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

Optimal. Leaf size=28 \[ -\frac {4 \sqrt [4]{x^3+1} \left (5 x^4-x^3-1\right )}{5 x^5} \]

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Rubi [A]  time = 0.12, antiderivative size = 47, normalized size of antiderivative = 1.68, number of steps used = 5, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1835, 1586, 449} \begin {gather*} -\frac {4 \sqrt [4]{x^3+1}}{x}+\frac {4 \sqrt [4]{x^3+1}}{5 x^5}+\frac {4 \sqrt [4]{x^3+1}}{5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(1 + x^3)^(1/4))/(5*x^5) + (4*(1 + x^3)^(1/4))/(5*x^2) - (4*(1 + x^3)^(1/4))/x

Rule 449

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] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[m, -1]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx &=\frac {4 \sqrt [4]{1+x^3}}{5 x^5}-\frac {1}{10} \int \frac {16 x^2-40 x^3+10 x^5-10 x^6}{x^5 \left (1+x^3\right )^{3/4}} \, dx\\ &=\frac {4 \sqrt [4]{1+x^3}}{5 x^5}-\frac {1}{10} \int \frac {16 x-40 x^2+10 x^4-10 x^5}{x^4 \left (1+x^3\right )^{3/4}} \, dx\\ &=\frac {4 \sqrt [4]{1+x^3}}{5 x^5}-\frac {1}{10} \int \frac {16-40 x+10 x^3-10 x^4}{x^3 \left (1+x^3\right )^{3/4}} \, dx\\ &=\frac {4 \sqrt [4]{1+x^3}}{5 x^5}+\frac {4 \sqrt [4]{1+x^3}}{5 x^2}+\frac {1}{40} \int \frac {160+40 x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx\\ &=\frac {4 \sqrt [4]{1+x^3}}{5 x^5}+\frac {4 \sqrt [4]{1+x^3}}{5 x^2}-\frac {4 \sqrt [4]{1+x^3}}{x}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 105, normalized size = 3.75 \begin {gather*} -x \, _2F_1\left (\frac {1}{3},\frac {3}{4};\frac {4}{3};-x^3\right )-\frac {4 \, _2F_1\left (-\frac {1}{3},\frac {3}{4};\frac {2}{3};-x^3\right )}{x}+\frac {4 \, _2F_1\left (-\frac {5}{3},\frac {3}{4};-\frac {2}{3};-x^3\right )}{5 x^5}+\frac {1}{2} x^2 \, _2F_1\left (\frac {2}{3},\frac {3}{4};\frac {5}{3};-x^3\right )+\frac {5 \, _2F_1\left (-\frac {2}{3},\frac {3}{4};\frac {1}{3};-x^3\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Hypergeometric2F1[-5/3, 3/4, -2/3, -x^3])/(5*x^5) + (5*Hypergeometric2F1[-2/3, 3/4, 1/3, -x^3])/(2*x^2) - (
4*Hypergeometric2F1[-1/3, 3/4, 2/3, -x^3])/x - x*Hypergeometric2F1[1/3, 3/4, 4/3, -x^3] + (x^2*Hypergeometric2
F1[2/3, 3/4, 5/3, -x^3])/2

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IntegrateAlgebraic [A]  time = 3.90, size = 28, normalized size = 1.00 \begin {gather*} -\frac {4 \sqrt [4]{1+x^3} \left (-1-x^3+5 x^4\right )}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 24, normalized size = 0.86 \begin {gather*} -\frac {4 \, {\left (5 \, x^{4} - x^{3} - 1\right )} {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5*(5*x^4 - x^3 - 1)*(x^3 + 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^{4} - x^{3} - 1\right )} {\left (x^{3} + 4\right )}}{{\left (x^{3} + 1\right )}^{\frac {3}{4}} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 25, normalized size = 0.89

method	result	size
trager	\(-\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}} \left (5 x^{4}-x^{3}-1\right )}{5 x^{5}}\)	\(25\)
risch	\(-\frac {4 \left (5 x^{7}-x^{6}+5 x^{4}-2 x^{3}-1\right )}{5 \left (x^{3}+1\right )^{\frac {3}{4}} x^{5}}\)	\(35\)
gosper	\(-\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (5 x^{4}-x^{3}-1\right )}{5 x^{5} \left (x^{3}+1\right )^{\frac {3}{4}}}\)	\(36\)
meijerg	\(\frac {\hypergeom \left (\left [\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {5}{3}\right ], -x^{3}\right ) x^{2}}{2}-\hypergeom \left (\left [\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {4}{3}\right ], -x^{3}\right ) x +\frac {5 \hypergeom \left (\left [-\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {1}{3}\right ], -x^{3}\right )}{2 x^{2}}+\frac {4 \hypergeom \left (\left [-\frac {5}{3}, \frac {3}{4}\right ], \left [-\frac {2}{3}\right ], -x^{3}\right )}{5 x^{5}}-\frac {4 \hypergeom \left (\left [-\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.77, size = 42, normalized size = 1.50 \begin {gather*} -\frac {4 \, {\left (5 \, x^{7} - x^{6} + 5 \, x^{4} - 2 \, x^{3} - 1\right )}}{5 \, {\left (x^{2} - x + 1\right )}^{\frac {3}{4}} {\left (x + 1\right )}^{\frac {3}{4}} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5*(5*x^7 - x^6 + 5*x^4 - 2*x^3 - 1)/((x^2 - x + 1)^(3/4)*(x + 1)^(3/4)*x^5)

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mupad [B]  time = 0.24, size = 39, normalized size = 1.39 \begin {gather*} \frac {4\,{\left (x^3+1\right )}^{1/4}+4\,x^3\,{\left (x^3+1\right )}^{1/4}-20\,x^4\,{\left (x^3+1\right )}^{1/4}}{5\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*(x^3 + 1)^(1/4) + 4*x^3*(x^3 + 1)^(1/4) - 20*x^4*(x^3 + 1)^(1/4))/(5*x^5)

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sympy [C]  time = 4.09, size = 167, normalized size = 5.96 \begin {gather*} \frac {x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{4} \\ \frac {5}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} - \frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{4} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {4 \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} - \frac {5 \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {3}{4} \\ \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} - \frac {4 \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {3}{4} \\ - \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{5} \Gamma \left (- \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*gamma(2/3)*hyper((2/3, 3/4), (5/3,), x**3*exp_polar(I*pi))/(3*gamma(5/3)) - x*gamma(1/3)*hyper((1/3, 3/4)
, (4/3,), x**3*exp_polar(I*pi))/(3*gamma(4/3)) + 4*gamma(-1/3)*hyper((-1/3, 3/4), (2/3,), x**3*exp_polar(I*pi)
)/(3*x*gamma(2/3)) - 5*gamma(-2/3)*hyper((-2/3, 3/4), (1/3,), x**3*exp_polar(I*pi))/(3*x**2*gamma(1/3)) - 4*ga
mma(-5/3)*hyper((-5/3, 3/4), (-2/3,), x**3*exp_polar(I*pi))/(3*x**5*gamma(-2/3))

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