3.1875 \(\int \frac{(1-2 x)^{3/2} (3+5 x)^2}{2+3 x} \, dx\)

Optimal. Leaf size=82 \[ \frac{25}{42} (1-2 x)^{7/2}-\frac{31}{18} (1-2 x)^{5/2}+\frac{2}{81} (1-2 x)^{3/2}+\frac{14}{81} \sqrt{1-2 x}-\frac{14}{81} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ) \]

[Out]

(14*Sqrt[1 - 2*x])/81 + (2*(1 - 2*x)^(3/2))/81 - (31*(1 - 2*x)^(5/2))/18 + (25*(1 - 2*x)^(7/2))/42 - (14*Sqrt[
7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/81

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Rubi [A]  time = 0.0287284, antiderivative size = 82, normalized size of antiderivative = 1., 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 = {88, 50, 63, 206} \[ \frac{25}{42} (1-2 x)^{7/2}-\frac{31}{18} (1-2 x)^{5/2}+\frac{2}{81} (1-2 x)^{3/2}+\frac{14}{81} \sqrt{1-2 x}-\frac{14}{81} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14*Sqrt[1 - 2*x])/81 + (2*(1 - 2*x)^(3/2))/81 - (31*(1 - 2*x)^(5/2))/18 + (25*(1 - 2*x)^(7/2))/42 - (14*Sqrt[
7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/81

Rule 88

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

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 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}{2+3 x} \, dx &=\int \left (\frac{155}{18} (1-2 x)^{3/2}-\frac{25}{6} (1-2 x)^{5/2}+\frac{(1-2 x)^{3/2}}{9 (2+3 x)}\right ) \, dx\\ &=-\frac{31}{18} (1-2 x)^{5/2}+\frac{25}{42} (1-2 x)^{7/2}+\frac{1}{9} \int \frac{(1-2 x)^{3/2}}{2+3 x} \, dx\\ &=\frac{2}{81} (1-2 x)^{3/2}-\frac{31}{18} (1-2 x)^{5/2}+\frac{25}{42} (1-2 x)^{7/2}+\frac{7}{27} \int \frac{\sqrt{1-2 x}}{2+3 x} \, dx\\ &=\frac{14}{81} \sqrt{1-2 x}+\frac{2}{81} (1-2 x)^{3/2}-\frac{31}{18} (1-2 x)^{5/2}+\frac{25}{42} (1-2 x)^{7/2}+\frac{49}{81} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=\frac{14}{81} \sqrt{1-2 x}+\frac{2}{81} (1-2 x)^{3/2}-\frac{31}{18} (1-2 x)^{5/2}+\frac{25}{42} (1-2 x)^{7/2}-\frac{49}{81} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{14}{81} \sqrt{1-2 x}+\frac{2}{81} (1-2 x)^{3/2}-\frac{31}{18} (1-2 x)^{5/2}+\frac{25}{42} (1-2 x)^{7/2}-\frac{14}{81} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0317668, size = 56, normalized size = 0.68 \[ \frac{3 \sqrt{1-2 x} \left (-2700 x^3+144 x^2+1853 x-527\right )-98 \sqrt{21} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{1701} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[1 - 2*x]*(-527 + 1853*x + 144*x^2 - 2700*x^3) - 98*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/1701

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Maple [A]  time = 0.006, size = 56, normalized size = 0.7 \begin{align*}{\frac{2}{81} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{31}{18} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}+{\frac{25}{42} \left ( 1-2\,x \right ) ^{{\frac{7}{2}}}}-{\frac{14\,\sqrt{21}}{243}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+{\frac{14}{81}\sqrt{1-2\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/81*(1-2*x)^(3/2)-31/18*(1-2*x)^(5/2)+25/42*(1-2*x)^(7/2)-14/243*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)
+14/81*(1-2*x)^(1/2)

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Maxima [A]  time = 3.90031, size = 99, normalized size = 1.21 \begin{align*} \frac{25}{42} \,{\left (-2 \, x + 1\right )}^{\frac{7}{2}} - \frac{31}{18} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} + \frac{2}{81} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{7}{243} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) + \frac{14}{81} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/42*(-2*x + 1)^(7/2) - 31/18*(-2*x + 1)^(5/2) + 2/81*(-2*x + 1)^(3/2) + 7/243*sqrt(21)*log(-(sqrt(21) - 3*sq
rt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 14/81*sqrt(-2*x + 1)

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Fricas [A]  time = 1.38551, size = 189, normalized size = 2.3 \begin{align*} \frac{7}{243} \, \sqrt{7} \sqrt{3} \log \left (\frac{\sqrt{7} \sqrt{3} \sqrt{-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - \frac{1}{567} \,{\left (2700 \, x^{3} - 144 \, x^{2} - 1853 \, x + 527\right )} \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/243*sqrt(7)*sqrt(3)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 1/567*(2700*x^3 - 144*x^2 -
1853*x + 527)*sqrt(-2*x + 1)

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Sympy [A]  time = 25.283, size = 114, normalized size = 1.39 \begin{align*} \frac{25 \left (1 - 2 x\right )^{\frac{7}{2}}}{42} - \frac{31 \left (1 - 2 x\right )^{\frac{5}{2}}}{18} + \frac{2 \left (1 - 2 x\right )^{\frac{3}{2}}}{81} + \frac{14 \sqrt{1 - 2 x}}{81} + \frac{98 \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 )}{81} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*(1 - 2*x)**(7/2)/42 - 31*(1 - 2*x)**(5/2)/18 + 2*(1 - 2*x)**(3/2)/81 + 14*sqrt(1 - 2*x)/81 + 98*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))/81

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Giac [A]  time = 2.38228, size = 122, normalized size = 1.49 \begin{align*} -\frac{25}{42} \,{\left (2 \, x - 1\right )}^{3} \sqrt{-2 \, x + 1} - \frac{31}{18} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} + \frac{2}{81} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{7}{243} \, \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{14}{81} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/42*(2*x - 1)^3*sqrt(-2*x + 1) - 31/18*(2*x - 1)^2*sqrt(-2*x + 1) + 2/81*(-2*x + 1)^(3/2) + 7/243*sqrt(21)*
log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 14/81*sqrt(-2*x + 1)