3.2556 \(\int \frac{5-x}{(3+2 x)^{5/2} (2+5 x+3 x^2)} \, dx\)

Optimal. Leaf size=68 \[ -\frac{198}{25 \sqrt{2 x+3}}-\frac{26}{15 (2 x+3)^{3/2}}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{102}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

-26/(15*(3 + 2*x)^(3/2)) - 198/(25*Sqrt[3 + 2*x]) + 12*ArcTanh[Sqrt[3 + 2*x]] - (102*Sqrt[3/5]*ArcTanh[Sqrt[3/
5]*Sqrt[3 + 2*x]])/25

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Rubi [A]  time = 0.0619013, antiderivative size = 68, 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 = {828, 826, 1166, 207} \[ -\frac{198}{25 \sqrt{2 x+3}}-\frac{26}{15 (2 x+3)^{3/2}}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{102}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-26/(15*(3 + 2*x)^(3/2)) - 198/(25*Sqrt[3 + 2*x]) + 12*ArcTanh[Sqrt[3 + 2*x]] - (102*Sqrt[3/5]*ArcTanh[Sqrt[3/
5]*Sqrt[3 + 2*x]])/25

Rule 828

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 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}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx &=-\frac{26}{15 (3+2 x)^{3/2}}+\frac{1}{5} \int \frac{-9-39 x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{26}{15 (3+2 x)^{3/2}}-\frac{198}{25 \sqrt{3+2 x}}+\frac{1}{25} \int \frac{-147-297 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{26}{15 (3+2 x)^{3/2}}-\frac{198}{25 \sqrt{3+2 x}}+\frac{2}{25} \operatorname{Subst}\left (\int \frac{597-297 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{26}{15 (3+2 x)^{3/2}}-\frac{198}{25 \sqrt{3+2 x}}+\frac{306}{25} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-36 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{26}{15 (3+2 x)^{3/2}}-\frac{198}{25 \sqrt{3+2 x}}+12 \tanh ^{-1}\left (\sqrt{3+2 x}\right )-\frac{102}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0895694, size = 65, normalized size = 0.96 \[ 12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{2 \left (2970 x+153 \sqrt{15} (2 x+3)^{3/2} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )+4780\right )}{375 (2 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

12*ArcTanh[Sqrt[3 + 2*x]] - (2*(4780 + 2970*x + 153*Sqrt[15]*(3 + 2*x)^(3/2)*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])
)/(375*(3 + 2*x)^(3/2))

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Maple [A]  time = 0.013, size = 62, normalized size = 0.9 \begin{align*} -{\frac{26}{15} \left ( 3+2\,x \right ) ^{-{\frac{3}{2}}}}-{\frac{198}{25}{\frac{1}{\sqrt{3+2\,x}}}}-{\frac{102\,\sqrt{15}}{125}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }+6\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\,\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+2*x)^(5/2)/(3*x^2+5*x+2),x)

[Out]

-26/15/(3+2*x)^(3/2)-198/25/(3+2*x)^(1/2)-102/125*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+6*ln(1+(3+2*x)^
(1/2))-6*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.44443, size = 101, normalized size = 1.49 \begin{align*} \frac{51}{125} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) - \frac{4 \,{\left (297 \, x + 478\right )}}{75 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}}} + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \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+2*x)^(5/2)/(3*x^2+5*x+2),x, algorithm="maxima")

[Out]

51/125*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 4/75*(297*x + 478)/(2*x + 3)
^(3/2) + 6*log(sqrt(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)

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Fricas [B]  time = 1.59618, size = 346, normalized size = 5.09 \begin{align*} \frac{153 \, \sqrt{5} \sqrt{3}{\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (-\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 2250 \,{\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) - 2250 \,{\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) - 20 \,{\left (297 \, x + 478\right )} \sqrt{2 \, x + 3}}{375 \,{\left (4 \, x^{2} + 12 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/375*(153*sqrt(5)*sqrt(3)*(4*x^2 + 12*x + 9)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) + 2250
*(4*x^2 + 12*x + 9)*log(sqrt(2*x + 3) + 1) - 2250*(4*x^2 + 12*x + 9)*log(sqrt(2*x + 3) - 1) - 20*(297*x + 478)
*sqrt(2*x + 3))/(4*x^2 + 12*x + 9)

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Sympy [A]  time = 41.6275, size = 114, normalized size = 1.68 \begin{align*} \frac{306 \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 )}{25} - 6 \log{\left (\sqrt{2 x + 3} - 1 \right )} + 6 \log{\left (\sqrt{2 x + 3} + 1 \right )} - \frac{198}{25 \sqrt{2 x + 3}} - \frac{26}{15 \left (2 x + 3\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

306*Piecewise((-sqrt(15)*acoth(sqrt(15)*sqrt(2*x + 3)/5)/15, 2*x + 3 > 5/3), (-sqrt(15)*atanh(sqrt(15)*sqrt(2*
x + 3)/5)/15, 2*x + 3 < 5/3))/25 - 6*log(sqrt(2*x + 3) - 1) + 6*log(sqrt(2*x + 3) + 1) - 198/(25*sqrt(2*x + 3)
) - 26/(15*(2*x + 3)**(3/2))

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Giac [A]  time = 1.12137, size = 107, normalized size = 1.57 \begin{align*} \frac{51}{125} \, \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{4 \,{\left (297 \, x + 478\right )}}{75 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}}} + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \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+2*x)^(5/2)/(3*x^2+5*x+2),x, algorithm="giac")

[Out]

51/125*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 4/75*(297*x + 478)/
(2*x + 3)^(3/2) + 6*log(sqrt(2*x + 3) + 1) - 6*log(abs(sqrt(2*x + 3) - 1))