3.19.66 \(\int \frac {(1-2 x)^{3/2} (3+5 x)}{(2+3 x)^3} \, dx\) [1866]

Optimal. Leaf size=81 \[ -\frac {71}{63} \sqrt {1-2 x}+\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}+\frac {71 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}} \]

[Out]

1/42*(1-2*x)^(5/2)/(2+3*x)^2-71/126*(1-2*x)^(3/2)/(2+3*x)+71/189*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-
71/63*(1-2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {79, 43, 52, 65, 212} \begin {gather*} \frac {(1-2 x)^{5/2}}{42 (3 x+2)^2}-\frac {71 (1-2 x)^{3/2}}{126 (3 x+2)}-\frac {71}{63} \sqrt {1-2 x}+\frac {71 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)^(3/2)*(3 + 5*x))/(2 + 3*x)^3,x]

[Out]

(-71*Sqrt[1 - 2*x])/63 + (1 - 2*x)^(5/2)/(42*(2 + 3*x)^2) - (71*(1 - 2*x)^(3/2))/(126*(2 + 3*x)) + (71*ArcTanh
[Sqrt[3/7]*Sqrt[1 - 2*x]])/(9*Sqrt[21])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{3/2} (3+5 x)}{(2+3 x)^3} \, dx &=\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}+\frac {71}{42} \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2} \, dx\\ &=\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}-\frac {71}{42} \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx\\ &=-\frac {71}{63} \sqrt {1-2 x}+\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}-\frac {71}{18} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {71}{63} \sqrt {1-2 x}+\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}+\frac {71}{18} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {71}{63} \sqrt {1-2 x}+\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}+\frac {71 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 58, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {1-2 x} \left (101+235 x+120 x^2\right )}{18 (2+3 x)^2}+\frac {71 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)^(3/2)*(3 + 5*x))/(2 + 3*x)^3,x]

[Out]

-1/18*(Sqrt[1 - 2*x]*(101 + 235*x + 120*x^2))/(2 + 3*x)^2 + (71*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(9*Sqrt[21])

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Maple [A]
time = 0.10, size = 57, normalized size = 0.70

method result size
risch \(\frac {240 x^{3}+350 x^{2}-33 x -101}{18 \left (2+3 x \right )^{2} \sqrt {1-2 x}}+\frac {71 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{189}\) \(51\)
derivativedivides \(-\frac {20 \sqrt {1-2 x}}{27}-\frac {4 \left (-\frac {25 \left (1-2 x \right )^{\frac {3}{2}}}{4}+\frac {511 \sqrt {1-2 x}}{36}\right )}{3 \left (-4-6 x \right )^{2}}+\frac {71 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{189}\) \(57\)
default \(-\frac {20 \sqrt {1-2 x}}{27}-\frac {4 \left (-\frac {25 \left (1-2 x \right )^{\frac {3}{2}}}{4}+\frac {511 \sqrt {1-2 x}}{36}\right )}{3 \left (-4-6 x \right )^{2}}+\frac {71 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{189}\) \(57\)
trager \(-\frac {\left (120 x^{2}+235 x +101\right ) \sqrt {1-2 x}}{18 \left (2+3 x \right )^{2}}+\frac {71 \RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \RootOf \left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{378}\) \(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-2*x)^(3/2)*(3+5*x)/(2+3*x)^3,x,method=_RETURNVERBOSE)

[Out]

-20/27*(1-2*x)^(1/2)-4/3*(-25/4*(1-2*x)^(3/2)+511/36*(1-2*x)^(1/2))/(-4-6*x)^2+71/189*arctanh(1/7*21^(1/2)*(1-
2*x)^(1/2))*21^(1/2)

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Maxima [A]
time = 0.49, size = 83, normalized size = 1.02 \begin {gather*} -\frac {71}{378} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {20}{27} \, \sqrt {-2 \, x + 1} + \frac {225 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 511 \, \sqrt {-2 \, x + 1}}{27 \, {\left (9 \, {\left (2 \, x - 1\right )}^{2} + 84 \, x + 7\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)/(2+3*x)^3,x, algorithm="maxima")

[Out]

-71/378*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 20/27*sqrt(-2*x + 1) + 1/
27*(225*(-2*x + 1)^(3/2) - 511*sqrt(-2*x + 1))/(9*(2*x - 1)^2 + 84*x + 7)

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Fricas [A]
time = 1.10, size = 75, normalized size = 0.93 \begin {gather*} \frac {71 \, \sqrt {21} {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (120 \, x^{2} + 235 \, x + 101\right )} \sqrt {-2 \, x + 1}}{378 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)/(2+3*x)^3,x, algorithm="fricas")

[Out]

1/378*(71*sqrt(21)*(9*x^2 + 12*x + 4)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(120*x^2 + 235*x
 + 101)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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Sympy [F(-1)] Timed out
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((1-2*x)**(3/2)*(3+5*x)/(2+3*x)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.49, size = 77, normalized size = 0.95 \begin {gather*} -\frac {71}{378} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {20}{27} \, \sqrt {-2 \, x + 1} + \frac {225 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 511 \, \sqrt {-2 \, x + 1}}{108 \, {\left (3 \, x + 2\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)/(2+3*x)^3,x, algorithm="giac")

[Out]

-71/378*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 20/27*sqrt(-2*x
+ 1) + 1/108*(225*(-2*x + 1)^(3/2) - 511*sqrt(-2*x + 1))/(3*x + 2)^2

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Mupad [B]
time = 0.06, size = 63, normalized size = 0.78 \begin {gather*} \frac {71\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{189}-\frac {20\,\sqrt {1-2\,x}}{27}-\frac {\frac {511\,\sqrt {1-2\,x}}{243}-\frac {25\,{\left (1-2\,x\right )}^{3/2}}{27}}{\frac {28\,x}{3}+{\left (2\,x-1\right )}^2+\frac {7}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - 2*x)^(3/2)*(5*x + 3))/(3*x + 2)^3,x)

[Out]

(71*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/189 - (20*(1 - 2*x)^(1/2))/27 - ((511*(1 - 2*x)^(1/2))/243 -
 (25*(1 - 2*x)^(3/2))/27)/((28*x)/3 + (2*x - 1)^2 + 7/9)

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