3.2563 \(\int \frac{5-x}{(3+2 x)^{3/2} (2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac{3 (47 x+37)}{5 \sqrt{2 x+3} \left (3 x^2+5 x+2\right )}-\frac{506}{25 \sqrt{2 x+3}}-34 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{1356}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

-506/(25*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(5*Sqrt[3 + 2*x]*(2 + 5*x + 3*x^2)) - 34*ArcTanh[Sqrt[3 + 2*x]] + (1
356*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25

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Rubi [A]  time = 0.0612812, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {822, 828, 826, 1166, 207} \[ -\frac{3 (47 x+37)}{5 \sqrt{2 x+3} \left (3 x^2+5 x+2\right )}-\frac{506}{25 \sqrt{2 x+3}}-34 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{1356}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-506/(25*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(5*Sqrt[3 + 2*x]*(2 + 5*x + 3*x^2)) - 34*ArcTanh[Sqrt[3 + 2*x]] + (1
356*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25

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 828

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((
e*f - d*g)*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d
+ e*x)^(m + 1)*Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x])/(a + b*x + c*x^2), 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] && FractionQ[m] && LtQ[m, -1]

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}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2} \, dx &=-\frac{3 (37+47 x)}{5 \sqrt{3+2 x} \left (2+5 x+3 x^2\right )}-\frac{1}{5} \int \frac{508+423 x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{506}{25 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 \sqrt{3+2 x} \left (2+5 x+3 x^2\right )}-\frac{1}{25} \int \frac{1184+759 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{506}{25 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 \sqrt{3+2 x} \left (2+5 x+3 x^2\right )}-\frac{2}{25} \operatorname{Subst}\left (\int \frac{91+759 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{506}{25 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 \sqrt{3+2 x} \left (2+5 x+3 x^2\right )}+102 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-\frac{4068}{25} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{506}{25 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 \sqrt{3+2 x} \left (2+5 x+3 x^2\right )}-34 \tanh ^{-1}\left (\sqrt{3+2 x}\right )+\frac{1356}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0873708, size = 85, normalized size = 1. \[ -\frac{3 (47 x+37)}{5 \sqrt{2 x+3} \left (3 x^2+5 x+2\right )}-\frac{506}{25 \sqrt{2 x+3}}-34 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{1356}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-506/(25*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(5*Sqrt[3 + 2*x]*(2 + 5*x + 3*x^2)) - 34*ArcTanh[Sqrt[3 + 2*x]] + (1
356*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25

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Maple [A]  time = 0.02, size = 95, normalized size = 1.1 \begin{align*} -{\frac{104}{25}{\frac{1}{\sqrt{3+2\,x}}}}-{\frac{102}{25}\sqrt{3+2\,x} \left ( 2\,x+{\frac{4}{3}} \right ) ^{-1}}+{\frac{1356\,\sqrt{15}}{125}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-6\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}-17\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}+17\,\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+2*x)^(3/2)/(3*x^2+5*x+2)^2,x)

[Out]

-104/25/(3+2*x)^(1/2)-102/25*(3+2*x)^(1/2)/(2*x+4/3)+1356/125*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-6/(
1+(3+2*x)^(1/2))-17*ln(1+(3+2*x)^(1/2))-6/(-1+(3+2*x)^(1/2))+17*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.46917, size = 144, normalized size = 1.69 \begin{align*} -\frac{678}{125} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) - \frac{2 \,{\left (759 \,{\left (2 \, x + 3\right )}^{2} - 2638 \, x - 3697\right )}}{25 \,{\left (3 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} - 8 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} + 5 \, \sqrt{2 \, x + 3}\right )}} - 17 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 17 \, \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+2*x)^(3/2)/(3*x^2+5*x+2)^2,x, algorithm="maxima")

[Out]

-678/125*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/25*(759*(2*x + 3)^2 - 26
38*x - 3697)/(3*(2*x + 3)^(5/2) - 8*(2*x + 3)^(3/2) + 5*sqrt(2*x + 3)) - 17*log(sqrt(2*x + 3) + 1) + 17*log(sq
rt(2*x + 3) - 1)

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Fricas [B]  time = 1.62298, size = 409, normalized size = 4.81 \begin{align*} \frac{678 \, \sqrt{5} \sqrt{3}{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )} \log \left (\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 2125 \,{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) + 2125 \,{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) - 5 \,{\left (1518 \, x^{2} + 3235 \, x + 1567\right )} \sqrt{2 \, x + 3}}{125 \,{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/125*(678*sqrt(5)*sqrt(3)*(6*x^3 + 19*x^2 + 19*x + 6)*log((sqrt(5)*sqrt(3)*sqrt(2*x + 3) + 3*x + 7)/(3*x + 2)
) - 2125*(6*x^3 + 19*x^2 + 19*x + 6)*log(sqrt(2*x + 3) + 1) + 2125*(6*x^3 + 19*x^2 + 19*x + 6)*log(sqrt(2*x +
3) - 1) - 5*(1518*x^2 + 3235*x + 1567)*sqrt(2*x + 3))/(6*x^3 + 19*x^2 + 19*x + 6)

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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+2*x)**(3/2)/(3*x**2+5*x+2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.10135, size = 150, normalized size = 1.76 \begin{align*} -\frac{678}{125} \, \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{2 \,{\left (759 \,{\left (2 \, x + 3\right )}^{2} - 2638 \, x - 3697\right )}}{25 \,{\left (3 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} - 8 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} + 5 \, \sqrt{2 \, x + 3}\right )}} - 17 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 17 \, \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+2*x)^(3/2)/(3*x^2+5*x+2)^2,x, algorithm="giac")

[Out]

-678/125*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/25*(759*(2*x +
3)^2 - 2638*x - 3697)/(3*(2*x + 3)^(5/2) - 8*(2*x + 3)^(3/2) + 5*sqrt(2*x + 3)) - 17*log(sqrt(2*x + 3) + 1) +
17*log(abs(sqrt(2*x + 3) - 1))