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

Optimal. Leaf size=94 \[ \frac {47 (1-2 x)^{5/2}}{294 (3 x+2)}-\frac {(1-2 x)^{5/2}}{126 (3 x+2)^2}+\frac {2873 (1-2 x)^{3/2}}{3969}+\frac {2873}{567} \sqrt {1-2 x}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}} \]

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Rubi [A]  time = 0.03, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {89, 78, 50, 63, 206} \begin {gather*} \frac {47 (1-2 x)^{5/2}}{294 (3 x+2)}-\frac {(1-2 x)^{5/2}}{126 (3 x+2)^2}+\frac {2873 (1-2 x)^{3/2}}{3969}+\frac {2873}{567} \sqrt {1-2 x}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2873*Sqrt[1 - 2*x])/567 + (2873*(1 - 2*x)^(3/2))/3969 - (1 - 2*x)^(5/2)/(126*(2 + 3*x)^2) + (47*(1 - 2*x)^(5/
2))/(294*(2 + 3*x)) - (2873*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(81*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 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 {(1-2 x)^{3/2} (3+5 x)^2}{(2+3 x)^3} \, dx &=-\frac {(1-2 x)^{5/2}}{126 (2+3 x)^2}+\frac {1}{126} \int \frac {(1-2 x)^{3/2} (559+1050 x)}{(2+3 x)^2} \, dx\\ &=-\frac {(1-2 x)^{5/2}}{126 (2+3 x)^2}+\frac {47 (1-2 x)^{5/2}}{294 (2+3 x)}+\frac {2873}{882} \int \frac {(1-2 x)^{3/2}}{2+3 x} \, dx\\ &=\frac {2873 (1-2 x)^{3/2}}{3969}-\frac {(1-2 x)^{5/2}}{126 (2+3 x)^2}+\frac {47 (1-2 x)^{5/2}}{294 (2+3 x)}+\frac {2873}{378} \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx\\ &=\frac {2873}{567} \sqrt {1-2 x}+\frac {2873 (1-2 x)^{3/2}}{3969}-\frac {(1-2 x)^{5/2}}{126 (2+3 x)^2}+\frac {47 (1-2 x)^{5/2}}{294 (2+3 x)}+\frac {2873}{162} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {2873}{567} \sqrt {1-2 x}+\frac {2873 (1-2 x)^{3/2}}{3969}-\frac {(1-2 x)^{5/2}}{126 (2+3 x)^2}+\frac {47 (1-2 x)^{5/2}}{294 (2+3 x)}-\frac {2873}{162} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {2873}{567} \sqrt {1-2 x}+\frac {2873 (1-2 x)^{3/2}}{3969}-\frac {(1-2 x)^{5/2}}{126 (2+3 x)^2}+\frac {47 (1-2 x)^{5/2}}{294 (2+3 x)}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 63, normalized size = 0.67 \begin {gather*} \frac {\sqrt {1-2 x} \left (-1800 x^3+5520 x^2+10195 x+3803\right )}{162 (3 x+2)^2}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(3803 + 10195*x + 5520*x^2 - 1800*x^3))/(162*(2 + 3*x)^2) - (2873*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*
x]])/(81*Sqrt[21])

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IntegrateAlgebraic [A]  time = 0.18, size = 79, normalized size = 0.84 \begin {gather*} \frac {\left (450 (1-2 x)^3+1410 (1-2 x)^2-14365 (1-2 x)+20111\right ) \sqrt {1-2 x}}{81 (3 (1-2 x)-7)^2}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((20111 - 14365*(1 - 2*x) + 1410*(1 - 2*x)^2 + 450*(1 - 2*x)^3)*Sqrt[1 - 2*x])/(81*(-7 + 3*(1 - 2*x))^2) - (28
73*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(81*Sqrt[21])

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fricas [A]  time = 1.71, size = 79, normalized size = 0.84 \begin {gather*} \frac {2873 \, \sqrt {21} {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (1800 \, x^{3} - 5520 \, x^{2} - 10195 \, x - 3803\right )} \sqrt {-2 \, x + 1}}{3402 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3402*(2873*sqrt(21)*(9*x^2 + 12*x + 4)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(1800*x^3 - 5
520*x^2 - 10195*x - 3803)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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giac [A]  time = 0.88, size = 86, normalized size = 0.91 \begin {gather*} \frac {50}{81} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {2873}{3402} \, \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 {130}{27} \, \sqrt {-2 \, x + 1} - \frac {435 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1001 \, \sqrt {-2 \, x + 1}}{324 \, {\left (3 \, x + 2\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50/81*(-2*x + 1)^(3/2) + 2873/3402*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*
x + 1))) + 130/27*sqrt(-2*x + 1) - 1/324*(435*(-2*x + 1)^(3/2) - 1001*sqrt(-2*x + 1))/(3*x + 2)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.70 \begin {gather*} -\frac {2873 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{1701}+\frac {50 \left (-2 x +1\right )^{\frac {3}{2}}}{81}+\frac {130 \sqrt {-2 x +1}}{27}+\frac {-\frac {145 \left (-2 x +1\right )^{\frac {3}{2}}}{27}+\frac {1001 \sqrt {-2 x +1}}{81}}{\left (-6 x -4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50/81*(-2*x+1)^(3/2)+130/27*(-2*x+1)^(1/2)+2/3*(-145/18*(-2*x+1)^(3/2)+1001/54*(-2*x+1)^(1/2))/(-6*x-4)^2-2873
/1701*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.21, size = 92, normalized size = 0.98 \begin {gather*} \frac {50}{81} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {2873}{3402} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {130}{27} \, \sqrt {-2 \, x + 1} - \frac {435 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1001 \, \sqrt {-2 \, x + 1}}{81 \, {\left (9 \, {\left (2 \, x - 1\right )}^{2} + 84 \, x + 7\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50/81*(-2*x + 1)^(3/2) + 2873/3402*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1)))
+ 130/27*sqrt(-2*x + 1) - 1/81*(435*(-2*x + 1)^(3/2) - 1001*sqrt(-2*x + 1))/(9*(2*x - 1)^2 + 84*x + 7)

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mupad [B]  time = 0.06, size = 73, normalized size = 0.78 \begin {gather*} \frac {130\,\sqrt {1-2\,x}}{27}+\frac {50\,{\left (1-2\,x\right )}^{3/2}}{81}+\frac {\frac {1001\,\sqrt {1-2\,x}}{729}-\frac {145\,{\left (1-2\,x\right )}^{3/2}}{243}}{\frac {28\,x}{3}+{\left (2\,x-1\right )}^2+\frac {7}{9}}+\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,2873{}\mathrm {i}}{1701} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*2873i)/1701 + (130*(1 - 2*x)^(1/2))/27 + (50*(1 - 2*x)^(3/2))/
81 + ((1001*(1 - 2*x)^(1/2))/729 - (145*(1 - 2*x)^(3/2))/243)/((28*x)/3 + (2*x - 1)^2 + 7/9)

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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/(2+3*x)**3,x)

[Out]

Timed out

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