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

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

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Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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*}

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Mathematica [A]  time = 0.04, size = 67, normalized size = 1.40 \begin {gather*} \frac {\sqrt {1-2 x} \left (-110 x+2 \sqrt {55} \sqrt {2 x-1} (5 x+3) \tan ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )+55\right )}{275 \left (10 x^2+x-3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(55 - 110*x + 2*Sqrt[55]*Sqrt[-1 + 2*x]*(3 + 5*x)*ArcTan[Sqrt[5/11]*Sqrt[-1 + 2*x]]))/(275*(-3
+ x + 10*x^2))

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IntegrateAlgebraic [A]  time = 0.10, size = 52, normalized size = 1.08 \begin {gather*} \frac {2 \sqrt {1-2 x}}{5 (5 (1-2 x)-11)}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.38, size = 54, normalized size = 1.12 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.23, size = 56, normalized size = 1.17 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.01, size = 36, normalized size = 0.75 \begin {gather*} \frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{275}+\frac {2 \sqrt {-2 x +1}}{25 \left (-2 x -\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.20, size = 53, normalized size = 1.10 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 1.17, size = 35, normalized size = 0.73 \begin {gather*} \frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{275}-\frac {2\,\sqrt {1-2\,x}}{25\,\left (2\,x+\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(1/2)/(5*x + 3)^2,x)

[Out]

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

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sympy [B]  time = 1.61, size = 175, normalized size = 3.65 \begin {gather*} \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 {gather*}

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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