3.2018 \(\int \frac{(3+5 x)^2}{\sqrt{1-2 x} (2+3 x)} \, dx\)

Optimal. Leaf size=54 \[ \frac{25}{18} (1-2 x)^{3/2}-\frac{155}{18} \sqrt{1-2 x}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{9 \sqrt{21}} \]

[Out]

(-155*Sqrt[1 - 2*x])/18 + (25*(1 - 2*x)^(3/2))/18 - (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(9*Sqrt[21])

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Rubi [A]  time = 0.019101, 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} \[ \frac{25}{18} (1-2 x)^{3/2}-\frac{155}{18} \sqrt{1-2 x}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{9 \sqrt{21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-155*Sqrt[1 - 2*x])/18 + (25*(1 - 2*x)^(3/2))/18 - (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(9*Sqrt[21])

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{(3+5 x)^2}{\sqrt{1-2 x} (2+3 x)} \, dx &=\int \left (\frac{155}{18 \sqrt{1-2 x}}-\frac{25}{6} \sqrt{1-2 x}+\frac{1}{9 \sqrt{1-2 x} (2+3 x)}\right ) \, dx\\ &=-\frac{155}{18} \sqrt{1-2 x}+\frac{25}{18} (1-2 x)^{3/2}+\frac{1}{9} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=-\frac{155}{18} \sqrt{1-2 x}+\frac{25}{18} (1-2 x)^{3/2}-\frac{1}{9} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{155}{18} \sqrt{1-2 x}+\frac{25}{18} (1-2 x)^{3/2}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{9 \sqrt{21}}\\ \end{align*}

Mathematica [A]  time = 0.0278732, size = 46, normalized size = 0.85 \[ -\frac{5}{9} \sqrt{1-2 x} (5 x+13)-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{9 \sqrt{21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Sqrt[1 - 2*x]*(13 + 5*x))/9 - (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(9*Sqrt[21])

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Maple [A]  time = 0.006, size = 38, normalized size = 0.7 \begin{align*}{\frac{25}{18} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{2\,\sqrt{21}}{189}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }-{\frac{155}{18}\sqrt{1-2\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/18*(1-2*x)^(3/2)-2/189*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-155/18*(1-2*x)^(1/2)

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Maxima [A]  time = 1.62914, size = 74, normalized size = 1.37 \begin{align*} \frac{25}{18} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1}{189} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) - \frac{155}{18} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/18*(-2*x + 1)^(3/2) + 1/189*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 15
5/18*sqrt(-2*x + 1)

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Fricas [A]  time = 1.77962, size = 135, normalized size = 2.5 \begin{align*} -\frac{5}{9} \,{\left (5 \, x + 13\right )} \sqrt{-2 \, x + 1} + \frac{1}{189} \, \sqrt{21} \log \left (\frac{3 \, x + \sqrt{21} \sqrt{-2 \, x + 1} - 5}{3 \, x + 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/9*(5*x + 13)*sqrt(-2*x + 1) + 1/189*sqrt(21)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2))

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Sympy [A]  time = 14.3985, size = 90, normalized size = 1.67 \begin{align*} \frac{25 \left (1 - 2 x\right )^{\frac{3}{2}}}{18} - \frac{155 \sqrt{1 - 2 x}}{18} + \frac{2 \left (\begin{cases} - \frac{\sqrt{21} \operatorname{acoth}{\left (\frac{\sqrt{21}}{3 \sqrt{1 - 2 x}} \right )}}{21} & \text{for}\: \frac{1}{1 - 2 x} > \frac{3}{7} \\- \frac{\sqrt{21} \operatorname{atanh}{\left (\frac{\sqrt{21}}{3 \sqrt{1 - 2 x}} \right )}}{21} & \text{for}\: \frac{1}{1 - 2 x} < \frac{3}{7} \end{cases}\right )}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*(1 - 2*x)**(3/2)/18 - 155*sqrt(1 - 2*x)/18 + 2*Piecewise((-sqrt(21)*acoth(sqrt(21)/(3*sqrt(1 - 2*x)))/21, 1
/(1 - 2*x) > 3/7), (-sqrt(21)*atanh(sqrt(21)/(3*sqrt(1 - 2*x)))/21, 1/(1 - 2*x) < 3/7))/9

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Giac [A]  time = 2.02442, size = 78, normalized size = 1.44 \begin{align*} \frac{25}{18} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1}{189} \, \sqrt{21} \log \left (\frac{{\left | -2 \, \sqrt{21} + 6 \, \sqrt{-2 \, x + 1} \right |}}{2 \,{\left (\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}\right )}}\right ) - \frac{155}{18} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/18*(-2*x + 1)^(3/2) + 1/189*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x +
1))) - 155/18*sqrt(-2*x + 1)