3.5.45 \(\int \frac {1+x^6}{x^6 (1+x^3) \sqrt [4]{x+x^4}} \, dx\)

Optimal. Leaf size=35 \[ \frac {4 \left (x^4+x\right )^{3/4} \left (53 x^6+8 x^3-3\right )}{63 x^6 \left (x^3+1\right )} \]

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Rubi [A]  time = 0.17, antiderivative size = 47, normalized size of antiderivative = 1.34, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2056, 1487, 453, 271, 264} \begin {gather*} \frac {212 x}{63 \sqrt [4]{x^4+x}}-\frac {4}{21 \sqrt [4]{x^4+x} x^5}+\frac {32}{63 \sqrt [4]{x^4+x} x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^6)/(x^6*(1 + x^3)*(x + x^4)^(1/4)),x]

[Out]

-4/(21*x^5*(x + x^4)^(1/4)) + 32/(63*x^2*(x + x^4)^(1/4)) + (212*x)/(63*(x + x^4)^(1/4))

Rule 264

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

Rule 271

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

Rule 453

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

Rule 1487

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(c^p
*(f*x)^(m + 2*n*p - n + 1)*(d + e*x^n)^(q + 1))/(e*f^(2*n*p - n + 1)*(m + 2*n*p + n*q + 1)), x] + Dist[1/(e*(m
+ 2*n*p + n*q + 1)), Int[(f*x)^m*(d + e*x^n)^q*ExpandToSum[e*(m + 2*n*p + n*q + 1)*((a + c*x^(2*n))^p - c^p*x
^(2*n*p)) - d*c^p*(m + 2*n*p - n + 1)*x^(2*n*p - n), x], x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2,
2*n] && IGtQ[n, 0] && IGtQ[p, 0] && GtQ[2*n*p, n - 1] &&  !IntegerQ[q] && NeQ[m + 2*n*p + n*q + 1, 0]

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^6}{x^6 \left (1+x^3\right ) \sqrt [4]{x+x^4}} \, dx &=\frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^3}\right ) \int \frac {1+x^6}{x^{25/4} \left (1+x^3\right )^{5/4}} \, dx}{\sqrt [4]{x+x^4}}\\ &=-\frac {1}{3 x^2 \sqrt [4]{x+x^4}}-\frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^3}\right ) \int \frac {-3+\frac {9 x^3}{4}}{x^{25/4} \left (1+x^3\right )^{5/4}} \, dx}{3 \sqrt [4]{x+x^4}}\\ &=-\frac {4}{21 x^5 \sqrt [4]{x+x^4}}-\frac {1}{3 x^2 \sqrt [4]{x+x^4}}-\frac {\left (53 \sqrt [4]{x} \sqrt [4]{1+x^3}\right ) \int \frac {1}{x^{13/4} \left (1+x^3\right )^{5/4}} \, dx}{28 \sqrt [4]{x+x^4}}\\ &=-\frac {4}{21 x^5 \sqrt [4]{x+x^4}}+\frac {32}{63 x^2 \sqrt [4]{x+x^4}}+\frac {\left (53 \sqrt [4]{x} \sqrt [4]{1+x^3}\right ) \int \frac {1}{\sqrt [4]{x} \left (1+x^3\right )^{5/4}} \, dx}{21 \sqrt [4]{x+x^4}}\\ &=-\frac {4}{21 x^5 \sqrt [4]{x+x^4}}+\frac {32}{63 x^2 \sqrt [4]{x+x^4}}+\frac {212 x}{63 \sqrt [4]{x+x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 0.80 \begin {gather*} \frac {4 \left (53 x^6+8 x^3-3\right )}{63 x^5 \sqrt [4]{x^4+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^6)/(x^6*(1 + x^3)*(x + x^4)^(1/4)),x]

[Out]

(4*(-3 + 8*x^3 + 53*x^6))/(63*x^5*(x + x^4)^(1/4))

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IntegrateAlgebraic [A]  time = 0.42, size = 35, normalized size = 1.00 \begin {gather*} \frac {4 \left (x+x^4\right )^{3/4} \left (-3+8 x^3+53 x^6\right )}{63 x^6 \left (1+x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^6)/(x^6*(1 + x^3)*(x + x^4)^(1/4)),x]

[Out]

(4*(x + x^4)^(3/4)*(-3 + 8*x^3 + 53*x^6))/(63*x^6*(1 + x^3))

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fricas [A]  time = 0.68, size = 30, normalized size = 0.86 \begin {gather*} \frac {4 \, {\left (53 \, x^{6} + 8 \, x^{3} - 3\right )} {\left (x^{4} + x\right )}^{\frac {3}{4}}}{63 \, {\left (x^{9} + x^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6+1)/x^6/(x^3+1)/(x^4+x)^(1/4),x, algorithm="fricas")

[Out]

4/63*(53*x^6 + 8*x^3 - 3)*(x^4 + x)^(3/4)/(x^9 + x^6)

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giac [A]  time = 0.28, size = 28, normalized size = 0.80 \begin {gather*} -\frac {4}{21} \, {\left (\frac {1}{x^{3}} + 1\right )}^{\frac {7}{4}} + \frac {8}{9} \, {\left (\frac {1}{x^{3}} + 1\right )}^{\frac {3}{4}} + \frac {8}{3 \, {\left (\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6+1)/x^6/(x^3+1)/(x^4+x)^(1/4),x, algorithm="giac")

[Out]

-4/21*(1/x^3 + 1)^(7/4) + 8/9*(1/x^3 + 1)^(3/4) + 8/3/(1/x^3 + 1)^(1/4)

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maple [A]  time = 0.09, size = 25, normalized size = 0.71

method	result	size
gosper	\(\frac {\frac {212}{63} x^{6}+\frac {32}{63} x^{3}-\frac {4}{21}}{\left (x^{4}+x \right )^{\frac {1}{4}} x^{5}}\)	\(25\)
risch	\(\frac {\frac {212}{63} x^{6}+\frac {32}{63} x^{3}-\frac {4}{21}}{x^{5} \left (x \left (x^{3}+1\right )\right )^{\frac {1}{4}}}\)	\(27\)
trager	\(\frac {4 \left (x^{4}+x \right )^{\frac {3}{4}} \left (53 x^{6}+8 x^{3}-3\right )}{63 x^{6} \left (x^{3}+1\right )}\)	\(32\)
meijerg	\(-\frac {4 \left (-32 x^{6}-8 x^{3}+3\right ) \left (x^{3}+1\right )^{\frac {3}{4}}}{21 \left (3 x^{3}+3\right ) x^{\frac {21}{4}}}+\frac {4 x^{\frac {3}{4}}}{3 \left (x^{3}+1\right )^{\frac {1}{4}}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6+1)/x^6/(x^3+1)/(x^4+x)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/63*(53*x^6+8*x^3-3)/(x^4+x)^(1/4)/x^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6} + 1}{{\left (x^{4} + x\right )}^{\frac {1}{4}} {\left (x^{3} + 1\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6+1)/x^6/(x^3+1)/(x^4+x)^(1/4),x, algorithm="maxima")

[Out]

integrate((x^6 + 1)/((x^4 + x)^(1/4)*(x^3 + 1)*x^6), x)

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mupad [B]  time = 0.30, size = 31, normalized size = 0.89 \begin {gather*} \frac {4\,{\left (x^4+x\right )}^{3/4}\,\left (53\,x^6+8\,x^3-3\right )}{63\,x^6\,\left (x^3+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6 + 1)/(x^6*(x^3 + 1)*(x + x^4)^(1/4)),x)

[Out]

(4*(x + x^4)^(3/4)*(8*x^3 + 53*x^6 - 3))/(63*x^6*(x^3 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} + 1\right ) \left (x^{4} - x^{2} + 1\right )}{x^{6} \sqrt [4]{x \left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**6+1)/x**6/(x**3+1)/(x**4+x)**(1/4),x)

[Out]

Integral((x**2 + 1)*(x**4 - x**2 + 1)/(x**6*(x*(x + 1)*(x**2 - x + 1))**(1/4)*(x + 1)*(x**2 - x + 1)), x)

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