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

Optimal. Leaf size=59 \[ \frac {(1-2 x)^{3/2}}{21 (3 x+2)}+\frac {8}{7} \sqrt {1-2 x}-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 50, 63, 206} \begin {gather*} \frac {(1-2 x)^{3/2}}{21 (3 x+2)}+\frac {8}{7} \sqrt {1-2 x}-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[1 - 2*x])/7 + (1 - 2*x)^(3/2)/(21*(2 + 3*x)) - (8*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/Sqrt[21]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 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 {\sqrt {1-2 x} (3+5 x)}{(2+3 x)^2} \, dx &=\frac {(1-2 x)^{3/2}}{21 (2+3 x)}+\frac {12}{7} \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx\\ &=\frac {8}{7} \sqrt {1-2 x}+\frac {(1-2 x)^{3/2}}{21 (2+3 x)}+4 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {8}{7} \sqrt {1-2 x}+\frac {(1-2 x)^{3/2}}{21 (2+3 x)}-4 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {8}{7} \sqrt {1-2 x}+\frac {(1-2 x)^{3/2}}{21 (2+3 x)}-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 55, normalized size = 0.93 \begin {gather*} \frac {7 \sqrt {1-2 x} (10 x+7)-8 \sqrt {21} (3 x+2) \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{63 x+42} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[1 - 2*x]*(7 + 10*x) - 8*Sqrt[21]*(2 + 3*x)*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(42 + 63*x)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.12, size = 59, normalized size = 1.00 \begin {gather*} \frac {2 (5 (1-2 x)-12) \sqrt {1-2 x}}{3 (3 (1-2 x)-7)}-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-12 + 5*(1 - 2*x))*Sqrt[1 - 2*x])/(3*(-7 + 3*(1 - 2*x))) - (8*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/Sqrt[21]

________________________________________________________________________________________

fricas [A]  time = 1.20, size = 59, normalized size = 1.00 \begin {gather*} \frac {4 \, \sqrt {21} {\left (3 \, x + 2\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 7 \, {\left (10 \, x + 7\right )} \sqrt {-2 \, x + 1}}{21 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(4*sqrt(21)*(3*x + 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 7*(10*x + 7)*sqrt(-2*x + 1))/(
3*x + 2)

________________________________________________________________________________________

giac [A]  time = 1.20, size = 65, normalized size = 1.10 \begin {gather*} \frac {4}{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 {10}{9} \, \sqrt {-2 \, x + 1} + \frac {\sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/21*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 10/9*sqrt(-2*x + 1)
 + 1/9*sqrt(-2*x + 1)/(3*x + 2)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 45, normalized size = 0.76 \begin {gather*} -\frac {8 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{21}+\frac {10 \sqrt {-2 x +1}}{9}-\frac {2 \sqrt {-2 x +1}}{27 \left (-2 x -\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10/9*(-2*x+1)^(1/2)-2/27*(-2*x+1)^(1/2)/(-4/3-2*x)-8/21*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

________________________________________________________________________________________

maxima [A]  time = 1.21, size = 62, normalized size = 1.05 \begin {gather*} \frac {4}{21} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {10}{9} \, \sqrt {-2 \, x + 1} + \frac {\sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/21*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 10/9*sqrt(-2*x + 1) + 1/9*sq
rt(-2*x + 1)/(3*x + 2)

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 44, normalized size = 0.75 \begin {gather*} \frac {2\,\sqrt {1-2\,x}}{27\,\left (2\,x+\frac {4}{3}\right )}-\frac {8\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{21}+\frac {10\,\sqrt {1-2\,x}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(1 - 2*x)^(1/2))/(27*(2*x + 4/3)) - (8*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/21 + (10*(1 - 2*x)^(1/
2))/9

________________________________________________________________________________________

sympy [A]  time = 112.41, size = 178, normalized size = 3.02 \begin {gather*} \frac {10 \sqrt {1 - 2 x}}{9} + \frac {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 )}{9} + \frac {74 \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 )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*sqrt(1 - 2*x)/9 + 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*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147, (x <= 1/2) & (x
> -2/3)))/9 + 74*Piecewise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 < -7/3), (-sqrt(21)*atanh(sq
rt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 > -7/3))/9

________________________________________________________________________________________