3.25.62 \(\int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx\) [2462]

Optimal. Leaf size=50 \[ -\frac {1}{2} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2 \sqrt {10}} \]

[Out]

11/20*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-1/2*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {52, 56, 222} \begin {gather*} \frac {11 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[3 + 5*x]/Sqrt[1 - 2*x],x]

[Out]

-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 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 {3+5 x}}{\sqrt {1-2 x}} \, dx &=-\frac {1}{2} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {1}{2} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{2 \sqrt {5}}\\ &=-\frac {1}{2} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 60, normalized size = 1.20 \begin {gather*} -\frac {1}{2} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {11 \tan ^{-1}\left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{\sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) - (11*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/Sqrt[10]

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Maple [A]
time = 0.07, size = 56, normalized size = 1.12

method	result	size
default	\(-\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}}{2}+\frac {11 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{40 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(56\)
risch	\(\frac {\sqrt {3+5 x}\, \left (-1+2 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{2 \sqrt {-\left (3+5 x \right ) \left (-1+2 x \right )}\, \sqrt {1-2 x}}+\frac {11 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{40 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(1-2*x)^(1/2)*(3+5*x)^(1/2)+11/40*10^(1/2)*arcsin(20/11*x+1/11)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)/(3+
5*x)^(1/2)

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Maxima [A]
time = 0.51, size = 29, normalized size = 0.58 \begin {gather*} \frac {11}{40} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {1}{2} \, \sqrt {-10 \, x^{2} - x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/40*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 1/2*sqrt(-10*x^2 - x + 3)

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Fricas [A]
time = 0.43, size = 57, normalized size = 1.14 \begin {gather*} -\frac {11}{40} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - \frac {1}{2} \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/40*sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 1/2*sqrt(5*x
+ 3)*sqrt(-2*x + 1)

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Sympy [C] Result contains complex when optimal does not.
time = 1.09, size = 139, normalized size = 2.78 \begin {gather*} \begin {cases} - \frac {5 i \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{\sqrt {10 x - 5}} + \frac {11 i \sqrt {x + \frac {3}{5}}}{2 \sqrt {10 x - 5}} - \frac {11 \sqrt {10} i \operatorname {acosh}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{20} & \text {for}\: \left |{x + \frac {3}{5}}\right | > \frac {11}{10} \\\frac {11 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{20} + \frac {5 \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{\sqrt {5 - 10 x}} - \frac {11 \sqrt {x + \frac {3}{5}}}{2 \sqrt {5 - 10 x}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-5*I*(x + 3/5)**(3/2)/sqrt(10*x - 5) + 11*I*sqrt(x + 3/5)/(2*sqrt(10*x - 5)) - 11*sqrt(10)*I*acosh(
sqrt(110)*sqrt(x + 3/5)/11)/20, Abs(x + 3/5) > 11/10), (11*sqrt(10)*asin(sqrt(110)*sqrt(x + 3/5)/11)/20 + 5*(x
+ 3/5)**(3/2)/sqrt(5 - 10*x) - 11*sqrt(x + 3/5)/(2*sqrt(5 - 10*x)), True))

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Giac [A]
time = 1.37, size = 40, normalized size = 0.80 \begin {gather*} \frac {1}{20} \, \sqrt {5} {\left (11 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - 2 \, \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*sqrt(5)*(11*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 2*sqrt(5*x + 3)*sqrt(-10*x + 5))

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Mupad [B]
time = 3.67, size = 186, normalized size = 3.72 \begin {gather*} \frac {11\,\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,\left (\sqrt {1-2\,x}-1\right )}{2\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}\right )}{10}-\frac {\frac {{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {2\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {8\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}}{\frac {4\,{\left (\sqrt {1-2\,x}-1\right )}^2}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {4}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*10^(1/2)*atan((10^(1/2)*((1 - 2*x)^(1/2) - 1))/(2*(3^(1/2) - (5*x + 3)^(1/2)))))/10 - (((1 - 2*x)^(1/2) -
1)^3/(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (2*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))) + (8*3^(1/
2)*((1 - 2*x)^(1/2) - 1)^2)/(5*(3^(1/2) - (5*x + 3)^(1/2))^2))/((4*((1 - 2*x)^(1/2) - 1)^2)/(5*(3^(1/2) - (5*x
+ 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + 4/25)

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