3.1922 \(\int \frac{(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=81 \[ -\frac{(1-2 x)^{5/2}}{110 (5 x+3)^2}-\frac{13 (1-2 x)^{3/2}}{110 (5 x+3)}-\frac{39}{275} \sqrt{1-2 x}+\frac{39 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}} \]

[Out]

(-39*Sqrt[1 - 2*x])/275 - (1 - 2*x)^(5/2)/(110*(3 + 5*x)^2) - (13*(1 - 2*x)^(3/2))/(110*(3 + 5*x)) + (39*ArcTa
nh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

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Rubi [A]  time = 0.0187935, antiderivative size = 81, normalized size of antiderivative = 1., 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 = {78, 47, 50, 63, 206} \[ -\frac{(1-2 x)^{5/2}}{110 (5 x+3)^2}-\frac{13 (1-2 x)^{3/2}}{110 (5 x+3)}-\frac{39}{275} \sqrt{1-2 x}+\frac{39 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-39*Sqrt[1 - 2*x])/275 - (1 - 2*x)^(5/2)/(110*(3 + 5*x)^2) - (13*(1 - 2*x)^(3/2))/(110*(3 + 5*x)) + (39*ArcTa
nh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

Rule 78

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] || LtQ
[p, n]))))

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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)}{(3+5 x)^3} \, dx &=-\frac{(1-2 x)^{5/2}}{110 (3+5 x)^2}+\frac{13}{22} \int \frac{(1-2 x)^{3/2}}{(3+5 x)^2} \, dx\\ &=-\frac{(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac{13 (1-2 x)^{3/2}}{110 (3+5 x)}-\frac{39}{110} \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx\\ &=-\frac{39}{275} \sqrt{1-2 x}-\frac{(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac{13 (1-2 x)^{3/2}}{110 (3+5 x)}-\frac{39}{50} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{39}{275} \sqrt{1-2 x}-\frac{(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac{13 (1-2 x)^{3/2}}{110 (3+5 x)}+\frac{39}{50} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{39}{275} \sqrt{1-2 x}-\frac{(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac{13 (1-2 x)^{3/2}}{110 (3+5 x)}+\frac{39 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}}\\ \end{align*}

Mathematica [C]  time = 0.0133043, size = 48, normalized size = 0.59 \[ -\frac{(1-2 x)^{5/2} \left (52 (5 x+3)^2 \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{5}{11} (1-2 x)\right )+121\right )}{13310 (5 x+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - 2*x)^(5/2)*(121 + 52*(3 + 5*x)^2*Hypergeometric2F1[2, 5/2, 7/2, (5*(1 - 2*x))/11]))/(13310*(3 + 5*x)^2)

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Maple [A]  time = 0.009, size = 57, normalized size = 0.7 \begin{align*} -{\frac{12}{125}\sqrt{1-2\,x}}-{\frac{4}{5\, \left ( -10\,x-6 \right ) ^{2}} \left ( -{\frac{61}{20} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{693}{100}\sqrt{1-2\,x}} \right ) }+{\frac{39\,\sqrt{55}}{1375}{\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)/(3+5*x)^3,x)

[Out]

-12/125*(1-2*x)^(1/2)-4/5*(-61/20*(1-2*x)^(3/2)+693/100*(1-2*x)^(1/2))/(-10*x-6)^2+39/1375*arctanh(1/11*55^(1/
2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.59548, size = 112, normalized size = 1.38 \begin{align*} -\frac{39}{2750} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{12}{125} \, \sqrt{-2 \, x + 1} + \frac{305 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 693 \, \sqrt{-2 \, x + 1}}{125 \,{\left (25 \,{\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-39/2750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 12/125*sqrt(-2*x + 1) +
1/125*(305*(-2*x + 1)^(3/2) - 693*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]  time = 1.57091, size = 211, normalized size = 2.6 \begin{align*} \frac{39 \, \sqrt{55}{\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac{5 \, x - \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \,{\left (120 \, x^{2} + 205 \, x + 82\right )} \sqrt{-2 \, x + 1}}{2750 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2750*(39*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(120*x^2 + 205
*x + 82)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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

[Out]

Timed out

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Giac [A]  time = 2.42057, size = 104, normalized size = 1.28 \begin{align*} -\frac{39}{2750} \, \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{12}{125} \, \sqrt{-2 \, x + 1} + \frac{305 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 693 \, \sqrt{-2 \, x + 1}}{500 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-39/2750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 12/125*sqrt(-2
*x + 1) + 1/500*(305*(-2*x + 1)^(3/2) - 693*sqrt(-2*x + 1))/(5*x + 3)^2