3.1850 \(\int \frac{\sqrt{1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=81 \[ -\frac{133 (1-2 x)^{3/2}}{6050 (5 x+3)}-\frac{(1-2 x)^{3/2}}{550 (5 x+3)^2}+\frac{409 \sqrt{1-2 x}}{3025}-\frac{409 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{275 \sqrt{55}} \]

[Out]

(409*Sqrt[1 - 2*x])/3025 - (1 - 2*x)^(3/2)/(550*(3 + 5*x)^2) - (133*(1 - 2*x)^(3/2))/(6050*(3 + 5*x)) - (409*A
rcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqrt[55])

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Rubi [A]  time = 0.0196173, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {89, 78, 50, 63, 206} \[ -\frac{133 (1-2 x)^{3/2}}{6050 (5 x+3)}-\frac{(1-2 x)^{3/2}}{550 (5 x+3)^2}+\frac{409 \sqrt{1-2 x}}{3025}-\frac{409 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{275 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(409*Sqrt[1 - 2*x])/3025 - (1 - 2*x)^(3/2)/(550*(3 + 5*x)^2) - (133*(1 - 2*x)^(3/2))/(6050*(3 + 5*x)) - (409*A
rcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqrt[55])

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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)^2}{(3+5 x)^3} \, dx &=-\frac{(1-2 x)^{3/2}}{550 (3+5 x)^2}+\frac{1}{550} \int \frac{\sqrt{1-2 x} (727+990 x)}{(3+5 x)^2} \, dx\\ &=-\frac{(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac{133 (1-2 x)^{3/2}}{6050 (3+5 x)}+\frac{409 \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx}{1210}\\ &=\frac{409 \sqrt{1-2 x}}{3025}-\frac{(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac{133 (1-2 x)^{3/2}}{6050 (3+5 x)}+\frac{409}{550} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=\frac{409 \sqrt{1-2 x}}{3025}-\frac{(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac{133 (1-2 x)^{3/2}}{6050 (3+5 x)}-\frac{409}{550} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{409 \sqrt{1-2 x}}{3025}-\frac{(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac{133 (1-2 x)^{3/2}}{6050 (3+5 x)}-\frac{409 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{275 \sqrt{55}}\\ \end{align*}

Mathematica [A]  time = 0.0377517, size = 58, normalized size = 0.72 \[ \frac{\sqrt{1-2 x} \left (1980 x^2+2245 x+632\right )}{550 (5 x+3)^2}-\frac{409 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{275 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(632 + 2245*x + 1980*x^2))/(550*(3 + 5*x)^2) - (409*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqr
t[55])

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Maple [A]  time = 0.01, size = 57, normalized size = 0.7 \begin{align*}{\frac{18}{125}\sqrt{1-2\,x}}+{\frac{2}{5\, \left ( -10\,x-6 \right ) ^{2}} \left ({\frac{131}{110} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{133}{50}\sqrt{1-2\,x}} \right ) }-{\frac{409\,\sqrt{55}}{15125}{\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)^2*(1-2*x)^(1/2)/(3+5*x)^3,x)

[Out]

18/125*(1-2*x)^(1/2)+2/5*(131/110*(1-2*x)^(3/2)-133/50*(1-2*x)^(1/2))/(-10*x-6)^2-409/15125*arctanh(1/11*55^(1
/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.50062, size = 112, normalized size = 1.38 \begin{align*} \frac{409}{30250} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{18}{125} \, \sqrt{-2 \, x + 1} + \frac{655 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 1463 \, \sqrt{-2 \, x + 1}}{1375 \,{\left (25 \,{\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

409/30250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/125*sqrt(-2*x + 1) +
1/1375*(655*(-2*x + 1)^(3/2) - 1463*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]  time = 1.53752, size = 217, normalized size = 2.68 \begin{align*} \frac{409 \, \sqrt{55}{\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \,{\left (1980 \, x^{2} + 2245 \, x + 632\right )} \sqrt{-2 \, x + 1}}{30250 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30250*(409*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(1980*x^2 +
2245*x + 632)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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Sympy [A]  time = 167.594, size = 326, normalized size = 4.02 \begin{align*} \frac{18 \sqrt{1 - 2 x}}{125} - \frac{256 \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 )}{125} + \frac{88 \left (\begin{cases} \frac{\sqrt{55} \left (\frac{3 \log{\left (\frac{\sqrt{55} \sqrt{1 - 2 x}}{11} - 1 \right )}}{16} - \frac{3 \log{\left (\frac{\sqrt{55} \sqrt{1 - 2 x}}{11} + 1 \right )}}{16} + \frac{3}{16 \left (\frac{\sqrt{55} \sqrt{1 - 2 x}}{11} + 1\right )} + \frac{1}{16 \left (\frac{\sqrt{55} \sqrt{1 - 2 x}}{11} + 1\right )^{2}} + \frac{3}{16 \left (\frac{\sqrt{55} \sqrt{1 - 2 x}}{11} - 1\right )} - \frac{1}{16 \left (\frac{\sqrt{55} \sqrt{1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text{for}\: x \leq \frac{1}{2} \wedge x > - \frac{3}{5} \end{cases}\right )}{125} + \frac{174 \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 )}{125} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

18*sqrt(1 - 2*x)/125 - 256*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)))/125 + 88*Piecewise((sqrt(55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1
- 2*x)/11 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)**2) + 3/(1
6*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (x <= 1/2) & (x > -3/5))
)/125 + 174*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))/125

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Giac [A]  time = 2.5569, size = 104, normalized size = 1.28 \begin{align*} \frac{409}{30250} \, \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{18}{125} \, \sqrt{-2 \, x + 1} + \frac{655 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 1463 \, \sqrt{-2 \, x + 1}}{5500 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

409/30250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/125*sqrt(-
2*x + 1) + 1/5500*(655*(-2*x + 1)^(3/2) - 1463*sqrt(-2*x + 1))/(5*x + 3)^2