3.5.88 \(\int \frac {(-4+x^5) (1+x^5)^{3/4} (2-x^4+2 x^5)}{x^{12}} \, dx\)

Optimal. Leaf size=38 \[ \frac {4 \left (x^5+1\right )^{3/4} \left (14 x^{10}-11 x^9+28 x^5-11 x^4+14\right )}{77 x^{11}} \]

________________________________________________________________________________________

Rubi [B]  time = 0.19, antiderivative size = 88, normalized size of antiderivative = 2.32, number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1820, 1825, 1835, 1586, 449} \begin {gather*} \frac {30 \left (x^5+1\right )^{3/4}}{11 x}+\frac {5 \left (x^5+1\right )^{3/4}}{11 x^6}-\frac {15 \left (x^5+1\right )^{3/4}}{14 x^2}+\frac {1}{154} \left (x^5+1\right )^{3/4} \left (\frac {112}{x^{11}}-\frac {88}{x^7}+\frac {154}{x^6}+\frac {77}{x^2}-\frac {308}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-4 + x^5)*(1 + x^5)^(3/4)*(2 - x^4 + 2*x^5))/x^12,x]

[Out]

((112/x^11 - 88/x^7 + 154/x^6 + 77/x^2 - 308/x)*(1 + x^5)^(3/4))/154 + (5*(1 + x^5)^(3/4))/(11*x^6) - (15*(1 +
x^5)^(3/4))/(14*x^2) + (30*(1 + x^5)^(3/4))/(11*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 1820

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

Rule 1825

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{u = IntHide[x^m*Pq, x]}, Simp[u*(a +
b*x^n)^p, x] - Dist[b*n*p, Int[x^(m + n)*(a + b*x^n)^(p - 1)*ExpandToSum[u/x^(m + 1), x], x], x]] /; FreeQ[{a
, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && GtQ[p, 0] && LtQ[m + Expon[Pq, x] + 1, 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^5\right ) \left (1+x^5\right )^{3/4} \left (2-x^4+2 x^5\right )}{x^{12}} \, dx &=\frac {1}{154} \left (\frac {112}{x^{11}}-\frac {88}{x^7}+\frac {154}{x^6}+\frac {77}{x^2}-\frac {308}{x}\right ) \left (1+x^5\right )^{3/4}-\frac {15}{4} \int \frac {\frac {8}{11}-\frac {4 x^4}{7}+x^5+\frac {x^9}{2}-2 x^{10}}{x^7 \sqrt [4]{1+x^5}} \, dx\\ &=\frac {1}{154} \left (\frac {112}{x^{11}}-\frac {88}{x^7}+\frac {154}{x^6}+\frac {77}{x^2}-\frac {308}{x}\right ) \left (1+x^5\right )^{3/4}+\frac {5 \left (1+x^5\right )^{3/4}}{11 x^6}+\frac {5}{16} \int \frac {\frac {48 x^3}{7}-\frac {96 x^4}{11}-6 x^8+24 x^9}{x^6 \sqrt [4]{1+x^5}} \, dx\\ &=\frac {1}{154} \left (\frac {112}{x^{11}}-\frac {88}{x^7}+\frac {154}{x^6}+\frac {77}{x^2}-\frac {308}{x}\right ) \left (1+x^5\right )^{3/4}+\frac {5 \left (1+x^5\right )^{3/4}}{11 x^6}+\frac {5}{16} \int \frac {\frac {48 x^2}{7}-\frac {96 x^3}{11}-6 x^7+24 x^8}{x^5 \sqrt [4]{1+x^5}} \, dx\\ &=\frac {1}{154} \left (\frac {112}{x^{11}}-\frac {88}{x^7}+\frac {154}{x^6}+\frac {77}{x^2}-\frac {308}{x}\right ) \left (1+x^5\right )^{3/4}+\frac {5 \left (1+x^5\right )^{3/4}}{11 x^6}+\frac {5}{16} \int \frac {\frac {48 x}{7}-\frac {96 x^2}{11}-6 x^6+24 x^7}{x^4 \sqrt [4]{1+x^5}} \, dx\\ &=\frac {1}{154} \left (\frac {112}{x^{11}}-\frac {88}{x^7}+\frac {154}{x^6}+\frac {77}{x^2}-\frac {308}{x}\right ) \left (1+x^5\right )^{3/4}+\frac {5 \left (1+x^5\right )^{3/4}}{11 x^6}+\frac {5}{16} \int \frac {\frac {48}{7}-\frac {96 x}{11}-6 x^5+24 x^6}{x^3 \sqrt [4]{1+x^5}} \, dx\\ &=\frac {1}{154} \left (\frac {112}{x^{11}}-\frac {88}{x^7}+\frac {154}{x^6}+\frac {77}{x^2}-\frac {308}{x}\right ) \left (1+x^5\right )^{3/4}+\frac {5 \left (1+x^5\right )^{3/4}}{11 x^6}-\frac {15 \left (1+x^5\right )^{3/4}}{14 x^2}-\frac {5}{64} \int \frac {\frac {384}{11}-96 x^5}{x^2 \sqrt [4]{1+x^5}} \, dx\\ &=\frac {1}{154} \left (\frac {112}{x^{11}}-\frac {88}{x^7}+\frac {154}{x^6}+\frac {77}{x^2}-\frac {308}{x}\right ) \left (1+x^5\right )^{3/4}+\frac {5 \left (1+x^5\right )^{3/4}}{11 x^6}-\frac {15 \left (1+x^5\right )^{3/4}}{14 x^2}+\frac {30 \left (1+x^5\right )^{3/4}}{11 x}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 106, normalized size = 2.79 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {3}{4},-\frac {1}{5};\frac {4}{5};-x^5\right )}{x}+\frac {8 \, _2F_1\left (-\frac {11}{5},-\frac {3}{4};-\frac {6}{5};-x^5\right )}{11 x^{11}}-\frac {4 \, _2F_1\left (-\frac {7}{5},-\frac {3}{4};-\frac {2}{5};-x^5\right )}{7 x^7}+\frac {\, _2F_1\left (-\frac {6}{5},-\frac {3}{4};-\frac {1}{5};-x^5\right )}{x^6}+\frac {\, _2F_1\left (-\frac {3}{4},-\frac {2}{5};\frac {3}{5};-x^5\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-4 + x^5)*(1 + x^5)^(3/4)*(2 - x^4 + 2*x^5))/x^12,x]

[Out]

(8*Hypergeometric2F1[-11/5, -3/4, -6/5, -x^5])/(11*x^11) - (4*Hypergeometric2F1[-7/5, -3/4, -2/5, -x^5])/(7*x^
7) + Hypergeometric2F1[-6/5, -3/4, -1/5, -x^5]/x^6 + Hypergeometric2F1[-3/4, -2/5, 3/5, -x^5]/(2*x^2) - (2*Hyp
ergeometric2F1[-3/4, -1/5, 4/5, -x^5])/x

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 2.60, size = 28, normalized size = 0.74 \begin {gather*} \frac {4 \left (1+x^5\right )^{7/4} \left (14-11 x^4+14 x^5\right )}{77 x^{11}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-4 + x^5)*(1 + x^5)^(3/4)*(2 - x^4 + 2*x^5))/x^12,x]

[Out]

(4*(1 + x^5)^(7/4)*(14 - 11*x^4 + 14*x^5))/(77*x^11)

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 34, normalized size = 0.89 \begin {gather*} \frac {4 \, {\left (14 \, x^{10} - 11 \, x^{9} + 28 \, x^{5} - 11 \, x^{4} + 14\right )} {\left (x^{5} + 1\right )}^{\frac {3}{4}}}{77 \, x^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-4)*(x^5+1)^(3/4)*(2*x^5-x^4+2)/x^12,x, algorithm="fricas")

[Out]

4/77*(14*x^10 - 11*x^9 + 28*x^5 - 11*x^4 + 14)*(x^5 + 1)^(3/4)/x^11

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{5} - x^{4} + 2\right )} {\left (x^{5} + 1\right )}^{\frac {3}{4}} {\left (x^{5} - 4\right )}}{x^{12}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-4)*(x^5+1)^(3/4)*(2*x^5-x^4+2)/x^12,x, algorithm="giac")

[Out]

integrate((2*x^5 - x^4 + 2)*(x^5 + 1)^(3/4)*(x^5 - 4)/x^12, x)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 35, normalized size = 0.92

method	result	size
trager	\(\frac {4 \left (x^{5}+1\right )^{\frac {3}{4}} \left (14 x^{10}-11 x^{9}+28 x^{5}-11 x^{4}+14\right )}{77 x^{11}}\)	\(35\)
gosper	\(\frac {4 \left (x^{4}-x^{3}+x^{2}-x +1\right ) \left (1+x \right ) \left (14 x^{5}-11 x^{4}+14\right ) \left (x^{5}+1\right )^{\frac {3}{4}}}{77 x^{11}}\)	\(44\)
risch	\(\frac {-\frac {4}{7} x^{14}-\frac {8}{7} x^{9}+\frac {8}{11} x^{15}+\frac {24}{11} x^{10}+\frac {24}{11} x^{5}+\frac {8}{11}-\frac {4}{7} x^{4}}{x^{11} \left (x^{5}+1\right )^{\frac {1}{4}}}\)	\(45\)
meijerg	\(\frac {\hypergeom \left (\left [-\frac {6}{5}, -\frac {3}{4}\right ], \left [-\frac {1}{5}\right ], -x^{5}\right )}{x^{6}}-\frac {4 \hypergeom \left (\left [-\frac {7}{5}, -\frac {3}{4}\right ], \left [-\frac {2}{5}\right ], -x^{5}\right )}{7 x^{7}}+\frac {8 \hypergeom \left (\left [-\frac {11}{5}, -\frac {3}{4}\right ], \left [-\frac {6}{5}\right ], -x^{5}\right )}{11 x^{11}}-\frac {2 \hypergeom \left (\left [-\frac {3}{4}, -\frac {1}{5}\right ], \left [\frac {4}{5}\right ], -x^{5}\right )}{x}+\frac {\hypergeom \left (\left [-\frac {3}{4}, -\frac {2}{5}\right ], \left [\frac {3}{5}\right ], -x^{5}\right )}{2 x^{2}}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5-4)*(x^5+1)^(3/4)*(2*x^5-x^4+2)/x^12,x,method=_RETURNVERBOSE)

[Out]

4/77*(x^5+1)^(3/4)*(14*x^10-11*x^9+28*x^5-11*x^4+14)/x^11

________________________________________________________________________________________

maxima [A]  time = 0.52, size = 50, normalized size = 1.32 \begin {gather*} \frac {4 \, {\left (14 \, x^{10} - 11 \, x^{9} + 28 \, x^{5} - 11 \, x^{4} + 14\right )} {\left (x^{4} - x^{3} + x^{2} - x + 1\right )}^{\frac {3}{4}} {\left (x + 1\right )}^{\frac {3}{4}}}{77 \, x^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-4)*(x^5+1)^(3/4)*(2*x^5-x^4+2)/x^12,x, algorithm="maxima")

[Out]

4/77*(14*x^10 - 11*x^9 + 28*x^5 - 11*x^4 + 14)*(x^4 - x^3 + x^2 - x + 1)^(3/4)*(x + 1)^(3/4)/x^11

________________________________________________________________________________________

mupad [B]  time = 0.70, size = 61, normalized size = 1.61 \begin {gather*} \frac {8\,{\left (x^5+1\right )}^{3/4}}{11\,x}-\frac {4\,{\left (x^5+1\right )}^{3/4}}{7\,x^2}+\frac {16\,{\left (x^5+1\right )}^{3/4}}{11\,x^6}-\frac {4\,{\left (x^5+1\right )}^{3/4}}{7\,x^7}+\frac {8\,{\left (x^5+1\right )}^{3/4}}{11\,x^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^5 + 1)^(3/4)*(x^5 - 4)*(2*x^5 - x^4 + 2))/x^12,x)

[Out]

(8*(x^5 + 1)^(3/4))/(11*x) - (4*(x^5 + 1)^(3/4))/(7*x^2) + (16*(x^5 + 1)^(3/4))/(11*x^6) - (4*(x^5 + 1)^(3/4))
/(7*x^7) + (8*(x^5 + 1)^(3/4))/(11*x^11)

________________________________________________________________________________________

sympy [C]  time = 6.57, size = 192, normalized size = 5.05 \begin {gather*} \frac {2 \Gamma \left (- \frac {1}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{5} \\ \frac {4}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x \Gamma \left (\frac {4}{5}\right )} - \frac {\Gamma \left (- \frac {2}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {2}{5} \\ \frac {3}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x^{2} \Gamma \left (\frac {3}{5}\right )} - \frac {6 \Gamma \left (- \frac {6}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {6}{5}, - \frac {3}{4} \\ - \frac {1}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x^{6} \Gamma \left (- \frac {1}{5}\right )} + \frac {4 \Gamma \left (- \frac {7}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{5}, - \frac {3}{4} \\ - \frac {2}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x^{7} \Gamma \left (- \frac {2}{5}\right )} - \frac {8 \Gamma \left (- \frac {11}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {11}{5}, - \frac {3}{4} \\ - \frac {6}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x^{11} \Gamma \left (- \frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**5-4)*(x**5+1)**(3/4)*(2*x**5-x**4+2)/x**12,x)

[Out]

2*gamma(-1/5)*hyper((-3/4, -1/5), (4/5,), x**5*exp_polar(I*pi))/(5*x*gamma(4/5)) - gamma(-2/5)*hyper((-3/4, -2
/5), (3/5,), x**5*exp_polar(I*pi))/(5*x**2*gamma(3/5)) - 6*gamma(-6/5)*hyper((-6/5, -3/4), (-1/5,), x**5*exp_p
olar(I*pi))/(5*x**6*gamma(-1/5)) + 4*gamma(-7/5)*hyper((-7/5, -3/4), (-2/5,), x**5*exp_polar(I*pi))/(5*x**7*ga
mma(-2/5)) - 8*gamma(-11/5)*hyper((-11/5, -3/4), (-6/5,), x**5*exp_polar(I*pi))/(5*x**11*gamma(-6/5))

________________________________________________________________________________________