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

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

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Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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 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 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*}

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Mathematica [A]  time = 0.04, size = 48, normalized size = 1.00 \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 52, normalized size = 1.08 \begin {gather*} \frac {2 \sqrt {1-2 x}}{55 (5 (1-2 x)-11)}-\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.56, size = 54, normalized size = 1.12 \begin {gather*} \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 {gather*}

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
)

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giac [A]  time = 1.28, size = 56, normalized size = 1.17 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.01, size = 36, normalized size = 0.75 \begin {gather*} -\frac {68 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{3025}+\frac {2 \sqrt {-2 x +1}}{275 \left (-2 x -\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.12, size = 53, normalized size = 1.10 \begin {gather*} \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 {gather*}

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)

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mupad [B]  time = 0.06, size = 35, normalized size = 0.73 \begin {gather*} -\frac {68\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{3025}-\frac {2\,\sqrt {1-2\,x}}{275\,\left (2\,x+\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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