3.1.77 \(\int \frac {(1+x^2) (3+x^2)}{x^6 \sqrt [4]{x+x^3}} \, dx\)

Optimal. Leaf size=16 \[ -\frac {4 \left (x^3+x\right )^{7/4}}{7 x^7} \]

________________________________________________________________________________________

Rubi [B]  time = 0.26, antiderivative size = 33, normalized size of antiderivative = 2.06, number of steps used = 14, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2052, 2025, 2011, 364} \begin {gather*} -\frac {4 \left (x^3+x\right )^{3/4}}{7 x^6}-\frac {4 \left (x^3+x\right )^{3/4}}{7 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)*(3 + x^2))/(x^6*(x + x^3)^(1/4)),x]

[Out]

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2052

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {\left (1+x^2\right ) \left (3+x^2\right )}{x^6 \sqrt [4]{x+x^3}} \, dx &=\int \left (\frac {3}{x^6 \sqrt [4]{x+x^3}}+\frac {4}{x^4 \sqrt [4]{x+x^3}}+\frac {1}{x^2 \sqrt [4]{x+x^3}}\right ) \, dx\\ &=3 \int \frac {1}{x^6 \sqrt [4]{x+x^3}} \, dx+4 \int \frac {1}{x^4 \sqrt [4]{x+x^3}} \, dx+\int \frac {1}{x^2 \sqrt [4]{x+x^3}} \, dx\\ &=-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^6}-\frac {16 \left (x+x^3\right )^{3/4}}{13 x^4}-\frac {4 \left (x+x^3\right )^{3/4}}{5 x^2}+\frac {1}{5} \int \frac {1}{\sqrt [4]{x+x^3}} \, dx-\frac {15}{7} \int \frac {1}{x^4 \sqrt [4]{x+x^3}} \, dx-\frac {28}{13} \int \frac {1}{x^2 \sqrt [4]{x+x^3}} \, dx\\ &=-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^6}-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^4}+\frac {12 \left (x+x^3\right )^{3/4}}{13 x^2}-\frac {28}{65} \int \frac {1}{\sqrt [4]{x+x^3}} \, dx+\frac {15}{13} \int \frac {1}{x^2 \sqrt [4]{x+x^3}} \, dx+\frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{1+x^2}} \, dx}{5 \sqrt [4]{x+x^3}}\\ &=-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^6}-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^4}+\frac {4 x \sqrt [4]{1+x^2} \, _2F_1\left (\frac {1}{4},\frac {3}{8};\frac {11}{8};-x^2\right )}{15 \sqrt [4]{x+x^3}}+\frac {3}{13} \int \frac {1}{\sqrt [4]{x+x^3}} \, dx-\frac {\left (28 \sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{1+x^2}} \, dx}{65 \sqrt [4]{x+x^3}}\\ &=-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^6}-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^4}-\frac {4 x \sqrt [4]{1+x^2} \, _2F_1\left (\frac {1}{4},\frac {3}{8};\frac {11}{8};-x^2\right )}{13 \sqrt [4]{x+x^3}}+\frac {\left (3 \sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{1+x^2}} \, dx}{13 \sqrt [4]{x+x^3}}\\ &=-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^6}-\frac {4 \left (x+x^3\right )^{3/4}}{7 x^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 21, normalized size = 1.31 \begin {gather*} -\frac {4 \left (x^2+1\right ) \left (x^3+x\right )^{3/4}}{7 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^2)*(3 + x^2))/(x^6*(x + x^3)^(1/4)),x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.33, size = 16, normalized size = 1.00 \begin {gather*} -\frac {4 \left (x+x^3\right )^{7/4}}{7 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 + x^2)*(3 + x^2))/(x^6*(x + x^3)^(1/4)),x]

[Out]

(-4*(x + x^3)^(7/4))/(7*x^7)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 17, normalized size = 1.06 \begin {gather*} -\frac {4 \, {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x^{2} + 1\right )}}{7 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^2+3)/x^6/(x^3+x)^(1/4),x, algorithm="fricas")

[Out]

-4/7*(x^3 + x)^(3/4)*(x^2 + 1)/x^6

________________________________________________________________________________________

giac [A]  time = 0.42, size = 11, normalized size = 0.69 \begin {gather*} -\frac {4}{7} \, {\left (\frac {1}{x} + \frac {1}{x^{3}}\right )}^{\frac {7}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^2+3)/x^6/(x^3+x)^(1/4),x, algorithm="giac")

[Out]

-4/7*(1/x + 1/x^3)^(7/4)

________________________________________________________________________________________

maple [A]  time = 0.10, size = 18, normalized size = 1.12

method	result	size
trager	\(-\frac {4 \left (x^{2}+1\right ) \left (x^{3}+x \right )^{\frac {3}{4}}}{7 x^{6}}\)	\(18\)
gosper	\(-\frac {4 \left (x^{2}+1\right )^{2}}{7 x^{5} \left (x^{3}+x \right )^{\frac {1}{4}}}\)	\(20\)
risch	\(-\frac {4 \left (x^{4}+2 x^{2}+1\right )}{7 x^{5} \left (\left (x^{2}+1\right ) x \right )^{\frac {1}{4}}}\)	\(25\)
meijerg	\(-\frac {4 \hypergeom \left (\left [-\frac {21}{8}, \frac {1}{4}\right ], \left [-\frac {13}{8}\right ], -x^{2}\right )}{7 x^{\frac {21}{4}}}-\frac {16 \hypergeom \left (\left [-\frac {13}{8}, \frac {1}{4}\right ], \left [-\frac {5}{8}\right ], -x^{2}\right )}{13 x^{\frac {13}{4}}}-\frac {4 \hypergeom \left (\left [-\frac {5}{8}, \frac {1}{4}\right ], \left [\frac {3}{8}\right ], -x^{2}\right )}{5 x^{\frac {5}{4}}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)*(x^2+3)/x^6/(x^3+x)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^2+3)/x^6/(x^3+x)^(1/4),x, algorithm="maxima")

[Out]

integrate((x^2 + 3)*(x^2 + 1)/((x^3 + x)^(1/4)*x^6), x)

________________________________________________________________________________________

mupad [B]  time = 0.35, size = 27, normalized size = 1.69 \begin {gather*} -\frac {4\,{\left (x^3+x\right )}^{3/4}+4\,x^2\,{\left (x^3+x\right )}^{3/4}}{7\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2 + 1)*(x^2 + 3))/(x^6*(x + x^3)^(1/4)),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} + 1\right ) \left (x^{2} + 3\right )}{x^{6} \sqrt [4]{x \left (x^{2} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+1)*(x**2+3)/x**6/(x**3+x)**(1/4),x)

[Out]

Integral((x**2 + 1)*(x**2 + 3)/(x**6*(x*(x**2 + 1))**(1/4)), x)

________________________________________________________________________________________