3.19.33 \(\int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)} \, dx\) [1833]

Optimal. Leaf size=55 \[ 2 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

-2/5*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+2/3*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {85, 65, 212} \begin {gather*} 2 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[(b*e - a*f)/(b*c
- a*d), Int[(e + f*x)^(p - 1)/(a + b*x), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[(e + f*x)^(p - 1)/(c + d*x
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[0, p, 1]

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 {\sqrt {1-2 x}}{(2+3 x) (3+5 x)} \, dx &=-\left (7 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\right )+11 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=7 \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-11 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=2 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2 \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 = 0.09, size = 55, normalized size = 1.00 \begin {gather*} 2 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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Maple [A]
time = 0.13, size = 38, normalized size = 0.69

method	result	size
derivativedivides	\(-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{5}+\frac {2 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}\)	\(38\)
default	\(-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{5}+\frac {2 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}\)	\(38\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{5}+\frac {\RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \RootOf \left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{3}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+2/3*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)

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Maxima [A]
time = 0.53, size = 73, normalized size = 1.33 \begin {gather*} \frac {1}{5} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {1}{3} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/3*sqrt(21)*log(-(sqrt(21) -
3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1)))

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Fricas [A]
time = 0.95, size = 74, normalized size = 1.35 \begin {gather*} \frac {1}{5} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {1}{3} \, \sqrt {7} \sqrt {3} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 1/3*sqrt(7)*sqrt(3)*log(-(sq
rt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2))

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Sympy [C] Result contains complex when optimal does not.
time = 3.86, size = 70, normalized size = 1.27 \begin {gather*} 2 \sqrt {2} i \sqrt {x - \frac {1}{2}} + \frac {2 \sqrt {55} i \operatorname {atan}{\left (\frac {\sqrt {110}}{10 \sqrt {x - \frac {1}{2}}} \right )}}{5} + \frac {2 \sqrt {21} i \operatorname {atan}{\left (\frac {\sqrt {42} \sqrt {x - \frac {1}{2}}}{7} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(2)*I*sqrt(x - 1/2) + 2*sqrt(55)*I*atan(sqrt(110)/(10*sqrt(x - 1/2)))/5 + 2*sqrt(21)*I*atan(sqrt(42)*sqr
t(x - 1/2)/7)/3

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (37) = 74\).
time = 1.43, size = 79, normalized size = 1.44 \begin {gather*} \frac {1}{5} \, \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 {1}{3} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/3*sqrt(21)*log(1/
2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1)))

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Mupad [B]
time = 0.09, size = 37, normalized size = 0.67 \begin {gather*} \frac {2\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/3 - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/5

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