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

Optimal. Leaf size=66 \[ -\frac {\sqrt {2 x+3} (35 x+29)}{3 x^2+5 x+2}-82 \tanh ^{-1}\left (\sqrt {2 x+3}\right )+\frac {316 \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}} \]

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Rubi [A]  time = 0.04, antiderivative size = 66, 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 = {820, 826, 1166, 207} \begin {gather*} -\frac {\sqrt {2 x+3} (35 x+29)}{3 x^2+5 x+2}-82 \tanh ^{-1}\left (\sqrt {2 x+3}\right )+\frac {316 \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[3 + 2*x]*(29 + 35*x))/(2 + 5*x + 3*x^2)) - 82*ArcTanh[Sqrt[3 + 2*x]] + (316*ArcTanh[Sqrt[3/5]*Sqrt[3 +
2*x]])/Sqrt[15]

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 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/
((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g*(2*a*e*m + b*d*(2*p + 3)) - f*
(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], 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] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p]
|| IntegersQ[2*m, 2*p])

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}}{\left (2+5 x+3 x^2\right )^2} \, dx &=-\frac {\sqrt {3+2 x} (29+35 x)}{2+5 x+3 x^2}-\int \frac {76+35 x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac {\sqrt {3+2 x} (29+35 x)}{2+5 x+3 x^2}-2 \operatorname {Subst}\left (\int \frac {47+35 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt {3+2 x}\right )\\ &=-\frac {\sqrt {3+2 x} (29+35 x)}{2+5 x+3 x^2}+246 \operatorname {Subst}\left (\int \frac {1}{-3+3 x^2} \, dx,x,\sqrt {3+2 x}\right )-316 \operatorname {Subst}\left (\int \frac {1}{-5+3 x^2} \, dx,x,\sqrt {3+2 x}\right )\\ &=-\frac {\sqrt {3+2 x} (29+35 x)}{2+5 x+3 x^2}-82 \tanh ^{-1}\left (\sqrt {3+2 x}\right )+\frac {316 \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )}{\sqrt {15}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 66, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {2 x+3} (35 x+29)}{3 x^2+5 x+2}-82 \tanh ^{-1}\left (\sqrt {2 x+3}\right )+\frac {316 \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[3 + 2*x]*(29 + 35*x))/(2 + 5*x + 3*x^2)) - 82*ArcTanh[Sqrt[3 + 2*x]] + (316*ArcTanh[Sqrt[3/5]*Sqrt[3 +
2*x]])/Sqrt[15]

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IntegrateAlgebraic [A]  time = 0.19, size = 83, normalized size = 1.26 \begin {gather*} -\frac {2 \left (35 (2 x+3)^{3/2}-47 \sqrt {2 x+3}\right )}{3 (2 x+3)^2-8 (2 x+3)+5}-82 \tanh ^{-1}\left (\sqrt {2 x+3}\right )+\frac {316 \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-47*Sqrt[3 + 2*x] + 35*(3 + 2*x)^(3/2)))/(5 - 8*(3 + 2*x) + 3*(3 + 2*x)^2) - 82*ArcTanh[Sqrt[3 + 2*x]] +
(316*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15]

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fricas [B]  time = 0.41, size = 113, normalized size = 1.71 \begin {gather*} \frac {158 \, \sqrt {15} {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\frac {\sqrt {15} \sqrt {2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 615 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) + 615 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 15 \, {\left (35 \, x + 29\right )} \sqrt {2 \, x + 3}}{15 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \end {gather*}

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)^2,x, algorithm="fricas")

[Out]

1/15*(158*sqrt(15)*(3*x^2 + 5*x + 2)*log((sqrt(15)*sqrt(2*x + 3) + 3*x + 7)/(3*x + 2)) - 615*(3*x^2 + 5*x + 2)
*log(sqrt(2*x + 3) + 1) + 615*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) - 1) - 15*(35*x + 29)*sqrt(2*x + 3))/(3*x^2
+ 5*x + 2)

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giac [A]  time = 0.18, size = 102, normalized size = 1.55 \begin {gather*} -\frac {158}{15} \, \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 \, {\left (35 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 47 \, \sqrt {2 \, x + 3}\right )}}{3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19} - 41 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 41 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \end {gather*}

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)^2,x, algorithm="giac")

[Out]

-158/15*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2*(35*(2*x + 3)^(3
/2) - 47*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 41*log(sqrt(2*x + 3) + 1) + 41*log(abs(sqrt(2*x + 3) - 1
))

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maple [A]  time = 0.02, size = 86, normalized size = 1.30 \begin {gather*} \frac {316 \sqrt {15}\, \arctanh \left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{15}+41 \ln \left (-1+\sqrt {2 x +3}\right )-41 \ln \left (\sqrt {2 x +3}+1\right )-\frac {34 \sqrt {2 x +3}}{3 \left (2 x +\frac {4}{3}\right )}-\frac {6}{\sqrt {2 x +3}+1}-\frac {6}{-1+\sqrt {2 x +3}} \end {gather*}

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)^2,x)

[Out]

-34/3*(2*x+3)^(1/2)/(2*x+4/3)+316/15*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))*15^(1/2)-6/((2*x+3)^(1/2)+1)-41*ln((2
*x+3)^(1/2)+1)-6/(-1+(2*x+3)^(1/2))+41*ln(-1+(2*x+3)^(1/2))

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maxima [A]  time = 1.12, size = 98, normalized size = 1.48 \begin {gather*} -\frac {158}{15} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {2 \, {\left (35 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 47 \, \sqrt {2 \, x + 3}\right )}}{3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19} - 41 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 41 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \end {gather*}

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)^2,x, algorithm="maxima")

[Out]

-158/15*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2*(35*(2*x + 3)^(3/2) - 47*
sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 41*log(sqrt(2*x + 3) + 1) + 41*log(sqrt(2*x + 3) - 1)

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mupad [B]  time = 2.40, size = 66, normalized size = 1.00 \begin {gather*} \frac {316\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{15}-\frac {\frac {94\,\sqrt {2\,x+3}}{3}-\frac {70\,{\left (2\,x+3\right )}^{3/2}}{3}}{\frac {16\,x}{3}-{\left (2\,x+3\right )}^2+\frac {19}{3}}-82\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right ) \end {gather*}

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)^2,x)

[Out]

(316*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/15 - ((94*(2*x + 3)^(1/2))/3 - (70*(2*x + 3)^(3/2))/3)/((16
*x)/3 - (2*x + 3)^2 + 19/3) - 82*atanh((2*x + 3)^(1/2))

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sympy [A]  time = 122.22, size = 212, normalized size = 3.21 \begin {gather*} 340 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: x \geq - \frac {3}{2} \wedge x < - \frac {2}{3} \end {cases}\right ) - 282 \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 ) + 41 \log {\left (\sqrt {2 x + 3} - 1 \right )} - 41 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {6}{\sqrt {2 x + 3} + 1} - \frac {6}{\sqrt {2 x + 3} - 1} \end {gather*}

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)**2,x)

[Out]

340*Piecewise((sqrt(15)*(-log(sqrt(15)*sqrt(2*x + 3)/5 - 1)/4 + log(sqrt(15)*sqrt(2*x + 3)/5 + 1)/4 - 1/(4*(sq
rt(15)*sqrt(2*x + 3)/5 + 1)) - 1/(4*(sqrt(15)*sqrt(2*x + 3)/5 - 1)))/75, (x >= -3/2) & (x < -2/3))) - 282*Piec
ewise((-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)) + 41*log(sqrt(2*x + 3) - 1) - 41*log(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) + 1) - 6/(sqr
t(2*x + 3) - 1)

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