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

Optimal. Leaf size=68 \[ \frac {\sqrt {1-2 x}}{110 (5 x+3)}-\frac {\sqrt {1-2 x}}{10 (5 x+3)^2}+\frac {\tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \]

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Rubi [A]  time = 0.01, antiderivative size = 68, normalized size of antiderivative = 1.00, 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, 51, 63, 206} \begin {gather*} \frac {\sqrt {1-2 x}}{110 (5 x+3)}-\frac {\sqrt {1-2 x}}{10 (5 x+3)^2}+\frac {\tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - 2*x]/(3 + 5*x)^3,x]

[Out]

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

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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)^3} \, dx &=-\frac {\sqrt {1-2 x}}{10 (3+5 x)^2}-\frac {1}{10} \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {\sqrt {1-2 x}}{10 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{110 (3+5 x)}-\frac {1}{110} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {\sqrt {1-2 x}}{10 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{110 (3+5 x)}+\frac {1}{110} \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}}{10 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{110 (3+5 x)}+\frac {\tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 30, normalized size = 0.44 \begin {gather*} -\frac {8 (1-2 x)^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};-\frac {5}{11} (2 x-1)\right )}{3993} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(1 - 2*x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, (-5*(-1 + 2*x))/11])/3993

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IntegrateAlgebraic [A]  time = 0.17, size = 61, normalized size = 0.90 \begin {gather*} \frac {\tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}}-\frac {(5 (1-2 x)+11) \sqrt {1-2 x}}{55 (5 (1-2 x)-11)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.55, size = 69, normalized size = 1.01 \begin {gather*} \frac {\sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (5 \, x - 8\right )} \sqrt {-2 \, x + 1}}{6050 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6050*(sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(5*x - 8)*sqrt(-2
*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.22, size = 68, normalized size = 1.00 \begin {gather*} -\frac {1}{6050} \, \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 {5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 11 \, \sqrt {-2 \, x + 1}}{220 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6050*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/220*(5*(-2*x
+ 1)^(3/2) + 11*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 48, normalized size = 0.71 \begin {gather*} \frac {\sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{3025}+\frac {-\frac {\left (-2 x +1\right )^{\frac {3}{2}}}{11}-\frac {\sqrt {-2 x +1}}{5}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

200*(-1/2200*(-2*x+1)^(3/2)-1/1000*(-2*x+1)^(1/2))/(-10*x-6)^2+1/3025*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55
^(1/2)

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maxima [A]  time = 1.17, size = 74, normalized size = 1.09 \begin {gather*} -\frac {1}{6050} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 11 \, \sqrt {-2 \, x + 1}}{55 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6050*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/55*(5*(-2*x + 1)^(3/2)
+ 11*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 1.19, size = 54, normalized size = 0.79 \begin {gather*} \frac {\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{3025}-\frac {\frac {\sqrt {1-2\,x}}{125}+\frac {{\left (1-2\,x\right )}^{3/2}}{275}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(1/2)/(5*x + 3)^3,x)

[Out]

(55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/3025 - ((1 - 2*x)^(1/2)/125 + (1 - 2*x)^(3/2)/275)/((44*x)/5 +
(2*x - 1)^2 + 11/25)

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sympy [B]  time = 2.73, size = 231, normalized size = 3.40 \begin {gather*} \begin {cases} \frac {\sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{3025} - \frac {\sqrt {2}}{550 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {3 \sqrt {2}}{500 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} - \frac {11 \sqrt {2}}{2500 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{2}}} & \text {for}\: \frac {11}{10 \left |{x + \frac {3}{5}}\right |} > 1 \\- \frac {\sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{3025} + \frac {\sqrt {2} i}{550 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {3 \sqrt {2} i}{500 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} + \frac {11 \sqrt {2} i}{2500 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{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)**3,x)

[Out]

Piecewise((sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/3025 - sqrt(2)/(550*sqrt(-1 + 11/(10*(x + 3/5)))*sqrt(
x + 3/5)) + 3*sqrt(2)/(500*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)) - 11*sqrt(2)/(2500*sqrt(-1 + 11/(10*
(x + 3/5)))*(x + 3/5)**(5/2)), 11/(10*Abs(x + 3/5)) > 1), (-sqrt(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5)))/3025
+ sqrt(2)*I/(550*sqrt(1 - 11/(10*(x + 3/5)))*sqrt(x + 3/5)) - 3*sqrt(2)*I/(500*sqrt(1 - 11/(10*(x + 3/5)))*(x
+ 3/5)**(3/2)) + 11*sqrt(2)*I/(2500*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(5/2)), True))

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