3.1913 \(\int \frac{(1-2 x)^{3/2}}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=63 \[ -\frac{(1-2 x)^{3/2}}{5 (5 x+3)}-\frac{6}{25} \sqrt{1-2 x}+\frac{6}{25} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(-6*Sqrt[1 - 2*x])/25 - (1 - 2*x)^(3/2)/(5*(3 + 5*x)) + (6*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/25

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Rubi [A]  time = 0.0131594, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {47, 50, 63, 206} \[ -\frac{(1-2 x)^{3/2}}{5 (5 x+3)}-\frac{6}{25} \sqrt{1-2 x}+\frac{6}{25} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(3/2)/(3 + 5*x)^2,x]

[Out]

(-6*Sqrt[1 - 2*x])/25 - (1 - 2*x)^(3/2)/(5*(3 + 5*x)) + (6*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/25

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}}{(3+5 x)^2} \, dx &=-\frac{(1-2 x)^{3/2}}{5 (3+5 x)}-\frac{3}{5} \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx\\ &=-\frac{6}{25} \sqrt{1-2 x}-\frac{(1-2 x)^{3/2}}{5 (3+5 x)}-\frac{33}{25} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{6}{25} \sqrt{1-2 x}-\frac{(1-2 x)^{3/2}}{5 (3+5 x)}+\frac{33}{25} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{6}{25} \sqrt{1-2 x}-\frac{(1-2 x)^{3/2}}{5 (3+5 x)}+\frac{6}{25} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [C]  time = 0.00499, size = 30, normalized size = 0.48 \[ -\frac{4}{605} (1-2 x)^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{5}{11} (1-2 x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(1 - 2*x)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, (5*(1 - 2*x))/11])/605

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Maple [A]  time = 0.009, size = 45, normalized size = 0.7 \begin{align*} -{\frac{4}{25}\sqrt{1-2\,x}}+{\frac{22}{125}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}+{\frac{6\,\sqrt{55}}{125}{\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)/(3+5*x)^2,x)

[Out]

-4/25*(1-2*x)^(1/2)+22/125*(1-2*x)^(1/2)/(-2*x-6/5)+6/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 3.64382, size = 84, normalized size = 1.33 \begin{align*} -\frac{3}{125} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{4}{25} \, \sqrt{-2 \, x + 1} - \frac{11 \, \sqrt{-2 \, x + 1}}{25 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 4/25*sqrt(-2*x + 1) - 11/2
5*sqrt(-2*x + 1)/(5*x + 3)

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Fricas [A]  time = 1.62916, size = 188, normalized size = 2.98 \begin{align*} \frac{3 \, \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 (20 \, x + 23\right )} \sqrt{-2 \, x + 1}}{125 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/125*(3*sqrt(11)*sqrt(5)*(5*x + 3)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) - 5*(20*x + 23
)*sqrt(-2*x + 1))/(5*x + 3)

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Sympy [B]  time = 1.96571, size = 238, normalized size = 3.78 \begin{align*} \begin{cases} \frac{6 \sqrt{55} \operatorname{acosh}{\left (\frac{\sqrt{110}}{10 \sqrt{x + \frac{3}{5}}} \right )}}{125} + \frac{4 \sqrt{2} \sqrt{x + \frac{3}{5}}}{25 \sqrt{-1 + \frac{11}{10 \left (x + \frac{3}{5}\right )}}} - \frac{11 \sqrt{2}}{125 \sqrt{-1 + \frac{11}{10 \left (x + \frac{3}{5}\right )}} \sqrt{x + \frac{3}{5}}} - \frac{121 \sqrt{2}}{1250 \sqrt{-1 + \frac{11}{10 \left (x + \frac{3}{5}\right )}} \left (x + \frac{3}{5}\right )^{\frac{3}{2}}} & \text{for}\: \frac{11}{10 \left |{x + \frac{3}{5}}\right |} > 1 \\- \frac{6 \sqrt{55} i \operatorname{asin}{\left (\frac{\sqrt{110}}{10 \sqrt{x + \frac{3}{5}}} \right )}}{125} - \frac{4 \sqrt{2} i \sqrt{x + \frac{3}{5}}}{25 \sqrt{1 - \frac{11}{10 \left (x + \frac{3}{5}\right )}}} + \frac{11 \sqrt{2} i}{125 \sqrt{1 - \frac{11}{10 \left (x + \frac{3}{5}\right )}} \sqrt{x + \frac{3}{5}}} + \frac{121 \sqrt{2} i}{1250 \sqrt{1 - \frac{11}{10 \left (x + \frac{3}{5}\right )}} \left (x + \frac{3}{5}\right )^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((6*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/125 + 4*sqrt(2)*sqrt(x + 3/5)/(25*sqrt(-1 + 11/(10*(
x + 3/5)))) - 11*sqrt(2)/(125*sqrt(-1 + 11/(10*(x + 3/5)))*sqrt(x + 3/5)) - 121*sqrt(2)/(1250*sqrt(-1 + 11/(10
*(x + 3/5)))*(x + 3/5)**(3/2)), 11/(10*Abs(x + 3/5)) > 1), (-6*sqrt(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5)))/1
25 - 4*sqrt(2)*I*sqrt(x + 3/5)/(25*sqrt(1 - 11/(10*(x + 3/5)))) + 11*sqrt(2)*I/(125*sqrt(1 - 11/(10*(x + 3/5))
)*sqrt(x + 3/5)) + 121*sqrt(2)*I/(1250*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)), True))

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Giac [A]  time = 2.25491, size = 88, normalized size = 1.4 \begin{align*} -\frac{3}{125} \, \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{4}{25} \, \sqrt{-2 \, x + 1} - \frac{11 \, \sqrt{-2 \, x + 1}}{25 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 4/25*sqrt(-2*x +
1) - 11/25*sqrt(-2*x + 1)/(5*x + 3)