3.2131 \(\int \frac{1}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\)

Optimal. Leaf size=83 \[ \frac{15}{1331 \sqrt{1-2 x}}-\frac{5}{242 \sqrt{1-2 x} (5 x+3)}-\frac{1}{22 \sqrt{1-2 x} (5 x+3)^2}-\frac{15 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{1331} \]

[Out]

15/(1331*Sqrt[1 - 2*x]) - 1/(22*Sqrt[1 - 2*x]*(3 + 5*x)^2) - 5/(242*Sqrt[1 - 2*x]*(3 + 5*x)) - (15*Sqrt[5/11]*
ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1331

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Rubi [A]  time = 0.0220152, antiderivative size = 90, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {51, 63, 206} \[ -\frac{75 \sqrt{1-2 x}}{2662 (5 x+3)}-\frac{25 \sqrt{1-2 x}}{242 (5 x+3)^2}+\frac{2}{11 \sqrt{1-2 x} (5 x+3)^2}-\frac{15 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{1331} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2/(11*Sqrt[1 - 2*x]*(3 + 5*x)^2) - (25*Sqrt[1 - 2*x])/(242*(3 + 5*x)^2) - (75*Sqrt[1 - 2*x])/(2662*(3 + 5*x))
- (15*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1331

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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}{(1-2 x)^{3/2} (3+5 x)^3} \, dx &=\frac{2}{11 \sqrt{1-2 x} (3+5 x)^2}+\frac{25}{11} \int \frac{1}{\sqrt{1-2 x} (3+5 x)^3} \, dx\\ &=\frac{2}{11 \sqrt{1-2 x} (3+5 x)^2}-\frac{25 \sqrt{1-2 x}}{242 (3+5 x)^2}+\frac{75}{242} \int \frac{1}{\sqrt{1-2 x} (3+5 x)^2} \, dx\\ &=\frac{2}{11 \sqrt{1-2 x} (3+5 x)^2}-\frac{25 \sqrt{1-2 x}}{242 (3+5 x)^2}-\frac{75 \sqrt{1-2 x}}{2662 (3+5 x)}+\frac{75 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{2662}\\ &=\frac{2}{11 \sqrt{1-2 x} (3+5 x)^2}-\frac{25 \sqrt{1-2 x}}{242 (3+5 x)^2}-\frac{75 \sqrt{1-2 x}}{2662 (3+5 x)}-\frac{75 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{2662}\\ &=\frac{2}{11 \sqrt{1-2 x} (3+5 x)^2}-\frac{25 \sqrt{1-2 x}}{242 (3+5 x)^2}-\frac{75 \sqrt{1-2 x}}{2662 (3+5 x)}-\frac{15 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{1331}\\ \end{align*}

Mathematica [C]  time = 0.0051925, size = 30, normalized size = 0.36 \[ \frac{8 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{5}{11} (1-2 x)\right )}{1331 \sqrt{1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Hypergeometric2F1[-1/2, 3, 1/2, (5*(1 - 2*x))/11])/(1331*Sqrt[1 - 2*x])

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Maple [A]  time = 0.01, size = 57, normalized size = 0.7 \begin{align*}{\frac{8}{1331}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{1000}{1331\, \left ( -10\,x-6 \right ) ^{2}} \left ({\frac{7}{40} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{99}{200}\sqrt{1-2\,x}} \right ) }-{\frac{15\,\sqrt{55}}{14641}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/1331/(1-2*x)^(1/2)+1000/1331*(7/40*(1-2*x)^(3/2)-99/200*(1-2*x)^(1/2))/(-10*x-6)^2-15/14641*arctanh(1/11*55^
(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.8369, size = 112, normalized size = 1.35 \begin{align*} \frac{15}{29282} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{375 \,{\left (2 \, x - 1\right )}^{2} + 2750 \, x - 407}{1331 \,{\left (25 \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - 110 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + 121 \, \sqrt{-2 \, x + 1}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/29282*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/1331*(375*(2*x - 1)^2
+ 2750*x - 407)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))

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Fricas [A]  time = 1.53028, size = 258, normalized size = 3.11 \begin{align*} \frac{15 \, \sqrt{11} \sqrt{5}{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 11 \,{\left (750 \, x^{2} + 625 \, x - 16\right )} \sqrt{-2 \, x + 1}}{29282 \,{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/29282*(15*sqrt(11)*sqrt(5)*(50*x^3 + 35*x^2 - 12*x - 9)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x
+ 3)) - 11*(750*x^2 + 625*x - 16)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)

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Sympy [A]  time = 3.88838, size = 231, normalized size = 2.78 \begin{align*} \begin{cases} - \frac{15 \sqrt{55} \operatorname{acosh}{\left (\frac{\sqrt{110}}{10 \sqrt{x + \frac{3}{5}}} \right )}}{14641} + \frac{15 \sqrt{2}}{2662 \sqrt{-1 + \frac{11}{10 \left (x + \frac{3}{5}\right )}} \sqrt{x + \frac{3}{5}}} - \frac{\sqrt{2}}{484 \sqrt{-1 + \frac{11}{10 \left (x + \frac{3}{5}\right )}} \left (x + \frac{3}{5}\right )^{\frac{3}{2}}} - \frac{\sqrt{2}}{1100 \sqrt{-1 + \frac{11}{10 \left (x + \frac{3}{5}\right )}} \left (x + \frac{3}{5}\right )^{\frac{5}{2}}} & \text{for}\: \frac{11}{10 \left |{x + \frac{3}{5}}\right |} > 1 \\\frac{15 \sqrt{55} i \operatorname{asin}{\left (\frac{\sqrt{110}}{10 \sqrt{x + \frac{3}{5}}} \right )}}{14641} - \frac{15 \sqrt{2} i}{2662 \sqrt{1 - \frac{11}{10 \left (x + \frac{3}{5}\right )}} \sqrt{x + \frac{3}{5}}} + \frac{\sqrt{2} i}{484 \sqrt{1 - \frac{11}{10 \left (x + \frac{3}{5}\right )}} \left (x + \frac{3}{5}\right )^{\frac{3}{2}}} + \frac{\sqrt{2} i}{1100 \sqrt{1 - \frac{11}{10 \left (x + \frac{3}{5}\right )}} \left (x + \frac{3}{5}\right )^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-15*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/14641 + 15*sqrt(2)/(2662*sqrt(-1 + 11/(10*(x + 3/5
)))*sqrt(x + 3/5)) - sqrt(2)/(484*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)) - sqrt(2)/(1100*sqrt(-1 + 11/
(10*(x + 3/5)))*(x + 3/5)**(5/2)), 11/(10*Abs(x + 3/5)) > 1), (15*sqrt(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5))
)/14641 - 15*sqrt(2)*I/(2662*sqrt(1 - 11/(10*(x + 3/5)))*sqrt(x + 3/5)) + sqrt(2)*I/(484*sqrt(1 - 11/(10*(x +
3/5)))*(x + 3/5)**(3/2)) + sqrt(2)*I/(1100*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(5/2)), True))

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Giac [A]  time = 2.09494, size = 104, normalized size = 1.25 \begin{align*} \frac{15}{29282} \, \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{8}{1331 \, \sqrt{-2 \, x + 1}} + \frac{5 \,{\left (35 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 99 \, \sqrt{-2 \, x + 1}\right )}}{5324 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/29282*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 8/1331/sqrt(-2
*x + 1) + 5/5324*(35*(-2*x + 1)^(3/2) - 99*sqrt(-2*x + 1))/(5*x + 3)^2