3.2052 \(\int \frac{2+3 x}{\sqrt{1-2 x} (3+5 x)^2} \, dx\)

Optimal. Leaf size=48 \[ -\frac{\sqrt{1-2 x}}{55 (5 x+3)}-\frac{68 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{55 \sqrt{55}} \]

[Out]

-Sqrt[1 - 2*x]/(55*(3 + 5*x)) - (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

________________________________________________________________________________________

Rubi [A]  time = 0.0089465, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {78, 63, 206} \[ -\frac{\sqrt{1-2 x}}{55 (5 x+3)}-\frac{68 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{55 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - 2*x]/(55*(3 + 5*x)) - (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

Rule 78

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

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}{\sqrt{1-2 x} (3+5 x)^2} \, dx &=-\frac{\sqrt{1-2 x}}{55 (3+5 x)}+\frac{34}{55} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{\sqrt{1-2 x}}{55 (3+5 x)}-\frac{34}{55} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{\sqrt{1-2 x}}{55 (3+5 x)}-\frac{68 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{55 \sqrt{55}}\\ \end{align*}

Mathematica [A]  time = 0.0175411, size = 48, normalized size = 1. \[ -\frac{\sqrt{1-2 x}}{55 (5 x+3)}-\frac{68 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{55 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - 2*x]/(55*(3 + 5*x)) - (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 36, normalized size = 0.8 \begin{align*}{\frac{2}{275}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{68\,\sqrt{55}}{3025}{\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)/(3+5*x)^2/(1-2*x)^(1/2),x)

[Out]

2/275*(1-2*x)^(1/2)/(-2*x-6/5)-68/3025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 1.48714, size = 72, normalized size = 1.5 \begin{align*} \frac{34}{3025} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{\sqrt{-2 \, x + 1}}{55 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

34/3025*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/55*sqrt(-2*x + 1)/(5*x
+ 3)

________________________________________________________________________________________

Fricas [A]  time = 1.62244, size = 153, normalized size = 3.19 \begin{align*} \frac{34 \, \sqrt{55}{\left (5 \, x + 3\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, \sqrt{-2 \, x + 1}}{3025 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3025*(34*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*sqrt(-2*x + 1))/(5*x + 3
)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Giac [A]  time = 1.73881, size = 76, normalized size = 1.58 \begin{align*} \frac{34}{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{\sqrt{-2 \, x + 1}}{55 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

34/3025*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/55*sqrt(-2*x
+ 1)/(5*x + 3)