\(\int \frac {1}{(a+b x^4)^{9/4}} \, dx\) [1171]

Optimal result

Integrand size = 11, antiderivative size = 39 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {x}{5 a \left (a+b x^4\right )^{5/4}}+\frac {4 x}{5 a^2 \sqrt [4]{a+b x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {198, 197} \[ \int \frac {1}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {4 x}{5 a^2 \sqrt [4]{a+b x^4}}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}} \]

[In]

Int[(a + b*x^4)^(-9/4),x]

[Out]

x/(5*a*(a + b*x^4)^(5/4)) + (4*x)/(5*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 198

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

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {5 a x+4 b x^5}{5 a^2 \left (a+b x^4\right )^{5/4}} \]

[In]

Integrate[(a + b*x^4)^(-9/4),x]

[Out]

(5*a*x + 4*b*x^5)/(5*a^2*(a + b*x^4)^(5/4))

Maple [A] (verified)

Time = 4.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {x \left (4 b \,x^{4}+5 a \right )}{5 \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{2}}\)	\(26\)
trager	\(\frac {x \left (4 b \,x^{4}+5 a \right )}{5 \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{2}}\)	\(26\)
pseudoelliptic	\(\frac {x \left (4 b \,x^{4}+5 a \right )}{5 \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{2}}\)	\(26\)

[In]

int(1/(b*x^4+a)^(9/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {{\left (4 \, b x^{5} + 5 \, a x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{5 \, {\left (a^{2} b^{2} x^{8} + 2 \, a^{3} b x^{4} + a^{4}\right )}} \]

[In]

integrate(1/(b*x^4+a)^(9/4),x, algorithm="fricas")

[Out]

1/5*(4*b*x^5 + 5*a*x)*(b*x^4 + a)^(3/4)/(a^2*b^2*x^8 + 2*a^3*b*x^4 + a^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (32) = 64\).

Time = 0.65 (sec) , antiderivative size = 126, normalized size of antiderivative = 3.23 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {5 a x \Gamma \left (\frac {1}{4}\right )}{16 a^{\frac {13}{4}} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right ) + 16 a^{\frac {9}{4}} b x^{4} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right )} + \frac {4 b x^{5} \Gamma \left (\frac {1}{4}\right )}{16 a^{\frac {13}{4}} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right ) + 16 a^{\frac {9}{4}} b x^{4} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right )} \]

[In]

integrate(1/(b*x**4+a)**(9/4),x)

[Out]

5*a*x*gamma(1/4)/(16*a**(13/4)*(1 + b*x**4/a)**(1/4)*gamma(9/4) + 16*a**(9/4)*b*x**4*(1 + b*x**4/a)**(1/4)*gam
ma(9/4)) + 4*b*x**5*gamma(1/4)/(16*a**(13/4)*(1 + b*x**4/a)**(1/4)*gamma(9/4) + 16*a**(9/4)*b*x**4*(1 + b*x**4
/a)**(1/4)*gamma(9/4))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4}} \, dx=-\frac {{\left (b - \frac {5 \, {\left (b x^{4} + a\right )}}{x^{4}}\right )} x^{5}}{5 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}} \]

[In]

integrate(1/(b*x^4+a)^(9/4),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{9/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate(1/(b*x^4+a)^(9/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(-9/4), x)

Mupad [B] (verification not implemented)

Time = 5.53 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {4\,x\,\left (b\,x^4+a\right )+a\,x}{5\,a^2\,{\left (b\,x^4+a\right )}^{5/4}} \]

[In]

int(1/(a + b*x^4)^(9/4),x)

[Out]

(4*x*(a + b*x^4) + a*x)/(5*a^2*(a + b*x^4)^(5/4))