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

Optimal. Leaf size=76 \[ \frac {(1-2 x)^{5/2}}{21 (3 x+2)}+\frac {76}{189} (1-2 x)^{3/2}+\frac {76}{27} \sqrt {1-2 x}-\frac {76}{27} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, 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)^{5/2}}{21 (3 x+2)}+\frac {76}{189} (1-2 x)^{3/2}+\frac {76}{27} \sqrt {1-2 x}-\frac {76}{27} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(76*Sqrt[1 - 2*x])/27 + (76*(1 - 2*x)^(3/2))/189 + (1 - 2*x)^(5/2)/(21*(2 + 3*x)) - (76*Sqrt[7/3]*ArcTanh[Sqrt
[3/7]*Sqrt[1 - 2*x]])/27

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 {(1-2 x)^{3/2} (3+5 x)}{(2+3 x)^2} \, dx &=\frac {(1-2 x)^{5/2}}{21 (2+3 x)}+\frac {38}{21} \int \frac {(1-2 x)^{3/2}}{2+3 x} \, dx\\ &=\frac {76}{189} (1-2 x)^{3/2}+\frac {(1-2 x)^{5/2}}{21 (2+3 x)}+\frac {38}{9} \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx\\ &=\frac {76}{27} \sqrt {1-2 x}+\frac {76}{189} (1-2 x)^{3/2}+\frac {(1-2 x)^{5/2}}{21 (2+3 x)}+\frac {266}{27} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {76}{27} \sqrt {1-2 x}+\frac {76}{189} (1-2 x)^{3/2}+\frac {(1-2 x)^{5/2}}{21 (2+3 x)}-\frac {266}{27} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {76}{27} \sqrt {1-2 x}+\frac {76}{189} (1-2 x)^{3/2}+\frac {(1-2 x)^{5/2}}{21 (2+3 x)}-\frac {76}{27} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 58, normalized size = 0.76 \begin {gather*} \frac {1}{81} \left (\frac {3 \sqrt {1-2 x} \left (-60 x^2+212 x+175\right )}{3 x+2}-76 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*Sqrt[1 - 2*x]*(175 + 212*x - 60*x^2))/(2 + 3*x) - 76*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/81

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IntegrateAlgebraic [A]  time = 0.12, size = 72, normalized size = 0.95 \begin {gather*} \frac {2 \left (15 (1-2 x)^2+76 (1-2 x)-266\right ) \sqrt {1-2 x}}{27 (3 (1-2 x)-7)}-\frac {76}{27} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-266 + 76*(1 - 2*x) + 15*(1 - 2*x)^2)*Sqrt[1 - 2*x])/(27*(-7 + 3*(1 - 2*x))) - (76*Sqrt[7/3]*ArcTanh[Sqrt[
3/7]*Sqrt[1 - 2*x]])/27

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fricas [A]  time = 1.39, size = 70, normalized size = 0.92 \begin {gather*} \frac {38 \, \sqrt {7} \sqrt {3} {\left (3 \, x + 2\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 3 \, {\left (60 \, x^{2} - 212 \, x - 175\right )} \sqrt {-2 \, x + 1}}{81 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/81*(38*sqrt(7)*sqrt(3)*(3*x + 2)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 3*(60*x^2 - 212
*x - 175)*sqrt(-2*x + 1))/(3*x + 2)

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giac [A]  time = 0.91, size = 74, normalized size = 0.97 \begin {gather*} \frac {10}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {38}{81} \, \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 {74}{27} \, \sqrt {-2 \, x + 1} + \frac {7 \, \sqrt {-2 \, x + 1}}{27 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10/27*(-2*x + 1)^(3/2) + 38/81*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x +
1))) + 74/27*sqrt(-2*x + 1) + 7/27*sqrt(-2*x + 1)/(3*x + 2)

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maple [A]  time = 0.01, size = 54, normalized size = 0.71 \begin {gather*} -\frac {76 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{81}+\frac {10 \left (-2 x +1\right )^{\frac {3}{2}}}{27}+\frac {74 \sqrt {-2 x +1}}{27}-\frac {14 \sqrt {-2 x +1}}{81 \left (-2 x -\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10/27*(-2*x+1)^(3/2)+74/27*(-2*x+1)^(1/2)-14/81*(-2*x+1)^(1/2)/(-2*x-4/3)-76/81*arctanh(1/7*21^(1/2)*(-2*x+1)^
(1/2))*21^(1/2)

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maxima [A]  time = 1.24, size = 71, normalized size = 0.93 \begin {gather*} \frac {10}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {38}{81} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {74}{27} \, \sqrt {-2 \, x + 1} + \frac {7 \, \sqrt {-2 \, x + 1}}{27 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10/27*(-2*x + 1)^(3/2) + 38/81*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 74
/27*sqrt(-2*x + 1) + 7/27*sqrt(-2*x + 1)/(3*x + 2)

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mupad [B]  time = 1.20, size = 55, normalized size = 0.72 \begin {gather*} \frac {14\,\sqrt {1-2\,x}}{81\,\left (2\,x+\frac {4}{3}\right )}+\frac {74\,\sqrt {1-2\,x}}{27}+\frac {10\,{\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 )\,76{}\mathrm {i}}{81} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*76i)/81 + (14*(1 - 2*x)^(1/2))/(81*(2*x + 4/3)) + (74*(1 - 2*x
)^(1/2))/27 + (10*(1 - 2*x)^(3/2))/27

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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