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

Optimal. Leaf size=93 \[ \frac {71 \sqrt {1-2 x}}{10 (5 x+3)}-\frac {11 \sqrt {1-2 x}}{10 (5 x+3)^2}+14 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2379 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}} \]

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Rubi [A]  time = 0.04, 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 {71 \sqrt {1-2 x}}{10 (5 x+3)}-\frac {11 \sqrt {1-2 x}}{10 (5 x+3)^2}+14 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2379 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*Sqrt[1 - 2*x])/(10*(3 + 5*x)^2) + (71*Sqrt[1 - 2*x])/(10*(3 + 5*x)) + 14*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[
1 - 2*x]] - (2379*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55])

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+5 x)^3} \, dx &=-\frac {11 \sqrt {1-2 x}}{10 (3+5 x)^2}-\frac {1}{10} \int \frac {92-107 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^2} \, dx\\ &=-\frac {11 \sqrt {1-2 x}}{10 (3+5 x)^2}+\frac {71 \sqrt {1-2 x}}{10 (3+5 x)}+\frac {1}{110} \int \frac {3828-2343 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac {11 \sqrt {1-2 x}}{10 (3+5 x)^2}+\frac {71 \sqrt {1-2 x}}{10 (3+5 x)}-147 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {2379}{10} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {11 \sqrt {1-2 x}}{10 (3+5 x)^2}+\frac {71 \sqrt {1-2 x}}{10 (3+5 x)}+147 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {2379}{10} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {11 \sqrt {1-2 x}}{10 (3+5 x)^2}+\frac {71 \sqrt {1-2 x}}{10 (3+5 x)}+14 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2379 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 78, normalized size = 0.84 \begin {gather*} \frac {\sqrt {1-2 x} (355 x+202)}{10 (5 x+3)^2}+14 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2379 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(202 + 355*x))/(10*(3 + 5*x)^2) + 14*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - (2379*ArcTanh[
Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55])

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IntegrateAlgebraic [A]  time = 0.27, size = 91, normalized size = 0.98 \begin {gather*} \frac {759 \sqrt {1-2 x}-355 (1-2 x)^{3/2}}{5 (5 (1-2 x)-11)^2}+14 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2379 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(759*Sqrt[1 - 2*x] - 355*(1 - 2*x)^(3/2))/(5*(-11 + 5*(1 - 2*x))^2) + 14*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2
*x]] - (2379*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55])

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fricas [A]  time = 1.32, size = 110, normalized size = 1.18 \begin {gather*} \frac {2379 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 3850 \, \sqrt {21} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 55 \, {\left (355 \, x + 202\right )} \sqrt {-2 \, x + 1}}{550 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/550*(2379*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 3850*sqrt(21)*(2
5*x^2 + 30*x + 9)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 55*(355*x + 202)*sqrt(-2*x + 1))/(25*x^
2 + 30*x + 9)

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giac [A]  time = 1.19, size = 107, normalized size = 1.15 \begin {gather*} \frac {2379}{550} \, \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 ) - 7 \, \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 {355 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 759 \, \sqrt {-2 \, x + 1}}{20 \, {\left (5 \, x + 3\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+5*x)^3,x, algorithm="giac")

[Out]

2379/550*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 7*sqrt(21)*log
(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/20*(355*(-2*x + 1)^(3/2) - 759*sqr
t(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.71 \begin {gather*} 14 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )-\frac {2379 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{275}+\frac {-71 \left (-2 x +1\right )^{\frac {3}{2}}+\frac {759 \sqrt {-2 x +1}}{5}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50*(-71/50*(-2*x+1)^(3/2)+759/250*(-2*x+1)^(1/2))/(-10*x-6)^2-2379/275*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*5
5^(1/2)+14*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.21, size = 110, normalized size = 1.18 \begin {gather*} \frac {2379}{550} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - 7 \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {355 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 759 \, \sqrt {-2 \, x + 1}}{5 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2379/550*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 7*sqrt(21)*log(-(sqrt(21
) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/5*(355*(-2*x + 1)^(3/2) - 759*sqrt(-2*x + 1))/(25*(2*
x - 1)^2 + 220*x + 11)

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mupad [B]  time = 1.21, size = 71, normalized size = 0.76 \begin {gather*} 14\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )-\frac {2379\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{275}+\frac {\frac {759\,\sqrt {1-2\,x}}{125}-\frac {71\,{\left (1-2\,x\right )}^{3/2}}{25}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

14*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7) - (2379*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/275 + (
(759*(1 - 2*x)^(1/2))/125 - (71*(1 - 2*x)^(3/2))/25)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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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+5*x)**3,x)

[Out]

Timed out

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