3.4.25 \(\int \frac {(4+x^3) (-1-x^3+x^4)}{x^8 \sqrt [4]{1+x^3}} \, dx\)

Optimal. Leaf size=28 \[ -\frac {4 \left (x^3+1\right )^{3/4} \left (7 x^4-3 x^3-3\right )}{21 x^7} \]

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Rubi [A]  time = 0.11, antiderivative size = 49, normalized size of antiderivative = 1.75, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1835, 1586, 446, 74} \begin {gather*} -\frac {4 \left (x^3+1\right )^{3/4}}{3 x^3}+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((4 + x^3)*(-1 - x^3 + x^4))/(x^8*(1 + x^3)^(1/4)),x]

[Out]

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

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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^8 \sqrt [4]{1+x^3}} \, dx &=\frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32 x^2-56 x^3+14 x^5-14 x^6}{x^7 \sqrt [4]{1+x^3}} \, dx\\ &=\frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32 x-56 x^2+14 x^4-14 x^5}{x^6 \sqrt [4]{1+x^3}} \, dx\\ &=\frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32-56 x+14 x^3-14 x^4}{x^5 \sqrt [4]{1+x^3}} \, dx\\ &=\frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}+\frac {1}{112} \int \frac {448+112 x^3}{x^4 \sqrt [4]{1+x^3}} \, dx\\ &=\frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}+\frac {1}{336} \operatorname {Subst}\left (\int \frac {448+112 x}{x^2 \sqrt [4]{1+x}} \, dx,x,x^3\right )\\ &=\frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}-\frac {4 \left (1+x^3\right )^{3/4}}{3 x^3}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 118, normalized size = 4.21 \begin {gather*} \frac {\, _2F_1\left (-\frac {1}{3},\frac {1}{4};\frac {2}{3};-x^3\right )}{x}+\frac {16}{9} \left (x^3+1\right )^{3/4} \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};x^3+1\right )+\frac {4 \, _2F_1\left (-\frac {7}{3},\frac {1}{4};-\frac {4}{3};-x^3\right )}{7 x^7}+\frac {5 \, _2F_1\left (-\frac {4}{3},\frac {1}{4};-\frac {1}{3};-x^3\right )}{4 x^4}+\frac {2}{3} \tan ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {2}{3} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((4 + x^3)*(-1 - x^3 + x^4))/(x^8*(1 + x^3)^(1/4)),x]

[Out]

(2*ArcTan[(1 + x^3)^(1/4)])/3 - (2*ArcTanh[(1 + x^3)^(1/4)])/3 + (4*Hypergeometric2F1[-7/3, 1/4, -4/3, -x^3])/
(7*x^7) + (5*Hypergeometric2F1[-4/3, 1/4, -1/3, -x^3])/(4*x^4) + Hypergeometric2F1[-1/3, 1/4, 2/3, -x^3]/x + (
16*(1 + x^3)^(3/4)*Hypergeometric2F1[3/4, 2, 7/4, 1 + x^3])/9

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IntegrateAlgebraic [A]  time = 0.59, size = 28, normalized size = 1.00 \begin {gather*} -\frac {4 \left (1+x^3\right )^{3/4} \left (-3-3 x^3+7 x^4\right )}{21 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((4 + x^3)*(-1 - x^3 + x^4))/(x^8*(1 + x^3)^(1/4)),x]

[Out]

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

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fricas [A]  time = 0.48, size = 24, normalized size = 0.86 \begin {gather*} -\frac {4 \, {\left (7 \, x^{4} - 3 \, x^{3} - 3\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{21 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+4)*(x^4-x^3-1)/x^8/(x^3+1)^(1/4),x, algorithm="fricas")

[Out]

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

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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 {1}{4}} x^{8}}\,{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^8/(x^3+1)^(1/4),x, algorithm="giac")

[Out]

integrate((x^4 - x^3 - 1)*(x^3 + 4)/((x^3 + 1)^(1/4)*x^8), x)

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maple [A]  time = 0.09, size = 25, normalized size = 0.89

method	result	size
trager	\(-\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}} \left (7 x^{4}-3 x^{3}-3\right )}{21 x^{7}}\)	\(25\)
risch	\(-\frac {4 \left (7 x^{7}-3 x^{6}+7 x^{4}-6 x^{3}-3\right )}{21 x^{7} \left (x^{3}+1\right )^{\frac {1}{4}}}\)	\(35\)
gosper	\(-\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (7 x^{4}-3 x^{3}-3\right )}{21 x^{7} \left (x^{3}+1\right )^{\frac {1}{4}}}\)	\(36\)
meijerg	\(\frac {4 \hypergeom \left (\left [-\frac {7}{3}, \frac {1}{4}\right ], \left [-\frac {4}{3}\right ], -x^{3}\right )}{7 x^{7}}+\frac {2 \sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {\pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right ) x^{3}}-\frac {\left (3-3 \ln \relax (2)-\frac {\pi }{2}+3 \ln \relax (x )\right ) \pi \sqrt {2}}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {5 \hypergeom \left (\left [1, 1, \frac {9}{4}\right ], \left [2, 3\right ], -x^{3}\right ) \pi \sqrt {2}\, x^{3}}{32 \Gamma \left (\frac {3}{4}\right )}\right )}{3 \pi }+\frac {5 \hypergeom \left (\left [-\frac {4}{3}, \frac {1}{4}\right ], \left [-\frac {1}{3}\right ], -x^{3}\right )}{4 x^{4}}+\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (\frac {\left (-3 \ln \relax (2)-\frac {\pi }{2}+3 \ln \relax (x )\right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}-\frac {\hypergeom \left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], -x^{3}\right ) \pi \sqrt {2}\, x^{3}}{4 \Gamma \left (\frac {3}{4}\right )}\right )}{6 \pi }+\frac {\hypergeom \left (\left [-\frac {1}{3}, \frac {1}{4}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+4)*(x^4-x^3-1)/x^8/(x^3+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.61, size = 42, normalized size = 1.50 \begin {gather*} -\frac {4 \, {\left (7 \, x^{7} - 3 \, x^{6} + 7 \, x^{4} - 6 \, x^{3} - 3\right )}}{21 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{4}} {\left (x + 1\right )}^{\frac {1}{4}} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+4)*(x^4-x^3-1)/x^8/(x^3+1)^(1/4),x, algorithm="maxima")

[Out]

-4/21*(7*x^7 - 3*x^6 + 7*x^4 - 6*x^3 - 3)/((x^2 - x + 1)^(1/4)*(x + 1)^(1/4)*x^7)

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mupad [B]  time = 0.14, size = 39, normalized size = 1.39 \begin {gather*} \frac {12\,{\left (x^3+1\right )}^{3/4}+12\,x^3\,{\left (x^3+1\right )}^{3/4}-28\,x^4\,{\left (x^3+1\right )}^{3/4}}{21\,x^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x^3 + 4)*(x^3 - x^4 + 1))/(x^8*(x^3 + 1)^(1/4)),x)

[Out]

(12*(x^3 + 1)^(3/4) + 12*x^3*(x^3 + 1)^(3/4) - 28*x^4*(x^3 + 1)^(3/4))/(21*x^7)

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sympy [C]  time = 4.80, size = 177, normalized size = 6.32 \begin {gather*} - \frac {\Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{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 {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{4} \\ - \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{4} \Gamma \left (- \frac {1}{3}\right )} - \frac {4 \Gamma \left (- \frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{3}, \frac {1}{4} \\ - \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{7} \Gamma \left (- \frac {4}{3}\right )} - \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} - \frac {4 \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {15}{4}} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+4)*(x**4-x**3-1)/x**8/(x**3+1)**(1/4),x)

[Out]

-gamma(-1/3)*hyper((-1/3, 1/4), (2/3,), x**3*exp_polar(I*pi))/(3*x*gamma(2/3)) - 5*gamma(-4/3)*hyper((-4/3, 1/
4), (-1/3,), x**3*exp_polar(I*pi))/(3*x**4*gamma(-1/3)) - 4*gamma(-7/3)*hyper((-7/3, 1/4), (-4/3,), x**3*exp_p
olar(I*pi))/(3*x**7*gamma(-4/3)) - gamma(1/4)*hyper((1/4, 1/4), (5/4,), exp_polar(I*pi)/x**3)/(3*x**(3/4)*gamm
a(5/4)) - 4*gamma(5/4)*hyper((1/4, 5/4), (9/4,), exp_polar(I*pi)/x**3)/(3*x**(15/4)*gamma(9/4))

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