\(\int \frac {1}{(4+x^2) \sqrt {1+4 x^2}} \, dx\) [239]

Optimal result

Integrand size = 19, antiderivative size = 31 \[ \int \frac {1}{\left (4+x^2\right ) \sqrt {1+4 x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {15} x}{2 \sqrt {1+4 x^2}}\right )}{2 \sqrt {15}} \]

[Out]

1/30*arctanh(1/2*x*15^(1/2)/(4*x^2+1)^(1/2))*15^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {385, 212} \[ \int \frac {1}{\left (4+x^2\right ) \sqrt {1+4 x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {15} x}{2 \sqrt {4 x^2+1}}\right )}{2 \sqrt {15}} \]

[In]

Int[1/((4 + x^2)*Sqrt[1 + 4*x^2]),x]

[Out]

ArcTanh[(Sqrt[15]*x)/(2*Sqrt[1 + 4*x^2])]/(2*Sqrt[15])

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 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}{4-15 x^2} \, dx,x,\frac {x}{\sqrt {1+4 x^2}}\right ) \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {15} x}{2 \sqrt {1+4 x^2}}\right )}{2 \sqrt {15}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\left (4+x^2\right ) \sqrt {1+4 x^2}} \, dx=\frac {\text {arctanh}\left (\frac {8+2 x^2-x \sqrt {1+4 x^2}}{2 \sqrt {15}}\right )}{2 \sqrt {15}} \]

[In]

Integrate[1/((4 + x^2)*Sqrt[1 + 4*x^2]),x]

[Out]

ArcTanh[(8 + 2*x^2 - x*Sqrt[1 + 4*x^2])/(2*Sqrt[15])]/(2*Sqrt[15])

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {\operatorname {arctanh}\left (\frac {x \sqrt {15}}{2 \sqrt {4 x^{2}+1}}\right ) \sqrt {15}}{30}\)	\(22\)
pseudoelliptic	\(\frac {\sqrt {15}\, \operatorname {arctanh}\left (\frac {2 \sqrt {4 x^{2}+1}\, \sqrt {15}}{15 x}\right )}{30}\)	\(24\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {31 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x^{2}+60 \sqrt {4 x^{2}+1}\, x +4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{x^{2}+4}\right )}{60}\)	\(50\)

[In]

int(1/(x^2+4)/(4*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/30*arctanh(1/2*x*15^(1/2)/(4*x^2+1)^(1/2))*15^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (21) = 42\).

Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {1}{\left (4+x^2\right ) \sqrt {1+4 x^2}} \, dx=\frac {1}{60} \, \sqrt {15} \log \left (\frac {961 \, x^{2} + 8 \, \sqrt {15} {\left (31 \, x^{2} + 4\right )} + 4 \, \sqrt {4 \, x^{2} + 1} {\left (31 \, \sqrt {15} x + 120 \, x\right )} + 124}{x^{2} + 4}\right ) \]

[In]

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

[Out]

1/60*sqrt(15)*log((961*x^2 + 8*sqrt(15)*(31*x^2 + 4) + 4*sqrt(4*x^2 + 1)*(31*sqrt(15)*x + 120*x) + 124)/(x^2 +
4))

Sympy [F]

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

[In]

integrate(1/(x**2+4)/(4*x**2+1)**(1/2),x)

[Out]

Integral(1/((x**2 + 4)*sqrt(4*x**2 + 1)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(1/(sqrt(4*x^2 + 1)*(x^2 + 4)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (21) = 42\).

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.84 \[ \int \frac {1}{\left (4+x^2\right ) \sqrt {1+4 x^2}} \, dx=-\frac {1}{60} \, \sqrt {15} \log \left (\frac {{\left (2 \, x - \sqrt {4 \, x^{2} + 1}\right )}^{2} - 8 \, \sqrt {15} + 31}{{\left (2 \, x - \sqrt {4 \, x^{2} + 1}\right )}^{2} + 8 \, \sqrt {15} + 31}\right ) \]

[In]

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

[Out]

-1/60*sqrt(15)*log(((2*x - sqrt(4*x^2 + 1))^2 - 8*sqrt(15) + 31)/((2*x - sqrt(4*x^2 + 1))^2 + 8*sqrt(15) + 31)
)

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.97 \[ \int \frac {1}{\left (4+x^2\right ) \sqrt {1+4 x^2}} \, dx=-\frac {\sqrt {15}\,\left (\ln \left (x-2{}\mathrm {i}\right )-\ln \left (x+\frac {\sqrt {15}\,\sqrt {x^2+\frac {1}{4}}}{4}-\frac {1}{8}{}\mathrm {i}\right )\right )}{60}+\frac {\sqrt {15}\,\left (\ln \left (x+2{}\mathrm {i}\right )-\ln \left (x-\frac {\sqrt {15}\,\sqrt {x^2+\frac {1}{4}}}{4}+\frac {1}{8}{}\mathrm {i}\right )\right )}{60} \]

[In]

int(1/((x^2 + 4)*(4*x^2 + 1)^(1/2)),x)

[Out]

(15^(1/2)*(log(x + 2i) - log(x - (15^(1/2)*(x^2 + 1/4)^(1/2))/4 + 1i/8)))/60 - (15^(1/2)*(log(x - 2i) - log(x
+ (15^(1/2)*(x^2 + 1/4)^(1/2))/4 - 1i/8)))/60