\(\int \frac {1+2 x^4}{x^4 (1+x^4)^{5/4}} \, dx\) [230]

Optimal result

Integrand size = 20, antiderivative size = 23 \[ \int \frac {1+2 x^4}{x^4 \left (1+x^4\right )^{5/4}} \, dx=\frac {-1+2 x^4}{3 x^3 \sqrt [4]{1+x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {464, 197} \[ \int \frac {1+2 x^4}{x^4 \left (1+x^4\right )^{5/4}} \, dx=\frac {2 x}{3 \sqrt [4]{x^4+1}}-\frac {1}{3 x^3 \sqrt [4]{x^4+1}} \]

[In]

Int[(1 + 2*x^4)/(x^4*(1 + x^4)^(5/4)),x]

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 x^3 \sqrt [4]{1+x^4}}+\frac {2}{3} \int \frac {1}{\left (1+x^4\right )^{5/4}} \, dx \\ & = -\frac {1}{3 x^3 \sqrt [4]{1+x^4}}+\frac {2 x}{3 \sqrt [4]{1+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x^4}{x^4 \left (1+x^4\right )^{5/4}} \, dx=\frac {-1+2 x^4}{3 x^3 \sqrt [4]{1+x^4}} \]

[In]

Integrate[(1 + 2*x^4)/(x^4*(1 + x^4)^(5/4)),x]

[Out]

(-1 + 2*x^4)/(3*x^3*(1 + x^4)^(1/4))

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
gosper	\(\frac {2 x^{4}-1}{3 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(20\)
trager	\(\frac {2 x^{4}-1}{3 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(20\)
risch	\(\frac {2 x^{4}-1}{3 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(20\)
pseudoelliptic	\(\frac {2 x^{4}-1}{3 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(20\)
meijerg	\(-\frac {4 x^{4}+1}{3 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}}+\frac {2 x}{\left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(31\)

[In]

int((2*x^4+1)/x^4/(x^4+1)^(5/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {1+2 x^4}{x^4 \left (1+x^4\right )^{5/4}} \, dx=\frac {{\left (2 \, x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {3}{4}}}{3 \, {\left (x^{7} + x^{3}\right )}} \]

[In]

integrate((2*x^4+1)/x^4/(x^4+1)^(5/4),x, algorithm="fricas")

[Out]

1/3*(2*x^4 - 1)*(x^4 + 1)^(3/4)/(x^7 + x^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (20) = 40\).

Time = 2.49 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.22 \[ \int \frac {1+2 x^4}{x^4 \left (1+x^4\right )^{5/4}} \, dx=\frac {4 x^{4} \left (x^{4} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{16 x^{7} \Gamma \left (\frac {5}{4}\right ) + 16 x^{3} \Gamma \left (\frac {5}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right )}{2 \sqrt [4]{x^{4} + 1} \Gamma \left (\frac {5}{4}\right )} + \frac {\left (x^{4} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{16 x^{7} \Gamma \left (\frac {5}{4}\right ) + 16 x^{3} \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate((2*x**4+1)/x**4/(x**4+1)**(5/4),x)

[Out]

4*x**4*(x**4 + 1)**(3/4)*gamma(-3/4)/(16*x**7*gamma(5/4) + 16*x**3*gamma(5/4)) + x*gamma(1/4)/(2*(x**4 + 1)**(
1/4)*gamma(5/4)) + (x**4 + 1)**(3/4)*gamma(-3/4)/(16*x**7*gamma(5/4) + 16*x**3*gamma(5/4))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1+2 x^4}{x^4 \left (1+x^4\right )^{5/4}} \, dx=\frac {x}{{\left (x^{4} + 1\right )}^{\frac {1}{4}}} - \frac {{\left (x^{4} + 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

integrate((2*x^4+1)/x^4/(x^4+1)^(5/4),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1+2 x^4}{x^4 \left (1+x^4\right )^{5/4}} \, dx=\int { \frac {2 \, x^{4} + 1}{{\left (x^{4} + 1\right )}^{\frac {5}{4}} x^{4}} \,d x } \]

[In]

integrate((2*x^4+1)/x^4/(x^4+1)^(5/4),x, algorithm="giac")

[Out]

integrate((2*x^4 + 1)/((x^4 + 1)^(5/4)*x^4), x)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x^4}{x^4 \left (1+x^4\right )^{5/4}} \, dx=\frac {2\,x^4-1}{3\,x^3\,{\left (x^4+1\right )}^{1/4}} \]

[In]

int((2*x^4 + 1)/(x^4*(x^4 + 1)^(5/4)),x)

[Out]

(2*x^4 - 1)/(3*x^3*(x^4 + 1)^(1/4))