\(\int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx\) [1006]

Optimal result

Integrand size = 15, antiderivative size = 44 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx=-\frac {\left (a+b x^4\right )^{5/4}}{9 a x^9}+\frac {4 b \left (a+b x^4\right )^{5/4}}{45 a^2 x^5} \]

[Out]

-1/9*(b*x^4+a)^(5/4)/a/x^9+4/45*b*(b*x^4+a)^(5/4)/a^2/x^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {277, 270} \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx=\frac {4 b \left (a+b x^4\right )^{5/4}}{45 a^2 x^5}-\frac {\left (a+b x^4\right )^{5/4}}{9 a x^9} \]

[In]

Int[(a + b*x^4)^(1/4)/x^10,x]

[Out]

-1/9*(a + b*x^4)^(5/4)/(a*x^9) + (4*b*(a + b*x^4)^(5/4))/(45*a^2*x^5)

Rule 270

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

Rule 277

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+b x^4\right )^{5/4}}{9 a x^9}-\frac {(4 b) \int \frac {\sqrt [4]{a+b x^4}}{x^6} \, dx}{9 a} \\ & = -\frac {\left (a+b x^4\right )^{5/4}}{9 a x^9}+\frac {4 b \left (a+b x^4\right )^{5/4}}{45 a^2 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx=\frac {\sqrt [4]{a+b x^4} \left (-5 a^2-a b x^4+4 b^2 x^8\right )}{45 a^2 x^9} \]

[In]

Integrate[(a + b*x^4)^(1/4)/x^10,x]

[Out]

((a + b*x^4)^(1/4)*(-5*a^2 - a*b*x^4 + 4*b^2*x^8))/(45*a^2*x^9)

Maple [A] (verified)

Time = 4.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.64

method	result	size
gosper	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}} \left (-4 b \,x^{4}+5 a \right )}{45 a^{2} x^{9}}\)	\(28\)
pseudoelliptic	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}} \left (-4 b \,x^{4}+5 a \right )}{45 a^{2} x^{9}}\)	\(28\)
trager	\(-\frac {\left (-4 b^{2} x^{8}+a b \,x^{4}+5 a^{2}\right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{45 a^{2} x^{9}}\)	\(38\)
risch	\(-\frac {\left (-4 b^{2} x^{8}+a b \,x^{4}+5 a^{2}\right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{45 a^{2} x^{9}}\)	\(38\)

[In]

int((b*x^4+a)^(1/4)/x^10,x,method=_RETURNVERBOSE)

[Out]

-1/45*(b*x^4+a)^(5/4)*(-4*b*x^4+5*a)/a^2/x^9

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx=\frac {{\left (4 \, b^{2} x^{8} - a b x^{4} - 5 \, a^{2}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{45 \, a^{2} x^{9}} \]

[In]

integrate((b*x^4+a)^(1/4)/x^10,x, algorithm="fricas")

[Out]

1/45*(4*b^2*x^8 - a*b*x^4 - 5*a^2)*(b*x^4 + a)^(1/4)/(a^2*x^9)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (37) = 74\).

Time = 0.66 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.48 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx=- \frac {5 \sqrt [4]{b} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{16 x^{8} \Gamma \left (- \frac {1}{4}\right )} - \frac {b^{\frac {5}{4}} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{16 a x^{4} \Gamma \left (- \frac {1}{4}\right )} + \frac {b^{\frac {9}{4}} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{4 a^{2} \Gamma \left (- \frac {1}{4}\right )} \]

[In]

integrate((b*x**4+a)**(1/4)/x**10,x)

[Out]

-5*b**(1/4)*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(16*x**8*gamma(-1/4)) - b**(5/4)*(a/(b*x**4) + 1)**(1/4)*gamma
(-9/4)/(16*a*x**4*gamma(-1/4)) + b**(9/4)*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(4*a**2*gamma(-1/4))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx=\frac {\frac {9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b}{x^{5}} - \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}}}{x^{9}}}{45 \, a^{2}} \]

[In]

integrate((b*x^4+a)^(1/4)/x^10,x, algorithm="maxima")

[Out]

1/45*(9*(b*x^4 + a)^(5/4)*b/x^5 - 5*(b*x^4 + a)^(9/4)/x^9)/a^2

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x^{10}} \,d x } \]

[In]

integrate((b*x^4+a)^(1/4)/x^10,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(1/4)/x^10, x)

Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{10}} \, dx=-\frac {{\left (b\,x^4+a\right )}^{1/4}\,\left (5\,a^2+a\,b\,x^4-4\,b^2\,x^8\right )}{45\,a^2\,x^9} \]

[In]

int((a + b*x^4)^(1/4)/x^10,x)

[Out]

-((a + b*x^4)^(1/4)*(5*a^2 - 4*b^2*x^8 + a*b*x^4))/(45*a^2*x^9)