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

Optimal. Leaf size=128 \[ -\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}+\frac {10551 x+9146}{50 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}+\frac {6853}{125 \sqrt {2 x+3}}+\frac {7451}{75 (2 x+3)^{3/2}}+310 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {45603}{125} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 8, 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 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}+\frac {10551 x+9146}{50 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}+\frac {6853}{125 \sqrt {2 x+3}}+\frac {7451}{75 (2 x+3)^{3/2}}+310 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {45603}{125} \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)^3),x]

[Out]

7451/(75*(3 + 2*x)^(3/2)) + 6853/(125*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(10*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)^2
) + (9146 + 10551*x)/(50*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)) + 310*ArcTanh[Sqrt[3 + 2*x]] - (45603*Sqrt[3/5]*Ar
cTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/125

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)^{5/2} \left (2+5 x+3 x^2\right )^3} \, dx &=-\frac {3 (37+47 x)}{10 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}-\frac {1}{10} \int \frac {1550+1269 x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac {3 (37+47 x)}{10 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}+\frac {9146+10551 x}{50 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}+\frac {1}{50} \int \frac {60505+52755 x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=\frac {7451}{75 (3+2 x)^{3/2}}-\frac {3 (37+47 x)}{10 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}+\frac {9146+10551 x}{50 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}+\frac {1}{250} \int \frac {150515+111765 x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=\frac {7451}{75 (3+2 x)^{3/2}}+\frac {6853}{125 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}+\frac {9146+10551 x}{50 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}+\frac {\int \frac {296545+102795 x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )} \, dx}{1250}\\ &=\frac {7451}{75 (3+2 x)^{3/2}}+\frac {6853}{125 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}+\frac {9146+10551 x}{50 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}+\frac {1}{625} \operatorname {Subst}\left (\int \frac {284705+102795 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt {3+2 x}\right )\\ &=\frac {7451}{75 (3+2 x)^{3/2}}+\frac {6853}{125 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}+\frac {9146+10551 x}{50 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-930 \operatorname {Subst}\left (\int \frac {1}{-3+3 x^2} \, dx,x,\sqrt {3+2 x}\right )+\frac {136809}{125} \operatorname {Subst}\left (\int \frac {1}{-5+3 x^2} \, dx,x,\sqrt {3+2 x}\right )\\ &=\frac {7451}{75 (3+2 x)^{3/2}}+\frac {6853}{125 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}+\frac {9146+10551 x}{50 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}+310 \tanh ^{-1}\left (\sqrt {3+2 x}\right )-\frac {45603}{125} \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 = 0.29, size = 91, normalized size = 0.71 \begin {gather*} \frac {\frac {5 \left (740124 x^5+4247856 x^4+9453447 x^3+10168583 x^2+5278129 x+1057511\right )}{(2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}-273618 \sqrt {15} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{3750}+310 \tanh ^{-1}\left (\sqrt {2 x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

310*ArcTanh[Sqrt[3 + 2*x]] + ((5*(1057511 + 5278129*x + 10168583*x^2 + 9453447*x^3 + 4247856*x^4 + 740124*x^5)
)/((3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)^2) - 273618*Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/3750

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IntegrateAlgebraic [A]  time = 0.30, size = 120, normalized size = 0.94 \begin {gather*} \frac {185031 (2 x+3)^5-651537 (2 x+3)^4+619101 (2 x+3)^3-10115 (2 x+3)^2-114080 (2 x+3)-10400}{375 (2 x+3)^{3/2} \left (3 (2 x+3)^2-8 (2 x+3)+5\right )^2}+310 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {45603}{125} \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)^3),x]

[Out]

(-10400 - 114080*(3 + 2*x) - 10115*(3 + 2*x)^2 + 619101*(3 + 2*x)^3 - 651537*(3 + 2*x)^4 + 185031*(3 + 2*x)^5)
/(375*(3 + 2*x)^(3/2)*(5 - 8*(3 + 2*x) + 3*(3 + 2*x)^2)^2) + 310*ArcTanh[Sqrt[3 + 2*x]] - (45603*Sqrt[3/5]*Arc
Tanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/125

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fricas [B]  time = 0.43, size = 220, normalized size = 1.72 \begin {gather*} \frac {136809 \, \sqrt {5} \sqrt {3} {\left (36 \, x^{6} + 228 \, x^{5} + 589 \, x^{4} + 794 \, x^{3} + 589 \, x^{2} + 228 \, x + 36\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 581250 \, {\left (36 \, x^{6} + 228 \, x^{5} + 589 \, x^{4} + 794 \, x^{3} + 589 \, x^{2} + 228 \, x + 36\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 581250 \, {\left (36 \, x^{6} + 228 \, x^{5} + 589 \, x^{4} + 794 \, x^{3} + 589 \, x^{2} + 228 \, x + 36\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) + 5 \, {\left (740124 \, x^{5} + 4247856 \, x^{4} + 9453447 \, x^{3} + 10168583 \, x^{2} + 5278129 \, x + 1057511\right )} \sqrt {2 \, x + 3}}{3750 \, {\left (36 \, x^{6} + 228 \, x^{5} + 589 \, x^{4} + 794 \, x^{3} + 589 \, x^{2} + 228 \, x + 36\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)^3,x, algorithm="fricas")

[Out]

1/3750*(136809*sqrt(5)*sqrt(3)*(36*x^6 + 228*x^5 + 589*x^4 + 794*x^3 + 589*x^2 + 228*x + 36)*log(-(sqrt(5)*sqr
t(3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) + 581250*(36*x^6 + 228*x^5 + 589*x^4 + 794*x^3 + 589*x^2 + 228*x + 36
)*log(sqrt(2*x + 3) + 1) - 581250*(36*x^6 + 228*x^5 + 589*x^4 + 794*x^3 + 589*x^2 + 228*x + 36)*log(sqrt(2*x +
3) - 1) + 5*(740124*x^5 + 4247856*x^4 + 9453447*x^3 + 10168583*x^2 + 5278129*x + 1057511)*sqrt(2*x + 3))/(36*
x^6 + 228*x^5 + 589*x^4 + 794*x^3 + 589*x^2 + 228*x + 36)

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giac [A]  time = 0.18, size = 134, normalized size = 1.05 \begin {gather*} \frac {45603}{1250} \, \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 {64 \, {\left (921 \, x + 1414\right )}}{1875 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}}} + \frac {396801 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 1551207 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 1922011 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 737605 \, \sqrt {2 \, x + 3}}{625 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 155 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 155 \, \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)^3,x, algorithm="giac")

[Out]

45603/1250*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 64/1875*(921*x
+ 1414)/(2*x + 3)^(3/2) + 1/625*(396801*(2*x + 3)^(7/2) - 1551207*(2*x + 3)^(5/2) + 1922011*(2*x + 3)^(3/2) -
737605*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 + 155*log(sqrt(2*x + 3) + 1) - 155*log(abs(sqrt(2*x + 3) -
1))

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maple [A]  time = 0.02, size = 142, normalized size = 1.11 \begin {gather*} -\frac {45603 \sqrt {15}\, \arctanh \left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{625}-155 \ln \left (-1+\sqrt {2 x +3}\right )+155 \ln \left (\sqrt {2 x +3}+1\right )+\frac {\frac {171801 \left (2 x +3\right )^{\frac {3}{2}}}{625}-\frac {60021 \sqrt {2 x +3}}{125}}{\left (6 x +4\right )^{2}}-\frac {3}{\left (\sqrt {2 x +3}+1\right )^{2}}+\frac {20}{\sqrt {2 x +3}+1}-\frac {416}{375 \left (2 x +3\right )^{\frac {3}{2}}}-\frac {9824}{625 \sqrt {2 x +3}}+\frac {3}{\left (-1+\sqrt {2 x +3}\right )^{2}}+\frac {20}{-1+\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)^3,x)

[Out]

4374/625*(707/18*(2*x+3)^(3/2)-1235/18*(2*x+3)^(1/2))/(6*x+4)^2-45603/625*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))*
15^(1/2)-3/((2*x+3)^(1/2)+1)^2+20/((2*x+3)^(1/2)+1)+155*ln((2*x+3)^(1/2)+1)-416/375/(2*x+3)^(3/2)-9824/625/(2*
x+3)^(1/2)+3/(-1+(2*x+3)^(1/2))^2+20/(-1+(2*x+3)^(1/2))-155*ln(-1+(2*x+3)^(1/2))

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maxima [A]  time = 0.98, size = 152, normalized size = 1.19 \begin {gather*} \frac {45603}{1250} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {185031 \, {\left (2 \, x + 3\right )}^{5} - 651537 \, {\left (2 \, x + 3\right )}^{4} + 619101 \, {\left (2 \, x + 3\right )}^{3} - 10115 \, {\left (2 \, x + 3\right )}^{2} - 228160 \, x - 352640}{375 \, {\left (9 \, {\left (2 \, x + 3\right )}^{\frac {11}{2}} - 48 \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} + 94 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 80 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 25 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}}\right )}} + 155 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 155 \, \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)^3,x, algorithm="maxima")

[Out]

45603/1250*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/375*(185031*(2*x + 3)^
5 - 651537*(2*x + 3)^4 + 619101*(2*x + 3)^3 - 10115*(2*x + 3)^2 - 228160*x - 352640)/(9*(2*x + 3)^(11/2) - 48*
(2*x + 3)^(9/2) + 94*(2*x + 3)^(7/2) - 80*(2*x + 3)^(5/2) + 25*(2*x + 3)^(3/2)) + 155*log(sqrt(2*x + 3) + 1) -
155*log(sqrt(2*x + 3) - 1)

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mupad [B]  time = 0.08, size = 118, normalized size = 0.92 \begin {gather*} 310\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {45603\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{625}-\frac {\frac {45632\,x}{675}+\frac {2023\,{\left (2\,x+3\right )}^2}{675}-\frac {68789\,{\left (2\,x+3\right )}^3}{375}+\frac {24131\,{\left (2\,x+3\right )}^4}{125}-\frac {6853\,{\left (2\,x+3\right )}^5}{125}+\frac {70528}{675}}{\frac {25\,{\left (2\,x+3\right )}^{3/2}}{9}-\frac {80\,{\left (2\,x+3\right )}^{5/2}}{9}+\frac {94\,{\left (2\,x+3\right )}^{7/2}}{9}-\frac {16\,{\left (2\,x+3\right )}^{9/2}}{3}+{\left (2\,x+3\right )}^{11/2}} \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)^3),x)

[Out]

310*atanh((2*x + 3)^(1/2)) - (45603*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/625 - ((45632*x)/675 + (2023
*(2*x + 3)^2)/675 - (68789*(2*x + 3)^3)/375 + (24131*(2*x + 3)^4)/125 - (6853*(2*x + 3)^5)/125 + 70528/675)/((
25*(2*x + 3)^(3/2))/9 - (80*(2*x + 3)^(5/2))/9 + (94*(2*x + 3)^(7/2))/9 - (16*(2*x + 3)^(9/2))/3 + (2*x + 3)^(
11/2))

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

[Out]

Timed out

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