\(\int \frac {(b+a x^4)^{3/4}}{x^4} \, dx\) [991]

Optimal result

Integrand size = 15, antiderivative size = 75 \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4} \, dx=-\frac {\left (b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {283, 246, 218, 212, 209} \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4} \, dx=\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )-\frac {\left (a x^4+b\right )^{3/4}}{3 x^3} \]

[In]

Int[(b + a*x^4)^(3/4)/x^4,x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 283

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

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4} \, dx=-\frac {\left (b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right ) \]

[In]

Integrate[(b + a*x^4)^(3/4)/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20

method	result	size
pseudoelliptic	\(\frac {3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) a^{\frac {3}{4}} x^{3}-6 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) a^{\frac {3}{4}} x^{3}-4 \left (a \,x^{4}+b \right )^{\frac {3}{4}}}{12 x^{3}}\)	\(90\)

[In]

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

[Out]

1/12*(3*ln((-a^(1/4)*x-(a*x^4+b)^(1/4))/(a^(1/4)*x-(a*x^4+b)^(1/4)))*a^(3/4)*x^3-6*arctan(1/a^(1/4)/x*(a*x^4+b
)^(1/4))*a^(3/4)*x^3-4*(a*x^4+b)^(3/4))/x^3

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.83 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.56 \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4} \, dx=\frac {b^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

b**(3/4)*gamma(-3/4)*hyper((-3/4, -3/4), (1/4,), a*x**4*exp_polar(I*pi)/b)/(4*x**3*gamma(1/4))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13 \[ \int \frac {\left (b+a x^4\right )^{3/4}}{x^4} \, dx=-\frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )} - \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate((a*x^4 + b)^(3/4)/x^4, x)

Mupad [F(-1)]

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

[In]

int((b + a*x^4)^(3/4)/x^4,x)

[Out]

int((b + a*x^4)^(3/4)/x^4, x)