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

Optimal. Leaf size=93 \[ \frac {65 \sqrt {1-2 x}}{6 (3 x+2)}+\frac {7 \sqrt {1-2 x}}{6 (3 x+2)^2}+\frac {2243 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{3 \sqrt {21}}-22 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 93, 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 = {98, 151, 156, 63, 206} \begin {gather*} \frac {65 \sqrt {1-2 x}}{6 (3 x+2)}+\frac {7 \sqrt {1-2 x}}{6 (3 x+2)^2}+\frac {2243 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{3 \sqrt {21}}-22 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[1 - 2*x])/(6*(2 + 3*x)^2) + (65*Sqrt[1 - 2*x])/(6*(2 + 3*x)) + (2243*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])
/(3*Sqrt[21]) - 22*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*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 98

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

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
 Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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}}{(2+3 x)^3 (3+5 x)} \, dx &=\frac {7 \sqrt {1-2 x}}{6 (2+3 x)^2}+\frac {1}{6} \int \frac {87-97 x}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)} \, dx\\ &=\frac {7 \sqrt {1-2 x}}{6 (2+3 x)^2}+\frac {65 \sqrt {1-2 x}}{6 (2+3 x)}+\frac {1}{42} \int \frac {3717-2275 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=\frac {7 \sqrt {1-2 x}}{6 (2+3 x)^2}+\frac {65 \sqrt {1-2 x}}{6 (2+3 x)}-\frac {2243}{6} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+605 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {7 \sqrt {1-2 x}}{6 (2+3 x)^2}+\frac {65 \sqrt {1-2 x}}{6 (2+3 x)}+\frac {2243}{6} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-605 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {7 \sqrt {1-2 x}}{6 (2+3 x)^2}+\frac {65 \sqrt {1-2 x}}{6 (2+3 x)}+\frac {2243 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{3 \sqrt {21}}-22 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 78, normalized size = 0.84 \begin {gather*} \frac {\sqrt {1-2 x} (195 x+137)}{6 (3 x+2)^2}+\frac {2243 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{3 \sqrt {21}}-22 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(137 + 195*x))/(6*(2 + 3*x)^2) + (2243*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(3*Sqrt[21]) - 22*Sqrt
[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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IntegrateAlgebraic [A]  time = 0.26, size = 91, normalized size = 0.98 \begin {gather*} \frac {469 \sqrt {1-2 x}-195 (1-2 x)^{3/2}}{3 (3 (1-2 x)-7)^2}+\frac {2243 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{3 \sqrt {21}}-22 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(469*Sqrt[1 - 2*x] - 195*(1 - 2*x)^(3/2))/(3*(-7 + 3*(1 - 2*x))^2) + (2243*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(
3*Sqrt[21]) - 22*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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fricas [A]  time = 1.37, size = 110, normalized size = 1.18 \begin {gather*} \frac {1386 \, \sqrt {55} {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 2243 \, \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 (195 \, x + 137\right )} \sqrt {-2 \, x + 1}}{126 \, {\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)/(2+3*x)^3/(3+5*x),x, algorithm="fricas")

[Out]

1/126*(1386*sqrt(55)*(9*x^2 + 12*x + 4)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 2243*sqrt(21)*(9*
x^2 + 12*x + 4)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 21*(195*x + 137)*sqrt(-2*x + 1))/(9*x^2 +
 12*x + 4)

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giac [A]  time = 1.06, size = 107, normalized size = 1.15 \begin {gather*} 11 \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {2243}{126} \, \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 {195 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 469 \, \sqrt {-2 \, x + 1}}{12 \, {\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)/(2+3*x)^3/(3+5*x),x, algorithm="giac")

[Out]

11*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2243/126*sqrt(21)*lo
g(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/12*(195*(-2*x + 1)^(3/2) - 469*sq
rt(-2*x + 1))/(3*x + 2)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.71 \begin {gather*} \frac {2243 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{63}-22 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )-\frac {18 \left (\frac {65 \left (-2 x +1\right )^{\frac {3}{2}}}{18}-\frac {469 \sqrt {-2 x +1}}{54}\right )}{\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)/(3*x+2)^3/(5*x+3),x)

[Out]

-22*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-18*(65/18*(-2*x+1)^(3/2)-469/54*(-2*x+1)^(1/2))/(-6*x-4)^2+
2243/63*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.19, size = 110, normalized size = 1.18 \begin {gather*} 11 \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2243}{126} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {195 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 469 \, \sqrt {-2 \, x + 1}}{3 \, {\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)/(2+3*x)^3/(3+5*x),x, algorithm="maxima")

[Out]

11*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2243/126*sqrt(21)*log(-(sqrt(2
1) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/3*(195*(-2*x + 1)^(3/2) - 469*sqrt(-2*x + 1))/(9*(2*
x - 1)^2 + 84*x + 7)

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mupad [B]  time = 0.09, size = 71, normalized size = 0.76 \begin {gather*} \frac {2243\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{63}-22\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )+\frac {\frac {469\,\sqrt {1-2\,x}}{27}-\frac {65\,{\left (1-2\,x\right )}^{3/2}}{9}}{\frac {28\,x}{3}+{\left (2\,x-1\right )}^2+\frac {7}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2243*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/63 - 22*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11) + ((
469*(1 - 2*x)^(1/2))/27 - (65*(1 - 2*x)^(3/2))/9)/((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)/(2+3*x)**3/(3+5*x),x)

[Out]

Timed out

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