3.2039 \(\int \frac{(2+3 x)^2}{\sqrt{1-2 x} (3+5 x)} \, dx\)

Optimal. Leaf size=54 \[ \frac{3}{10} (1-2 x)^{3/2}-\frac{111}{50} \sqrt{1-2 x}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}} \]

[Out]

(-111*Sqrt[1 - 2*x])/50 + (3*(1 - 2*x)^(3/2))/10 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

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Rubi [A]  time = 0.022043, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {88, 63, 206} \[ \frac{3}{10} (1-2 x)^{3/2}-\frac{111}{50} \sqrt{1-2 x}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-111*Sqrt[1 - 2*x])/50 + (3*(1 - 2*x)^(3/2))/10 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

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 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{(2+3 x)^2}{\sqrt{1-2 x} (3+5 x)} \, dx &=\int \left (\frac{111}{50 \sqrt{1-2 x}}-\frac{9}{10} \sqrt{1-2 x}+\frac{1}{25 \sqrt{1-2 x} (3+5 x)}\right ) \, dx\\ &=-\frac{111}{50} \sqrt{1-2 x}+\frac{3}{10} (1-2 x)^{3/2}+\frac{1}{25} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{111}{50} \sqrt{1-2 x}+\frac{3}{10} (1-2 x)^{3/2}-\frac{1}{25} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{111}{50} \sqrt{1-2 x}+\frac{3}{10} (1-2 x)^{3/2}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}}\\ \end{align*}

Mathematica [A]  time = 0.0335851, size = 46, normalized size = 0.85 \[ -\frac{3}{25} \sqrt{1-2 x} (5 x+16)-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{25 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[1 - 2*x]*(16 + 5*x))/25 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

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Maple [A]  time = 0.006, size = 38, normalized size = 0.7 \begin{align*}{\frac{3}{10} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{2\,\sqrt{55}}{1375}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) }-{\frac{111}{50}\sqrt{1-2\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(1-2*x)^(3/2)-2/1375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-111/50*(1-2*x)^(1/2)

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Maxima [A]  time = 1.51723, size = 74, normalized size = 1.37 \begin{align*} \frac{3}{10} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1}{1375} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{111}{50} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(-2*x + 1)^(3/2) + 1/1375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 11
1/50*sqrt(-2*x + 1)

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Fricas [A]  time = 1.64776, size = 138, normalized size = 2.56 \begin{align*} -\frac{3}{25} \,{\left (5 \, x + 16\right )} \sqrt{-2 \, x + 1} + \frac{1}{1375} \, \sqrt{55} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/25*(5*x + 16)*sqrt(-2*x + 1) + 1/1375*sqrt(55)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3))

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Sympy [A]  time = 14.3353, size = 90, normalized size = 1.67 \begin{align*} \frac{3 \left (1 - 2 x\right )^{\frac{3}{2}}}{10} - \frac{111 \sqrt{1 - 2 x}}{50} + \frac{2 \left (\begin{cases} - \frac{\sqrt{55} \operatorname{acoth}{\left (\frac{\sqrt{55}}{5 \sqrt{1 - 2 x}} \right )}}{55} & \text{for}\: \frac{1}{1 - 2 x} > \frac{5}{11} \\- \frac{\sqrt{55} \operatorname{atanh}{\left (\frac{\sqrt{55}}{5 \sqrt{1 - 2 x}} \right )}}{55} & \text{for}\: \frac{1}{1 - 2 x} < \frac{5}{11} \end{cases}\right )}{25} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(1 - 2*x)**(3/2)/10 - 111*sqrt(1 - 2*x)/50 + 2*Piecewise((-sqrt(55)*acoth(sqrt(55)/(5*sqrt(1 - 2*x)))/55, 1/
(1 - 2*x) > 5/11), (-sqrt(55)*atanh(sqrt(55)/(5*sqrt(1 - 2*x)))/55, 1/(1 - 2*x) < 5/11))/25

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Giac [A]  time = 2.69079, size = 78, normalized size = 1.44 \begin{align*} \frac{3}{10} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1}{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{111}{50} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(-2*x + 1)^(3/2) + 1/1375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x +
1))) - 111/50*sqrt(-2*x + 1)