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

Optimal. Leaf size=68 \[ \frac {\sqrt {1-2 x}}{3 x+2}+\frac {68 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}}-2 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {99, 156, 63, 206} \begin {gather*} \frac {\sqrt {1-2 x}}{3 x+2}+\frac {68 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}}-2 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - 2*x]/(2 + 3*x) + (68*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/Sqrt[21] - 2*Sqrt[55]*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 99

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

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)} \, dx &=\frac {\sqrt {1-2 x}}{2+3 x}-\int \frac {-8+5 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{2+3 x}-34 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+55 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{2+3 x}+34 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-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}}{2+3 x}+\frac {68 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}}-2 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 81, normalized size = 1.19 \begin {gather*} \frac {68 \sqrt {21} (3 x+2) \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+21 \left (\sqrt {1-2 x}-2 \sqrt {55} (3 x+2) \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )}{63 x+42} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(68*Sqrt[21]*(2 + 3*x)*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + 21*(Sqrt[1 - 2*x] - 2*Sqrt[55]*(2 + 3*x)*ArcTanh[Sqr
t[5/11]*Sqrt[1 - 2*x]]))/(42 + 63*x)

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IntegrateAlgebraic [A]  time = 0.16, size = 73, normalized size = 1.07 \begin {gather*} -\frac {2 \sqrt {1-2 x}}{3 (1-2 x)-7}+\frac {68 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}}-2 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - 2*x])/(-7 + 3*(1 - 2*x)) + (68*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/Sqrt[21] - 2*Sqrt[55]*ArcTanh[Sq
rt[5/11]*Sqrt[1 - 2*x]]

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fricas [A]  time = 1.38, size = 90, normalized size = 1.32 \begin {gather*} \frac {21 \, \sqrt {55} {\left (3 \, x + 2\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 34 \, \sqrt {21} {\left (3 \, x + 2\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, \sqrt {-2 \, x + 1}}{21 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(21*sqrt(55)*(3*x + 2)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 34*sqrt(21)*(3*x + 2)*log((3*
x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 21*sqrt(-2*x + 1))/(3*x + 2)

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giac [A]  time = 1.23, size = 93, normalized size = 1.37 \begin {gather*} \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 {34}{21} \, \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 ) + \frac {\sqrt {-2 \, x + 1}}{3 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 34/21*sqrt(21)*log(1/2*
abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + sqrt(-2*x + 1)/(3*x + 2)

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maple [A]  time = 0.01, size = 54, normalized size = 0.79 \begin {gather*} \frac {68 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{21}-2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )-\frac {2 \sqrt {-2 x +1}}{3 \left (-2 x -\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-2/3*(-2*x+1)^(1/2)/(-2*x-4/3)+68/21*arctanh(1/7*21^(1/2)*(-2
*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.17, size = 87, normalized size = 1.28 \begin {gather*} \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {34}{21} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {\sqrt {-2 \, x + 1}}{3 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 34/21*sqrt(21)*log(-(sqrt(21) - 3
*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + sqrt(-2*x + 1)/(3*x + 2)

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mupad [B]  time = 1.19, size = 53, normalized size = 0.78 \begin {gather*} \frac {68\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{21}-2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )+\frac {2\,\sqrt {1-2\,x}}{3\,\left (2\,x+\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(68*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/21 - 2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11) + (2*(1
- 2*x)^(1/2))/(3*(2*x + 4/3))

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sympy [A]  time = 55.36, size = 230, normalized size = 3.38 \begin {gather*} 28 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: x \leq \frac {1}{2} \wedge x > - \frac {2}{3} \end {cases}\right ) - 66 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 < - \frac {7}{3} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 > - \frac {7}{3} \end {cases}\right ) + 110 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

28*Piecewise((sqrt(21)*(-log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/4 + log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/4 - 1/(4*(sqr
t(21)*sqrt(1 - 2*x)/7 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147, (x <= 1/2) & (x > -2/3))) - 66*Piecew
ise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 < -7/3), (-sqrt(21)*atanh(sqrt(21)*sqrt(1 - 2*x)/7)
/21, 2*x - 1 > -7/3)) + 110*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt
(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))

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