\(\int \frac {1}{x^4 (a+b x^4)^{5/4}} \, dx\) [1158]

Optimal result

Integrand size = 15, antiderivative size = 42 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{3 a x^3 \sqrt [4]{a+b x^4}}-\frac {4 b x}{3 a^2 \sqrt [4]{a+b x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, 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, 197} \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {4 b x}{3 a^2 \sqrt [4]{a+b x^4}}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^4}} \]

[In]

Int[1/(x^4*(a + b*x^4)^(5/4)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{5/4}} \, dx=\frac {-a-4 b x^4}{3 a^2 x^3 \sqrt [4]{a+b x^4}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.62

method	result	size
gosper	\(-\frac {4 b \,x^{4}+a}{3 x^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}}\)	\(26\)
trager	\(-\frac {4 b \,x^{4}+a}{3 x^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}}\)	\(26\)
pseudoelliptic	\(-\frac {4 b \,x^{4}+a}{3 x^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}}\)	\(26\)
risch	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{3 a^{2} x^{3}}-\frac {b x}{a^{2} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {{\left (4 \, b x^{4} + a\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{3 \, {\left (a^{2} b x^{7} + a^{3} x^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.62 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{5/4}} \, dx=\frac {\Gamma \left (- \frac {3}{4}\right )}{16 a \sqrt [4]{b} x^{4} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (\frac {5}{4}\right )} + \frac {b^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{4 a^{2} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate(1/x**4/(b*x**4+a)**(5/4),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {b x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{3 \, a^{2} x^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {4\,b\,x^4+a}{3\,a^2\,x^3\,{\left (b\,x^4+a\right )}^{1/4}} \]

[In]

int(1/(x^4*(a + b*x^4)^(5/4)),x)

[Out]

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