^3 \sqrt [4]{1+x^4}+\frac {1}{8} x^7 \sqrt [4]{1+x^4}-\frac {3}{32} \int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx\\ &=\frac {1}{32} x^3 \sqrt [4]{1+x^4}+\frac {1}{8} x^7 \sqrt [4]{1+x^4}-\frac {3}{32} \operatorname {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {1}{32} x^3 \sqrt [4]{1+x^4}+\frac {1}{8} x^7 \sqrt [4]{1+x^4}-\frac {3}{64} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {3}{64} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {1}{32} x^3 \sqrt [4]{1+x^4}+\frac {1}{8} x^7 \sqrt [4]{1+x^4}+\frac {3}{64} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3}{64} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.62 \begin {gather*} \frac {1}{8} x^3 \left (\left (x^4+1\right )^{5/4}-\, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-x^4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6*(1 + x^4)^(1/4),x]

[Out]

(x^3*((1 + x^4)^(5/4) - Hypergeometric2F1[-1/4, 3/4, 7/4, -x^4]))/8

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IntegrateAlgebraic [A]  time = 0.18, size = 55, normalized size = 1.00 \begin {gather*} \frac {1}{32} \sqrt [4]{1+x^4} \left (x^3+4 x^7\right )+\frac {3}{64} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3}{64} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^6*(1 + x^4)^(1/4),x]

[Out]

((1 + x^4)^(1/4)*(x^3 + 4*x^7))/32 + (3*ArcTan[x/(1 + x^4)^(1/4)])/64 - (3*ArcTanh[x/(1 + x^4)^(1/4)])/64

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fricas [A]  time = 0.47, size = 68, normalized size = 1.24 \begin {gather*} \frac {1}{32} \, {\left (4 \, x^{7} + x^{3}\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}} - \frac {3}{64} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{128} \, \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {3}{128} \, \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(x^4+1)^(1/4),x, algorithm="fricas")

[Out]

1/32*(4*x^7 + x^3)*(x^4 + 1)^(1/4) - 3/64*arctan((x^4 + 1)^(1/4)/x) - 3/128*log((x + (x^4 + 1)^(1/4))/x) + 3/1
28*log(-(x - (x^4 + 1)^(1/4))/x)

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giac [A]  time = 0.17, size = 81, normalized size = 1.47 \begin {gather*} \frac {1}{32} \, x^{8} {\left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}} {\left (\frac {1}{x^{4}} + 1\right )}}{x} + \frac {3 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right )} - \frac {3}{64} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{128} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {3}{128} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(x^4+1)^(1/4),x, algorithm="giac")

[Out]

1/32*x^8*((x^4 + 1)^(1/4)*(1/x^4 + 1)/x + 3*(x^4 + 1)^(1/4)/x) - 3/64*arctan((x^4 + 1)^(1/4)/x) - 3/128*log((x
^4 + 1)^(1/4)/x + 1) + 3/128*log((x^4 + 1)^(1/4)/x - 1)

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maple [C]  time = 1.90, size = 17, normalized size = 0.31

method	result	size
meijerg	\(\frac {x^{7} \hypergeom \left (\left [-\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], -x^{4}\right )}{7}\)	\(17\)
risch	\(\frac {x^{3} \left (4 x^{4}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}}}{32}-\frac {x^{3} \hypergeom \left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{4}\right )}{32}\)	\(37\)
trager	\(\frac {x^{3} \left (4 x^{4}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}}}{32}-\frac {3 \ln \left (2 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 x^{2} \sqrt {x^{4}+1}+2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}+2 x^{4}+1\right )}{128}+\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}\, x^{2}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{128}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(x^4+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/7*x^7*hypergeom([-1/4,7/4],[11/4],-x^4)

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maxima [B]  time = 0.42, size = 99, normalized size = 1.80 \begin {gather*} -\frac {\frac {3 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + \frac {{\left (x^{4} + 1\right )}^{\frac {5}{4}}}{x^{5}}}{32 \, {\left (\frac {2 \, {\left (x^{4} + 1\right )}}{x^{4}} - \frac {{\left (x^{4} + 1\right )}^{2}}{x^{8}} - 1\right )}} - \frac {3}{64} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{128} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {3}{128} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(x^4+1)^(1/4),x, algorithm="maxima")

[Out]

-1/32*(3*(x^4 + 1)^(1/4)/x + (x^4 + 1)^(5/4)/x^5)/(2*(x^4 + 1)/x^4 - (x^4 + 1)^2/x^8 - 1) - 3/64*arctan((x^4 +
1)^(1/4)/x) - 3/128*log((x^4 + 1)^(1/4)/x + 1) + 3/128*log((x^4 + 1)^(1/4)/x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^6\,{\left (x^4+1\right )}^{1/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(x^4 + 1)^(1/4),x)

[Out]

int(x^6*(x^4 + 1)^(1/4), x)

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sympy [C]  time = 1.04, size = 31, normalized size = 0.56 \begin {gather*} \frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(x**4+1)**(1/4),x)

[Out]

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

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