\(\int \frac {(1+x^2) (3+x^2)}{x^6 \sqrt [4]{x+x^3}} \, dx\) [77]

Optimal result

Integrand size = 23, antiderivative size = 16 \[ \int \frac {\left (1+x^2\right ) \left (3+x^2\right )}{x^6 \sqrt [4]{x+x^3}} \, dx=-\frac {4 \left (x+x^3\right )^{7/4}}{7 x^7} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(33\) vs. \(2(16)=32\).

Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.06, number of steps used = 14, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2077, 2050, 2036, 371} \[ \int \frac {\left (1+x^2\right ) \left (3+x^2\right )}{x^6 \sqrt [4]{x+x^3}} \, dx=-\frac {4 \left (x^3+x\right )^{3/4}}{7 x^6}-\frac {4 \left (x^3+x\right )^{3/4}}{7 x^4} \]

[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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2036

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 2050

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 2077

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*} \text {integral}& = \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} \operatorname {Hypergeometric2F1}\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} \operatorname {Hypergeometric2F1}\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] (verified)

Time = 10.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {\left (1+x^2\right ) \left (3+x^2\right )}{x^6 \sqrt [4]{x+x^3}} \, dx=-\frac {4 \left (1+x^2\right ) \left (x+x^3\right )^{3/4}}{7 x^6} \]

[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)

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 18, normalized size of antiderivative = 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	\(-\frac {4 \operatorname {hypergeom}\left (\left [-\frac {21}{8}, \frac {1}{4}\right ], \left [-\frac {13}{8}\right ], -x^{2}\right )}{7 x^{\frac {21}{4}}}-\frac {16 \operatorname {hypergeom}\left (\left [-\frac {13}{8}, \frac {1}{4}\right ], \left [-\frac {5}{8}\right ], -x^{2}\right )}{13 x^{\frac {13}{4}}}-\frac {4 \operatorname {hypergeom}\left (\left [-\frac {5}{8}, \frac {1}{4}\right ], \left [\frac {3}{8}\right ], -x^{2}\right )}{5 x^{\frac {5}{4}}}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {\left (1+x^2\right ) \left (3+x^2\right )}{x^6 \sqrt [4]{x+x^3}} \, dx=-\frac {4 \, {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x^{2} + 1\right )}}{7 \, x^{6}} \]

[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

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int \frac {\left (1+x^2\right ) \left (3+x^2\right )}{x^6 \sqrt [4]{x+x^3}} \, dx=-\frac {4}{7} \, {\left (\frac {1}{x} + \frac {1}{x^{3}}\right )}^{\frac {7}{4}} \]

[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)

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {\left (1+x^2\right ) \left (3+x^2\right )}{x^6 \sqrt [4]{x+x^3}} \, dx=-\frac {4\,{\left (x^3+x\right )}^{3/4}+4\,x^2\,{\left (x^3+x\right )}^{3/4}}{7\,x^6} \]

[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)