\(\int \frac {1}{(a+b x^4) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx\) [1000]

Optimal result

Integrand size = 33, antiderivative size = 40 \[ \int \frac {1}{\left (a+b x^4\right ) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {c x^2+d \sqrt {a+b x^4}}}\right )}{a \sqrt {c}} \]

[Out]

arctanh(x*c^(1/2)/(c*x^2+d*(b*x^4+a)^(1/2))^(1/2))/a/c^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2153, 212} \[ \int \frac {1}{\left (a+b x^4\right ) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {d \sqrt {a+b x^4}+c x^2}}\right )}{a \sqrt {c}} \]

[In]

Int[1/((a + b*x^4)*Sqrt[c*x^2 + d*Sqrt[a + b*x^4]]),x]

[Out]

ArcTanh[(Sqrt[c]*x)/Sqrt[c*x^2 + d*Sqrt[a + b*x^4]]]/(a*Sqrt[c])

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 2153

Int[1/(((a_) + (b_.)*(x_)^(n_.))*Sqrt[(c_.)*(x_)^2 + (d_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.)]), x_Symbol] :> Dis
t[1/a, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[c*x^2 + d*(a + b*x^n)^(2/n)]], x] /; FreeQ[{a, b, c, d, n}, x] &
& EqQ[p, 2/n]

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

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (a+b x^4\right ) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c x^2+d \sqrt {a+b x^4}}}{\sqrt {c} x}\right )}{a \sqrt {c}} \]

[In]

Integrate[1/((a + b*x^4)*Sqrt[c*x^2 + d*Sqrt[a + b*x^4]]),x]

[Out]

ArcTanh[Sqrt[c*x^2 + d*Sqrt[a + b*x^4]]/(Sqrt[c]*x)]/(a*Sqrt[c])

Maple [F]

\[\int \frac {1}{\left (b \,x^{4}+a \right ) \sqrt {c \,x^{2}+d \sqrt {b \,x^{4}+a}}}d x\]

[In]

int(1/(b*x^4+a)/(c*x^2+d*(b*x^4+a)^(1/2))^(1/2),x)

[Out]

int(1/(b*x^4+a)/(c*x^2+d*(b*x^4+a)^(1/2))^(1/2),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right ) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x^4+a)/(c*x^2+d*(b*x^4+a)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\left (a+b x^4\right ) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx=\int \frac {1}{\left (a + b x^{4}\right ) \sqrt {c x^{2} + d \sqrt {a + b x^{4}}}}\, dx \]

[In]

integrate(1/(b*x**4+a)/(c*x**2+d*(b*x**4+a)**(1/2))**(1/2),x)

[Out]

Integral(1/((a + b*x**4)*sqrt(c*x**2 + d*sqrt(a + b*x**4))), x)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^4\right ) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )} \sqrt {c x^{2} + \sqrt {b x^{4} + a} d}} \,d x } \]

[In]

integrate(1/(b*x^4+a)/(c*x^2+d*(b*x^4+a)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)*sqrt(c*x^2 + sqrt(b*x^4 + a)*d)), x)

Giac [F]

\[ \int \frac {1}{\left (a+b x^4\right ) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )} \sqrt {c x^{2} + \sqrt {b x^{4} + a} d}} \,d x } \]

[In]

integrate(1/(b*x^4+a)/(c*x^2+d*(b*x^4+a)^(1/2))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)*sqrt(c*x^2 + sqrt(b*x^4 + a)*d)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right ) \sqrt {c x^2+d \sqrt {a+b x^4}}} \, dx=\int \frac {1}{\left (b\,x^4+a\right )\,\sqrt {d\,\sqrt {b\,x^4+a}+c\,x^2}} \,d x \]

[In]

int(1/((a + b*x^4)*(d*(a + b*x^4)^(1/2) + c*x^2)^(1/2)),x)

[Out]

int(1/((a + b*x^4)*(d*(a + b*x^4)^(1/2) + c*x^2)^(1/2)), x)