3.2.51 \(\int \frac {(-1+x^4) \sqrt [4]{x^3+x^5}}{x^4} \, dx\)

Optimal. Leaf size=18 \[ \frac {4 \left (x^5+x^3\right )^{9/4}}{9 x^9} \]

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Rubi [B]  time = 0.18, antiderivative size = 53, normalized size of antiderivative = 2.94, number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2052, 2004, 2032, 364, 2020, 2025} \begin {gather*} \frac {4}{9} \sqrt [4]{x^5+x^3} x+\frac {8 \sqrt [4]{x^5+x^3}}{9 x}+\frac {4 \sqrt [4]{x^5+x^3}}{9 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(x^3 + x^5)^(1/4))/(9*x^3) + (8*(x^3 + x^5)^(1/4))/(9*x) + (4*x*(x^3 + x^5)^(1/4))/9

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 2004

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

Rule 2020

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

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 (-1+x^4\right ) \sqrt [4]{x^3+x^5}}{x^4} \, dx &=\int \left (\sqrt [4]{x^3+x^5}-\frac {\sqrt [4]{x^3+x^5}}{x^4}\right ) \, dx\\ &=\int \sqrt [4]{x^3+x^5} \, dx-\int \frac {\sqrt [4]{x^3+x^5}}{x^4} \, dx\\ &=\frac {4 \sqrt [4]{x^3+x^5}}{9 x^3}+\frac {4}{9} x \sqrt [4]{x^3+x^5}-\frac {2}{9} \int \frac {x}{\left (x^3+x^5\right )^{3/4}} \, dx+\frac {2}{9} \int \frac {x^3}{\left (x^3+x^5\right )^{3/4}} \, dx\\ &=\frac {4 \sqrt [4]{x^3+x^5}}{9 x^3}+\frac {8 \sqrt [4]{x^3+x^5}}{9 x}+\frac {4}{9} x \sqrt [4]{x^3+x^5}-\frac {2}{9} \int \frac {x^3}{\left (x^3+x^5\right )^{3/4}} \, dx+\frac {\left (2 x^{9/4} \left (1+x^2\right )^{3/4}\right ) \int \frac {x^{3/4}}{\left (1+x^2\right )^{3/4}} \, dx}{9 \left (x^3+x^5\right )^{3/4}}\\ &=\frac {4 \sqrt [4]{x^3+x^5}}{9 x^3}+\frac {8 \sqrt [4]{x^3+x^5}}{9 x}+\frac {4}{9} x \sqrt [4]{x^3+x^5}+\frac {8 x^4 \left (1+x^2\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {7}{8};\frac {15}{8};-x^2\right )}{63 \left (x^3+x^5\right )^{3/4}}-\frac {\left (2 x^{9/4} \left (1+x^2\right )^{3/4}\right ) \int \frac {x^{3/4}}{\left (1+x^2\right )^{3/4}} \, dx}{9 \left (x^3+x^5\right )^{3/4}}\\ &=\frac {4 \sqrt [4]{x^3+x^5}}{9 x^3}+\frac {8 \sqrt [4]{x^3+x^5}}{9 x}+\frac {4}{9} x \sqrt [4]{x^3+x^5}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 65, normalized size = 3.61 \begin {gather*} \frac {4 \sqrt [4]{x^5+x^3} \left (7 \, _2F_1\left (-\frac {9}{8},-\frac {1}{4};-\frac {1}{8};-x^2\right )+9 x^4 \, _2F_1\left (-\frac {1}{4},\frac {7}{8};\frac {15}{8};-x^2\right )\right )}{63 x^3 \sqrt [4]{x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(x^3 + x^5)^(1/4)*(7*Hypergeometric2F1[-9/8, -1/4, -1/8, -x^2] + 9*x^4*Hypergeometric2F1[-1/4, 7/8, 15/8, -
x^2]))/(63*x^3*(1 + x^2)^(1/4))

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IntegrateAlgebraic [A]  time = 0.09, size = 18, normalized size = 1.00 \begin {gather*} \frac {4 \left (x^3+x^5\right )^{9/4}}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(x^3 + x^5)^(9/4))/(9*x^9)

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fricas [A]  time = 0.44, size = 24, normalized size = 1.33 \begin {gather*} \frac {4 \, {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{4} + 2 \, x^{2} + 1\right )}}{9 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/9*(x^5 + x^3)^(1/4)*(x^4 + 2*x^2 + 1)/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 22, normalized size = 1.22

method	result	size
gosper	\(\frac {4 \left (x^{2}+1\right )^{2} \left (x^{5}+x^{3}\right )^{\frac {1}{4}}}{9 x^{3}}\)	\(22\)
trager	\(\frac {4 \left (x^{4}+2 x^{2}+1\right ) \left (x^{5}+x^{3}\right )^{\frac {1}{4}}}{9 x^{3}}\)	\(25\)
meijerg	\(\frac {4 \hypergeom \left (\left [-\frac {9}{8}, -\frac {1}{4}\right ], \left [-\frac {1}{8}\right ], -x^{2}\right )}{9 x^{\frac {9}{4}}}+\frac {4 \hypergeom \left (\left [-\frac {1}{4}, \frac {7}{8}\right ], \left [\frac {15}{8}\right ], -x^{2}\right ) x^{\frac {7}{4}}}{7}\)	\(34\)
risch	\(\frac {4 \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} \left (x^{6}+3 x^{4}+3 x^{2}+1\right )}{9 x^{3} \left (x^{2}+1\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/9*(x^2+1)^2*(x^5+x^3)^(1/4)/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.27, size = 41, normalized size = 2.28 \begin {gather*} \frac {4\,x\,{\left (x^5+x^3\right )}^{1/4}}{9}+\frac {8\,{\left (x^5+x^3\right )}^{1/4}}{9\,x}+\frac {4\,{\left (x^5+x^3\right )}^{1/4}}{9\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x*(x^3 + x^5)^(1/4))/9 + (8*(x^3 + x^5)^(1/4))/(9*x) + (4*(x^3 + x^5)^(1/4))/(9*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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