3.2553 \(\int \frac {(5-x) \sqrt {3+2 x}}{2+5 x+3 x^2} \, dx\)

Optimal. Leaf size=55 \[ -\frac {2}{3} \sqrt {2 x+3}+12 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {34}{3} \sqrt {\frac {5}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right ) \]

[Out]

12*arctanh((3+2*x)^(1/2))-34/9*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-2/3*(3+2*x)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 55, normalized size of antiderivative = 1.00, 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 = {824, 826, 1166, 207} \[ -\frac {2}{3} \sqrt {2 x+3}+12 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {34}{3} \sqrt {\frac {5}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), 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] && FractionQ[m] && GtQ[m, 0]

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]

Rubi steps

\begin {align*} \int \frac {(5-x) \sqrt {3+2 x}}{2+5 x+3 x^2} \, dx &=-\frac {2}{3} \sqrt {3+2 x}+\frac {1}{3} \int \frac {49+31 x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac {2}{3} \sqrt {3+2 x}+\frac {2}{3} \operatorname {Subst}\left (\int \frac {5+31 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt {3+2 x}\right )\\ &=-\frac {2}{3} \sqrt {3+2 x}-36 \operatorname {Subst}\left (\int \frac {1}{-3+3 x^2} \, dx,x,\sqrt {3+2 x}\right )+\frac {170}{3} \operatorname {Subst}\left (\int \frac {1}{-5+3 x^2} \, dx,x,\sqrt {3+2 x}\right )\\ &=-\frac {2}{3} \sqrt {3+2 x}+12 \tanh ^{-1}\left (\sqrt {3+2 x}\right )-\frac {34}{3} \sqrt {\frac {5}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 1.00 \[ -\frac {2}{3} \sqrt {2 x+3}+12 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {34}{3} \sqrt {\frac {5}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 71, normalized size = 1.29 \[ \frac {17}{9} \, \sqrt {5} \sqrt {3} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) - \frac {2}{3} \, \sqrt {2 \, x + 3} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/9*sqrt(5)*sqrt(3)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) - 2/3*sqrt(2*x + 3) + 6*log(sqr
t(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)

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giac [A]  time = 0.17, size = 74, normalized size = 1.35 \[ \frac {17}{9} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {2}{3} \, \sqrt {2 \, x + 3} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/9*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/3*sqrt(2*x + 3) + 6
*log(sqrt(2*x + 3) + 1) - 6*log(abs(sqrt(2*x + 3) - 1))

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maple [A]  time = 0.01, size = 53, normalized size = 0.96 \[ -\frac {34 \sqrt {15}\, \arctanh \left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{9}-6 \ln \left (-1+\sqrt {2 x +3}\right )+6 \ln \left (\sqrt {2 x +3}+1\right )-\frac {2 \sqrt {2 x +3}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(2*x+3)^(1/2)-34/9*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))*15^(1/2)+6*ln((2*x+3)^(1/2)+1)-6*ln(-1+(2*x+3)^(1/
2))

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maxima [A]  time = 1.30, size = 70, normalized size = 1.27 \[ \frac {17}{9} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {2}{3} \, \sqrt {2 \, x + 3} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/9*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/3*sqrt(2*x + 3) + 6*log(sqrt
(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)

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mupad [B]  time = 0.07, size = 38, normalized size = 0.69 \[ 12\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {34\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{9}-\frac {2\,\sqrt {2\,x+3}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*atanh((2*x + 3)^(1/2)) - (34*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/9 - (2*(2*x + 3)^(1/2))/3

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sympy [A]  time = 8.20, size = 102, normalized size = 1.85 \[ - \frac {2 \sqrt {2 x + 3}}{3} + \frac {170 \left (\begin {cases} - \frac {\sqrt {15} \operatorname {acoth}{\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} \right )}}{15} & \text {for}\: 2 x + 3 > \frac {5}{3} \\- \frac {\sqrt {15} \operatorname {atanh}{\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} \right )}}{15} & \text {for}\: 2 x + 3 < \frac {5}{3} \end {cases}\right )}{3} - 6 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 6 \log {\left (\sqrt {2 x + 3} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(2*x + 3)/3 + 170*Piecewise((-sqrt(15)*acoth(sqrt(15)*sqrt(2*x + 3)/5)/15, 2*x + 3 > 5/3), (-sqrt(15)*a
tanh(sqrt(15)*sqrt(2*x + 3)/5)/15, 2*x + 3 < 5/3))/3 - 6*log(sqrt(2*x + 3) - 1) + 6*log(sqrt(2*x + 3) + 1)

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