3.2.51 \(\int \frac {(-1+x^4) \sqrt [4]{x^3+x^5}}{x^4} \, dx\) [151]

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

[Out]

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

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(18)=36\).
time = 0.12, 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 = {2077, 2029, 2057, 371, 2045, 2050} \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 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 2029

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 2045

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 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 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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 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*} \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 [A]
time = 0.20, size = 25, normalized size = 1.39 \begin {gather*} \frac {4 \left (1+x^2\right )^2 \sqrt [4]{x^3+x^5}}{9 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.30, 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	\(\frac {4 \hypergeom \left (\left [-\frac {9}{8}, -\frac {1}{4}\right ], \left [-\frac {1}{8}\right ], -x^{2}\right )}{9 x^{\frac {9}{4}}}+\frac {4 x^{\frac {7}{4}} \hypergeom \left (\left [-\frac {1}{4}, \frac {7}{8}\right ], \left [\frac {15}{8}\right ], -x^{2}\right )}{7}\)	\(34\)
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*} \text {Failed to integrate} \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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Fricas [A]
time = 0.35, 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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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