3.1832 \(\int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\)

Optimal. Leaf size=43 \[ \frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 43, 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 = {50, 63, 206} \[ \frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - 2*x]/(3 + 5*x),x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.00 \[ \frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - 2*x]/(3 + 5*x),x]

[Out]

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

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fricas [A]  time = 0.92, size = 46, normalized size = 1.07 \[ \frac {1}{25} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {2}{5} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 2/5*sqrt(-2*x + 1)

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giac [A]  time = 1.27, size = 49, normalized size = 1.14 \[ \frac {1}{25} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {2}{5} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 29, normalized size = 0.67 \[ -\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{25}+\frac {2 \sqrt {-2 x +1}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.37, size = 46, normalized size = 1.07 \[ \frac {1}{25} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2}{5} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2/5*sqrt(-2*x + 1)

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mupad [B]  time = 0.06, size = 28, normalized size = 0.65 \[ \frac {2\,\sqrt {1-2\,x}}{5}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(1/2)/(5*x + 3),x)

[Out]

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

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sympy [A]  time = 1.27, size = 107, normalized size = 2.49 \[ \begin {cases} \frac {2 \sqrt {5} i \sqrt {10 x - 5}}{25} + \frac {2 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{25} & \text {for}\: \frac {10 \left |{x + \frac {3}{5}}\right |}{11} > 1 \\\frac {2 \sqrt {5} \sqrt {5 - 10 x}}{25} + \frac {\sqrt {55} \log {\left (x + \frac {3}{5} \right )}}{25} - \frac {2 \sqrt {55} \log {\left (\sqrt {\frac {5}{11} - \frac {10 x}{11}} + 1 \right )}}{25} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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