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

Optimal. Leaf size=102 \[ -\frac {68 \sqrt {1-2 x}}{3 (5 x+3)}+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)}-134 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+138 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.04, 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 = {98, 151, 156, 63, 206} \begin {gather*} -\frac {68 \sqrt {1-2 x}}{3 (5 x+3)}+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)}-134 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+138 \sqrt {\frac {11}{5}} \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)^2*(3 + 5*x)^2),x]

[Out]

(-68*Sqrt[1 - 2*x])/(3*(3 + 5*x)) + (7*Sqrt[1 - 2*x])/(3*(2 + 3*x)*(3 + 5*x)) - 134*Sqrt[7/3]*ArcTanh[Sqrt[3/7
]*Sqrt[1 - 2*x]] + 138*Sqrt[11/5]*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)^2 (3+5 x)^2} \, dx &=\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)}+\frac {1}{3} \int \frac {89-101 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^2} \, dx\\ &=-\frac {68 \sqrt {1-2 x}}{3 (3+5 x)}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)}-\frac {1}{33} \int \frac {3663-2244 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac {68 \sqrt {1-2 x}}{3 (3+5 x)}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)}+469 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx-759 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {68 \sqrt {1-2 x}}{3 (3+5 x)}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)}-469 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )+759 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {68 \sqrt {1-2 x}}{3 (3+5 x)}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)}-134 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+138 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 105, normalized size = 1.03 \begin {gather*} \frac {-670 \sqrt {21} \left (15 x^2+19 x+6\right ) \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+414 \sqrt {55} \left (15 x^2+19 x+6\right ) \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-15 \sqrt {1-2 x} (68 x+43)}{15 (3 x+2) (5 x+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*Sqrt[1 - 2*x]*(43 + 68*x) - 670*Sqrt[21]*(6 + 19*x + 15*x^2)*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + 414*Sqrt[
55]*(6 + 19*x + 15*x^2)*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15*(2 + 3*x)*(3 + 5*x))

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IntegrateAlgebraic [A]  time = 0.22, size = 95, normalized size = 0.93 \begin {gather*} \frac {4 \sqrt {1-2 x} (34 (1-2 x)-77)}{15 (1-2 x)^2-68 (1-2 x)+77}-134 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+138 \sqrt {\frac {11}{5}} \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)^2*(3 + 5*x)^2),x]

[Out]

(4*(-77 + 34*(1 - 2*x))*Sqrt[1 - 2*x])/(77 - 68*(1 - 2*x) + 15*(1 - 2*x)^2) - 134*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*
Sqrt[1 - 2*x]] + 138*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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fricas [A]  time = 1.53, size = 122, normalized size = 1.20 \begin {gather*} \frac {207 \, \sqrt {11} \sqrt {5} {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 335 \, \sqrt {7} \sqrt {3} {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 15 \, {\left (68 \, x + 43\right )} \sqrt {-2 \, x + 1}}{15 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(207*sqrt(11)*sqrt(5)*(15*x^2 + 19*x + 6)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 3
35*sqrt(7)*sqrt(3)*(15*x^2 + 19*x + 6)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 15*(68*x +
43)*sqrt(-2*x + 1))/(15*x^2 + 19*x + 6)

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giac [A]  time = 1.06, size = 116, normalized size = 1.14 \begin {gather*} -\frac {69}{5} \, \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 {67}{3} \, \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 {4 \, {\left (34 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 77 \, \sqrt {-2 \, x + 1}\right )}}{15 \, {\left (2 \, x - 1\right )}^{2} + 136 \, x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-69/5*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 67/3*sqrt(21)*log
(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4*(34*(-2*x + 1)^(3/2) - 77*sqrt(-2*
x + 1))/(15*(2*x - 1)^2 + 136*x + 9)

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maple [A]  time = 0.01, size = 70, normalized size = 0.69 \begin {gather*} -\frac {134 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{3}+\frac {138 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{5}+\frac {22 \sqrt {-2 x +1}}{5 \left (-2 x -\frac {6}{5}\right )}+\frac {14 \sqrt {-2 x +1}}{3 \left (-2 x -\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

22/5*(-2*x+1)^(1/2)/(-2*x-6/5)+138/5*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)+14/3*(-2*x+1)^(1/2)/(-2*x-
4/3)-134/3*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.14, size = 110, normalized size = 1.08 \begin {gather*} -\frac {69}{5} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {67}{3} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {4 \, {\left (34 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 77 \, \sqrt {-2 \, x + 1}\right )}}{15 \, {\left (2 \, x - 1\right )}^{2} + 136 \, x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-69/5*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 67/3*sqrt(21)*log(-(sqrt(21
) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4*(34*(-2*x + 1)^(3/2) - 77*sqrt(-2*x + 1))/(15*(2*x -
1)^2 + 136*x + 9)

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mupad [B]  time = 1.21, size = 72, normalized size = 0.71 \begin {gather*} \frac {138\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{5}-\frac {134\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3}-\frac {\frac {308\,\sqrt {1-2\,x}}{15}-\frac {136\,{\left (1-2\,x\right )}^{3/2}}{15}}{\frac {136\,x}{15}+{\left (2\,x-1\right )}^2+\frac {3}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(138*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/5 - (134*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/3 -
((308*(1 - 2*x)^(1/2))/15 - (136*(1 - 2*x)^(3/2))/15)/((136*x)/15 + (2*x - 1)^2 + 3/5)

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

[Out]

Timed out

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