3.1809 \(\int \frac {\sqrt {1-2 x} (3+5 x)^2}{(2+3 x)^2} \, dx\)

Optimal. Leaf size=74 \[ -\frac {(1-2 x)^{3/2}}{63 (3 x+2)}-\frac {25}{27} (1-2 x)^{3/2}-\frac {142}{189} \sqrt {1-2 x}+\frac {142 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}} \]

[Out]

-25/27*(1-2*x)^(3/2)-1/63*(1-2*x)^(3/2)/(2+3*x)+142/567*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-142/189*(
1-2*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {89, 80, 50, 63, 206} \[ -\frac {(1-2 x)^{3/2}}{63 (3 x+2)}-\frac {25}{27} (1-2 x)^{3/2}-\frac {142}{189} \sqrt {1-2 x}+\frac {142 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-142*Sqrt[1 - 2*x])/189 - (25*(1 - 2*x)^(3/2))/27 - (1 - 2*x)^(3/2)/(63*(2 + 3*x)) + (142*ArcTanh[Sqrt[3/7]*S
qrt[1 - 2*x]])/(27*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 80

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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}{(2+3 x)^2} \, dx &=-\frac {(1-2 x)^{3/2}}{63 (2+3 x)}+\frac {1}{63} \int \frac {\sqrt {1-2 x} (279+525 x)}{2+3 x} \, dx\\ &=-\frac {25}{27} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{63 (2+3 x)}-\frac {71}{63} \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx\\ &=-\frac {142}{189} \sqrt {1-2 x}-\frac {25}{27} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{63 (2+3 x)}-\frac {71}{27} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {142}{189} \sqrt {1-2 x}-\frac {25}{27} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{63 (2+3 x)}+\frac {71}{27} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {142}{189} \sqrt {1-2 x}-\frac {25}{27} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{63 (2+3 x)}+\frac {142 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 55, normalized size = 0.74 \[ \frac {\sqrt {1-2 x} \left (150 x^2-35 x-91\right )}{81 x+54}+\frac {142 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(-91 - 35*x + 150*x^2))/(54 + 81*x) + (142*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(27*Sqrt[21])

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fricas [A]  time = 0.99, size = 65, normalized size = 0.88 \[ \frac {71 \, \sqrt {21} {\left (3 \, x + 2\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, {\left (150 \, x^{2} - 35 \, x - 91\right )} \sqrt {-2 \, x + 1}}{567 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/567*(71*sqrt(21)*(3*x + 2)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 21*(150*x^2 - 35*x - 91)*sqr
t(-2*x + 1))/(3*x + 2)

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giac [A]  time = 1.23, size = 74, normalized size = 1.00 \[ -\frac {25}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {71}{567} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {20}{27} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{27 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/27*(-2*x + 1)^(3/2) - 71/567*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x
+ 1))) - 20/27*sqrt(-2*x + 1) - 1/27*sqrt(-2*x + 1)/(3*x + 2)

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maple [A]  time = 0.01, size = 54, normalized size = 0.73 \[ \frac {142 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{567}-\frac {25 \left (-2 x +1\right )^{\frac {3}{2}}}{27}-\frac {20 \sqrt {-2 x +1}}{27}+\frac {2 \sqrt {-2 x +1}}{81 \left (-2 x -\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/27*(-2*x+1)^(3/2)-20/27*(-2*x+1)^(1/2)+2/81*(-2*x+1)^(1/2)/(-2*x-4/3)+142/567*arctanh(1/7*21^(1/2)*(-2*x+1
)^(1/2))*21^(1/2)

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maxima [A]  time = 1.22, size = 71, normalized size = 0.96 \[ -\frac {25}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {71}{567} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {20}{27} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{27 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/27*(-2*x + 1)^(3/2) - 71/567*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) -
20/27*sqrt(-2*x + 1) - 1/27*sqrt(-2*x + 1)/(3*x + 2)

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mupad [B]  time = 1.19, size = 55, normalized size = 0.74 \[ -\frac {2\,\sqrt {1-2\,x}}{81\,\left (2\,x+\frac {4}{3}\right )}-\frac {20\,\sqrt {1-2\,x}}{27}-\frac {25\,{\left (1-2\,x\right )}^{3/2}}{27}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,142{}\mathrm {i}}{567} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*142i)/567 - (2*(1 - 2*x)^(1/2))/(81*(2*x + 4/3)) - (20*(1 -
2*x)^(1/2))/27 - (25*(1 - 2*x)^(3/2))/27

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sympy [A]  time = 121.34, size = 192, normalized size = 2.59 \[ - \frac {25 \left (1 - 2 x\right )^{\frac {3}{2}}}{27} - \frac {20 \sqrt {1 - 2 x}}{27} - \frac {28 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: x \leq \frac {1}{2} \wedge x > - \frac {2}{3} \end {cases}\right )}{27} - \frac {16 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 < - \frac {7}{3} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 > - \frac {7}{3} \end {cases}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25*(1 - 2*x)**(3/2)/27 - 20*sqrt(1 - 2*x)/27 - 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)))/27 - 16*Piecewise((-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))/3

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