\(\int \frac {1}{(1+8 x+3 x^2)^{5/2}} \, dx\) [285]

Optimal result

Integrand size = 14, antiderivative size = 47 \[ \int \frac {1}{\left (1+8 x+3 x^2\right )^{5/2}} \, dx=-\frac {4+3 x}{39 \left (1+8 x+3 x^2\right )^{3/2}}+\frac {2 (4+3 x)}{169 \sqrt {1+8 x+3 x^2}} \]

[Out]

1/39*(-4-3*x)/(3*x^2+8*x+1)^(3/2)+2/169*(4+3*x)/(3*x^2+8*x+1)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {628, 627} \[ \int \frac {1}{\left (1+8 x+3 x^2\right )^{5/2}} \, dx=\frac {2 (3 x+4)}{169 \sqrt {3 x^2+8 x+1}}-\frac {3 x+4}{39 \left (3 x^2+8 x+1\right )^{3/2}} \]

[In]

Int[(1 + 8*x + 3*x^2)^(-5/2),x]

[Out]

-1/39*(4 + 3*x)/(1 + 8*x + 3*x^2)^(3/2) + (2*(4 + 3*x))/(169*Sqrt[1 + 8*x + 3*x^2])

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {4+3 x}{39 \left (1+8 x+3 x^2\right )^{3/2}}-\frac {2}{13} \int \frac {1}{\left (1+8 x+3 x^2\right )^{3/2}} \, dx \\ & = -\frac {4+3 x}{39 \left (1+8 x+3 x^2\right )^{3/2}}+\frac {2 (4+3 x)}{169 \sqrt {1+8 x+3 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (1+8 x+3 x^2\right )^{5/2}} \, dx=\frac {(4+3 x) \left (-7+48 x+18 x^2\right )}{507 \left (1+8 x+3 x^2\right )^{3/2}} \]

[In]

Integrate[(1 + 8*x + 3*x^2)^(-5/2),x]

[Out]

((4 + 3*x)*(-7 + 48*x + 18*x^2))/(507*(1 + 8*x + 3*x^2)^(3/2))

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.64

method	result	size
gosper	\(\frac {54 x^{3}+216 x^{2}+171 x -28}{507 \left (3 x^{2}+8 x +1\right )^{\frac {3}{2}}}\)	\(30\)
trager	\(\frac {54 x^{3}+216 x^{2}+171 x -28}{507 \left (3 x^{2}+8 x +1\right )^{\frac {3}{2}}}\)	\(30\)
risch	\(\frac {54 x^{3}+216 x^{2}+171 x -28}{507 \left (3 x^{2}+8 x +1\right )^{\frac {3}{2}}}\)	\(30\)
default	\(-\frac {6 x +8}{78 \left (3 x^{2}+8 x +1\right )^{\frac {3}{2}}}+\frac {6 x +8}{169 \sqrt {3 x^{2}+8 x +1}}\)	\(40\)

[In]

int(1/(3*x^2+8*x+1)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/507*(54*x^3+216*x^2+171*x-28)/(3*x^2+8*x+1)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.55 \[ \int \frac {1}{\left (1+8 x+3 x^2\right )^{5/2}} \, dx=-\frac {252 \, x^{4} + 1344 \, x^{3} + 1960 \, x^{2} - {\left (54 \, x^{3} + 216 \, x^{2} + 171 \, x - 28\right )} \sqrt {3 \, x^{2} + 8 \, x + 1} + 448 \, x + 28}{507 \, {\left (9 \, x^{4} + 48 \, x^{3} + 70 \, x^{2} + 16 \, x + 1\right )}} \]

[In]

integrate(1/(3*x^2+8*x+1)^(5/2),x, algorithm="fricas")

[Out]

-1/507*(252*x^4 + 1344*x^3 + 1960*x^2 - (54*x^3 + 216*x^2 + 171*x - 28)*sqrt(3*x^2 + 8*x + 1) + 448*x + 28)/(9
*x^4 + 48*x^3 + 70*x^2 + 16*x + 1)

Sympy [F]

\[ \int \frac {1}{\left (1+8 x+3 x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (3 x^{2} + 8 x + 1\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(3*x**2+8*x+1)**(5/2),x)

[Out]

Integral((3*x**2 + 8*x + 1)**(-5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (1+8 x+3 x^2\right )^{5/2}} \, dx=\frac {6 \, x}{169 \, \sqrt {3 \, x^{2} + 8 \, x + 1}} + \frac {8}{169 \, \sqrt {3 \, x^{2} + 8 \, x + 1}} - \frac {x}{13 \, {\left (3 \, x^{2} + 8 \, x + 1\right )}^{\frac {3}{2}}} - \frac {4}{39 \, {\left (3 \, x^{2} + 8 \, x + 1\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(3*x^2+8*x+1)^(5/2),x, algorithm="maxima")

[Out]

6/169*x/sqrt(3*x^2 + 8*x + 1) + 8/169/sqrt(3*x^2 + 8*x + 1) - 1/13*x/(3*x^2 + 8*x + 1)^(3/2) - 4/39/(3*x^2 + 8
*x + 1)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (1+8 x+3 x^2\right )^{5/2}} \, dx=\frac {9 \, {\left (6 \, {\left (x + 4\right )} x + 19\right )} x - 28}{507 \, {\left (3 \, x^{2} + 8 \, x + 1\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(3*x^2+8*x+1)^(5/2),x, algorithm="giac")

[Out]

1/507*(9*(6*(x + 4)*x + 19)*x - 28)/(3*x^2 + 8*x + 1)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\left (1+8 x+3 x^2\right )^{5/2}} \, dx=\frac {\left (12\,x+16\right )\,\left (72\,x^2+192\,x-28\right )}{8112\,{\left (3\,x^2+8\,x+1\right )}^{3/2}} \]

[In]

int(1/(8*x + 3*x^2 + 1)^(5/2),x)

[Out]

((12*x + 16)*(192*x + 72*x^2 - 28))/(8112*(8*x + 3*x^2 + 1)^(3/2))