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

Optimal. Leaf size=115 \[ -\frac {3 (47 x+37)}{10 \sqrt {2 x+3} \left (3 x^2+5 x+2\right )^2}+\frac {2229 x+1888}{10 \sqrt {2 x+3} \left (3 x^2+5 x+2\right )}+\frac {2667}{25 \sqrt {2 x+3}}+402 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {12717}{25} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {822, 828, 826, 1166, 207} \begin {gather*} -\frac {3 (47 x+37)}{10 \sqrt {2 x+3} \left (3 x^2+5 x+2\right )^2}+\frac {2229 x+1888}{10 \sqrt {2 x+3} \left (3 x^2+5 x+2\right )}+\frac {2667}{25 \sqrt {2 x+3}}+402 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {12717}{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)^(3/2)*(2 + 5*x + 3*x^2)^3),x]

[Out]

2667/(25*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(10*Sqrt[3 + 2*x]*(2 + 5*x + 3*x^2)^2) + (1888 + 2229*x)/(10*Sqrt[3
+ 2*x]*(2 + 5*x + 3*x^2)) + 402*ArcTanh[Sqrt[3 + 2*x]] - (12717*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 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], 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] && LtQ[p, -1] && (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 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)^{3/2} \left (2+5 x+3 x^2\right )^3} \, dx &=-\frac {3 (37+47 x)}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2}-\frac {1}{10} \int \frac {1328+987 x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac {3 (37+47 x)}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2}+\frac {1888+2229 x}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )}+\frac {1}{50} \int \frac {43485+33435 x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=\frac {2667}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2}+\frac {1888+2229 x}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )}+\frac {1}{250} \int \frac {90255+40005 x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=\frac {2667}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2}+\frac {1888+2229 x}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )}+\frac {1}{125} \operatorname {Subst}\left (\int \frac {60495+40005 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt {3+2 x}\right )\\ &=\frac {2667}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2}+\frac {1888+2229 x}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )}-1206 \operatorname {Subst}\left (\int \frac {1}{-3+3 x^2} \, dx,x,\sqrt {3+2 x}\right )+\frac {38151}{25} \operatorname {Subst}\left (\int \frac {1}{-5+3 x^2} \, dx,x,\sqrt {3+2 x}\right )\\ &=\frac {2667}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2}+\frac {1888+2229 x}{10 \sqrt {3+2 x} \left (2+5 x+3 x^2\right )}+402 \tanh ^{-1}\left (\sqrt {3+2 x}\right )-\frac {12717}{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 = 0.14, size = 86, normalized size = 0.75 \begin {gather*} \frac {1}{50} \left (\frac {48006 x^4+193455 x^3+281403 x^2+175465 x+39661}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )^2}+20100 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-25434 \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((39661 + 175465*x + 281403*x^2 + 193455*x^3 + 48006*x^4)/(Sqrt[3 + 2*x]*(2 + 5*x + 3*x^2)^2) + 20100*ArcTanh[
Sqrt[3 + 2*x]] - 25434*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/50

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IntegrateAlgebraic [A]  time = 0.31, size = 111, normalized size = 0.97 \begin {gather*} \frac {24003 (2 x+3)^4-94581 (2 x+3)^3+117873 (2 x+3)^2-44015 (2 x+3)-2080}{25 \sqrt {2 x+3} \left (3 (2 x+3)^2-8 (2 x+3)+5\right )^2}+402 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {12717}{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)^(3/2)*(2 + 5*x + 3*x^2)^3),x]

[Out]

(-2080 - 44015*(3 + 2*x) + 117873*(3 + 2*x)^2 - 94581*(3 + 2*x)^3 + 24003*(3 + 2*x)^4)/(25*Sqrt[3 + 2*x]*(5 -
8*(3 + 2*x) + 3*(3 + 2*x)^2)^2) + 402*ArcTanh[Sqrt[3 + 2*x]] - (12717*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x
]])/25

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fricas [B]  time = 0.41, size = 195, normalized size = 1.70 \begin {gather*} \frac {12717 \, \sqrt {5} \sqrt {3} {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 50250 \, {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 50250 \, {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) + 5 \, {\left (48006 \, x^{4} + 193455 \, x^{3} + 281403 \, x^{2} + 175465 \, x + 39661\right )} \sqrt {2 \, x + 3}}{250 \, {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/250*(12717*sqrt(5)*sqrt(3)*(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x
+ 3) - 3*x - 7)/(3*x + 2)) + 50250*(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12)*log(sqrt(2*x + 3) + 1) -
50250*(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12)*log(sqrt(2*x + 3) - 1) + 5*(48006*x^4 + 193455*x^3 + 2
81403*x^2 + 175465*x + 39661)*sqrt(2*x + 3))/(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12)

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giac [A]  time = 0.18, size = 129, normalized size = 1.12 \begin {gather*} \frac {12717}{250} \, \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 {416}{125 \, \sqrt {2 \, x + 3}} + \frac {123759 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 492873 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 628469 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 253355 \, \sqrt {2 \, x + 3}}{125 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 201 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 201 \, \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)^(3/2)/(3*x^2+5*x+2)^3,x, algorithm="giac")

[Out]

12717/250*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 416/125/sqrt(2*x
+ 3) + 1/125*(123759*(2*x + 3)^(7/2) - 492873*(2*x + 3)^(5/2) + 628469*(2*x + 3)^(3/2) - 253355*sqrt(2*x + 3)
)/(3*(2*x + 3)^2 - 16*x - 19)^2 + 201*log(sqrt(2*x + 3) + 1) - 201*log(abs(sqrt(2*x + 3) - 1))

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maple [A]  time = 0.02, size = 133, normalized size = 1.16 \begin {gather*} -\frac {12717 \sqrt {15}\, \arctanh \left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{125}-201 \ln \left (-1+\sqrt {2 x +3}\right )+201 \ln \left (\sqrt {2 x +3}+1\right )+\frac {\frac {51759 \left (2 x +3\right )^{\frac {3}{2}}}{125}-\frac {18171 \sqrt {2 x +3}}{25}}{\left (6 x +4\right )^{2}}+\frac {32}{\sqrt {2 x +3}+1}-\frac {3}{\left (\sqrt {2 x +3}+1\right )^{2}}-\frac {416}{125 \sqrt {2 x +3}}+\frac {32}{-1+\sqrt {2 x +3}}+\frac {3}{\left (-1+\sqrt {2 x +3}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

486/125*(213/2*(2*x+3)^(3/2)-3365/18*(2*x+3)^(1/2))/(6*x+4)^2-12717/125*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))*15
^(1/2)+32/((2*x+3)^(1/2)+1)-3/((2*x+3)^(1/2)+1)^2+201*ln((2*x+3)^(1/2)+1)-416/125/(2*x+3)^(1/2)+32/(-1+(2*x+3)
^(1/2))+3/(-1+(2*x+3)^(1/2))^2-201*ln(-1+(2*x+3)^(1/2))

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maxima [A]  time = 0.99, size = 143, normalized size = 1.24 \begin {gather*} \frac {12717}{250} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {24003 \, {\left (2 \, x + 3\right )}^{4} - 94581 \, {\left (2 \, x + 3\right )}^{3} + 117873 \, {\left (2 \, x + 3\right )}^{2} - 88030 \, x - 134125}{25 \, {\left (9 \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} - 48 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} + 94 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} - 80 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} + 25 \, \sqrt {2 \, x + 3}\right )}} + 201 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 201 \, \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)^(3/2)/(3*x^2+5*x+2)^3,x, algorithm="maxima")

[Out]

12717/250*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/25*(24003*(2*x + 3)^4 -
94581*(2*x + 3)^3 + 117873*(2*x + 3)^2 - 88030*x - 134125)/(9*(2*x + 3)^(9/2) - 48*(2*x + 3)^(7/2) + 94*(2*x
+ 3)^(5/2) - 80*(2*x + 3)^(3/2) + 25*sqrt(2*x + 3)) + 201*log(sqrt(2*x + 3) + 1) - 201*log(sqrt(2*x + 3) - 1)

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mupad [B]  time = 2.40, size = 109, normalized size = 0.95 \begin {gather*} 402\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {12717\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{125}-\frac {\frac {17606\,x}{45}-\frac {13097\,{\left (2\,x+3\right )}^2}{25}+\frac {10509\,{\left (2\,x+3\right )}^3}{25}-\frac {2667\,{\left (2\,x+3\right )}^4}{25}+\frac {5365}{9}}{\frac {25\,\sqrt {2\,x+3}}{9}-\frac {80\,{\left (2\,x+3\right )}^{3/2}}{9}+\frac {94\,{\left (2\,x+3\right )}^{5/2}}{9}-\frac {16\,{\left (2\,x+3\right )}^{7/2}}{3}+{\left (2\,x+3\right )}^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

402*atanh((2*x + 3)^(1/2)) - (12717*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/125 - ((17606*x)/45 - (13097
*(2*x + 3)^2)/25 + (10509*(2*x + 3)^3)/25 - (2667*(2*x + 3)^4)/25 + 5365/9)/((25*(2*x + 3)^(1/2))/9 - (80*(2*x
+ 3)^(3/2))/9 + (94*(2*x + 3)^(5/2))/9 - (16*(2*x + 3)^(7/2))/3 + (2*x + 3)^(9/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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