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

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

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Rubi [B]  time = 0.12, antiderivative size = 37, normalized size of antiderivative = 2.06, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2052, 2016, 2014} \begin {gather*} -\frac {4 \left (x^4+x^3\right )^{5/4}}{9 x^6}-\frac {4 \left (x^4+x^3\right )^{5/4}}{9 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(x^3 + x^4)^(5/4))/(9*x^6) - (4*(x^3 + x^4)^(5/4))/(9*x^5)

Rule 2014

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*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

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, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
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 {(1+x) \sqrt [4]{x^3+x^4}}{x^4} \, dx &=\int \left (\frac {\sqrt [4]{x^3+x^4}}{x^4}+\frac {\sqrt [4]{x^3+x^4}}{x^3}\right ) \, dx\\ &=\int \frac {\sqrt [4]{x^3+x^4}}{x^4} \, dx+\int \frac {\sqrt [4]{x^3+x^4}}{x^3} \, dx\\ &=-\frac {4 \left (x^3+x^4\right )^{5/4}}{9 x^6}-\frac {4 \left (x^3+x^4\right )^{5/4}}{5 x^5}-\frac {4}{9} \int \frac {\sqrt [4]{x^3+x^4}}{x^3} \, dx\\ &=-\frac {4 \left (x^3+x^4\right )^{5/4}}{9 x^6}-\frac {4 \left (x^3+x^4\right )^{5/4}}{9 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} -\frac {4 \left (x^3 (x+1)\right )^{9/4}}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(x^3*(1 + x))^(9/4))/(9*x^9)

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IntegrateAlgebraic [A]  time = 0.20, size = 18, normalized size = 1.00 \begin {gather*} -\frac {4 \left (x^3+x^4\right )^{9/4}}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(x^3 + x^4)^(9/4))/(9*x^9)

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fricas [A]  time = 0.45, size = 22, normalized size = 1.22 \begin {gather*} -\frac {4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{2} + 2 \, x + 1\right )}}{9 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.00, size = 9, normalized size = 0.50 \begin {gather*} -\frac {4}{9} \, {\left (\frac {1}{x} + 1\right )}^{\frac {9}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/9*(1/x + 1)^(9/4)

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maple [A]  time = 0.06, size = 20, normalized size = 1.11

method	result	size
gosper	\(-\frac {4 \left (1+x \right )^{2} \left (x^{4}+x^{3}\right )^{\frac {1}{4}}}{9 x^{3}}\)	\(20\)
trager	\(-\frac {4 \left (x^{2}+2 x +1\right ) \left (x^{4}+x^{3}\right )^{\frac {1}{4}}}{9 x^{3}}\)	\(23\)
meijerg	\(-\frac {4 \left (-\frac {4}{5} x^{2}+\frac {1}{5} x +1\right ) \left (1+x \right )^{\frac {1}{4}}}{9 x^{\frac {9}{4}}}-\frac {4 \left (1+x \right )^{\frac {5}{4}}}{5 x^{\frac {5}{4}}}\)	\(32\)
risch	\(-\frac {4 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}} \left (x^{3}+3 x^{2}+3 x +1\right )}{9 \left (1+x \right ) x^{3}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.24, size = 43, normalized size = 2.39 \begin {gather*} -\frac {8\,x\,{\left (x^4+x^3\right )}^{1/4}+4\,{\left (x^4+x^3\right )}^{1/4}+4\,x^2\,{\left (x^4+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^4)^(1/4)*(x + 1))/x^4,x)

[Out]

-(8*x*(x^3 + x^4)^(1/4) + 4*(x^3 + x^4)^(1/4) + 4*x^2*(x^3 + x^4)^(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 + 1\right )} \left (x + 1\right )}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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