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

Optimal. Leaf size=75 \[ -\frac {10 (1-2 x)^{3/2}}{33 (5 x+3)^{3/2}}+\frac {6 \sqrt {1-2 x}}{\sqrt {5 x+3}}-6 \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {96, 94, 93, 204} \begin {gather*} -\frac {10 (1-2 x)^{3/2}}{33 (5 x+3)^{3/2}}+\frac {6 \sqrt {1-2 x}}{\sqrt {5 x+3}}-6 \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*(1 - 2*x)^(3/2))/(33*(3 + 5*x)^(3/2)) + (6*Sqrt[1 - 2*x])/Sqrt[3 + 5*x] - 6*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/
(Sqrt[7]*Sqrt[3 + 5*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 94

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[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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) (3+5 x)^{5/2}} \, dx &=-\frac {10 (1-2 x)^{3/2}}{33 (3+5 x)^{3/2}}-3 \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac {10 (1-2 x)^{3/2}}{33 (3+5 x)^{3/2}}+\frac {6 \sqrt {1-2 x}}{\sqrt {3+5 x}}+21 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {10 (1-2 x)^{3/2}}{33 (3+5 x)^{3/2}}+\frac {6 \sqrt {1-2 x}}{\sqrt {3+5 x}}+42 \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {10 (1-2 x)^{3/2}}{33 (3+5 x)^{3/2}}+\frac {6 \sqrt {1-2 x}}{\sqrt {3+5 x}}-6 \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 60, normalized size = 0.80 \begin {gather*} \frac {2 \sqrt {1-2 x} (505 x+292)}{33 (5 x+3)^{3/2}}-6 \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 - 2*x]*(292 + 505*x))/(33*(3 + 5*x)^(3/2)) - 6*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])]

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IntegrateAlgebraic [A]  time = 0.10, size = 78, normalized size = 1.04 \begin {gather*} -\frac {2}{33} \left (\frac {5 (1-2 x)^{3/2}}{(5 x+3)^{3/2}}-\frac {99 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-6 \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*((5*(1 - 2*x)^(3/2))/(3 + 5*x)^(3/2) - (99*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]))/33 - 6*Sqrt[7]*ArcTan[Sqrt[1 - 2
*x]/(Sqrt[7]*Sqrt[3 + 5*x])]

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fricas [A]  time = 1.33, size = 86, normalized size = 1.15 \begin {gather*} -\frac {99 \, \sqrt {7} {\left (25 \, x^{2} + 30 \, x + 9\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 ) - 2 \, {\left (505 \, x + 292\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{33 \, {\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)^(1/2)/(2+3*x)/(3+5*x)^(5/2),x, algorithm="fricas")

[Out]

-1/33*(99*sqrt(7)*(25*x^2 + 30*x + 9)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x
- 3)) - 2*(505*x + 292)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [B]  time = 1.50, size = 195, normalized size = 2.60 \begin {gather*} \frac {1}{2640} \, \sqrt {5} {\left (792 \, \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 )} - \sqrt {2} {\left ({\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} - \frac {792 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {3168 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2640*sqrt(5)*(792*sqrt(70)*sqrt(2)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) -
sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - sqrt(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)))^3 - 792*(sqrt(2)*sqrt(-10*x + 5) -
sqrt(22))/sqrt(5*x + 3) + 3168*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))))

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maple [B]  time = 0.01, size = 147, normalized size = 1.96 \begin {gather*} \frac {\left (2475 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+2970 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+1010 \sqrt {-10 x^{2}-x +3}\, x +891 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+584 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{33 \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)/(5*x+3)^(5/2),x)

[Out]

1/33*(2475*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+2970*7^(1/2)*x*arctan(1/14*(37*x+20)
*7^(1/2)/(-10*x^2-x+3)^(1/2))+891*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+1010*(-10*x^2-x+3
)^(1/2)*x+584*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)/(-10*x^2-x+3)^(1/2)/(5*x+3)^(3/2)

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maxima [A]  time = 1.26, size = 87, normalized size = 1.16 \begin {gather*} 3 \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {404 \, x}{33 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1054}{165 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {44 \, x}{15 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {22}{15 \, {\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)/(3+5*x)^(5/2),x, algorithm="maxima")

[Out]

3*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 404/33*x/sqrt(-10*x^2 - x + 3) + 1054/165/sqrt(-
10*x^2 - x + 3) + 44/15*x/(-10*x^2 - x + 3)^(3/2) - 22/15/(-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 )\,{\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)*(5*x + 3)^(5/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {1 - 2 x}}{\left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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