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

Optimal. Leaf size=28 \[ -\frac {4 \left (x^3+x\right )^{3/4} \left (7 x^3+3 x^2+3\right )}{21 x^6} \]

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Rubi [A]  time = 0.36, antiderivative size = 49, normalized size of antiderivative = 1.75, number of steps used = 21, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2052, 2025, 2011, 364, 2032} \begin {gather*} -\frac {4 \left (x^3+x\right )^{3/4}}{3 x^3}-\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[((3 + x^2)*(1 + x^2 + x^3))/(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) - (4*(x + x^3)^(3/4))/(3*x^3)

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

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 (3+x^2\right ) \left (1+x^2+x^3\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 {3}{x^3 \sqrt [4]{x+x^3}}+\frac {1}{x^2 \sqrt [4]{x+x^3}}+\frac {1}{x \sqrt [4]{x+x^3}}\right ) \, dx\\ &=3 \int \frac {1}{x^6 \sqrt [4]{x+x^3}} \, dx+3 \int \frac {1}{x^3 \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+\int \frac {1}{x \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}}{3 x^3}-\frac {4 \left (x+x^3\right )^{3/4}}{5 x^2}-\frac {4 \left (x+x^3\right )^{3/4}}{x}+\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+5 \int \frac {x}{\sqrt [4]{x+x^3}} \, dx-\int \frac {1}{x \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 {4 \left (x+x^3\right )^{3/4}}{3 x^3}+\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-5 \int \frac {x}{\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 {\left (5 \sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {x^{3/4}}{\sqrt [4]{1+x^2}} \, dx}{\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 \left (x+x^3\right )^{3/4}}{3 x^3}+\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 {20 x^2 \sqrt [4]{1+x^2} \, _2F_1\left (\frac {1}{4},\frac {7}{8};\frac {15}{8};-x^2\right )}{7 \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 {\left (5 \sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {x^{3/4}}{\sqrt [4]{1+x^2}} \, dx}{\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 \left (x+x^3\right )^{3/4}}{3 x^3}-\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}-\frac {4 \left (x+x^3\right )^{3/4}}{3 x^3}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 124, normalized size = 4.43 \begin {gather*} -\frac {4 \sqrt [4]{x^2+1} \left (7 x^2 \left (60 \, _2F_1\left (-\frac {13}{8},\frac {1}{4};-\frac {5}{8};-x^2\right )+13 x \left (5 \, _2F_1\left (-\frac {9}{8},\frac {1}{4};-\frac {1}{8};-x^2\right )+3 x \left (\, _2F_1\left (-\frac {5}{8},\frac {1}{4};\frac {3}{8};-x^2\right )+5 x \, _2F_1\left (-\frac {1}{8},\frac {1}{4};\frac {7}{8};-x^2\right )\right )\right )\right )+195 \, _2F_1\left (-\frac {21}{8},\frac {1}{4};-\frac {13}{8};-x^2\right )\right )}{1365 x^5 \sqrt [4]{x^3+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(1 + x^2)^(1/4)*(195*Hypergeometric2F1[-21/8, 1/4, -13/8, -x^2] + 7*x^2*(60*Hypergeometric2F1[-13/8, 1/4,
-5/8, -x^2] + 13*x*(5*Hypergeometric2F1[-9/8, 1/4, -1/8, -x^2] + 3*x*(Hypergeometric2F1[-5/8, 1/4, 3/8, -x^2]
+ 5*x*Hypergeometric2F1[-1/8, 1/4, 7/8, -x^2])))))/(1365*x^5*(x + x^3)^(1/4))

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IntegrateAlgebraic [A]  time = 0.70, size = 28, normalized size = 1.00 \begin {gather*} -\frac {4 \left (x+x^3\right )^{3/4} \left (3+3 x^2+7 x^3\right )}{21 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 24, normalized size = 0.86 \begin {gather*} -\frac {4 \, {\left (7 \, x^{3} + 3 \, x^{2} + 3\right )} {\left (x^{3} + x\right )}^{\frac {3}{4}}}{21 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.56, size = 23, normalized size = 0.82 \begin {gather*} -\frac {4}{7} \, {\left (\frac {1}{x} + \frac {1}{x^{3}}\right )}^{\frac {7}{4}} - \frac {4}{3} \, {\left (\frac {1}{x} + \frac {1}{x^{3}}\right )}^{\frac {3}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

method	result	size
trager	\(-\frac {4 \left (x^{3}+x \right )^{\frac {3}{4}} \left (7 x^{3}+3 x^{2}+3\right )}{21 x^{6}}\)	\(25\)
gosper	\(-\frac {4 \left (x^{2}+1\right ) \left (7 x^{3}+3 x^{2}+3\right )}{21 x^{5} \left (x^{3}+x \right )^{\frac {1}{4}}}\)	\(30\)
risch	\(-\frac {4 \left (7 x^{5}+3 x^{4}+7 x^{3}+6 x^{2}+3\right )}{21 x^{5} \left (\left (x^{2}+1\right ) x \right )^{\frac {1}{4}}}\)	\(37\)
meijerg	\(-\frac {4 \hypergeom \left (\left [-\frac {21}{8}, \frac {1}{4}\right ], \left [-\frac {13}{8}\right ], -x^{2}\right )}{7 x^{\frac {21}{4}}}-\frac {4 \hypergeom \left (\left [-\frac {9}{8}, \frac {1}{4}\right ], \left [-\frac {1}{8}\right ], -x^{2}\right )}{3 x^{\frac {9}{4}}}-\frac {16 \hypergeom \left (\left [-\frac {13}{8}, \frac {1}{4}\right ], \left [-\frac {5}{8}\right ], -x^{2}\right )}{13 x^{\frac {13}{4}}}-\frac {4 \hypergeom \left (\left [-\frac {1}{8}, \frac {1}{4}\right ], \left [\frac {7}{8}\right ], -x^{2}\right )}{x^{\frac {1}{4}}}-\frac {4 \hypergeom \left (\left [-\frac {5}{8}, \frac {1}{4}\right ], \left [\frac {3}{8}\right ], -x^{2}\right )}{5 x^{\frac {5}{4}}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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mupad [B]  time = 0.24, size = 39, normalized size = 1.39 \begin {gather*} -\frac {12\,{\left (x^3+x\right )}^{3/4}+12\,x^2\,{\left (x^3+x\right )}^{3/4}+28\,x^3\,{\left (x^3+x\right )}^{3/4}}{21\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(12*(x + x^3)^(3/4) + 12*x^2*(x + x^3)^(3/4) + 28*x^3*(x + x^3)^(3/4))/(21*x^6)

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

[Out]

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

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