\(\int \frac {(b+a x^4)^{3/4}}{x^4 (2 b+a x^4)} \, dx\) [1476]

Optimal result

Integrand size = 26, antiderivative size = 104 \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx=-\frac {\left (b+a x^4\right )^{3/4}}{6 b x^3}+\frac {a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} b}+\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} b} \]

[Out]

-1/6*(a*x^4+b)^(3/4)/b/x^3+1/8*a^(3/4)*arctan(1/2*a^(1/4)*x*2^(3/4)/(a*x^4+b)^(1/4))*2^(1/4)/b+1/8*a^(3/4)*arc
tanh(1/2*a^(1/4)*x*2^(3/4)/(a*x^4+b)^(1/4))*2^(1/4)/b

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.44, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {525, 524} \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx=-\frac {\left (a x^4+b\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},\frac {a x^4}{2 \left (a x^4+b\right )}\right )}{6 b x^3} \]

[In]

Int[(b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)),x]

[Out]

-1/6*((b + a*x^4)^(3/4)*Hypergeometric2F1[-3/4, 1, 1/4, (a*x^4)/(2*(b + a*x^4))])/(b*x^3)

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b+a x^4\right )^{3/4} \int \frac {\left (1+\frac {a x^4}{b}\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \\ & = -\frac {\left (b+a x^4\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},\frac {a x^4}{2 \left (b+a x^4\right )}\right )}{6 b x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx=-\frac {\left (b+a x^4\right )^{3/4}}{6 b x^3}+\frac {a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} b}+\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} b} \]

[In]

Integrate[(b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)),x]

[Out]

-1/6*(b + a*x^4)^(3/4)/(b*x^3) + (a^(3/4)*ArcTan[(a^(1/4)*x)/(2^(1/4)*(b + a*x^4)^(1/4))])/(4*2^(3/4)*b) + (a^
(3/4)*ArcTanh[(a^(1/4)*x)/(2^(1/4)*(b + a*x^4)^(1/4))])/(4*2^(3/4)*b)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11

method	result	size
pseudoelliptic	\(-\frac {\left (\arctan \left (\frac {2^{\frac {1}{4}} \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) a \sqrt {2}\, x^{3}-\frac {\ln \left (\frac {-2^{\frac {3}{4}} a^{\frac {1}{4}} x -2 \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{2^{\frac {3}{4}} a^{\frac {1}{4}} x -2 \left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) a \sqrt {2}\, x^{3}}{2}+\frac {4 \left (a \,x^{4}+b \right )^{\frac {3}{4}} 2^{\frac {1}{4}} a^{\frac {1}{4}}}{3}\right ) 2^{\frac {3}{4}}}{16 a^{\frac {1}{4}} x^{3} b}\)	\(115\)

[In]

int((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x,method=_RETURNVERBOSE)

[Out]

-1/16/a^(1/4)*(arctan(1/a^(1/4)/x*2^(1/4)*(a*x^4+b)^(1/4))*a*2^(1/2)*x^3-1/2*ln((-2^(3/4)*a^(1/4)*x-2*(a*x^4+b
)^(1/4))/(2^(3/4)*a^(1/4)*x-2*(a*x^4+b)^(1/4)))*a*2^(1/2)*x^3+4/3*(a*x^4+b)^(3/4)*2^(1/4)*a^(1/4))*2^(3/4)/x^3
/b

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 34.11 (sec) , antiderivative size = 565, normalized size of antiderivative = 5.43 \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx=\frac {3 \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {a^{3}}{b^{4}}} + 8 \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} b^{3} x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} a^{2} x + \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (3 \, a^{2} b x^{4} + 2 \, a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) + 3 i \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {a^{3}}{b^{4}}} + 8 i \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} b^{3} x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} a^{2} x + \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (-3 i \, a^{2} b x^{4} - 2 i \, a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) - 3 i \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {a^{3}}{b^{4}}} - 8 i \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} b^{3} x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} a^{2} x + \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (3 i \, a^{2} b x^{4} + 2 i \, a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) - 3 \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {a^{3}}{b^{4}}} - 8 \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} b^{3} x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} a^{2} x - \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (3 \, a^{2} b x^{4} + 2 \, a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) - 8 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}}}{48 \, b x^{3}} \]

[In]

integrate((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x, algorithm="fricas")

[Out]

1/48*(3*(1/8)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(1/2*(2*sqrt(1/2)*(a*x^4 + b)^(1/4)*a*b^2*x^3*sqrt(a^3/b^4) + 8*(
1/8)^(3/4)*sqrt(a*x^4 + b)*b^3*x^2*(a^3/b^4)^(3/4) + 2*(a*x^4 + b)^(3/4)*a^2*x + (1/8)^(1/4)*(3*a^2*b*x^4 + 2*
a*b^2)*(a^3/b^4)^(1/4))/(a*x^4 + 2*b)) + 3*I*(1/8)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(-1/2*(2*sqrt(1/2)*(a*x^4 +
b)^(1/4)*a*b^2*x^3*sqrt(a^3/b^4) + 8*I*(1/8)^(3/4)*sqrt(a*x^4 + b)*b^3*x^2*(a^3/b^4)^(3/4) - 2*(a*x^4 + b)^(3/
4)*a^2*x + (1/8)^(1/4)*(-3*I*a^2*b*x^4 - 2*I*a*b^2)*(a^3/b^4)^(1/4))/(a*x^4 + 2*b)) - 3*I*(1/8)^(1/4)*b*x^3*(a
^3/b^4)^(1/4)*log(-1/2*(2*sqrt(1/2)*(a*x^4 + b)^(1/4)*a*b^2*x^3*sqrt(a^3/b^4) - 8*I*(1/8)^(3/4)*sqrt(a*x^4 + b
)*b^3*x^2*(a^3/b^4)^(3/4) - 2*(a*x^4 + b)^(3/4)*a^2*x + (1/8)^(1/4)*(3*I*a^2*b*x^4 + 2*I*a*b^2)*(a^3/b^4)^(1/4
))/(a*x^4 + 2*b)) - 3*(1/8)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(1/2*(2*sqrt(1/2)*(a*x^4 + b)^(1/4)*a*b^2*x^3*sqrt(
a^3/b^4) - 8*(1/8)^(3/4)*sqrt(a*x^4 + b)*b^3*x^2*(a^3/b^4)^(3/4) + 2*(a*x^4 + b)^(3/4)*a^2*x - (1/8)^(1/4)*(3*
a^2*b*x^4 + 2*a*b^2)*(a^3/b^4)^(1/4))/(a*x^4 + 2*b)) - 8*(a*x^4 + b)^(3/4))/(b*x^3)

Sympy [F]

\[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx=\int \frac {\left (a x^{4} + b\right )^{\frac {3}{4}}}{x^{4} \left (a x^{4} + 2 b\right )}\, dx \]

[In]

integrate((a*x**4+b)**(3/4)/x**4/(a*x**4+2*b),x)

[Out]

Integral((a*x**4 + b)**(3/4)/(x**4*(a*x**4 + 2*b)), x)

Maxima [F]

\[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx=\int { \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}}}{{\left (a x^{4} + 2 \, b\right )} x^{4}} \,d x } \]

[In]

integrate((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x, algorithm="maxima")

[Out]

integrate((a*x^4 + b)^(3/4)/((a*x^4 + 2*b)*x^4), x)

Giac [F]

\[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx=\int { \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}}}{{\left (a x^{4} + 2 \, b\right )} x^{4}} \,d x } \]

[In]

integrate((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x, algorithm="giac")

[Out]

integrate((a*x^4 + b)^(3/4)/((a*x^4 + 2*b)*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx=\int \frac {{\left (a\,x^4+b\right )}^{3/4}}{x^4\,\left (a\,x^4+2\,b\right )} \,d x \]

[In]

int((b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)),x)

[Out]

int((b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)), x)