3.1843 \(\int \frac{\sqrt{1-2 x}}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=48 \[ \frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{5 \sqrt{55}}-\frac{\sqrt{1-2 x}}{5 (5 x+3)} \]

[Out]

-Sqrt[1 - 2*x]/(5*(3 + 5*x)) + (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55])

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Rubi [A]  time = 0.0087964, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {47, 63, 206} \[ \frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{5 \sqrt{55}}-\frac{\sqrt{1-2 x}}{5 (5 x+3)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - 2*x]/(3 + 5*x)^2,x]

[Out]

-Sqrt[1 - 2*x]/(5*(3 + 5*x)) + (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55])

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 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{\sqrt{1-2 x}}{(3+5 x)^2} \, dx &=-\frac{\sqrt{1-2 x}}{5 (3+5 x)}-\frac{1}{5} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{\sqrt{1-2 x}}{5 (3+5 x)}+\frac{1}{5} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{\sqrt{1-2 x}}{5 (3+5 x)}+\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{5 \sqrt{55}}\\ \end{align*}

Mathematica [A]  time = 0.0365168, size = 46, normalized size = 0.96 \[ \frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{5 \sqrt{55}}-\frac{\sqrt{1-2 x}}{25 x+15} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]/(15 + 25*x)) + (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55])

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Maple [A]  time = 0.007, size = 36, normalized size = 0.8 \begin{align*}{\frac{2}{25}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}+{\frac{2\,\sqrt{55}}{275}{\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)^(1/2)/(3+5*x)^2,x)

[Out]

2/25*(1-2*x)^(1/2)/(-2*x-6/5)+2/275*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.62286, size = 72, normalized size = 1.5 \begin{align*} -\frac{1}{275} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{\sqrt{-2 \, x + 1}}{5 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/275*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/5*sqrt(-2*x + 1)/(5*x +
3)

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Fricas [A]  time = 1.60335, size = 147, normalized size = 3.06 \begin{align*} \frac{\sqrt{55}{\left (5 \, x + 3\right )} \log \left (\frac{5 \, x - \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, \sqrt{-2 \, x + 1}}{275 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/275*(sqrt(55)*(5*x + 3)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*sqrt(-2*x + 1))/(5*x + 3)

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Sympy [B]  time = 1.64611, size = 175, normalized size = 3.65 \begin{align*} \begin{cases} \frac{2 \sqrt{55} \operatorname{acosh}{\left (\frac{\sqrt{110}}{10 \sqrt{x + \frac{3}{5}}} \right )}}{275} + \frac{\sqrt{2}}{25 \sqrt{-1 + \frac{11}{10 \left (x + \frac{3}{5}\right )}} \sqrt{x + \frac{3}{5}}} - \frac{11 \sqrt{2}}{250 \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{2 \sqrt{55} i \operatorname{asin}{\left (\frac{\sqrt{110}}{10 \sqrt{x + \frac{3}{5}}} \right )}}{275} - \frac{\sqrt{2} i}{25 \sqrt{1 - \frac{11}{10 \left (x + \frac{3}{5}\right )}} \sqrt{x + \frac{3}{5}}} + \frac{11 \sqrt{2} i}{250 \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)**(1/2)/(3+5*x)**2,x)

[Out]

Piecewise((2*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/275 + sqrt(2)/(25*sqrt(-1 + 11/(10*(x + 3/5)))*sqrt(
x + 3/5)) - 11*sqrt(2)/(250*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)), 11/(10*Abs(x + 3/5)) > 1), (-2*sqr
t(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5)))/275 - sqrt(2)*I/(25*sqrt(1 - 11/(10*(x + 3/5)))*sqrt(x + 3/5)) + 11
*sqrt(2)*I/(250*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)), True))

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Giac [A]  time = 1.83545, size = 76, normalized size = 1.58 \begin{align*} -\frac{1}{275} \, \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{\sqrt{-2 \, x + 1}}{5 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/275*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/5*sqrt(-2*x +
1)/(5*x + 3)