\(\int \frac {1}{(1+2 x^2)^{5/2}} \, dx\) [468]

Optimal result

Integrand size = 11, antiderivative size = 33 \[ \int \frac {1}{\left (1+2 x^2\right )^{5/2}} \, dx=\frac {x}{3 \left (1+2 x^2\right )^{3/2}}+\frac {2 x}{3 \sqrt {1+2 x^2}} \]

[Out]

1/3*x/(2*x^2+1)^(3/2)+2/3*x/(2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {198, 197} \[ \int \frac {1}{\left (1+2 x^2\right )^{5/2}} \, dx=\frac {2 x}{3 \sqrt {2 x^2+1}}+\frac {x}{3 \left (2 x^2+1\right )^{3/2}} \]

[In]

Int[(1 + 2*x^2)^(-5/2),x]

[Out]

x/(3*(1 + 2*x^2)^(3/2)) + (2*x)/(3*Sqrt[1 + 2*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

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

Rubi steps \begin{align*} \text {integral}& = \frac {x}{3 \left (1+2 x^2\right )^{3/2}}+\frac {2}{3} \int \frac {1}{\left (1+2 x^2\right )^{3/2}} \, dx \\ & = \frac {x}{3 \left (1+2 x^2\right )^{3/2}}+\frac {2 x}{3 \sqrt {1+2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (1+2 x^2\right )^{5/2}} \, dx=\frac {3 x+4 x^3}{3 \left (1+2 x^2\right )^{3/2}} \]

[In]

Integrate[(1 + 2*x^2)^(-5/2),x]

[Out]

(3*x + 4*x^3)/(3*(1 + 2*x^2)^(3/2))

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.61

method	result	size
gosper	\(\frac {x \left (4 x^{2}+3\right )}{3 \left (2 x^{2}+1\right )^{\frac {3}{2}}}\)	\(20\)
trager	\(\frac {x \left (4 x^{2}+3\right )}{3 \left (2 x^{2}+1\right )^{\frac {3}{2}}}\)	\(20\)
meijerg	\(\frac {x \left (4 x^{2}+3\right )}{3 \left (2 x^{2}+1\right )^{\frac {3}{2}}}\)	\(20\)
risch	\(\frac {x \left (4 x^{2}+3\right )}{3 \left (2 x^{2}+1\right )^{\frac {3}{2}}}\)	\(20\)
pseudoelliptic	\(\frac {4 x^{3}+3 x}{3 \left (2 x^{2}+1\right )^{\frac {3}{2}}}\)	\(21\)
default	\(\frac {x}{3 \left (2 x^{2}+1\right )^{\frac {3}{2}}}+\frac {2 x}{3 \sqrt {2 x^{2}+1}}\)	\(26\)

[In]

int(1/(2*x^2+1)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*x*(4*x^2+3)/(2*x^2+1)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (1+2 x^2\right )^{5/2}} \, dx=\frac {{\left (4 \, x^{3} + 3 \, x\right )} \sqrt {2 \, x^{2} + 1}}{3 \, {\left (4 \, x^{4} + 4 \, x^{2} + 1\right )}} \]

[In]

integrate(1/(2*x^2+1)^(5/2),x, algorithm="fricas")

[Out]

1/3*(4*x^3 + 3*x)*sqrt(2*x^2 + 1)/(4*x^4 + 4*x^2 + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (27) = 54\).

Time = 0.86 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.85 \[ \int \frac {1}{\left (1+2 x^2\right )^{5/2}} \, dx=\frac {4 x^{3}}{6 x^{2} \sqrt {2 x^{2} + 1} + 3 \sqrt {2 x^{2} + 1}} + \frac {3 x}{6 x^{2} \sqrt {2 x^{2} + 1} + 3 \sqrt {2 x^{2} + 1}} \]

[In]

integrate(1/(2*x**2+1)**(5/2),x)

[Out]

4*x**3/(6*x**2*sqrt(2*x**2 + 1) + 3*sqrt(2*x**2 + 1)) + 3*x/(6*x**2*sqrt(2*x**2 + 1) + 3*sqrt(2*x**2 + 1))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (1+2 x^2\right )^{5/2}} \, dx=\frac {2 \, x}{3 \, \sqrt {2 \, x^{2} + 1}} + \frac {x}{3 \, {\left (2 \, x^{2} + 1\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(2*x^2+1)^(5/2),x, algorithm="maxima")

[Out]

2/3*x/sqrt(2*x^2 + 1) + 1/3*x/(2*x^2 + 1)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\left (1+2 x^2\right )^{5/2}} \, dx=\frac {{\left (4 \, x^{2} + 3\right )} x}{3 \, {\left (2 \, x^{2} + 1\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(2*x^2+1)^(5/2),x, algorithm="giac")

[Out]

1/3*(4*x^2 + 3)*x/(2*x^2 + 1)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.00 \[ \int \frac {1}{\left (1+2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {x^2+\frac {1}{2}}\,1{}\mathrm {i}}{24\,\left (-x^2+1{}\mathrm {i}\,\sqrt {2}\,x+\frac {1}{2}\right )}+\frac {\sqrt {x^2+\frac {1}{2}}\,1{}\mathrm {i}}{24\,\left (x^2+1{}\mathrm {i}\,\sqrt {2}\,x-\frac {1}{2}\right )}+\frac {\sqrt {2}\,\sqrt {x^2+\frac {1}{2}}}{6\,\left (x-\frac {\sqrt {2}\,1{}\mathrm {i}}{2}\right )}+\frac {\sqrt {2}\,\sqrt {x^2+\frac {1}{2}}}{6\,\left (x+\frac {\sqrt {2}\,1{}\mathrm {i}}{2}\right )} \]

[In]

int(1/(2*x^2 + 1)^(5/2),x)

[Out]

((x^2 + 1/2)^(1/2)*1i)/(24*(2^(1/2)*x*1i - x^2 + 1/2)) + ((x^2 + 1/2)^(1/2)*1i)/(24*(2^(1/2)*x*1i + x^2 - 1/2)
) + (2^(1/2)*(x^2 + 1/2)^(1/2))/(6*(x - (2^(1/2)*1i)/2)) + (2^(1/2)*(x^2 + 1/2)^(1/2))/(6*(x + (2^(1/2)*1i)/2)
)