3.2108 \(\int \frac{(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)} \, dx\)

Optimal. Leaf size=54 \[ \frac{9}{10} \sqrt{1-2 x}+\frac{49}{22 \sqrt{1-2 x}}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{55 \sqrt{55}} \]

[Out]

49/(22*Sqrt[1 - 2*x]) + (9*Sqrt[1 - 2*x])/10 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

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Rubi [A]  time = 0.0236737, 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 = {87, 63, 206} \[ \frac{9}{10} \sqrt{1-2 x}+\frac{49}{22 \sqrt{1-2 x}}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{55 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

49/(22*Sqrt[1 - 2*x]) + (9*Sqrt[1 - 2*x])/10 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

Rule 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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}{(1-2 x)^{3/2} (3+5 x)} \, dx &=\int \left (\frac{49}{22 (1-2 x)^{3/2}}-\frac{9}{10 \sqrt{1-2 x}}+\frac{1}{55 \sqrt{1-2 x} (3+5 x)}\right ) \, dx\\ &=\frac{49}{22 \sqrt{1-2 x}}+\frac{9}{10} \sqrt{1-2 x}+\frac{1}{55} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=\frac{49}{22 \sqrt{1-2 x}}+\frac{9}{10} \sqrt{1-2 x}-\frac{1}{55} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{49}{22 \sqrt{1-2 x}}+\frac{9}{10} \sqrt{1-2 x}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{55 \sqrt{55}}\\ \end{align*}

Mathematica [C]  time = 0.0168831, size = 37, normalized size = 0.69 \[ \frac{2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{5}{11} (1-2 x)\right )-495 x+858}{275 \sqrt{1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(858 - 495*x + 2*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 - 2*x))/11])/(275*Sqrt[1 - 2*x])

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Maple [A]  time = 0.007, size = 38, normalized size = 0.7 \begin{align*} -{\frac{2\,\sqrt{55}}{3025}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) }+{\frac{49}{22}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{9}{10}\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)^(3/2)/(3+5*x),x)

[Out]

-2/3025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+49/22/(1-2*x)^(1/2)+9/10*(1-2*x)^(1/2)

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Maxima [A]  time = 3.17041, size = 74, normalized size = 1.37 \begin{align*} \frac{1}{3025} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{9}{10} \, \sqrt{-2 \, x + 1} + \frac{49}{22 \, \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)^(3/2)/(3+5*x),x, algorithm="maxima")

[Out]

1/3025*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 9/10*sqrt(-2*x + 1) + 49/2
2/sqrt(-2*x + 1)

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Fricas [A]  time = 1.60525, size = 166, normalized size = 3.07 \begin{align*} \frac{\sqrt{55}{\left (2 \, x - 1\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \,{\left (99 \, x - 172\right )} \sqrt{-2 \, x + 1}}{3025 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3025*(sqrt(55)*(2*x - 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(99*x - 172)*sqrt(-2*x + 1)
)/(2*x - 1)

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Sympy [A]  time = 23.9308, size = 90, normalized size = 1.67 \begin{align*} \frac{9 \sqrt{1 - 2 x}}{10} + \frac{2 \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 )}{55} + \frac{49}{22 \sqrt{1 - 2 x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9*sqrt(1 - 2*x)/10 + 2*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))/55 + 49/(22*sqrt(1 - 2*x))

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Giac [A]  time = 2.70696, size = 78, normalized size = 1.44 \begin{align*} \frac{1}{3025} \, \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{9}{10} \, \sqrt{-2 \, x + 1} + \frac{49}{22 \, \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)^(3/2)/(3+5*x),x, algorithm="giac")

[Out]

1/3025*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 9/10*sqrt(-2*x +
1) + 49/22/sqrt(-2*x + 1)