3.1845 \(\int \frac{\sqrt{1-2 x}}{(2+3 x)^2 (3+5 x)^2} \, dx\)

Optimal. Leaf size=97 \[ -\frac{10 \sqrt{1-2 x}}{5 x+3}+\frac{\sqrt{1-2 x}}{(3 x+2) (5 x+3)}-138 \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )+134 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(-10*Sqrt[1 - 2*x])/(3 + 5*x) + Sqrt[1 - 2*x]/((2 + 3*x)*(3 + 5*x)) - 138*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 -
2*x]] + 134*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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Rubi [A]  time = 0.0314253, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {99, 151, 156, 63, 206} \[ -\frac{10 \sqrt{1-2 x}}{5 x+3}+\frac{\sqrt{1-2 x}}{(3 x+2) (5 x+3)}-138 \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )+134 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[1 - 2*x])/(3 + 5*x) + Sqrt[1 - 2*x]/((2 + 3*x)*(3 + 5*x)) - 138*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 -
2*x]] + 134*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*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 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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 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}}{(2+3 x)^2 (3+5 x)^2} \, dx &=\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)}-\int \frac{-13+15 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^2} \, dx\\ &=-\frac{10 \sqrt{1-2 x}}{3+5 x}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)}+\frac{1}{11} \int \frac{-539+330 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac{10 \sqrt{1-2 x}}{3+5 x}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)}+207 \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx-335 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{10 \sqrt{1-2 x}}{3+5 x}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)}-207 \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )+335 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{10 \sqrt{1-2 x}}{3+5 x}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)}-138 \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )+134 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0567936, size = 105, normalized size = 1.08 \[ \frac{-1518 \sqrt{21} \left (15 x^2+19 x+6\right ) \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )+938 \sqrt{55} \left (15 x^2+19 x+6\right ) \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-77 \sqrt{1-2 x} (30 x+19)}{77 (3 x+2) (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-77*Sqrt[1 - 2*x]*(19 + 30*x) - 1518*Sqrt[21]*(6 + 19*x + 15*x^2)*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + 938*Sqrt
[55]*(6 + 19*x + 15*x^2)*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(77*(2 + 3*x)*(3 + 5*x))

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Maple [A]  time = 0.012, size = 70, normalized size = 0.7 \begin{align*} 2\,{\frac{\sqrt{1-2\,x}}{-2\,x-4/3}}-{\frac{138\,\sqrt{21}}{7}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+2\,{\frac{\sqrt{1-2\,x}}{-2\,x-6/5}}+{\frac{134\,\sqrt{55}}{11}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1-2*x)^(1/2)/(-2*x-4/3)-138/7*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+2*(1-2*x)^(1/2)/(-2*x-6/5)+134/1
1*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.5261, size = 149, normalized size = 1.54 \begin{align*} -\frac{67}{11} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{69}{7} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) + \frac{4 \,{\left (15 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 34 \, \sqrt{-2 \, x + 1}\right )}}{15 \,{\left (2 \, x - 1\right )}^{2} + 136 \, x + 9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-67/11*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 69/7*sqrt(21)*log(-(sqrt(2
1) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4*(15*(-2*x + 1)^(3/2) - 34*sqrt(-2*x + 1))/(15*(2*x -
1)^2 + 136*x + 9)

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Fricas [A]  time = 1.65473, size = 352, normalized size = 3.63 \begin{align*} \frac{469 \, \sqrt{11} \sqrt{5}{\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (-\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 759 \, \sqrt{7} \sqrt{3}{\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (\frac{\sqrt{7} \sqrt{3} \sqrt{-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 77 \,{\left (30 \, x + 19\right )} \sqrt{-2 \, x + 1}}{77 \,{\left (15 \, x^{2} + 19 \, x + 6\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/77*(469*sqrt(11)*sqrt(5)*(15*x^2 + 19*x + 6)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 7
59*sqrt(7)*sqrt(3)*(15*x^2 + 19*x + 6)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 77*(30*x +
19)*sqrt(-2*x + 1))/(15*x^2 + 19*x + 6)

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Sympy [A]  time = 51.6882, size = 328, normalized size = 3.38 \begin{align*} - 84 \left (\begin{cases} \frac{\sqrt{21} \left (- \frac{\log{\left (\frac{\sqrt{21} \sqrt{1 - 2 x}}{7} - 1 \right )}}{4} + \frac{\log{\left (\frac{\sqrt{21} \sqrt{1 - 2 x}}{7} + 1 \right )}}{4} - \frac{1}{4 \left (\frac{\sqrt{21} \sqrt{1 - 2 x}}{7} + 1\right )} - \frac{1}{4 \left (\frac{\sqrt{21} \sqrt{1 - 2 x}}{7} - 1\right )}\right )}{147} & \text{for}\: x \leq \frac{1}{2} \wedge x > - \frac{2}{3} \end{cases}\right ) - 220 \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 ) + 408 \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 ) - 680 \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-84*Piecewise((sqrt(21)*(-log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/4 + log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/4 - 1/(4*(sq
rt(21)*sqrt(1 - 2*x)/7 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147, (x <= 1/2) & (x > -2/3))) - 220*Piec
ewise((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))) + 408*Piecewis
e((-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)/2
1, 2*x - 1 > -7/3)) - 680*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt(5
5)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))

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Giac [A]  time = 1.81764, size = 157, normalized size = 1.62 \begin{align*} -\frac{67}{11} \, \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{69}{7} \, \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{4 \,{\left (15 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 34 \, \sqrt{-2 \, x + 1}\right )}}{15 \,{\left (2 \, x - 1\right )}^{2} + 136 \, x + 9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-67/11*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 69/7*sqrt(21)*lo
g(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4*(15*(-2*x + 1)^(3/2) - 34*sqrt(-2
*x + 1))/(15*(2*x - 1)^2 + 136*x + 9)