3.2554 \(\int \frac{5-x}{\sqrt{3+2 x} (2+5 x+3 x^2)} \, dx\)

Optimal. Leaf size=38 \[ 12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{34 \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )}{\sqrt{15}} \]

[Out]

12*ArcTanh[Sqrt[3 + 2*x]] - (34*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15]

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Rubi [A]  time = 0.0304311, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {826, 1166, 207} \[ 12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{34 \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )}{\sqrt{15}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

12*ArcTanh[Sqrt[3 + 2*x]] - (34*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15]

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{5-x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{13-x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=34 \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-36 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=12 \tanh ^{-1}\left (\sqrt{3+2 x}\right )-\frac{34 \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )}{\sqrt{15}}\\ \end{align*}

Mathematica [A]  time = 0.0139014, size = 38, normalized size = 1. \[ 12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{34 \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )}{\sqrt{15}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

12*ArcTanh[Sqrt[3 + 2*x]] - (34*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15]

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Maple [A]  time = 0.01, size = 44, normalized size = 1.2 \begin{align*} -{\frac{34\,\sqrt{15}}{15}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }+6\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\,\ln \left ( -1+\sqrt{3+2\,x} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-34/15*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+6*ln(1+(3+2*x)^(1/2))-6*ln(-1+(3+2*x)^(1/2))

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Maxima [B]  time = 1.46203, size = 82, normalized size = 2.16 \begin{align*} \frac{17}{15} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt{2 \, x + 3} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/15*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 6*log(sqrt(2*x + 3) + 1) - 6*
log(sqrt(2*x + 3) - 1)

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Fricas [A]  time = 1.5823, size = 163, normalized size = 4.29 \begin{align*} \frac{17}{15} \, \sqrt{15} \log \left (-\frac{\sqrt{15} \sqrt{2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt{2 \, x + 3} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/15*sqrt(15)*log(-(sqrt(15)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) + 6*log(sqrt(2*x + 3) + 1) - 6*log(sqrt(2*x
+ 3) - 1)

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Sympy [A]  time = 32.7741, size = 95, normalized size = 2.5 \begin{align*} 34 \left (\begin{cases} - \frac{\sqrt{15} \operatorname{acoth}{\left (\frac{\sqrt{15}}{3 \sqrt{2 x + 3}} \right )}}{15} & \text{for}\: \frac{1}{2 x + 3} > \frac{3}{5} \\- \frac{\sqrt{15} \operatorname{atanh}{\left (\frac{\sqrt{15}}{3 \sqrt{2 x + 3}} \right )}}{15} & \text{for}\: \frac{1}{2 x + 3} < \frac{3}{5} \end{cases}\right ) - 6 \log{\left (-1 + \frac{1}{\sqrt{2 x + 3}} \right )} + 6 \log{\left (1 + \frac{1}{\sqrt{2 x + 3}} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

34*Piecewise((-sqrt(15)*acoth(sqrt(15)/(3*sqrt(2*x + 3)))/15, 1/(2*x + 3) > 3/5), (-sqrt(15)*atanh(sqrt(15)/(3
*sqrt(2*x + 3)))/15, 1/(2*x + 3) < 3/5)) - 6*log(-1 + 1/sqrt(2*x + 3)) + 6*log(1 + 1/sqrt(2*x + 3))

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Giac [B]  time = 1.10009, size = 88, normalized size = 2.32 \begin{align*} \frac{17}{15} \, \sqrt{15} \log \left (\frac{{\left | -2 \, \sqrt{15} + 6 \, \sqrt{2 \, x + 3} \right |}}{2 \,{\left (\sqrt{15} + 3 \, \sqrt{2 \, x + 3}\right )}}\right ) + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt{2 \, x + 3} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/15*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 6*log(sqrt(2*x + 3)
+ 1) - 6*log(abs(sqrt(2*x + 3) - 1))