3.2042 \(\int \frac {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\)

Optimal. Leaf size=55 \[ 2 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

2/7*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-2/11*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {86, 63, 206} \[ 2 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 2*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

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 {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx &=-\left (3 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\right )+5 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=3 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-5 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=2 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 55, normalized size = 1.00 \[ 2 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 2*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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fricas [A]  time = 1.15, size = 74, normalized size = 1.35 \[ \frac {1}{11} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {1}{7} \, \sqrt {7} \sqrt {3} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 1/7*sqrt(7)*sqrt(3)*log(-(s
qrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2))

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giac [B]  time = 1.17, size = 79, normalized size = 1.44 \[ \frac {1}{11} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {1}{7} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/7*sqrt(21)*log(1
/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1)))

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maple [A]  time = 0.01, size = 38, normalized size = 0.69 \[ \frac {2 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{7}-\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)-2/11*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.26, size = 73, normalized size = 1.33 \[ \frac {1}{11} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {1}{7} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/7*sqrt(21)*log(-(sqrt(21)
- 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1)))

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mupad [B]  time = 1.26, size = 37, normalized size = 0.67 \[ \frac {2\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{7}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/7 - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/11

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sympy [A]  time = 14.90, size = 131, normalized size = 2.38 \[ - 6 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21}}{3 \sqrt {1 - 2 x}} \right )}}{21} & \text {for}\: \frac {1}{1 - 2 x} > \frac {3}{7} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21}}{3 \sqrt {1 - 2 x}} \right )}}{21} & \text {for}\: \frac {1}{1 - 2 x} < \frac {3}{7} \end {cases}\right ) + 10 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{55} & \text {for}\: \frac {1}{1 - 2 x} > \frac {5}{11} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{55} & \text {for}\: \frac {1}{1 - 2 x} < \frac {5}{11} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*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)) + 10*Piecewise((-sqrt(55)*acoth(sqrt(55)/(5*sqrt(1 - 2*x)))/55, 1/(1
- 2*x) > 5/11), (-sqrt(55)*atanh(sqrt(55)/(5*sqrt(1 - 2*x)))/55, 1/(1 - 2*x) < 5/11))

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