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

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

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Rubi [A]  time = 0.06, antiderivative size = 68, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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 ) \end {gather*}

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 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 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 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 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}{(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*}

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Mathematica [A]  time = 0.08, size = 65, normalized size = 0.96 \begin {gather*} 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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.17, size = 64, normalized size = 0.94 \begin {gather*} -\frac {2 (297 (2 x+3)+65)}{75 (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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(65 + 297*(3 + 2*x)))/(75*(3 + 2*x)^(3/2)) + 12*ArcTanh[Sqrt[3 + 2*x]] - (102*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*
Sqrt[3 + 2*x]])/25

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fricas [B]  time = 0.42, size = 120, normalized size = 1.76 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.17, size = 79, normalized size = 1.16 \begin {gather*} \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 {gather*}

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))

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maple [A]  time = 0.01, size = 62, normalized size = 0.91 \begin {gather*} -\frac {102 \sqrt {15}\, \arctanh \left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{125}-6 \ln \left (-1+\sqrt {2 x +3}\right )+6 \ln \left (\sqrt {2 x +3}+1\right )-\frac {26}{15 \left (2 x +3\right )^{\frac {3}{2}}}-\frac {198}{25 \sqrt {2 x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.21, size = 75, normalized size = 1.10 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.07, size = 43, normalized size = 0.63 \begin {gather*} 12\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {\frac {396\,x}{25}+\frac {1912}{75}}{{\left (2\,x+3\right )}^{3/2}}-\frac {102\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*atanh((2*x + 3)^(1/2)) - ((396*x)/25 + 1912/75)/(2*x + 3)^(3/2) - (102*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(
1/2))/5))/125

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sympy [A]  time = 105.82, size = 114, normalized size = 1.68 \begin {gather*} \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 {gather*}

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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