3.1454 \(\int \frac{1}{(2+3 x) \sqrt{1+5 x}} \, dx\)

Optimal. Leaf size=25 \[ \frac{2 \tan ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{5 x+1}\right )}{\sqrt{21}} \]

[Out]

(2*ArcTan[Sqrt[3/7]*Sqrt[1 + 5*x]])/Sqrt[21]

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Rubi [A]  time = 0.0086591, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {63, 203} \[ \frac{2 \tan ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{5 x+1}\right )}{\sqrt{21}} \]

Antiderivative was successfully verified.

[In]

Int[1/((2 + 3*x)*Sqrt[1 + 5*x]),x]

[Out]

(2*ArcTan[Sqrt[3/7]*Sqrt[1 + 5*x]])/Sqrt[21]

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(2+3 x) \sqrt{1+5 x}} \, dx &=\frac{2}{5} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{5}+\frac{3 x^2}{5}} \, dx,x,\sqrt{1+5 x}\right )\\ &=\frac{2 \tan ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1+5 x}\right )}{\sqrt{21}}\\ \end{align*}

Mathematica [A]  time = 0.0088877, size = 25, normalized size = 1. \[ \frac{2 \tan ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{5 x+1}\right )}{\sqrt{21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[Sqrt[3/7]*Sqrt[1 + 5*x]])/Sqrt[21]

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Maple [A]  time = 0.005, size = 19, normalized size = 0.8 \begin{align*}{\frac{2\,\sqrt{21}}{21}\arctan \left ({\frac{\sqrt{21}}{7}\sqrt{1+5\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*arctan(1/7*21^(1/2)*(1+5*x)^(1/2))*21^(1/2)

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Maxima [A]  time = 1.44277, size = 24, normalized size = 0.96 \begin{align*} \frac{2}{21} \, \sqrt{21} \arctan \left (\frac{1}{7} \, \sqrt{21} \sqrt{5 \, x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*sqrt(21)*arctan(1/7*sqrt(21)*sqrt(5*x + 1))

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Fricas [A]  time = 2.05966, size = 68, normalized size = 2.72 \begin{align*} \frac{2}{21} \, \sqrt{21} \arctan \left (\frac{1}{7} \, \sqrt{21} \sqrt{5 \, x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*sqrt(21)*arctan(1/7*sqrt(21)*sqrt(5*x + 1))

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Sympy [A]  time = 0.991192, size = 61, normalized size = 2.44 \begin{align*} \begin{cases} \frac{2 \sqrt{21} i \operatorname{acosh}{\left (\frac{\sqrt{105}}{15 \sqrt{x + \frac{2}{3}}} \right )}}{21} & \text{for}\: \frac{7}{15 \left |{x + \frac{2}{3}}\right |} > 1 \\- \frac{2 \sqrt{21} \operatorname{asin}{\left (\frac{\sqrt{105}}{15 \sqrt{x + \frac{2}{3}}} \right )}}{21} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*sqrt(21)*I*acosh(sqrt(105)/(15*sqrt(x + 2/3)))/21, 7/(15*Abs(x + 2/3)) > 1), (-2*sqrt(21)*asin(sq
rt(105)/(15*sqrt(x + 2/3)))/21, True))

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Giac [A]  time = 1.07503, size = 24, normalized size = 0.96 \begin{align*} \frac{2}{21} \, \sqrt{21} \arctan \left (\frac{1}{7} \, \sqrt{21} \sqrt{5 \, x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*sqrt(21)*arctan(1/7*sqrt(21)*sqrt(5*x + 1))