3.2051 \(\int \frac{(2+3 x)^2}{\sqrt{1-2 x} (3+5 x)^2} \, dx\)

Optimal. Leaf size=61 \[ -\frac{9}{25} \sqrt{1-2 x}-\frac{\sqrt{1-2 x}}{275 (5 x+3)}-\frac{134 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{275 \sqrt{55}} \]

[Out]

(-9*Sqrt[1 - 2*x])/25 - Sqrt[1 - 2*x]/(275*(3 + 5*x)) - (134*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqrt[55])

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Rubi [A]  time = 0.0140236, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {89, 80, 63, 206} \[ -\frac{9}{25} \sqrt{1-2 x}-\frac{\sqrt{1-2 x}}{275 (5 x+3)}-\frac{134 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{275 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*Sqrt[1 - 2*x])/25 - Sqrt[1 - 2*x]/(275*(3 + 5*x)) - (134*ArcTanh[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 80

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

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)^2} \, dx &=-\frac{\sqrt{1-2 x}}{275 (3+5 x)}+\frac{1}{275} \int \frac{364+495 x}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{9}{25} \sqrt{1-2 x}-\frac{\sqrt{1-2 x}}{275 (3+5 x)}+\frac{67}{275} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{9}{25} \sqrt{1-2 x}-\frac{\sqrt{1-2 x}}{275 (3+5 x)}-\frac{67}{275} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{9}{25} \sqrt{1-2 x}-\frac{\sqrt{1-2 x}}{275 (3+5 x)}-\frac{134 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{275 \sqrt{55}}\\ \end{align*}

Mathematica [A]  time = 0.0276832, size = 53, normalized size = 0.87 \[ -\frac{\sqrt{1-2 x} (495 x+298)}{275 (5 x+3)}-\frac{134 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{275 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]*(298 + 495*x))/(275*(3 + 5*x)) - (134*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqrt[55])

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Maple [A]  time = 0.009, size = 45, normalized size = 0.7 \begin{align*} -{\frac{9}{25}\sqrt{1-2\,x}}+{\frac{2}{1375}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{134\,\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/(3+5*x)^2/(1-2*x)^(1/2),x)

[Out]

-9/25*(1-2*x)^(1/2)+2/1375*(1-2*x)^(1/2)/(-2*x-6/5)-134/15125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.49958, size = 84, normalized size = 1.38 \begin{align*} \frac{67}{15125} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{9}{25} \, \sqrt{-2 \, x + 1} - \frac{\sqrt{-2 \, x + 1}}{275 \,{\left (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)^2/(1-2*x)^(1/2),x, algorithm="maxima")

[Out]

67/15125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 9/25*sqrt(-2*x + 1) - 1/
275*sqrt(-2*x + 1)/(5*x + 3)

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Fricas [A]  time = 1.61941, size = 173, normalized size = 2.84 \begin{align*} \frac{67 \, \sqrt{55}{\left (5 \, x + 3\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \,{\left (495 \, x + 298\right )} \sqrt{-2 \, x + 1}}{15125 \,{\left (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)^2/(1-2*x)^(1/2),x, algorithm="fricas")

[Out]

1/15125*(67*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(495*x + 298)*sqrt(-2*x
+ 1))/(5*x + 3)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.75371, size = 88, normalized size = 1.44 \begin{align*} \frac{67}{15125} \, \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}{25} \, \sqrt{-2 \, x + 1} - \frac{\sqrt{-2 \, x + 1}}{275 \,{\left (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)^2/(1-2*x)^(1/2),x, algorithm="giac")

[Out]

67/15125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 9/25*sqrt(-2*x
+ 1) - 1/275*sqrt(-2*x + 1)/(5*x + 3)