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

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

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Rubi [A]  time = 0.03, antiderivative size = 103, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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 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*}

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Mathematica [A]  time = 0.04, size = 93, normalized size = 0.90 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.14, size = 104, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {1-2 x} \left (\frac {50 (1-2 x)^2}{(5 x+3)^2}-\frac {1630 (1-2 x)}{5 x+3}-17127\right )}{33 \sqrt {5 x+3} \left (\frac {1-2 x}{5 x+3}+7\right )}-\frac {519 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/33*(Sqrt[1 - 2*x]*(-17127 + (50*(1 - 2*x)^2)/(3 + 5*x)^2 - (1630*(1 - 2*x))/(3 + 5*x)))/(Sqrt[3 + 5*x]*(7 +
(1 - 2*x)/(3 + 5*x))) - (519*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

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fricas [A]  time = 1.42, size = 101, normalized size = 0.98 \begin {gather*} -\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 {gather*}

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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giac [B]  time = 1.93, size = 318, normalized size = 3.09 \begin {gather*} -\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 {gather*}

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))

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maple [B]  time = 0.02, size = 202, normalized size = 1.96 \begin {gather*} \frac {\left (1284525 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+2397780 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+523950 \sqrt {-10 x^{2}-x +3}\, x^{2}+1490049 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+652120 \sqrt {-10 x^{2}-x +3}\, x +308286 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+202342 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{462 \left (3 x +2\right ) \sqrt {-10 x^{2}-x +3}\, \left (5 x +3\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.19, size = 121, normalized size = 1.17 \begin {gather*} \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 {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {1-2\,x}}{{\left (3\,x+2\right )}^2\,{\left (5\,x+3\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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