3.25.100 \(\int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\) [2500]

Optimal. Leaf size=77 \[ -\frac {\tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{2 \sqrt {3}}+\frac {13 \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{2 \sqrt {5}} \]

[Out]

-1/6*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+13/10*arctanh(1/10*(7+8*x)*5^(1/2)/(3*x^2+5*x+2)
^(1/2))*5^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 77, 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 = {857, 635, 212, 738} \begin {gather*} \frac {13 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {5}}-\frac {\tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])]/Sqrt[3] + (13*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 +
5*x + 3*x^2])])/(2*Sqrt[5])

Rule 212

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

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

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

Rule 857

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

Rubi steps

\begin {align*} \int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx &=-\left (\frac {1}{2} \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx\right )+\frac {13}{2} \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\left (13 \text {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-7-8 x}{\sqrt {2+5 x+3 x^2}}\right )\right )-\text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{2 \sqrt {3}}+\frac {13 \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{2 \sqrt {5}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 61, normalized size = 0.79 \begin {gather*} \frac {13 \tanh ^{-1}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )}{\sqrt {5}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(13*ArcTanh[Sqrt[2/5 + x + (3*x^2)/5]/(1 + x)])/Sqrt[5] - ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)]/Sqrt[3]

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Maple [A]
time = 0.12, size = 61, normalized size = 0.79

method	result	size
default	\(-\frac {\ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{6}-\frac {13 \sqrt {5}\, \arctanh \left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{10}\)	\(61\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \RootOf \left (\textit {\_Z}^{2}-3\right ) x +5 \RootOf \left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{6}+\frac {13 \RootOf \left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {8 \RootOf \left (\textit {\_Z}^{2}-5\right ) x +7 \RootOf \left (\textit {\_Z}^{2}-5\right )+10 \sqrt {3 x^{2}+5 x +2}}{3+2 x}\right )}{10}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)-13/10*5^(1/2)*arctanh(2/5*(-7/2-4*x)*5^(1/2)/(12*(x
+3/2)^2-16*x-19)^(1/2))

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Maxima [A]
time = 0.50, size = 70, normalized size = 0.91 \begin {gather*} -\frac {1}{6} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) - \frac {13}{10} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) - 13/10*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/
abs(2*x + 3) + 5/2/abs(2*x + 3) - 2)

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Fricas [A]
time = 3.73, size = 90, normalized size = 1.17 \begin {gather*} \frac {1}{12} \, \sqrt {3} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + \frac {13}{20} \, \sqrt {5} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49) + 13/20*sqrt(5)*log((4*sqrt
(5)*sqrt(3*x^2 + 5*x + 2)*(8*x + 7) + 124*x^2 + 212*x + 89)/(4*x^2 + 12*x + 9))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x}{2 x \sqrt {3 x^{2} + 5 x + 2} + 3 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {5}{2 x \sqrt {3 x^{2} + 5 x + 2} + 3 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x/(2*x*sqrt(3*x**2 + 5*x + 2) + 3*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-5/(2*x*sqrt(3*x**2 + 5*x +
2) + 3*sqrt(3*x**2 + 5*x + 2)), x)

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Giac [A]
time = 3.96, size = 107, normalized size = 1.39 \begin {gather*} \frac {13}{10} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) + \frac {1}{6} \, \sqrt {3} \log \left ({\left | -6 \, \sqrt {3} x - 5 \, \sqrt {3} + 6 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/10*sqrt(5)*log(abs(-4*sqrt(3)*x - 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))/abs(-4*sqrt(3)*x + 2*sqr
t(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))) + 1/6*sqrt(3)*log(abs(-6*sqrt(3)*x - 5*sqrt(3) + 6*sqrt(3*x^2 + 5
*x + 2)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {x-5}{\left (2\,x+3\right )\,\sqrt {3\,x^2+5\,x+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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