Integrand size = 24, antiderivative size = 52 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+3 x^2}} \, dx=-\frac {\text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{2 \sqrt {3}}-\frac {13 \text {arctanh}\left (\frac {4-9 x}{\sqrt {35} \sqrt {2+3 x^2}}\right )}{2 \sqrt {35}} \] Output:
-1/6*arcsinh(1/2*x*6^(1/2))*3^(1/2)-13/70*35^(1/2)*arctanh(1/35*(4-9*x)*35 ^(1/2)/(3*x^2+2)^(1/2))
Time = 0.51 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.42 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+3 x^2}} \, dx=\frac {13 \text {arctanh}\left (\frac {3 \sqrt {3}+2 \sqrt {3} x-2 \sqrt {2+3 x^2}}{\sqrt {35}}\right )}{\sqrt {35}}+\frac {\log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right )}{2 \sqrt {3}} \] Input:
Integrate[(5 - x)/((3 + 2*x)*Sqrt[2 + 3*x^2]),x]
Output:
(13*ArcTanh[(3*Sqrt[3] + 2*Sqrt[3]*x - 2*Sqrt[2 + 3*x^2])/Sqrt[35]])/Sqrt[ 35] + Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^2]]/(2*Sqrt[3])
Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {719, 222, 488, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {5-x}{(2 x+3) \sqrt {3 x^2+2}} \, dx\) |
\(\Big \downarrow \) 719 |
\(\displaystyle \frac {13}{2} \int \frac {1}{(2 x+3) \sqrt {3 x^2+2}}dx-\frac {1}{2} \int \frac {1}{\sqrt {3 x^2+2}}dx\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {13}{2} \int \frac {1}{(2 x+3) \sqrt {3 x^2+2}}dx-\frac {\text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{2 \sqrt {3}}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle -\frac {13}{2} \int \frac {1}{35-\frac {(4-9 x)^2}{3 x^2+2}}d\frac {4-9 x}{\sqrt {3 x^2+2}}-\frac {\text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{2 \sqrt {3}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{2 \sqrt {3}}-\frac {13 \text {arctanh}\left (\frac {4-9 x}{\sqrt {35} \sqrt {3 x^2+2}}\right )}{2 \sqrt {35}}\) |
Input:
Int[(5 - x)/((3 + 2*x)*Sqrt[2 + 3*x^2]),x]
Output:
-1/2*ArcSinh[Sqrt[3/2]*x]/Sqrt[3] - (13*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/(2*Sqrt[35])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Time = 0.87 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85
method | result | size |
default | \(-\frac {\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{6}-\frac {13 \sqrt {35}\, \operatorname {arctanh}\left (\frac {2 \left (4-9 x \right ) \sqrt {35}}{35 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-36 x -19}}\right )}{70}\) | \(44\) |
trager | \(-\frac {13 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) \ln \left (-\frac {9 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right )-35 \sqrt {3 x^{2}+2}}{2 x +3}\right )}{70}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{6}\) | \(79\) |
Input:
int((5-x)/(2*x+3)/(3*x^2+2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/6*arcsinh(1/2*6^(1/2)*x)*3^(1/2)-13/70*35^(1/2)*arctanh(2/35*(4-9*x)*35 ^(1/2)/(12*(x+3/2)^2-36*x-19)^(1/2))
Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.46 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+3 x^2}} \, dx=\frac {1}{12} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + \frac {13}{140} \, \sqrt {35} \log \left (-\frac {\sqrt {35} \sqrt {3 \, x^{2} + 2} {\left (9 \, x - 4\right )} + 93 \, x^{2} - 36 \, x + 43}{4 \, x^{2} + 12 \, x + 9}\right ) \] Input:
integrate((5-x)/(3+2*x)/(3*x^2+2)^(1/2),x, algorithm="fricas")
Output:
1/12*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + 13/140*sqrt(35)* log(-(sqrt(35)*sqrt(3*x^2 + 2)*(9*x - 4) + 93*x^2 - 36*x + 43)/(4*x^2 + 12 *x + 9))
\[ \int \frac {5-x}{(3+2 x) \sqrt {2+3 x^2}} \, dx=- \int \frac {x}{2 x \sqrt {3 x^{2} + 2} + 3 \sqrt {3 x^{2} + 2}}\, dx - \int \left (- \frac {5}{2 x \sqrt {3 x^{2} + 2} + 3 \sqrt {3 x^{2} + 2}}\right )\, dx \] Input:
integrate((5-x)/(3+2*x)/(3*x**2+2)**(1/2),x)
Output:
-Integral(x/(2*x*sqrt(3*x**2 + 2) + 3*sqrt(3*x**2 + 2)), x) - Integral(-5/ (2*x*sqrt(3*x**2 + 2) + 3*sqrt(3*x**2 + 2)), x)
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+3 x^2}} \, dx=-\frac {1}{6} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {13}{70} \, \sqrt {35} \operatorname {arsinh}\left (\frac {3 \, \sqrt {6} x}{2 \, {\left | 2 \, x + 3 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 3 \right |}}\right ) \] Input:
integrate((5-x)/(3+2*x)/(3*x^2+2)^(1/2),x, algorithm="maxima")
Output:
-1/6*sqrt(3)*arcsinh(1/2*sqrt(6)*x) + 13/70*sqrt(35)*arcsinh(3/2*sqrt(6)*x /abs(2*x + 3) - 2/3*sqrt(6)/abs(2*x + 3))
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (38) = 76\).
Time = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.73 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+3 x^2}} \, dx=\frac {1}{6} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) + \frac {13}{70} \, \sqrt {35} \log \left (-\frac {{\left | -2 \, \sqrt {3} x - \sqrt {35} - 3 \, \sqrt {3} + 2 \, \sqrt {3 \, x^{2} + 2} \right |}}{2 \, \sqrt {3} x - \sqrt {35} + 3 \, \sqrt {3} - 2 \, \sqrt {3 \, x^{2} + 2}}\right ) \] Input:
integrate((5-x)/(3+2*x)/(3*x^2+2)^(1/2),x, algorithm="giac")
Output:
1/6*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 13/70*sqrt(35)*log(-abs(-2 *sqrt(3)*x - sqrt(35) - 3*sqrt(3) + 2*sqrt(3*x^2 + 2))/(2*sqrt(3)*x - sqrt (35) + 3*sqrt(3) - 2*sqrt(3*x^2 + 2)))
Time = 5.94 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+3 x^2}} \, dx=\frac {\sqrt {35}\,\left (26\,\ln \left (x+\frac {3}{2}\right )-26\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {35}\,\sqrt {x^2+\frac {2}{3}}}{9}-\frac {4}{9}\right )\right )}{140}-\frac {\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {2}\,\sqrt {3}\,x}{2}\right )}{6} \] Input:
int(-(x - 5)/((2*x + 3)*(3*x^2 + 2)^(1/2)),x)
Output:
(35^(1/2)*(26*log(x + 3/2) - 26*log(x - (3^(1/2)*35^(1/2)*(x^2 + 2/3)^(1/2 ))/9 - 4/9)))/140 - (3^(1/2)*asinh((2^(1/2)*3^(1/2)*x)/2))/6
Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.29 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+3 x^2}} \, dx=\frac {13 \sqrt {35}\, \mathit {atan} \left (\frac {2 \sqrt {3 x^{2}+2}\, i +2 \sqrt {3}\, i x}{\sqrt {35}-3 \sqrt {3}}\right ) i}{70}+\frac {13 \sqrt {35}\, \mathrm {log}\left (4 \sqrt {3 x^{2}+2}\, \sqrt {3}\, x +3 \sqrt {105}+12 x^{2}-27\right )}{140}-\frac {13 \sqrt {35}\, \mathrm {log}\left (\frac {2 \sqrt {3 x^{2}+2}+\sqrt {35}+2 \sqrt {3}\, x +3 \sqrt {3}}{\sqrt {2}}\right )}{70}-\frac {\sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right )}{6} \] Input:
int((5-x)/(3+2*x)/(3*x^2+2)^(1/2),x)
Output:
(78*sqrt(35)*atan((2*sqrt(3*x**2 + 2)*i + 2*sqrt(3)*i*x)/(sqrt(35) - 3*sqr t(3)))*i + 39*sqrt(35)*log(4*sqrt(3*x**2 + 2)*sqrt(3)*x + 3*sqrt(105) + 12 *x**2 - 27) - 78*sqrt(35)*log((2*sqrt(3*x**2 + 2) + sqrt(35) + 2*sqrt(3)*x + 3*sqrt(3))/sqrt(2)) - 70*sqrt(3)*log((sqrt(3*x**2 + 2) + sqrt(3)*x)/sqr t(2)))/420