3.1911 \(\int \frac{(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=89 \[ -\frac{(1-2 x)^{5/2}}{275 (5 x+3)}-\frac{9}{125} (1-2 x)^{5/2}+\frac{42 (1-2 x)^{3/2}}{1375}+\frac{126}{625} \sqrt{1-2 x}-\frac{126}{625} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(126*Sqrt[1 - 2*x])/625 + (42*(1 - 2*x)^(3/2))/1375 - (9*(1 - 2*x)^(5/2))/125 - (1 - 2*x)^(5/2)/(275*(3 + 5*x)
) - (126*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/625

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Rubi [A]  time = 0.0248406, 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} \[ -\frac{(1-2 x)^{5/2}}{275 (5 x+3)}-\frac{9}{125} (1-2 x)^{5/2}+\frac{42 (1-2 x)^{3/2}}{1375}+\frac{126}{625} \sqrt{1-2 x}-\frac{126}{625} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(126*Sqrt[1 - 2*x])/625 + (42*(1 - 2*x)^(3/2))/1375 - (9*(1 - 2*x)^(5/2))/125 - (1 - 2*x)^(5/2)/(275*(3 + 5*x)
) - (126*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/625

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} (2+3 x)^2}{(3+5 x)^2} \, dx &=-\frac{(1-2 x)^{5/2}}{275 (3+5 x)}+\frac{1}{275} \int \frac{(1-2 x)^{3/2} (360+495 x)}{3+5 x} \, dx\\ &=-\frac{9}{125} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{275 (3+5 x)}+\frac{63}{275} \int \frac{(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=\frac{42 (1-2 x)^{3/2}}{1375}-\frac{9}{125} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{275 (3+5 x)}+\frac{63}{125} \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx\\ &=\frac{126}{625} \sqrt{1-2 x}+\frac{42 (1-2 x)^{3/2}}{1375}-\frac{9}{125} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{275 (3+5 x)}+\frac{693}{625} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=\frac{126}{625} \sqrt{1-2 x}+\frac{42 (1-2 x)^{3/2}}{1375}-\frac{9}{125} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{275 (3+5 x)}-\frac{693}{625} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{126}{625} \sqrt{1-2 x}+\frac{42 (1-2 x)^{3/2}}{1375}-\frac{9}{125} (1-2 x)^{5/2}-\frac{(1-2 x)^{5/2}}{275 (3+5 x)}-\frac{126}{625} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0356058, size = 63, normalized size = 0.71 \[ \frac{\frac{5 \sqrt{1-2 x} \left (-900 x^3+160 x^2+935 x+298\right )}{5 x+3}-126 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3125} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(298 + 935*x + 160*x^2 - 900*x^3))/(3 + 5*x) - 126*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]
])/3125

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Maple [A]  time = 0.01, size = 63, normalized size = 0.7 \begin{align*} -{\frac{9}{125} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}+{\frac{4}{125} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{128}{625}\sqrt{1-2\,x}}+{\frac{22}{3125}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{126\,\sqrt{55}}{3125}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/125*(1-2*x)^(5/2)+4/125*(1-2*x)^(3/2)+128/625*(1-2*x)^(1/2)+22/3125*(1-2*x)^(1/2)/(-2*x-6/5)-126/3125*arcta
nh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.89637, size = 108, normalized size = 1.21 \begin{align*} -\frac{9}{125} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} + \frac{4}{125} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{63}{3125} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{128}{625} \, \sqrt{-2 \, x + 1} - \frac{11 \, \sqrt{-2 \, x + 1}}{625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/125*(-2*x + 1)^(5/2) + 4/125*(-2*x + 1)^(3/2) + 63/3125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(5
5) + 5*sqrt(-2*x + 1))) + 128/625*sqrt(-2*x + 1) - 11/625*sqrt(-2*x + 1)/(5*x + 3)

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Fricas [A]  time = 1.5757, size = 219, normalized size = 2.46 \begin{align*} \frac{63 \, \sqrt{11} \sqrt{5}{\left (5 \, x + 3\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 5 \,{\left (900 \, x^{3} - 160 \, x^{2} - 935 \, x - 298\right )} \sqrt{-2 \, x + 1}}{3125 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3125*(63*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 5*(900*x^3
- 160*x^2 - 935*x - 298)*sqrt(-2*x + 1))/(5*x + 3)

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

[Out]

Timed out

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Giac [A]  time = 1.96852, size = 122, normalized size = 1.37 \begin{align*} -\frac{9}{125} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} + \frac{4}{125} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{63}{3125} \, \sqrt{55} \log \left (\frac{{\left | -2 \, \sqrt{55} + 10 \, \sqrt{-2 \, x + 1} \right |}}{2 \,{\left (\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}\right )}}\right ) + \frac{128}{625} \, \sqrt{-2 \, x + 1} - \frac{11 \, \sqrt{-2 \, x + 1}}{625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/125*(2*x - 1)^2*sqrt(-2*x + 1) + 4/125*(-2*x + 1)^(3/2) + 63/3125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqr
t(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 128/625*sqrt(-2*x + 1) - 11/625*sqrt(-2*x + 1)/(5*x + 3)