\(\int \frac {1}{(a-(a-b) x^4) \sqrt [4]{a+b x^4}} \, dx\) [217]

Optimal result

Integrand size = 26, antiderivative size = 57 \[ \int \frac {1}{\left (a-(a-b) x^4\right ) \sqrt [4]{a+b x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}\right )}{2 a^{5/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}\right )}{2 a^{5/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {385, 218, 212, 209} \[ \int \frac {1}{\left (a-(a-b) x^4\right ) \sqrt [4]{a+b x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}\right )}{2 a^{5/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}\right )}{2 a^{5/4}} \]

[In]

Int[1/((a - (a - b)*x^4)*(a + b*x^4)^(1/4)),x]

[Out]

ArcTan[(a^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*a^(5/4)) + ArcTanh[(a^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*a^(5/4))

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 385

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

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a-(a-b) x^4\right ) \sqrt [4]{a+b x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}\right )}{2 a^{5/4}} \]

[In]

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

[Out]

(ArcTan[(a^(1/4)*x)/(a + b*x^4)^(1/4)] + ArcTanh[(a^(1/4)*x)/(a + b*x^4)^(1/4)])/(2*a^(5/4))

Maple [A] (verified)

Time = 4.87 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(\frac {-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\ln \left (\frac {-a^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )}{4 a^{\frac {5}{4}}}\)	\(65\)

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a-(a-b) x^4\right ) \sqrt [4]{a+b x^4}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

-Integral(1/(a*x**4*(a + b*x**4)**(1/4) - a*(a + b*x**4)**(1/4) - b*x**4*(a + b*x**4)**(1/4)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a-(a-b) x^4\right ) \sqrt [4]{a+b x^4}} \, dx=\int \frac {1}{{\left (b\,x^4+a\right )}^{1/4}\,\left (a-x^4\,\left (a-b\right )\right )} \,d x \]

[In]

int(1/((a + b*x^4)^(1/4)*(a - x^4*(a - b))),x)

[Out]

int(1/((a + b*x^4)^(1/4)*(a - x^4*(a - b))), x)