3.2510 \(\int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=74 \[ -\frac {404 \sqrt {1-2 x}}{9075 \sqrt {5 x+3}}-\frac {2 \sqrt {1-2 x}}{825 (5 x+3)^{3/2}}+\frac {9}{25} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ) \]

[Out]

9/125*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/825*(1-2*x)^(1/2)/(3+5*x)^(3/2)-404/9075*(1-2*x)^(1/2)/(3
+5*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 74, 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 = {89, 78, 54, 216} \[ -\frac {404 \sqrt {1-2 x}}{9075 \sqrt {5 x+3}}-\frac {2 \sqrt {1-2 x}}{825 (5 x+3)^{3/2}}+\frac {9}{25} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - 2*x])/(825*(3 + 5*x)^(3/2)) - (404*Sqrt[1 - 2*x])/(9075*Sqrt[3 + 5*x]) + (9*Sqrt[2/5]*ArcSin[Sqrt
[2/11]*Sqrt[3 + 5*x]])/25

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 78

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin {align*} \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx &=-\frac {2 \sqrt {1-2 x}}{825 (3+5 x)^{3/2}}+\frac {2}{825} \int \frac {\frac {1093}{2}+\frac {1485 x}{2}}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx\\ &=-\frac {2 \sqrt {1-2 x}}{825 (3+5 x)^{3/2}}-\frac {404 \sqrt {1-2 x}}{9075 \sqrt {3+5 x}}+\frac {9}{25} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {2 \sqrt {1-2 x}}{825 (3+5 x)^{3/2}}-\frac {404 \sqrt {1-2 x}}{9075 \sqrt {3+5 x}}+\frac {18 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{25 \sqrt {5}}\\ &=-\frac {2 \sqrt {1-2 x}}{825 (3+5 x)^{3/2}}-\frac {404 \sqrt {1-2 x}}{9075 \sqrt {3+5 x}}+\frac {9}{25} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 64, normalized size = 0.86 \[ \frac {\sqrt {1-2 x} \left (-\frac {10 (1010 x+617)}{(5 x+3)^{3/2}}-\frac {3267 \sqrt {10} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{\sqrt {2 x-1}}\right )}{45375} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*((-10*(617 + 1010*x))/(3 + 5*x)^(3/2) - (3267*Sqrt[10]*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]])/Sqrt
[-1 + 2*x]))/45375

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fricas [A]  time = 0.93, size = 92, normalized size = 1.24 \[ -\frac {3267 \, \sqrt {5} \sqrt {2} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (1010 \, x + 617\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{90750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/90750*(3267*sqrt(5)*sqrt(2)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-
2*x + 1)/(10*x^2 + x - 3)) + 20*(1010*x + 617)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [B]  time = 1.32, size = 139, normalized size = 1.88 \[ -\frac {1}{726000} \, \sqrt {10} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{{\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {1620 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}}\right )} + \frac {9}{125} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {405 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{45375 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/726000*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 1620*(sqrt(2)*sqrt(-10*x + 5) - s
qrt(22))/sqrt(5*x + 3)) + 9/125*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/45375*sqrt(10)*(5*x + 3)^(3/2
)*(405*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3

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maple [A]  time = 0.01, size = 96, normalized size = 1.30 \[ \frac {\left (81675 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+98010 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-20200 \sqrt {-10 x^{2}-x +3}\, x +29403 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-12340 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{90750 \sqrt {-10 x^{2}-x +3}\, \left (5 x +3\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90750*(81675*10^(1/2)*x^2*arcsin(20/11*x+1/11)+98010*10^(1/2)*x*arcsin(20/11*x+1/11)+29403*10^(1/2)*arcsin(2
0/11*x+1/11)-20200*(-10*x^2-x+3)^(1/2)*x-12340*(-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 = 1.31, size = 62, normalized size = 0.84 \[ \frac {9}{250} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{825 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {404 \, \sqrt {-10 \, x^{2} - x + 3}}{9075 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/250*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 2/825*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) - 404/9075*sqrt
(-10*x^2 - x + 3)/(5*x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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