3.168 \(\int \frac {-4+7 x}{(5+2 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=43 \[ -\frac {19 x+39}{28 \left (3 x^2+2 x+5\right )}-\frac {19 \tan ^{-1}\left (\frac {3 x+1}{\sqrt {14}}\right )}{28 \sqrt {14}} \]

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Rubi [A]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {638, 618, 204} \[ -\frac {19 x+39}{28 \left (3 x^2+2 x+5\right )}-\frac {19 \tan ^{-1}\left (\frac {3 x+1}{\sqrt {14}}\right )}{28 \sqrt {14}} \]

Antiderivative was successfully verified.

[In]

Int[(-4 + 7*x)/(5 + 2*x + 3*x^2)^2,x]

[Out]

-(39 + 19*x)/(28*(5 + 2*x + 3*x^2)) - (19*ArcTan[(1 + 3*x)/Sqrt[14]])/(28*Sqrt[14])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

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

Rule 638

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)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), 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*} \int \frac {-4+7 x}{\left (5+2 x+3 x^2\right )^2} \, dx &=-\frac {39+19 x}{28 \left (5+2 x+3 x^2\right )}-\frac {19}{28} \int \frac {1}{5+2 x+3 x^2} \, dx\\ &=-\frac {39+19 x}{28 \left (5+2 x+3 x^2\right )}+\frac {19}{14} \operatorname {Subst}\left (\int \frac {1}{-56-x^2} \, dx,x,2+6 x\right )\\ &=-\frac {39+19 x}{28 \left (5+2 x+3 x^2\right )}-\frac {19 \tan ^{-1}\left (\frac {1+3 x}{\sqrt {14}}\right )}{28 \sqrt {14}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 43, normalized size = 1.00 \[ \frac {-19 x-39}{28 \left (3 x^2+2 x+5\right )}-\frac {19 \tan ^{-1}\left (\frac {3 x+1}{\sqrt {14}}\right )}{28 \sqrt {14}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 7*x)/(5 + 2*x + 3*x^2)^2,x]

[Out]

(-39 - 19*x)/(28*(5 + 2*x + 3*x^2)) - (19*ArcTan[(1 + 3*x)/Sqrt[14]])/(28*Sqrt[14])

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IntegrateAlgebraic [A]  time = 0.05, size = 46, normalized size = 1.07 \[ \frac {-19 x-39}{28 \left (3 x^2+2 x+5\right )}-\frac {19 \tan ^{-1}\left (\frac {3 x}{\sqrt {14}}+\frac {1}{\sqrt {14}}\right )}{28 \sqrt {14}} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-4 + 7*x)/(5 + 2*x + 3*x^2)^2,x]

[Out]

(-39 - 19*x)/(28*(5 + 2*x + 3*x^2)) - (19*ArcTan[1/Sqrt[14] + (3*x)/Sqrt[14]])/(28*Sqrt[14])

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fricas [A]  time = 1.17, size = 45, normalized size = 1.05 \[ -\frac {19 \, \sqrt {14} {\left (3 \, x^{2} + 2 \, x + 5\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (3 \, x + 1\right )}\right ) + 266 \, x + 546}{392 \, {\left (3 \, x^{2} + 2 \, x + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4+7*x)/(3*x^2+2*x+5)^2,x, algorithm="fricas")

[Out]

-1/392*(19*sqrt(14)*(3*x^2 + 2*x + 5)*arctan(1/14*sqrt(14)*(3*x + 1)) + 266*x + 546)/(3*x^2 + 2*x + 5)

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giac [A]  time = 0.85, size = 36, normalized size = 0.84 \[ -\frac {19}{392} \, \sqrt {14} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (3 \, x + 1\right )}\right ) - \frac {19 \, x + 39}{28 \, {\left (3 \, x^{2} + 2 \, x + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4+7*x)/(3*x^2+2*x+5)^2,x, algorithm="giac")

[Out]

-19/392*sqrt(14)*arctan(1/14*sqrt(14)*(3*x + 1)) - 1/28*(19*x + 39)/(3*x^2 + 2*x + 5)

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maple [A]  time = 0.51, size = 34, normalized size = 0.79

method	result	size
risch	\(\frac {-\frac {19 x}{84}-\frac {13}{28}}{x^{2}+\frac {2}{3} x +\frac {5}{3}}-\frac {19 \arctan \left (\frac {\left (1+3 x \right ) \sqrt {14}}{14}\right ) \sqrt {14}}{392}\)	\(34\)
default	\(\frac {-38 x -78}{168 x^{2}+112 x +280}-\frac {19 \sqrt {14}\, \arctan \left (\frac {\left (6 x +2\right ) \sqrt {14}}{28}\right )}{392}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4+7*x)/(3*x^2+2*x+5)^2,x,method=_RETURNVERBOSE)

[Out]

(-19/84*x-13/28)/(x^2+2/3*x+5/3)-19/392*arctan(1/14*(1+3*x)*14^(1/2))*14^(1/2)

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maxima [A]  time = 0.97, size = 36, normalized size = 0.84 \[ -\frac {19}{392} \, \sqrt {14} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (3 \, x + 1\right )}\right ) - \frac {19 \, x + 39}{28 \, {\left (3 \, x^{2} + 2 \, x + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4+7*x)/(3*x^2+2*x+5)^2,x, algorithm="maxima")

[Out]

-19/392*sqrt(14)*arctan(1/14*sqrt(14)*(3*x + 1)) - 1/28*(19*x + 39)/(3*x^2 + 2*x + 5)

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mupad [B]  time = 0.19, size = 36, normalized size = 0.84 \[ -\frac {\frac {19\,x}{84}+\frac {13}{28}}{x^2+\frac {2\,x}{3}+\frac {5}{3}}-\frac {19\,\sqrt {14}\,\mathrm {atan}\left (\frac {3\,\sqrt {14}\,x}{14}+\frac {\sqrt {14}}{14}\right )}{392} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((7*x - 4)/(2*x + 3*x^2 + 5)^2,x)

[Out]

- ((19*x)/84 + 13/28)/((2*x)/3 + x^2 + 5/3) - (19*14^(1/2)*atan((3*14^(1/2)*x)/14 + 14^(1/2)/14))/392

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sympy [A]  time = 0.14, size = 42, normalized size = 0.98 \[ \frac {- 19 x - 39}{84 x^{2} + 56 x + 140} - \frac {19 \sqrt {14} \operatorname {atan}{\left (\frac {3 \sqrt {14} x}{14} + \frac {\sqrt {14}}{14} \right )}}{392} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4+7*x)/(3*x**2+2*x+5)**2,x)

[Out]

(-19*x - 39)/(84*x**2 + 56*x + 140) - 19*sqrt(14)*atan(3*sqrt(14)*x/14 + sqrt(14)/14)/392

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