3.24.21 \(\int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx\) [2321]

Optimal. Leaf size=74 \[ -\frac {2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}-\frac {6 \sqrt {1-2 x}}{25 \sqrt {3+5 x}}-\frac {6}{25} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \]

[Out]

-2/165*(1-2*x)^(3/2)/(3+5*x)^(3/2)-6/125*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-6/25*(1-2*x)^(1/2)/(3+5*
x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {79, 49, 56, 222} \begin {gather*} -\frac {6}{25} \sqrt {\frac {2}{5}} \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{165 (5 x+3)^{3/2}}-\frac {6 \sqrt {1-2 x}}{25 \sqrt {5 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - 2*x)^(3/2))/(165*(3 + 5*x)^(3/2)) - (6*Sqrt[1 - 2*x])/(25*Sqrt[3 + 5*x]) - (6*Sqrt[2/5]*ArcSin[Sqrt[2
/11]*Sqrt[3 + 5*x]])/25

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

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 79

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] || L
tQ[p, n]))))

Rule 222

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 {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx &=-\frac {2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}+\frac {3}{5} \int \frac {\sqrt {1-2 x}}{(3+5 x)^{3/2}} \, dx\\ &=-\frac {2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}-\frac {6 \sqrt {1-2 x}}{25 \sqrt {3+5 x}}-\frac {6}{25} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}-\frac {6 \sqrt {1-2 x}}{25 \sqrt {3+5 x}}-\frac {12 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{25 \sqrt {5}}\\ &=-\frac {2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}-\frac {6 \sqrt {1-2 x}}{25 \sqrt {3+5 x}}-\frac {6}{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 = 0.10, size = 61, normalized size = 0.82 \begin {gather*} -\frac {2 \sqrt {1-2 x} (302+485 x)}{825 (3+5 x)^{3/2}}+\frac {6}{25} \sqrt {\frac {2}{5}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\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]*(302 + 485*x))/(825*(3 + 5*x)^(3/2)) + (6*Sqrt[2/5]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/2
5

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Maple [A]
time = 0.14, size = 96, normalized size = 1.30

method	result	size
default	\(-\frac {\left (2475 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+2970 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +891 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+4850 x \sqrt {-10 x^{2}-x +3}+3020 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{4125 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4125*(2475*10^(1/2)*arcsin(20/11*x+1/11)*x^2+2970*10^(1/2)*arcsin(20/11*x+1/11)*x+891*10^(1/2)*arcsin(20/11
*x+1/11)+4850*x*(-10*x^2-x+3)^(1/2)+3020*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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Maxima [A]
time = 0.55, size = 48, normalized size = 0.65 \begin {gather*} -\frac {4 \, \sqrt {-10 \, x^{2} - x + 3}}{15 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {8 \, \sqrt {-10 \, x^{2} - x + 3}}{165 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) + 8/165*sqrt(-10*x^2 - x + 3)/(5*x + 3)

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Fricas [A]
time = 0.80, size = 92, normalized size = 1.24 \begin {gather*} \frac {99 \, \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 ) - 10 \, {\left (485 \, x + 302\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4125 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4125*(99*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)) - 10*(485*x + 302)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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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((2+3*x)*(1-2*x)**(1/2)/(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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (51) = 102\).
time = 2.07, size = 139, normalized size = 1.88 \begin {gather*} -\frac {1}{66000} \, \sqrt {10} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{{\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {780 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}}\right )} - \frac {6}{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 {195 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{4125 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/66000*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 780*(sqrt(2)*sqrt(-10*x + 5) - sqr
t(22))/sqrt(5*x + 3)) - 6/125*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/4125*sqrt(10)*(5*x + 3)^(3/2)*(
195*(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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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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