3.21.53 \(\int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx\) [2053]

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

[Out]

-2/605*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/11*(1-2*x)^(1/2)/(3+5*x)

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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 = {44, 65, 212} \begin {gather*} -\frac {\sqrt {1-2 x}}{11 (5 x+3)}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - 2*x]*(3 + 5*x)^2),x]

[Out]

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[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*} \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x}}{11 (3+5 x)}+\frac {1}{11} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {\sqrt {1-2 x}}{11 (3+5 x)}-\frac {1}{11} \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}}{11 (3+5 x)}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 46, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {1-2 x}}{33+55 x}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]/(33 + 55*x)) - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(11*Sqrt[55])

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Maple [A]
time = 0.11, size = 36, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {2 \sqrt {1-2 x}}{11 \left (-6-10 x \right )}-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}\)	\(36\)
default	\(\frac {2 \sqrt {1-2 x}}{11 \left (-6-10 x \right )}-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}\)	\(36\)
risch	\(\frac {-1+2 x}{11 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}\)	\(41\)
trager	\(-\frac {\sqrt {1-2 x}}{11 \left (3+5 x \right )}+\frac {\RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{605}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*(1-2*x)^(1/2)/(-6-10*x)-2/605*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]
time = 0.48, size = 53, normalized size = 1.10 \begin {gather*} \frac {1}{605} \, \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}}{11 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/605*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/11*sqrt(-2*x + 1)/(5*x +
3)

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Fricas [A]
time = 1.31, size = 53, normalized size = 1.10 \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}}{605 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/605*(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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Sympy [C] Result contains complex when optimal does not.
time = 1.09, size = 143, normalized size = 2.98 \begin {gather*} \begin {cases} - \frac {2 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{605} + \frac {\sqrt {2}}{55 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {\sqrt {2}}{50 \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 {\sqrt {2} i \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{55 \sqrt {x + \frac {3}{5}}} + \frac {2 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{605} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/605 + sqrt(2)/(55*sqrt(-1 + 11/(10*(x + 3/5)))*sqrt
(x + 3/5)) - sqrt(2)/(50*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)), 1/Abs(x + 3/5) > 10/11), (-sqrt(2)*I*
sqrt(1 - 11/(10*(x + 3/5)))/(55*sqrt(x + 3/5)) + 2*sqrt(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5)))/605, True))

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Giac [A]
time = 1.85, size = 56, normalized size = 1.17 \begin {gather*} \frac {1}{605} \, \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}}{11 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/605*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/11*sqrt(-2*x +
1)/(5*x + 3)

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Mupad [B]
time = 0.05, size = 35, normalized size = 0.73 \begin {gather*} -\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{605}-\frac {2\,\sqrt {1-2\,x}}{55\,\left (2\,x+\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((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))/605 - (2*(1 - 2*x)^(1/2))/(55*(2*x + 6/5))

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