\(\int \frac {5-x}{(3+2 x)^{3/2} (2+5 x+3 x^2)} \, dx\) [2555]

Optimal result

Integrand size = 27, antiderivative size = 55 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=-\frac {26}{5 \sqrt {3+2 x}}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {34}{5} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]

[Out]

12*arctanh((3+2*x)^(1/2))-34/25*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-26/5/(3+2*x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {842, 840, 1180, 213} \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=12 \text {arctanh}\left (\sqrt {2 x+3}\right )-\frac {34}{5} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )-\frac {26}{5 \sqrt {2 x+3}} \]

[In]

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

[Out]

-26/(5*Sqrt[3 + 2*x]) + 12*ArcTanh[Sqrt[3 + 2*x]] - (34*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/5

Rule 213

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

Rule 840

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 842

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 1180

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*} \text {integral}& = -\frac {26}{5 \sqrt {3+2 x}}+\frac {1}{5} \int \frac {-9-39 x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )} \, dx \\ & = -\frac {26}{5 \sqrt {3+2 x}}+\frac {2}{5} \text {Subst}\left (\int \frac {99-39 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt {3+2 x}\right ) \\ & = -\frac {26}{5 \sqrt {3+2 x}}+\frac {102}{5} \text {Subst}\left (\int \frac {1}{-5+3 x^2} \, dx,x,\sqrt {3+2 x}\right )-36 \text {Subst}\left (\int \frac {1}{-3+3 x^2} \, dx,x,\sqrt {3+2 x}\right ) \\ & = -\frac {26}{5 \sqrt {3+2 x}}+12 \tanh ^{-1}\left (\sqrt {3+2 x}\right )-\frac {34}{5} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=-\frac {26}{5 \sqrt {3+2 x}}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {34}{5} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]

[In]

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

[Out]

-26/(5*Sqrt[3 + 2*x]) + 12*ArcTanh[Sqrt[3 + 2*x]] - (34*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/5

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(-\frac {26}{5 \sqrt {3+2 x}}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )-\frac {34 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{25}\)	\(53\)
default	\(-\frac {26}{5 \sqrt {3+2 x}}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )-\frac {34 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{25}\)	\(53\)
risch	\(-\frac {26}{5 \sqrt {3+2 x}}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )-\frac {34 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{25}\)	\(53\)
pseudoelliptic	\(-\frac {2 \left (17 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}\, \sqrt {3+2 x}+75 \ln \left (\sqrt {3+2 x}-1\right ) \sqrt {3+2 x}-75 \ln \left (\sqrt {3+2 x}+1\right ) \sqrt {3+2 x}+65\right )}{25 \sqrt {3+2 x}}\)	\(75\)
trager	\(-\frac {26}{5 \sqrt {3+2 x}}-6 \ln \left (\frac {-2-x +\sqrt {3+2 x}}{1+x}\right )-\frac {17 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{25}\)	\(76\)

[In]

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

[Out]

-26/5/(3+2*x)^(1/2)-6*ln((3+2*x)^(1/2)-1)+6*ln((3+2*x)^(1/2)+1)-34/25*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(
1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (38) = 76\).

Time = 0.37 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.73 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {17 \, \sqrt {5} \sqrt {3} {\left (2 \, x + 3\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 150 \, {\left (2 \, x + 3\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 150 \, {\left (2 \, x + 3\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 130 \, \sqrt {2 \, x + 3}}{25 \, {\left (2 \, x + 3\right )}} \]

[In]

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

[Out]

1/25*(17*sqrt(5)*sqrt(3)*(2*x + 3)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) + 150*(2*x + 3)*l
og(sqrt(2*x + 3) + 1) - 150*(2*x + 3)*log(sqrt(2*x + 3) - 1) - 130*sqrt(2*x + 3))/(2*x + 3)

Sympy [A] (verification not implemented)

Time = 3.41 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.45 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {17 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{25} - 6 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 6 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {26}{5 \sqrt {2 x + 3}} \]

[In]

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

[Out]

17*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(15)/3))/25 - 6*log(sqrt(2*x + 3) - 1)
+ 6*log(sqrt(2*x + 3) + 1) - 26/(5*sqrt(2*x + 3))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.27 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {17}{25} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {26}{5 \, \sqrt {2 \, x + 3}} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]

[In]

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

[Out]

17/25*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 26/5/sqrt(2*x + 3) + 6*log(sq
rt(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.35 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {17}{25} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {26}{5 \, \sqrt {2 \, x + 3}} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]

[In]

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

[Out]

17/25*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 26/5/sqrt(2*x + 3) +
6*log(sqrt(2*x + 3) + 1) - 6*log(abs(sqrt(2*x + 3) - 1))

Mupad [B] (verification not implemented)

Time = 11.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=12\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {34\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{25}-\frac {26}{5\,\sqrt {2\,x+3}} \]

[In]

int(-(x - 5)/((2*x + 3)^(3/2)*(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))/25 - 26/(5*(2*x + 3)^(1/2))