\(\int \frac {1}{x \sqrt [4]{x^3+x^4}} \, dx\) [143]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {1}{x \sqrt [4]{x^3+x^4}} \, dx=-\frac {4 \left (x^3+x^4\right )^{3/4}}{3 x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2039} \[ \int \frac {1}{x \sqrt [4]{x^3+x^4}} \, dx=-\frac {4 \left (x^4+x^3\right )^{3/4}}{3 x^3} \]

[In]

Int[1/(x*(x^3 + x^4)^(1/4)),x]

[Out]

(-4*(x^3 + x^4)^(3/4))/(3*x^3)

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])

Rubi steps \begin{align*} \text {integral}& = -\frac {4 \left (x^3+x^4\right )^{3/4}}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt [4]{x^3+x^4}} \, dx=-\frac {4 (1+x)}{3 \sqrt [4]{x^3 (1+x)}} \]

[In]

Integrate[1/(x*(x^3 + x^4)^(1/4)),x]

[Out]

(-4*(1 + x))/(3*(x^3*(1 + x))^(1/4))

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61

method	result	size
meijerg	\(-\frac {4 \left (1+x \right )^{\frac {3}{4}}}{3 x^{\frac {3}{4}}}\)	\(11\)
gosper	\(-\frac {4 \left (1+x \right )}{3 \left (x^{4}+x^{3}\right )^{\frac {1}{4}}}\)	\(15\)
trager	\(-\frac {4 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(15\)
risch	\(-\frac {4 \left (1+x \right )}{3 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}\)	\(15\)
pseudoelliptic	\(-\frac {4 \left (x^{3} \left (1+x \right )\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(15\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \sqrt [4]{x^3+x^4}} \, dx=-\frac {4 \, {\left (x^{4} + x^{3}\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

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

[Out]

-4/3*(x^4 + x^3)^(3/4)/x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(1/((x^4 + x^3)^(1/4)*x), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x \sqrt [4]{x^3+x^4}} \, dx=-\frac {4}{3} \, {\left (\frac {1}{x} + 1\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

-4/3*(1/x + 1)^(3/4)

Mupad [B] (verification not implemented)

Time = 5.17 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \sqrt [4]{x^3+x^4}} \, dx=-\frac {4\,{\left (x^4+x^3\right )}^{3/4}}{3\,x^3} \]

[In]

int(1/(x*(x^3 + x^4)^(1/4)),x)

[Out]

-(4*(x^3 + x^4)^(3/4))/(3*x^3)