3.1842 \(\int \frac{\sqrt{1-2 x} (2+3 x)}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=61 \[ -\frac{(1-2 x)^{3/2}}{55 (5 x+3)}+\frac{64}{275} \sqrt{1-2 x}-\frac{64 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}} \]

[Out]

(64*Sqrt[1 - 2*x])/275 - (1 - 2*x)^(3/2)/(55*(3 + 5*x)) - (64*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

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Rubi [A]  time = 0.0129792, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {78, 50, 63, 206} \[ -\frac{(1-2 x)^{3/2}}{55 (5 x+3)}+\frac{64}{275} \sqrt{1-2 x}-\frac{64 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*Sqrt[1 - 2*x])/275 - (1 - 2*x)^(3/2)/(55*(3 + 5*x)) - (64*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

Rule 78

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{\sqrt{1-2 x} (2+3 x)}{(3+5 x)^2} \, dx &=-\frac{(1-2 x)^{3/2}}{55 (3+5 x)}+\frac{32}{55} \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx\\ &=\frac{64}{275} \sqrt{1-2 x}-\frac{(1-2 x)^{3/2}}{55 (3+5 x)}+\frac{32}{25} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=\frac{64}{275} \sqrt{1-2 x}-\frac{(1-2 x)^{3/2}}{55 (3+5 x)}-\frac{32}{25} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{64}{275} \sqrt{1-2 x}-\frac{(1-2 x)^{3/2}}{55 (3+5 x)}-\frac{64 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}}\\ \end{align*}

Mathematica [A]  time = 0.0230207, size = 53, normalized size = 0.87 \[ \frac{\sqrt{1-2 x} (30 x+17)}{25 (5 x+3)}-\frac{64 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(17 + 30*x))/(25*(3 + 5*x)) - (64*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

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Maple [A]  time = 0.009, size = 45, normalized size = 0.7 \begin{align*}{\frac{6}{25}\sqrt{1-2\,x}}+{\frac{2}{125}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{64\,\sqrt{55}}{1375}{\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((2+3*x)*(1-2*x)^(1/2)/(3+5*x)^2,x)

[Out]

6/25*(1-2*x)^(1/2)+2/125*(1-2*x)^(1/2)/(-2*x-6/5)-64/1375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 2.01089, size = 84, normalized size = 1.38 \begin{align*} \frac{32}{1375} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{6}{25} \, \sqrt{-2 \, x + 1} - \frac{\sqrt{-2 \, x + 1}}{25 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/1375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 6/25*sqrt(-2*x + 1) - 1/2
5*sqrt(-2*x + 1)/(5*x + 3)

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Fricas [A]  time = 1.55594, size = 169, normalized size = 2.77 \begin{align*} \frac{32 \, \sqrt{55}{\left (5 \, x + 3\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \,{\left (30 \, x + 17\right )} \sqrt{-2 \, x + 1}}{1375 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1375*(32*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(30*x + 17)*sqrt(-2*x +
1))/(5*x + 3)

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Sympy [B]  time = 50.9999, size = 178, normalized size = 2.92 \begin{align*} \frac{6 \sqrt{1 - 2 x}}{25} - \frac{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 )}{25} + \frac{62 \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 )}{25} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*sqrt(1 - 2*x)/25 - 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)))/25 + 62*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt(55)*a
tanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))/25

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Giac [A]  time = 1.32683, size = 88, normalized size = 1.44 \begin{align*} \frac{32}{1375} \, \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{6}{25} \, \sqrt{-2 \, x + 1} - \frac{\sqrt{-2 \, x + 1}}{25 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/1375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 6/25*sqrt(-2*x
+ 1) - 1/25*sqrt(-2*x + 1)/(5*x + 3)