3.2562 \(\int \frac{5-x}{\sqrt{3+2 x} (2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=72 \[ -\frac{3 \sqrt{2 x+3} (47 x+37)}{5 \left (3 x^2+5 x+2\right )}-58 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{384}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

(-3*Sqrt[3 + 2*x]*(37 + 47*x))/(5*(2 + 5*x + 3*x^2)) - 58*ArcTanh[Sqrt[3 + 2*x]] + (384*Sqrt[3/5]*ArcTanh[Sqrt
[3/5]*Sqrt[3 + 2*x]])/5

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Rubi [A]  time = 0.0450731, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {822, 826, 1166, 207} \[ -\frac{3 \sqrt{2 x+3} (47 x+37)}{5 \left (3 x^2+5 x+2\right )}-58 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{384}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[3 + 2*x]*(37 + 47*x))/(5*(2 + 5*x + 3*x^2)) - 58*ArcTanh[Sqrt[3 + 2*x]] + (384*Sqrt[3/5]*ArcTanh[Sqrt
[3/5]*Sqrt[3 + 2*x]])/5

Rule 822

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

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 )^2} \, dx &=-\frac{3 \sqrt{3+2 x} (37+47 x)}{5 \left (2+5 x+3 x^2\right )}-\frac{1}{5} \int \frac{286+141 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{3 \sqrt{3+2 x} (37+47 x)}{5 \left (2+5 x+3 x^2\right )}-\frac{2}{5} \operatorname{Subst}\left (\int \frac{149+141 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{3 \sqrt{3+2 x} (37+47 x)}{5 \left (2+5 x+3 x^2\right )}+174 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-\frac{1152}{5} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{3 \sqrt{3+2 x} (37+47 x)}{5 \left (2+5 x+3 x^2\right )}-58 \tanh ^{-1}\left (\sqrt{3+2 x}\right )+\frac{384}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0700324, size = 72, normalized size = 1. \[ -\frac{3 \sqrt{2 x+3} (47 x+37)}{5 \left (3 x^2+5 x+2\right )}-58 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{384}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[3 + 2*x]*(37 + 47*x))/(5*(2 + 5*x + 3*x^2)) - 58*ArcTanh[Sqrt[3 + 2*x]] + (384*Sqrt[3/5]*ArcTanh[Sqrt
[3/5]*Sqrt[3 + 2*x]])/5

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Maple [A]  time = 0.019, size = 86, normalized size = 1.2 \begin{align*} -{\frac{34}{5}\sqrt{3+2\,x} \left ( 2\,x+{\frac{4}{3}} \right ) ^{-1}}+{\frac{384\,\sqrt{15}}{25}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-6\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}-29\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}+29\,\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)^2/(3+2*x)^(1/2),x)

[Out]

-34/5*(3+2*x)^(1/2)/(2*x+4/3)+384/25*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-6/(1+(3+2*x)^(1/2))-29*ln(1+
(3+2*x)^(1/2))-6/(-1+(3+2*x)^(1/2))+29*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.48588, size = 132, normalized size = 1.83 \begin{align*} -\frac{192}{25} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) - \frac{6 \,{\left (47 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 67 \, \sqrt{2 \, x + 3}\right )}}{5 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 29 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 29 \, \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)^2/(3+2*x)^(1/2),x, algorithm="maxima")

[Out]

-192/25*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 6/5*(47*(2*x + 3)^(3/2) - 6
7*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 29*log(sqrt(2*x + 3) + 1) + 29*log(sqrt(2*x + 3) - 1)

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Fricas [B]  time = 1.55725, size = 332, normalized size = 4.61 \begin{align*} \frac{192 \, \sqrt{5} \sqrt{3}{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 725 \,{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) + 725 \,{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) - 15 \,{\left (47 \, x + 37\right )} \sqrt{2 \, x + 3}}{25 \,{\left (3 \, x^{2} + 5 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*(192*sqrt(5)*sqrt(3)*(3*x^2 + 5*x + 2)*log((sqrt(5)*sqrt(3)*sqrt(2*x + 3) + 3*x + 7)/(3*x + 2)) - 725*(3*
x^2 + 5*x + 2)*log(sqrt(2*x + 3) + 1) + 725*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) - 1) - 15*(47*x + 37)*sqrt(2*x
+ 3))/(3*x^2 + 5*x + 2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.08791, size = 138, normalized size = 1.92 \begin{align*} -\frac{192}{25} \, \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 ) - \frac{6 \,{\left (47 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 67 \, \sqrt{2 \, x + 3}\right )}}{5 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 29 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 29 \, \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)^2/(3+2*x)^(1/2),x, algorithm="giac")

[Out]

-192/25*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 6/5*(47*(2*x + 3)^
(3/2) - 67*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 29*log(sqrt(2*x + 3) + 1) + 29*log(abs(sqrt(2*x + 3) -
1))