\(\int \frac {(1+x) \sqrt [4]{x^3+x^4}}{x^4} \, dx\) [144]

   Optimal result
   Rubi [B] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 18, antiderivative size = 18 \[ \int \frac {(1+x) \sqrt [4]{x^3+x^4}}{x^4} \, dx=-\frac {4 \left (x^3+x^4\right )^{9/4}}{9 x^9} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(37\) vs. \(2(18)=36\).

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.06, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2077, 2041, 2039} \[ \int \frac {(1+x) \sqrt [4]{x^3+x^4}}{x^4} \, dx=-\frac {4 \left (x^4+x^3\right )^{5/4}}{9 x^6}-\frac {4 \left (x^4+x^3\right )^{5/4}}{9 x^5} \]

[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 2039

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] &&
 NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

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 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 {\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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {(1+x) \sqrt [4]{x^3+x^4}}{x^4} \, dx=-\frac {4 \left (x^3 (1+x)\right )^{9/4}}{9 x^9} \]

[In]

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

[Out]

(-4*(x^3*(1 + x))^(9/4))/(9*x^9)

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method result size
gosper \(-\frac {4 \left (1+x \right )^{2} \left (x^{4}+x^{3}\right )^{\frac {1}{4}}}{9 x^{3}}\) \(20\)
pseudoelliptic \(-\frac {4 \left (1+x \right )^{2} \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{9 x^{3}}\) \(20\)
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 \(-\frac {4 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}} \left (x^{3}+3 x^{2}+3 x +1\right )}{9 \left (1+x \right ) x^{3}}\) \(33\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {(1+x) \sqrt [4]{x^3+x^4}}{x^4} \, dx=-\frac {4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{2} + 2 \, x + 1\right )}}{9 \, x^{3}} \]

[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

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50 \[ \int \frac {(1+x) \sqrt [4]{x^3+x^4}}{x^4} \, dx=-\frac {4}{9} \, {\left (\frac {1}{x} + 1\right )}^{\frac {9}{4}} \]

[In]

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

[Out]

-4/9*(1/x + 1)^(9/4)

Mupad [B] (verification not implemented)

Time = 5.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.39 \[ \int \frac {(1+x) \sqrt [4]{x^3+x^4}}{x^4} \, dx=-\frac {8\,x\,{\left (x^4+x^3\right )}^{1/4}+4\,{\left (x^4+x^3\right )}^{1/4}+4\,x^2\,{\left (x^4+x^3\right )}^{1/4}}{9\,x^3} \]

[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)