\(\int \frac {1+3 x}{(1-8 x+2 x^2)^{5/2}} \, dx\) [481]

Optimal result

Integrand size = 20, antiderivative size = 47 \[ \int \frac {1+3 x}{\left (1-8 x+2 x^2\right )^{5/2}} \, dx=\frac {1-2 x}{6 \left (1-8 x+2 x^2\right )^{3/2}}-\frac {2 (2-x)}{21 \sqrt {1-8 x+2 x^2}} \]

[Out]

1/6*(1-2*x)/(2*x^2-8*x+1)^(3/2)-2/21*(2-x)/(2*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {652, 627} \[ \int \frac {1+3 x}{\left (1-8 x+2 x^2\right )^{5/2}} \, dx=\frac {1-2 x}{6 \left (2 x^2-8 x+1\right )^{3/2}}-\frac {2 (2-x)}{21 \sqrt {2 x^2-8 x+1}} \]

[In]

Int[(1 + 3*x)/(1 - 8*x + 2*x^2)^(5/2),x]

[Out]

(1 - 2*x)/(6*(1 - 8*x + 2*x^2)^(3/2)) - (2*(2 - x))/(21*Sqrt[1 - 8*x + 2*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 652

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

Rubi steps \begin{align*} \text {integral}& = \frac {1-2 x}{6 \left (1-8 x+2 x^2\right )^{3/2}}-\frac {2}{3} \int \frac {1}{\left (1-8 x+2 x^2\right )^{3/2}} \, dx \\ & = \frac {1-2 x}{6 \left (1-8 x+2 x^2\right )^{3/2}}-\frac {2 (2-x)}{21 \sqrt {1-8 x+2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.70 \[ \int \frac {1+3 x}{\left (1-8 x+2 x^2\right )^{5/2}} \, dx=\frac {-1+54 x-48 x^2+8 x^3}{42 \left (1-8 x+2 x^2\right )^{3/2}} \]

[In]

Integrate[(1 + 3*x)/(1 - 8*x + 2*x^2)^(5/2),x]

[Out]

(-1 + 54*x - 48*x^2 + 8*x^3)/(42*(1 - 8*x + 2*x^2)^(3/2))

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.64

method	result	size
gosper	\(\frac {8 x^{3}-48 x^{2}+54 x -1}{42 \left (2 x^{2}-8 x +1\right )^{\frac {3}{2}}}\)	\(30\)
trager	\(\frac {8 x^{3}-48 x^{2}+54 x -1}{42 \left (2 x^{2}-8 x +1\right )^{\frac {3}{2}}}\)	\(30\)
risch	\(\frac {8 x^{3}-48 x^{2}+54 x -1}{42 \left (2 x^{2}-8 x +1\right )^{\frac {3}{2}}}\)	\(30\)
default	\(-\frac {4 x -8}{12 \left (2 x^{2}-8 x +1\right )^{\frac {3}{2}}}+\frac {4 x -8}{42 \sqrt {2 x^{2}-8 x +1}}-\frac {1}{2 \left (2 x^{2}-8 x +1\right )^{\frac {3}{2}}}\)	\(54\)

[In]

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

[Out]

1/42*(8*x^3-48*x^2+54*x-1)/(2*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 \[ \int \frac {1+3 x}{\left (1-8 x+2 x^2\right )^{5/2}} \, dx=-\frac {4 \, x^{4} - 32 \, x^{3} + 68 \, x^{2} - {\left (8 \, x^{3} - 48 \, x^{2} + 54 \, x - 1\right )} \sqrt {2 \, x^{2} - 8 \, x + 1} - 16 \, x + 1}{42 \, {\left (4 \, x^{4} - 32 \, x^{3} + 68 \, x^{2} - 16 \, x + 1\right )}} \]

[In]

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

[Out]

-1/42*(4*x^4 - 32*x^3 + 68*x^2 - (8*x^3 - 48*x^2 + 54*x - 1)*sqrt(2*x^2 - 8*x + 1) - 16*x + 1)/(4*x^4 - 32*x^3
+ 68*x^2 - 16*x + 1)

Sympy [F]

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

[In]

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

[Out]

Integral((3*x + 1)/(2*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 \[ \int \frac {1+3 x}{\left (1-8 x+2 x^2\right )^{5/2}} \, dx=\frac {2 \, x}{21 \, \sqrt {2 \, x^{2} - 8 \, x + 1}} - \frac {4}{21 \, \sqrt {2 \, x^{2} - 8 \, x + 1}} - \frac {x}{3 \, {\left (2 \, x^{2} - 8 \, x + 1\right )}^{\frac {3}{2}}} + \frac {1}{6 \, {\left (2 \, x^{2} - 8 \, x + 1\right )}^{\frac {3}{2}}} \]

[In]

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

[Out]

2/21*x/sqrt(2*x^2 - 8*x + 1) - 4/21/sqrt(2*x^2 - 8*x + 1) - 1/3*x/(2*x^2 - 8*x + 1)^(3/2) + 1/6/(2*x^2 - 8*x +
1)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.57 \[ \int \frac {1+3 x}{\left (1-8 x+2 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (4 \, {\left (x - 6\right )} x + 27\right )} x - 1}{42 \, {\left (2 \, x^{2} - 8 \, x + 1\right )}^{\frac {3}{2}}} \]

[In]

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

[Out]

1/42*(2*(4*(x - 6)*x + 27)*x - 1)/(2*x^2 - 8*x + 1)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.62 \[ \int \frac {1+3 x}{\left (1-8 x+2 x^2\right )^{5/2}} \, dx=\frac {8\,x^3-48\,x^2+54\,x-1}{42\,{\left (2\,x^2-8\,x+1\right )}^{3/2}} \]

[In]

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

[Out]

(54*x - 48*x^2 + 8*x^3 - 1)/(42*(2*x^2 - 8*x + 1)^(3/2))