3.1844 \(\int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx\)

Optimal. Leaf size=69 \[ -\frac {\sqrt {1-2 x}}{5 x+3}-2 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}} \]

[Out]

-2*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+68/55*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-(1-2*x)^(1
/2)/(3+5*x)

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Rubi [A]  time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {99, 156, 63, 206} \[ -\frac {\sqrt {1-2 x}}{5 x+3}-2 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]/(3 + 5*x)) - 2*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x
]])/Sqrt[55]

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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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}}{(2+3 x) (3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x}}{3+5 x}+\int \frac {-5+3 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac {\sqrt {1-2 x}}{3+5 x}+21 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx-34 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {\sqrt {1-2 x}}{3+5 x}-21 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )+34 \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}}{3+5 x}-2 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 69, normalized size = 1.00 \[ -\frac {\sqrt {1-2 x}}{5 x+3}-2 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]/(3 + 5*x)) - 2*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x
]])/Sqrt[55]

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fricas [A]  time = 0.99, size = 90, normalized size = 1.30 \[ \frac {34 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, \sqrt {21} {\left (5 \, x + 3\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 55 \, \sqrt {-2 \, x + 1}}{55 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/55*(34*sqrt(55)*(5*x + 3)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*sqrt(21)*(5*x + 3)*log((3*
x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 55*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 1.26, size = 94, normalized size = 1.36 \[ -\frac {34}{55} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {\sqrt {-2 \, x + 1}}{5 \, x + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-34/55*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + sqrt(21)*log(1/2
*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 54, normalized size = 0.78 \[ -2 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )+\frac {68 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{55}+\frac {2 \sqrt {-2 x +1}}{5 \left (-2 x -\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(-2*x+1)^(1/2)/(-2*x-6/5)+68/55*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-2*arctanh(1/7*21^(1/2)*(-2*
x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.32, size = 88, normalized size = 1.28 \[ -\frac {34}{55} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {\sqrt {-2 \, x + 1}}{5 \, x + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-34/55*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + sqrt(21)*log(-(sqrt(21) -
3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - sqrt(-2*x + 1)/(5*x + 3)

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mupad [B]  time = 1.18, size = 53, normalized size = 0.77 \[ \frac {68\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{55}-2\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )-\frac {2\,\sqrt {1-2\,x}}{5\,\left (2\,x+\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(68*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/55 - 2*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7) - (2*(1
- 2*x)^(1/2))/(5*(2*x + 6/5))

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sympy [A]  time = 55.05, size = 230, normalized size = 3.33 \[ - 44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: x \leq \frac {1}{2} \wedge x > - \frac {3}{5} \end {cases}\right ) + 42 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 < - \frac {7}{3} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 > - \frac {7}{3} \end {cases}\right ) - 70 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-44*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(
sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (x <= 1/2) & (x > -3/5))) + 42*P
iecewise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 < -7/3), (-sqrt(21)*atanh(sqrt(21)*sqrt(1 - 2*
x)/7)/21, 2*x - 1 > -7/3)) - 70*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-
sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))

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