3.2571 \(\int \frac{5-x}{\sqrt{3+2 x} (2+5 x+3 x^2)^3} \, dx\)

Optimal. Leaf size=102 \[ -\frac{3 \sqrt{2 x+3} (47 x+37)}{10 \left (3 x^2+5 x+2\right )^2}+\frac{\sqrt{2 x+3} (11739 x+9734)}{50 \left (3 x^2+5 x+2\right )}+542 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{17463}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

(-3*Sqrt[3 + 2*x]*(37 + 47*x))/(10*(2 + 5*x + 3*x^2)^2) + (Sqrt[3 + 2*x]*(9734 + 11739*x))/(50*(2 + 5*x + 3*x^
2)) + 542*ArcTanh[Sqrt[3 + 2*x]] - (17463*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25

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Rubi [A]  time = 0.063425, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {822, 826, 1166, 207} \[ -\frac{3 \sqrt{2 x+3} (47 x+37)}{10 \left (3 x^2+5 x+2\right )^2}+\frac{\sqrt{2 x+3} (11739 x+9734)}{50 \left (3 x^2+5 x+2\right )}+542 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{17463}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[3 + 2*x]*(37 + 47*x))/(10*(2 + 5*x + 3*x^2)^2) + (Sqrt[3 + 2*x]*(9734 + 11739*x))/(50*(2 + 5*x + 3*x^
2)) + 542*ArcTanh[Sqrt[3 + 2*x]] - (17463*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 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 )^3} \, dx &=-\frac{3 \sqrt{3+2 x} (37+47 x)}{10 \left (2+5 x+3 x^2\right )^2}-\frac{1}{10} \int \frac{1106+705 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{3 \sqrt{3+2 x} (37+47 x)}{10 \left (2+5 x+3 x^2\right )^2}+\frac{\sqrt{3+2 x} (9734+11739 x)}{50 \left (2+5 x+3 x^2\right )}+\frac{1}{50} \int \frac{25289+11739 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{3 \sqrt{3+2 x} (37+47 x)}{10 \left (2+5 x+3 x^2\right )^2}+\frac{\sqrt{3+2 x} (9734+11739 x)}{50 \left (2+5 x+3 x^2\right )}+\frac{1}{25} \operatorname{Subst}\left (\int \frac{15361+11739 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)}{10 \left (2+5 x+3 x^2\right )^2}+\frac{\sqrt{3+2 x} (9734+11739 x)}{50 \left (2+5 x+3 x^2\right )}-1626 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )+\frac{52389}{25} \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)}{10 \left (2+5 x+3 x^2\right )^2}+\frac{\sqrt{3+2 x} (9734+11739 x)}{50 \left (2+5 x+3 x^2\right )}+542 \tanh ^{-1}\left (\sqrt{3+2 x}\right )-\frac{17463}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.108466, size = 81, normalized size = 0.79 \[ \frac{1}{50} \left (\frac{\sqrt{2 x+3} \left (35217 x^3+87897 x^2+71443 x+18913\right )}{\left (3 x^2+5 x+2\right )^2}+27100 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-34926 \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[3 + 2*x]*(18913 + 71443*x + 87897*x^2 + 35217*x^3))/(2 + 5*x + 3*x^2)^2 + 27100*ArcTanh[Sqrt[3 + 2*x]]
- 34926*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/50

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Maple [A]  time = 0.018, size = 124, normalized size = 1.2 \begin{align*} 486\,{\frac{1}{ \left ( 6\,x+4 \right ) ^{2}} \left ({\frac{571\, \left ( 3+2\,x \right ) ^{3/2}}{450}}-{\frac{121\,\sqrt{3+2\,x}}{54}} \right ) }-{\frac{17463\,\sqrt{15}}{125}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-3\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-2}+44\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}+271\,\ln \left ( 1+\sqrt{3+2\,x} \right ) +3\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-2}+44\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}-271\,\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/(3+2*x)^(1/2),x)

[Out]

486*(571/450*(3+2*x)^(3/2)-121/54*(3+2*x)^(1/2))/(6*x+4)^2-17463/125*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1
/2)-3/(1+(3+2*x)^(1/2))^2+44/(1+(3+2*x)^(1/2))+271*ln(1+(3+2*x)^(1/2))+3/(-1+(3+2*x)^(1/2))^2+44/(-1+(3+2*x)^(
1/2))-271*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.50646, size = 181, normalized size = 1.77 \begin{align*} \frac{17463}{250} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) + \frac{35217 \,{\left (2 \, x + 3\right )}^{\frac{7}{2}} - 141159 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + 181867 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 74725 \, \sqrt{2 \, x + 3}}{25 \,{\left (9 \,{\left (2 \, x + 3\right )}^{4} - 48 \,{\left (2 \, x + 3\right )}^{3} + 94 \,{\left (2 \, x + 3\right )}^{2} - 160 \, x - 215\right )}} + 271 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 271 \, \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/(3+2*x)^(1/2),x, algorithm="maxima")

[Out]

17463/250*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/25*(35217*(2*x + 3)^(7/
2) - 141159*(2*x + 3)^(5/2) + 181867*(2*x + 3)^(3/2) - 74725*sqrt(2*x + 3))/(9*(2*x + 3)^4 - 48*(2*x + 3)^3 +
94*(2*x + 3)^2 - 160*x - 215) + 271*log(sqrt(2*x + 3) + 1) - 271*log(sqrt(2*x + 3) - 1)

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Fricas [B]  time = 1.91248, size = 485, normalized size = 4.75 \begin{align*} \frac{17463 \, \sqrt{5} \sqrt{3}{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (-\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 67750 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) - 67750 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) + 5 \,{\left (35217 \, x^{3} + 87897 \, x^{2} + 71443 \, x + 18913\right )} \sqrt{2 \, x + 3}}{250 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/250*(17463*sqrt(5)*sqrt(3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x -
7)/(3*x + 2)) + 67750*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(sqrt(2*x + 3) + 1) - 67750*(9*x^4 + 30*x^3 + 37
*x^2 + 20*x + 4)*log(sqrt(2*x + 3) - 1) + 5*(35217*x^3 + 87897*x^2 + 71443*x + 18913)*sqrt(2*x + 3))/(9*x^4 +
30*x^3 + 37*x^2 + 20*x + 4)

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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)**3/(3+2*x)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.08575, size = 162, normalized size = 1.59 \begin{align*} \frac{17463}{250} \, \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{35217 \,{\left (2 \, x + 3\right )}^{\frac{7}{2}} - 141159 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + 181867 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 74725 \, \sqrt{2 \, x + 3}}{25 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 271 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 271 \, \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/(3+2*x)^(1/2),x, algorithm="giac")

[Out]

17463/250*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/25*(35217*(2*x
+ 3)^(7/2) - 141159*(2*x + 3)^(5/2) + 181867*(2*x + 3)^(3/2) - 74725*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 1
9)^2 + 271*log(sqrt(2*x + 3) + 1) - 271*log(abs(sqrt(2*x + 3) - 1))