3.2324 \(\int \frac{\sqrt{1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=103 \[ \frac{2495 \sqrt{1-2 x}}{33 \sqrt{5 x+3}}-\frac{25 \sqrt{1-2 x}}{3 (5 x+3)^{3/2}}+\frac{\sqrt{1-2 x}}{(3 x+2) (5 x+3)^{3/2}}-\frac{519 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{\sqrt{7}} \]

[Out]

(-25*Sqrt[1 - 2*x])/(3*(3 + 5*x)^(3/2)) + Sqrt[1 - 2*x]/((2 + 3*x)*(3 + 5*x)^(3/2)) + (2495*Sqrt[1 - 2*x])/(33
*Sqrt[3 + 5*x]) - (519*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

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Rubi [A]  time = 0.0296518, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {99, 152, 12, 93, 204} \[ \frac{2495 \sqrt{1-2 x}}{33 \sqrt{5 x+3}}-\frac{25 \sqrt{1-2 x}}{3 (5 x+3)^{3/2}}+\frac{\sqrt{1-2 x}}{(3 x+2) (5 x+3)^{3/2}}-\frac{519 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{\sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*Sqrt[1 - 2*x])/(3*(3 + 5*x)^(3/2)) + Sqrt[1 - 2*x]/((2 + 3*x)*(3 + 5*x)^(3/2)) + (2495*Sqrt[1 - 2*x])/(33
*Sqrt[3 + 5*x]) - (519*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 152

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] && IntegersQ[2*m, 2*n, 2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx &=\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)^{3/2}}-\int \frac{-\frac{31}{2}+20 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx\\ &=-\frac{25 \sqrt{1-2 x}}{3 (3+5 x)^{3/2}}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)^{3/2}}+\frac{2}{33} \int \frac{-\frac{3509}{4}+825 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac{25 \sqrt{1-2 x}}{3 (3+5 x)^{3/2}}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)^{3/2}}+\frac{2495 \sqrt{1-2 x}}{33 \sqrt{3+5 x}}-\frac{4}{363} \int -\frac{188397}{8 \sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{25 \sqrt{1-2 x}}{3 (3+5 x)^{3/2}}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)^{3/2}}+\frac{2495 \sqrt{1-2 x}}{33 \sqrt{3+5 x}}+\frac{519}{2} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{25 \sqrt{1-2 x}}{3 (3+5 x)^{3/2}}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)^{3/2}}+\frac{2495 \sqrt{1-2 x}}{33 \sqrt{3+5 x}}+519 \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )\\ &=-\frac{25 \sqrt{1-2 x}}{3 (3+5 x)^{3/2}}+\frac{\sqrt{1-2 x}}{(2+3 x) (3+5 x)^{3/2}}+\frac{2495 \sqrt{1-2 x}}{33 \sqrt{3+5 x}}-\frac{519 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{\sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0409544, size = 93, normalized size = 0.9 \[ \frac{7 \sqrt{1-2 x} \left (37425 x^2+46580 x+14453\right )-17127 \sqrt{7} \sqrt{5 x+3} \left (15 x^2+19 x+6\right ) \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{231 (3 x+2) (5 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[1 - 2*x]*(14453 + 46580*x + 37425*x^2) - 17127*Sqrt[7]*Sqrt[3 + 5*x]*(6 + 19*x + 15*x^2)*ArcTan[Sqrt[1
- 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(231*(2 + 3*x)*(3 + 5*x)^(3/2))

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Maple [B]  time = 0.013, size = 202, normalized size = 2. \begin{align*}{\frac{1}{924+1386\,x} \left ( 1284525\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{3}+2397780\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}+1490049\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+523950\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+308286\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +652120\,x\sqrt{-10\,{x}^{2}-x+3}+202342\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}} \left ( 3+5\,x \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/462*(1284525*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^3+2397780*7^(1/2)*arctan(1/14*(37*
x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+1490049*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+52
3950*x^2*(-10*x^2-x+3)^(1/2)+308286*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+652120*x*(-10*x
^2-x+3)^(1/2)+202342*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(2+3*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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Maxima [A]  time = 1.59784, size = 163, normalized size = 1.58 \begin{align*} \frac{519}{14} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) - \frac{4990 \, x}{33 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{2605}{33 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{38 \, x}{{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} + \frac{49}{9 \,{\left (3 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}} x + 2 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}\right )}} - \frac{185}{9 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

519/14*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 4990/33*x/sqrt(-10*x^2 - x + 3) + 2605/33/s
qrt(-10*x^2 - x + 3) + 38*x/(-10*x^2 - x + 3)^(3/2) + 49/9/(3*(-10*x^2 - x + 3)^(3/2)*x + 2*(-10*x^2 - x + 3)^
(3/2)) - 185/9/(-10*x^2 - x + 3)^(3/2)

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Fricas [A]  time = 1.51992, size = 309, normalized size = 3. \begin{align*} -\frac{17127 \, \sqrt{7}{\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \arctan \left (\frac{\sqrt{7}{\left (37 \, x + 20\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{14 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \,{\left (37425 \, x^{2} + 46580 \, x + 14453\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{462 \,{\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/462*(17127*sqrt(7)*(75*x^3 + 140*x^2 + 87*x + 18)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x +
1)/(10*x^2 + x - 3)) - 14*(37425*x^2 + 46580*x + 14453)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(75*x^3 + 140*x^2 + 87*
x + 18)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 3.0895, size = 429, normalized size = 4.17 \begin{align*} -\frac{1}{18480} \, \sqrt{5}{\left (35 \, \sqrt{2}{\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}^{3} - 68508 \, \sqrt{70} \sqrt{2}{\left (\pi + 2 \, \arctan \left (-\frac{\sqrt{70} \sqrt{5 \, x + 3}{\left (\frac{{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}\right )\right )} - 55440 \, \sqrt{2}{\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )} - \frac{3659040 \, \sqrt{2}{\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}}{{\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}^{2} + 280}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18480*sqrt(5)*(35*sqrt(2)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sq
rt(-10*x + 5) - sqrt(22)))^3 - 68508*sqrt(70)*sqrt(2)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*s
qrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 55440*sqrt(2)*((sqrt(2)*s
qrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))) - 3659040*sqr
t(2)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)
))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
)^2 + 280))