3.1876 \(\int \frac{(1-2 x)^{3/2} (3+5 x)^2}{(2+3 x)^2} \, dx\)

Optimal. Leaf size=89 \[ -\frac{(1-2 x)^{5/2}}{63 (3 x+2)}-\frac{5}{9} (1-2 x)^{5/2}-\frac{146}{567} (1-2 x)^{3/2}-\frac{146}{81} \sqrt{1-2 x}+\frac{146}{81} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ) \]

[Out]

(-146*Sqrt[1 - 2*x])/81 - (146*(1 - 2*x)^(3/2))/567 - (5*(1 - 2*x)^(5/2))/9 - (1 - 2*x)^(5/2)/(63*(2 + 3*x)) +
(146*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/81

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Rubi [A]  time = 0.0245972, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {89, 80, 50, 63, 206} \[ -\frac{(1-2 x)^{5/2}}{63 (3 x+2)}-\frac{5}{9} (1-2 x)^{5/2}-\frac{146}{567} (1-2 x)^{3/2}-\frac{146}{81} \sqrt{1-2 x}+\frac{146}{81} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-146*Sqrt[1 - 2*x])/81 - (146*(1 - 2*x)^(3/2))/567 - (5*(1 - 2*x)^(5/2))/9 - (1 - 2*x)^(5/2)/(63*(2 + 3*x)) +
(146*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/81

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

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

Rule 50

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 63

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 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}{(2+3 x)^2} \, dx &=-\frac{(1-2 x)^{5/2}}{63 (2+3 x)}+\frac{1}{63} \int \frac{(1-2 x)^{3/2} (277+525 x)}{2+3 x} \, dx\\ &=-\frac{5}{9} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{63 (2+3 x)}-\frac{73}{63} \int \frac{(1-2 x)^{3/2}}{2+3 x} \, dx\\ &=-\frac{146}{567} (1-2 x)^{3/2}-\frac{5}{9} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{63 (2+3 x)}-\frac{73}{27} \int \frac{\sqrt{1-2 x}}{2+3 x} \, dx\\ &=-\frac{146}{81} \sqrt{1-2 x}-\frac{146}{567} (1-2 x)^{3/2}-\frac{5}{9} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{63 (2+3 x)}-\frac{511}{81} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=-\frac{146}{81} \sqrt{1-2 x}-\frac{146}{567} (1-2 x)^{3/2}-\frac{5}{9} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{63 (2+3 x)}+\frac{511}{81} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{146}{81} \sqrt{1-2 x}-\frac{146}{567} (1-2 x)^{3/2}-\frac{5}{9} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{63 (2+3 x)}+\frac{146}{81} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0345369, size = 63, normalized size = 0.71 \[ \frac{1}{243} \left (146 \sqrt{21} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{3 \sqrt{1-2 x} \left (540 x^3-300 x^2+187 x+425\right )}{3 x+2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*Sqrt[1 - 2*x]*(425 + 187*x - 300*x^2 + 540*x^3))/(2 + 3*x) + 146*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]
])/243

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Maple [A]  time = 0.01, size = 63, normalized size = 0.7 \begin{align*} -{\frac{5}{9} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}-{\frac{20}{81} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{16}{9}\sqrt{1-2\,x}}+{\frac{14}{243}\sqrt{1-2\,x} \left ( -2\,x-{\frac{4}{3}} \right ) ^{-1}}+{\frac{146\,\sqrt{21}}{243}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/9*(1-2*x)^(5/2)-20/81*(1-2*x)^(3/2)-16/9*(1-2*x)^(1/2)+14/243*(1-2*x)^(1/2)/(-2*x-4/3)+146/243*arctanh(1/7*
21^(1/2)*(1-2*x)^(1/2))*21^(1/2)

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Maxima [A]  time = 1.54562, size = 108, normalized size = 1.21 \begin{align*} -\frac{5}{9} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - \frac{20}{81} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - \frac{73}{243} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) - \frac{16}{9} \, \sqrt{-2 \, x + 1} - \frac{7 \, \sqrt{-2 \, x + 1}}{81 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/9*(-2*x + 1)^(5/2) - 20/81*(-2*x + 1)^(3/2) - 73/243*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21)
+ 3*sqrt(-2*x + 1))) - 16/9*sqrt(-2*x + 1) - 7/81*sqrt(-2*x + 1)/(3*x + 2)

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Fricas [A]  time = 1.4291, size = 216, normalized size = 2.43 \begin{align*} \frac{73 \, \sqrt{7} \sqrt{3}{\left (3 \, x + 2\right )} \log \left (-\frac{\sqrt{7} \sqrt{3} \sqrt{-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) - 3 \,{\left (540 \, x^{3} - 300 \, x^{2} + 187 \, x + 425\right )} \sqrt{-2 \, x + 1}}{243 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/243*(73*sqrt(7)*sqrt(3)*(3*x + 2)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) - 3*(540*x^3 -
300*x^2 + 187*x + 425)*sqrt(-2*x + 1))/(3*x + 2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 2.62776, size = 122, normalized size = 1.37 \begin{align*} -\frac{5}{9} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} - \frac{20}{81} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - \frac{73}{243} \, \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{16}{9} \, \sqrt{-2 \, x + 1} - \frac{7 \, \sqrt{-2 \, x + 1}}{81 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/9*(2*x - 1)^2*sqrt(-2*x + 1) - 20/81*(-2*x + 1)^(3/2) - 73/243*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2
*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 16/9*sqrt(-2*x + 1) - 7/81*sqrt(-2*x + 1)/(3*x + 2)