3.1830 \(\int \frac{\sqrt{1-2 x} (2+3 x)^2}{3+5 x} \, dx\)

Optimal. Leaf size=69 \[ \frac{9}{50} (1-2 x)^{5/2}-\frac{37}{50} (1-2 x)^{3/2}+\frac{2}{125} \sqrt{1-2 x}-\frac{2}{125} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(2*Sqrt[1 - 2*x])/125 - (37*(1 - 2*x)^(3/2))/50 + (9*(1 - 2*x)^(5/2))/50 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sq
rt[1 - 2*x]])/125

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Rubi [A]  time = 0.0240918, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {88, 50, 63, 206} \[ \frac{9}{50} (1-2 x)^{5/2}-\frac{37}{50} (1-2 x)^{3/2}+\frac{2}{125} \sqrt{1-2 x}-\frac{2}{125} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 - 2*x])/125 - (37*(1 - 2*x)^(3/2))/50 + (9*(1 - 2*x)^(5/2))/50 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sq
rt[1 - 2*x]])/125

Rule 88

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

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} \, dx &=\int \left (\frac{111}{50} \sqrt{1-2 x}-\frac{9}{10} (1-2 x)^{3/2}+\frac{\sqrt{1-2 x}}{25 (3+5 x)}\right ) \, dx\\ &=-\frac{37}{50} (1-2 x)^{3/2}+\frac{9}{50} (1-2 x)^{5/2}+\frac{1}{25} \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx\\ &=\frac{2}{125} \sqrt{1-2 x}-\frac{37}{50} (1-2 x)^{3/2}+\frac{9}{50} (1-2 x)^{5/2}+\frac{11}{125} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=\frac{2}{125} \sqrt{1-2 x}-\frac{37}{50} (1-2 x)^{3/2}+\frac{9}{50} (1-2 x)^{5/2}-\frac{11}{125} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{2}{125} \sqrt{1-2 x}-\frac{37}{50} (1-2 x)^{3/2}+\frac{9}{50} (1-2 x)^{5/2}-\frac{2}{125} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.027363, size = 51, normalized size = 0.74 \[ \frac{1}{625} \left (5 \sqrt{1-2 x} \left (90 x^2+95 x-68\right )-2 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Sqrt[1 - 2*x]*(-68 + 95*x + 90*x^2) - 2*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/625

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Maple [A]  time = 0.006, size = 47, normalized size = 0.7 \begin{align*} -{\frac{37}{50} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{9}{50} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}-{\frac{2\,\sqrt{55}}{625}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) }+{\frac{2}{125}\sqrt{1-2\,x}} \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),x)

[Out]

-37/50*(1-2*x)^(3/2)+9/50*(1-2*x)^(5/2)-2/625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+2/125*(1-2*x)^(1/2
)

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Maxima [A]  time = 2.20614, size = 86, normalized size = 1.25 \begin{align*} \frac{9}{50} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - \frac{37}{50} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1}{625} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{2}{125} \, \sqrt{-2 \, x + 1} \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),x, algorithm="maxima")

[Out]

9/50*(-2*x + 1)^(5/2) - 37/50*(-2*x + 1)^(3/2) + 1/625*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) +
5*sqrt(-2*x + 1))) + 2/125*sqrt(-2*x + 1)

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Fricas [A]  time = 1.59393, size = 171, normalized size = 2.48 \begin{align*} \frac{1}{625} \, \sqrt{11} \sqrt{5} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac{1}{125} \,{\left (90 \, x^{2} + 95 \, x - 68\right )} \sqrt{-2 \, x + 1} \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),x, algorithm="fricas")

[Out]

1/625*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 1/125*(90*x^2 + 95*x - 68)
*sqrt(-2*x + 1)

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Sympy [A]  time = 4.48347, size = 102, normalized size = 1.48 \begin{align*} \frac{9 \left (1 - 2 x\right )^{\frac{5}{2}}}{50} - \frac{37 \left (1 - 2 x\right )^{\frac{3}{2}}}{50} + \frac{2 \sqrt{1 - 2 x}}{125} + \frac{22 \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),x)

[Out]

9*(1 - 2*x)**(5/2)/50 - 37*(1 - 2*x)**(3/2)/50 + 2*sqrt(1 - 2*x)/125 + 22*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 = 1.77202, size = 100, normalized size = 1.45 \begin{align*} \frac{9}{50} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} - \frac{37}{50} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1}{625} \, \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{2}{125} \, \sqrt{-2 \, x + 1} \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),x, algorithm="giac")

[Out]

9/50*(2*x - 1)^2*sqrt(-2*x + 1) - 37/50*(-2*x + 1)^(3/2) + 1/625*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2
*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2/125*sqrt(-2*x + 1)