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

Optimal. Leaf size=102 \[ -\frac {(1-2 x)^{7/2}}{63 (3 x+2)}-\frac {25}{63} (1-2 x)^{7/2}-\frac {10}{63} (1-2 x)^{5/2}-\frac {50}{81} (1-2 x)^{3/2}-\frac {350}{81} \sqrt {1-2 x}+\frac {350}{81} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \begin {gather*} -\frac {(1-2 x)^{7/2}}{63 (3 x+2)}-\frac {25}{63} (1-2 x)^{7/2}-\frac {10}{63} (1-2 x)^{5/2}-\frac {50}{81} (1-2 x)^{3/2}-\frac {350}{81} \sqrt {1-2 x}+\frac {350}{81} \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)^(5/2)*(3 + 5*x)^2)/(2 + 3*x)^2,x]

[Out]

(-350*Sqrt[1 - 2*x])/81 - (50*(1 - 2*x)^(3/2))/81 - (10*(1 - 2*x)^(5/2))/63 - (25*(1 - 2*x)^(7/2))/63 - (1 - 2
*x)^(7/2)/(63*(2 + 3*x)) + (350*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/81

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 {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^2} \, dx &=-\frac {(1-2 x)^{7/2}}{63 (2+3 x)}+\frac {1}{63} \int \frac {(1-2 x)^{5/2} (275+525 x)}{2+3 x} \, dx\\ &=-\frac {25}{63} (1-2 x)^{7/2}-\frac {(1-2 x)^{7/2}}{63 (2+3 x)}-\frac {25}{21} \int \frac {(1-2 x)^{5/2}}{2+3 x} \, dx\\ &=-\frac {10}{63} (1-2 x)^{5/2}-\frac {25}{63} (1-2 x)^{7/2}-\frac {(1-2 x)^{7/2}}{63 (2+3 x)}-\frac {25}{9} \int \frac {(1-2 x)^{3/2}}{2+3 x} \, dx\\ &=-\frac {50}{81} (1-2 x)^{3/2}-\frac {10}{63} (1-2 x)^{5/2}-\frac {25}{63} (1-2 x)^{7/2}-\frac {(1-2 x)^{7/2}}{63 (2+3 x)}-\frac {175}{27} \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx\\ &=-\frac {350}{81} \sqrt {1-2 x}-\frac {50}{81} (1-2 x)^{3/2}-\frac {10}{63} (1-2 x)^{5/2}-\frac {25}{63} (1-2 x)^{7/2}-\frac {(1-2 x)^{7/2}}{63 (2+3 x)}-\frac {1225}{81} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {350}{81} \sqrt {1-2 x}-\frac {50}{81} (1-2 x)^{3/2}-\frac {10}{63} (1-2 x)^{5/2}-\frac {25}{63} (1-2 x)^{7/2}-\frac {(1-2 x)^{7/2}}{63 (2+3 x)}+\frac {1225}{81} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {350}{81} \sqrt {1-2 x}-\frac {50}{81} (1-2 x)^{3/2}-\frac {10}{63} (1-2 x)^{5/2}-\frac {25}{63} (1-2 x)^{7/2}-\frac {(1-2 x)^{7/2}}{63 (2+3 x)}+\frac {350}{81} \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 = 0.06, size = 70, normalized size = 0.69 \begin {gather*} \frac {\sqrt {1-2 x} \left (5400 x^4-5508 x^3+1002 x^2-4471 x-6239\right )}{567 (3 x+2)}+\frac {350}{81} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(-6239 - 4471*x + 1002*x^2 - 5508*x^3 + 5400*x^4))/(567*(2 + 3*x)) + (350*Sqrt[7/3]*ArcTanh[Sqr
t[3/7]*Sqrt[1 - 2*x]])/81

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IntegrateAlgebraic [A]  time = 0.12, size = 90, normalized size = 0.88 \begin {gather*} \frac {350}{81} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {\left (675 (1-2 x)^4-1323 (1-2 x)^3+420 (1-2 x)^2+4900 (1-2 x)-17150\right ) \sqrt {1-2 x}}{567 (3 (1-2 x)-7)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/567*((-17150 + 4900*(1 - 2*x) + 420*(1 - 2*x)^2 - 1323*(1 - 2*x)^3 + 675*(1 - 2*x)^4)*Sqrt[1 - 2*x])/(-7 +
3*(1 - 2*x)) + (350*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/81

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fricas [A]  time = 1.42, size = 81, normalized size = 0.79 \begin {gather*} \frac {1225 \, \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 (5400 \, x^{4} - 5508 \, x^{3} + 1002 \, x^{2} - 4471 \, x - 6239\right )} \sqrt {-2 \, x + 1}}{1701 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1701*(1225*sqrt(7)*sqrt(3)*(3*x + 2)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 3*(5400*x^
4 - 5508*x^3 + 1002*x^2 - 4471*x - 6239)*sqrt(-2*x + 1))/(3*x + 2)

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giac [A]  time = 0.95, size = 106, normalized size = 1.04 \begin {gather*} \frac {25}{63} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {4}{27} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {16}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {175}{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 {1036}{243} \, \sqrt {-2 \, x + 1} - \frac {49 \, \sqrt {-2 \, x + 1}}{243 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/63*(2*x - 1)^3*sqrt(-2*x + 1) - 4/27*(2*x - 1)^2*sqrt(-2*x + 1) - 16/27*(-2*x + 1)^(3/2) - 175/243*sqrt(21)
*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1036/243*sqrt(-2*x + 1) - 49/243
*sqrt(-2*x + 1)/(3*x + 2)

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maple [A]  time = 0.01, size = 72, normalized size = 0.71 \begin {gather*} \frac {350 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{243}-\frac {25 \left (-2 x +1\right )^{\frac {7}{2}}}{63}-\frac {4 \left (-2 x +1\right )^{\frac {5}{2}}}{27}-\frac {16 \left (-2 x +1\right )^{\frac {3}{2}}}{27}-\frac {1036 \sqrt {-2 x +1}}{243}+\frac {98 \sqrt {-2 x +1}}{729 \left (-2 x -\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/63*(-2*x+1)^(7/2)-4/27*(-2*x+1)^(5/2)-16/27*(-2*x+1)^(3/2)-1036/243*(-2*x+1)^(1/2)+98/729*(-2*x+1)^(1/2)/(
-2*x-4/3)+350/243*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.15, size = 89, normalized size = 0.87 \begin {gather*} -\frac {25}{63} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {4}{27} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {16}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {175}{243} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {1036}{243} \, \sqrt {-2 \, x + 1} - \frac {49 \, \sqrt {-2 \, x + 1}}{243 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/63*(-2*x + 1)^(7/2) - 4/27*(-2*x + 1)^(5/2) - 16/27*(-2*x + 1)^(3/2) - 175/243*sqrt(21)*log(-(sqrt(21) - 3
*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1036/243*sqrt(-2*x + 1) - 49/243*sqrt(-2*x + 1)/(3*x + 2)

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mupad [B]  time = 1.18, size = 73, normalized size = 0.72 \begin {gather*} -\frac {98\,\sqrt {1-2\,x}}{729\,\left (2\,x+\frac {4}{3}\right )}-\frac {1036\,\sqrt {1-2\,x}}{243}-\frac {16\,{\left (1-2\,x\right )}^{3/2}}{27}-\frac {4\,{\left (1-2\,x\right )}^{5/2}}{27}-\frac {25\,{\left (1-2\,x\right )}^{7/2}}{63}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,350{}\mathrm {i}}{243} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*350i)/243 - (98*(1 - 2*x)^(1/2))/(729*(2*x + 4/3)) - (1036*(
1 - 2*x)^(1/2))/243 - (16*(1 - 2*x)^(3/2))/27 - (4*(1 - 2*x)^(5/2))/27 - (25*(1 - 2*x)^(7/2))/63

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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)**(5/2)*(3+5*x)**2/(2+3*x)**2,x)

[Out]

Timed out

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