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

Optimal. Leaf size=68 \[ -\frac {3 \sqrt {1-2 x}}{242 (5 x+3)}-\frac {\sqrt {1-2 x}}{22 (5 x+3)^2}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{121 \sqrt {55}} \]

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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 = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {51, 63, 206} \begin {gather*} -\frac {3 \sqrt {1-2 x}}{242 (5 x+3)}-\frac {\sqrt {1-2 x}}{22 (5 x+3)^2}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{121 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - 2*x]*(3 + 5*x)^3),x]

[Out]

-Sqrt[1 - 2*x]/(22*(3 + 5*x)^2) - (3*Sqrt[1 - 2*x])/(242*(3 + 5*x)) - (3*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(1
21*Sqrt[55])

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 {1}{\sqrt {1-2 x} (3+5 x)^3} \, dx &=-\frac {\sqrt {1-2 x}}{22 (3+5 x)^2}+\frac {3}{22} \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {\sqrt {1-2 x}}{22 (3+5 x)^2}-\frac {3 \sqrt {1-2 x}}{242 (3+5 x)}+\frac {3}{242} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {\sqrt {1-2 x}}{22 (3+5 x)^2}-\frac {3 \sqrt {1-2 x}}{242 (3+5 x)}-\frac {3}{242} \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}}{22 (3+5 x)^2}-\frac {3 \sqrt {1-2 x}}{242 (3+5 x)}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{121 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 30, normalized size = 0.44 \begin {gather*} -\frac {8 \sqrt {1-2 x} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};-\frac {5}{11} (2 x-1)\right )}{1331} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*Sqrt[1 - 2*x]*Hypergeometric2F1[1/2, 3, 3/2, (-5*(-1 + 2*x))/11])/1331

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IntegrateAlgebraic [A]  time = 0.11, size = 61, normalized size = 0.90 \begin {gather*} \frac {5 (3 (1-2 x)-11) \sqrt {1-2 x}}{121 (5 (1-2 x)-11)^2}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{121 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(-11 + 3*(1 - 2*x))*Sqrt[1 - 2*x])/(121*(-11 + 5*(1 - 2*x))^2) - (3*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(121
*Sqrt[55])

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fricas [A]  time = 1.32, size = 69, normalized size = 1.01 \begin {gather*} \frac {3 \, \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 ) - 275 \, {\left (3 \, x + 4\right )} \sqrt {-2 \, x + 1}}{13310 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13310*(3*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 275*(3*x + 4)*sqr
t(-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 {3}{13310} \, \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 (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}}{484 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/13310*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 5/484*(3*(-2*x
+ 1)^(3/2) - 11*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 52, normalized size = 0.76 \begin {gather*} -\frac {3 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{6655}-\frac {2 \sqrt {-2 x +1}}{11 \left (-10 x -6\right )^{2}}+\frac {3 \sqrt {-2 x +1}}{121 \left (-10 x -6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/11*(-2*x+1)^(1/2)/(-10*x-6)^2+3/121*(-2*x+1)^(1/2)/(-10*x-6)-3/6655*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*5
5^(1/2)

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maxima [A]  time = 1.39, size = 74, normalized size = 1.09 \begin {gather*} \frac {3}{13310} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {5 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}}{121 \, {\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/(3+5*x)^3/(1-2*x)^(1/2),x, algorithm="maxima")

[Out]

3/13310*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 5/121*(3*(-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 = 0.06, size = 54, normalized size = 0.79 \begin {gather*} -\frac {3\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{6655}-\frac {\frac {\sqrt {1-2\,x}}{55}-\frac {3\,{\left (1-2\,x\right )}^{3/2}}{605}}{\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/((1 - 2*x)^(1/2)*(5*x + 3)^3),x)

[Out]

- (3*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/6655 - ((1 - 2*x)^(1/2)/55 - (3*(1 - 2*x)^(3/2))/605)/((44
*x)/5 + (2*x - 1)^2 + 11/25)

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sympy [A]  time = 3.06, size = 231, normalized size = 3.40 \begin {gather*} \begin {cases} - \frac {3 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{6655} + \frac {3 \sqrt {2}}{1210 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {\sqrt {2}}{1100 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} - \frac {\sqrt {2}}{500 \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 {3 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{6655} - \frac {3 \sqrt {2} i}{1210 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {\sqrt {2} i}{1100 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} + \frac {\sqrt {2} i}{500 \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/(3+5*x)**3/(1-2*x)**(1/2),x)

[Out]

Piecewise((-3*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/6655 + 3*sqrt(2)/(1210*sqrt(-1 + 11/(10*(x + 3/5)))
*sqrt(x + 3/5)) - sqrt(2)/(1100*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)) - sqrt(2)/(500*sqrt(-1 + 11/(10
*(x + 3/5)))*(x + 3/5)**(5/2)), 11/(10*Abs(x + 3/5)) > 1), (3*sqrt(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5)))/66
55 - 3*sqrt(2)*I/(1210*sqrt(1 - 11/(10*(x + 3/5)))*sqrt(x + 3/5)) + sqrt(2)*I/(1100*sqrt(1 - 11/(10*(x + 3/5))
)*(x + 3/5)**(3/2)) + sqrt(2)*I/(500*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(5/2)), True))

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