\(\int \frac {1}{(1+x) \sqrt [4]{x^3+x^4}} \, dx\) [198]

Optimal result

Integrand size = 17, antiderivative size = 21 \[ \int \frac {1}{(1+x) \sqrt [4]{x^3+x^4}} \, dx=\frac {4 \left (x^3+x^4\right )^{3/4}}{x^2 (1+x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2081, 37} \[ \int \frac {1}{(1+x) \sqrt [4]{x^3+x^4}} \, dx=\frac {4 x}{\sqrt [4]{x^4+x^3}} \]

[In]

Int[1/((1 + x)*(x^3 + x^4)^(1/4)),x]

[Out]

(4*x)/(x^3 + x^4)^(1/4)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{3/4} \sqrt [4]{1+x}\right ) \int \frac {1}{x^{3/4} (1+x)^{5/4}} \, dx}{\sqrt [4]{x^3+x^4}} \\ & = \frac {4 x}{\sqrt [4]{x^3+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(1+x) \sqrt [4]{x^3+x^4}} \, dx=\frac {4 x}{\sqrt [4]{x^3 (1+x)}} \]

[In]

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

[Out]

(4*x)/(x^3*(1 + x))^(1/4)

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52

method	result	size
meijerg	\(\frac {4 x^{\frac {1}{4}}}{\left (1+x \right )^{\frac {1}{4}}}\)	\(11\)
gosper	\(\frac {4 x}{\left (x^{4}+x^{3}\right )^{\frac {1}{4}}}\)	\(13\)
risch	\(\frac {4 x}{\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}\)	\(13\)
pseudoelliptic	\(\frac {4 x}{\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}\)	\(13\)
trager	\(\frac {4 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}}{x^{2} \left (1+x \right )}\)	\(20\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1+x) \sqrt [4]{x^3+x^4}} \, dx=\frac {4 \, {\left (x^{4} + x^{3}\right )}^{\frac {3}{4}}}{x^{3} + x^{2}} \]

[In]

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

[Out]

4*(x^4 + x^3)^(3/4)/(x^3 + x^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.43 \[ \int \frac {1}{(1+x) \sqrt [4]{x^3+x^4}} \, dx=\frac {4}{{\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

4/(1/x + 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 5.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(1+x) \sqrt [4]{x^3+x^4}} \, dx=\frac {4\,{\left (x^4+x^3\right )}^{3/4}}{x^2\,\left (x+1\right )} \]

[In]

int(1/((x^3 + x^4)^(1/4)*(x + 1)),x)

[Out]

(4*(x^3 + x^4)^(3/4))/(x^2*(x + 1))