3.3.86 \(\int \frac {(-4+x^5) (1+x^4+x^5)}{x^6 (1+x^5)^{3/4}} \, dx\)

Optimal. Leaf size=26 \[ \frac {4 \sqrt [4]{x^5+1} \left (x^5+5 x^4+1\right )}{5 x^5} \]

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Rubi [A]  time = 0.10, antiderivative size = 44, normalized size of antiderivative = 1.69, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1835, 1585, 12, 261} \begin {gather*} \frac {4 \sqrt [4]{x^5+1}}{x}+\frac {4 \sqrt [4]{x^5+1}}{5 x^5}+\frac {4}{5} \sqrt [4]{x^5+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-4 + x^5)*(1 + x^4 + x^5))/(x^6*(1 + x^5)^(3/4)),x]

[Out]

(4*(1 + x^5)^(1/4))/5 + (4*(1 + x^5)^(1/4))/(5*x^5) + (4*(1 + x^5)^(1/4))/x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

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

Rule 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rule 1835

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{Pq0 = Coeff[Pq, x, 0]}, Simp[(Pq
0*(c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(2*a*c*(m + 1)), Int[(c*x)^(m + 1)*ExpandToSum
[(2*a*(m + 1)*(Pq - Pq0))/x - 2*b*Pq0*(m + n*(p + 1) + 1)*x^(n - 1), x]*(a + b*x^n)^p, x], x] /; NeQ[Pq0, 0]]
/; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[m, -1] && LeQ[n - 1, Expon[Pq, x]]

Rubi steps

\begin {align*} \int \frac {\left (-4+x^5\right ) \left (1+x^4+x^5\right )}{x^6 \left (1+x^5\right )^{3/4}} \, dx &=\frac {4 \sqrt [4]{1+x^5}}{5 x^5}-\frac {1}{10} \int \frac {40 x^3-10 x^8-10 x^9}{x^5 \left (1+x^5\right )^{3/4}} \, dx\\ &=\frac {4 \sqrt [4]{1+x^5}}{5 x^5}-\frac {1}{10} \int \frac {40-10 x^5-10 x^6}{x^2 \left (1+x^5\right )^{3/4}} \, dx\\ &=\frac {4 \sqrt [4]{1+x^5}}{5 x^5}+\frac {4 \sqrt [4]{1+x^5}}{x}+\frac {1}{20} \int \frac {20 x^4}{\left (1+x^5\right )^{3/4}} \, dx\\ &=\frac {4 \sqrt [4]{1+x^5}}{5 x^5}+\frac {4 \sqrt [4]{1+x^5}}{x}+\int \frac {x^4}{\left (1+x^5\right )^{3/4}} \, dx\\ &=\frac {4}{5} \sqrt [4]{1+x^5}+\frac {4 \sqrt [4]{1+x^5}}{5 x^5}+\frac {4 \sqrt [4]{1+x^5}}{x}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 107, normalized size = 4.12 \begin {gather*} -\frac {16}{5} \sqrt [4]{x^5+1} \, _2F_1\left (\frac {1}{4},2;\frac {5}{4};x^5+1\right )+\frac {4 \, _2F_1\left (-\frac {1}{5},\frac {3}{4};\frac {4}{5};-x^5\right )}{x}+\frac {1}{4} x^4 \, _2F_1\left (\frac {3}{4},\frac {4}{5};\frac {9}{5};-x^5\right )+\frac {4}{5} \sqrt [4]{x^5+1}+\frac {6}{5} \left (\tan ^{-1}\left (\sqrt [4]{x^5+1}\right )+\tanh ^{-1}\left (\sqrt [4]{x^5+1}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-4 + x^5)*(1 + x^4 + x^5))/(x^6*(1 + x^5)^(3/4)),x]

[Out]

(4*(1 + x^5)^(1/4))/5 + (6*(ArcTan[(1 + x^5)^(1/4)] + ArcTanh[(1 + x^5)^(1/4)]))/5 + (4*Hypergeometric2F1[-1/5
, 3/4, 4/5, -x^5])/x - (16*(1 + x^5)^(1/4)*Hypergeometric2F1[1/4, 2, 5/4, 1 + x^5])/5 + (x^4*Hypergeometric2F1
[3/4, 4/5, 9/5, -x^5])/4

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IntegrateAlgebraic [A]  time = 1.88, size = 26, normalized size = 1.00 \begin {gather*} \frac {4 \sqrt [4]{1+x^5} \left (1+5 x^4+x^5\right )}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-4 + x^5)*(1 + x^4 + x^5))/(x^6*(1 + x^5)^(3/4)),x]

[Out]

(4*(1 + x^5)^(1/4)*(1 + 5*x^4 + x^5))/(5*x^5)

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fricas [A]  time = 0.47, size = 22, normalized size = 0.85 \begin {gather*} \frac {4 \, {\left (x^{5} + 5 \, x^{4} + 1\right )} {\left (x^{5} + 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-4)*(x^5+x^4+1)/x^6/(x^5+1)^(3/4),x, algorithm="fricas")

[Out]

4/5*(x^5 + 5*x^4 + 1)*(x^5 + 1)^(1/4)/x^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{5} + x^{4} + 1\right )} {\left (x^{5} - 4\right )}}{{\left (x^{5} + 1\right )}^{\frac {3}{4}} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-4)*(x^5+x^4+1)/x^6/(x^5+1)^(3/4),x, algorithm="giac")

[Out]

integrate((x^5 + x^4 + 1)*(x^5 - 4)/((x^5 + 1)^(3/4)*x^6), x)

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maple [A]  time = 0.12, size = 23, normalized size = 0.88

method	result	size
trager	\(\frac {4 \left (x^{5}+1\right )^{\frac {1}{4}} \left (x^{5}+5 x^{4}+1\right )}{5 x^{5}}\)	\(23\)
risch	\(\frac {\frac {8}{5} x^{5}+\frac {4}{5}+4 x^{9}+4 x^{4}+\frac {4}{5} x^{10}}{\left (x^{5}+1\right )^{\frac {3}{4}} x^{5}}\)	\(33\)
gosper	\(\frac {4 \left (x^{5}+5 x^{4}+1\right ) \left (1+x \right ) \left (x^{4}-x^{3}+x^{2}-x +1\right )}{5 \left (x^{5}+1\right )^{\frac {3}{4}} x^{5}}\)	\(42\)
meijerg	\(\frac {\hypergeom \left (\left [\frac {3}{4}, 1\right ], \relax [2], -x^{5}\right ) x^{5}}{5}+\frac {\hypergeom \left (\left [\frac {3}{4}, \frac {4}{5}\right ], \left [\frac {9}{5}\right ], -x^{5}\right ) x^{4}}{4}-\frac {3 \left (\left (-3 \ln \relax (2)+\frac {\pi }{2}+5 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )-\frac {3 \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{5}\right ) \Gamma \left (\frac {3}{4}\right ) x^{5}}{4}\right )}{5 \Gamma \left (\frac {3}{4}\right )}-\frac {4 \left (-\frac {\Gamma \left (\frac {3}{4}\right )}{x^{5}}-\frac {3 \left (\frac {1}{3}-3 \ln \relax (2)+\frac {\pi }{2}+5 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )}{4}+\frac {21 \hypergeom \left (\left [1, 1, \frac {11}{4}\right ], \left [2, 3\right ], -x^{5}\right ) \Gamma \left (\frac {3}{4}\right ) x^{5}}{32}\right )}{5 \Gamma \left (\frac {3}{4}\right )}+\frac {4 \hypergeom \left (\left [-\frac {1}{5}, \frac {3}{4}\right ], \left [\frac {4}{5}\right ], -x^{5}\right )}{x}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5-4)*(x^5+x^4+1)/x^6/(x^5+1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

4/5*(x^5+1)^(1/4)*(x^5+5*x^4+1)/x^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4 \, {\left (x^{5} + 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} - \frac {6}{5} \, \arctan \left ({\left (x^{5} + 1\right )}^{\frac {1}{4}}\right ) + \int \frac {{\left (x^{6} + x^{5} - 3 \, x - 4\right )} {\left (x^{4} - x^{3} + x^{2} - x + 1\right )}^{\frac {1}{4}} {\left (x + 1\right )}^{\frac {1}{4}}}{x^{7} + x^{2}}\,{d x} - \frac {3}{5} \, \log \left ({\left (x^{5} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {3}{5} \, \log \left ({\left (x^{5} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-4)*(x^5+x^4+1)/x^6/(x^5+1)^(3/4),x, algorithm="maxima")

[Out]

4/5*(x^5 + 1)^(1/4)/x^5 - 6/5*arctan((x^5 + 1)^(1/4)) + integrate((x^6 + x^5 - 3*x - 4)*(x^4 - x^3 + x^2 - x +
1)^(1/4)*(x + 1)^(1/4)/(x^7 + x^2), x) - 3/5*log((x^5 + 1)^(1/4) + 1) + 3/5*log((x^5 + 1)^(1/4) - 1)

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mupad [B]  time = 0.18, size = 27, normalized size = 1.04 \begin {gather*} \frac {4\,{\left (x^5+1\right )}^{5/4}+20\,x^4\,{\left (x^5+1\right )}^{1/4}}{5\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^5 - 4)*(x^4 + x^5 + 1))/(x^6*(x^5 + 1)^(3/4)),x)

[Out]

(4*(x^5 + 1)^(5/4) + 20*x^4*(x^5 + 1)^(1/4))/(5*x^5)

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sympy [C]  time = 4.41, size = 143, normalized size = 5.50 \begin {gather*} \frac {x^{4} \Gamma \left (\frac {4}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {4}{5} \\ \frac {9}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 \Gamma \left (\frac {9}{5}\right )} + \frac {4 \sqrt [4]{x^{5} + 1}}{5} - \frac {4 \Gamma \left (- \frac {1}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{5}, \frac {3}{4} \\ \frac {4}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x \Gamma \left (\frac {4}{5}\right )} + \frac {3 \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{5}}} \right )}}{5 x^{\frac {15}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {4 \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{5}}} \right )}}{5 x^{\frac {35}{4}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**5-4)*(x**5+x**4+1)/x**6/(x**5+1)**(3/4),x)

[Out]

x**4*gamma(4/5)*hyper((3/4, 4/5), (9/5,), x**5*exp_polar(I*pi))/(5*gamma(9/5)) + 4*(x**5 + 1)**(1/4)/5 - 4*gam
ma(-1/5)*hyper((-1/5, 3/4), (4/5,), x**5*exp_polar(I*pi))/(5*x*gamma(4/5)) + 3*gamma(3/4)*hyper((3/4, 3/4), (7
/4,), exp_polar(I*pi)/x**5)/(5*x**(15/4)*gamma(7/4)) + 4*gamma(7/4)*hyper((3/4, 7/4), (11/4,), exp_polar(I*pi)
/x**5)/(5*x**(35/4)*gamma(11/4))

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