\(\int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx\) [1843]

Optimal result

Integrand size = 17, antiderivative size = 48 \[ \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x}}{5 (3+5 x)}+\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}} \]

[Out]

2/275*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/5*(1-2*x)^(1/2)/(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {43, 65, 212} \[ \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx=\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {\sqrt {1-2 x}}{5 (5 x+3)} \]

[In]

Int[Sqrt[1 - 2*x]/(3 + 5*x)^2,x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps \begin{align*} \text {integral}& = -\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} \text {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] (verified)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x}}{15+25 x}+\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}} \]

[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])

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {2 \sqrt {1-2 x}}{25 \left (-\frac {6}{5}-2 x \right )}+\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{275}\)	\(36\)
default	\(\frac {2 \sqrt {1-2 x}}{25 \left (-\frac {6}{5}-2 x \right )}+\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{275}\)	\(36\)
risch	\(\frac {-1+2 x}{5 \left (3+5 x \right ) \sqrt {1-2 x}}+\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{275}\)	\(41\)
pseudoelliptic	\(\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}-55 \sqrt {1-2 x}}{825+1375 x}\)	\(42\)
trager	\(-\frac {\sqrt {1-2 x}}{5 \left (3+5 x \right )}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{275}\)	\(62\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx=\frac {\sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, \sqrt {-2 \, x + 1}}{275 \, {\left (5 \, x + 3\right )}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.95 (sec) , antiderivative size = 175, normalized size of antiderivative = 3.65 \[ \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx=\begin {cases} \frac {2 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{275} + \frac {\sqrt {2}}{25 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {11 \sqrt {2}}{250 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} & \text {for}\: \frac {1}{\left |{x + \frac {3}{5}}\right |} > \frac {10}{11} \\- \frac {2 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{275} - \frac {\sqrt {2} i}{25 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {11 \sqrt {2} i}{250 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[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)), 1/Abs(x + 3/5) > 10/11), (-2*sqrt(
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*s
qrt(2)*I/(250*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx=-\frac {1}{275} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {\sqrt {-2 \, x + 1}}{5 \, {\left (5 \, x + 3\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx=-\frac {1}{275} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {\sqrt {-2 \, x + 1}}{5 \, {\left (5 \, x + 3\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx=\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{275}-\frac {2\,\sqrt {1-2\,x}}{25\,\left (2\,x+\frac {6}{5}\right )} \]

[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))