\(\int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx\) [1557]

Optimal result

Integrand size = 19, antiderivative size = 26 \[ \int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx=\sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{21}} \sqrt {3+5 x}\right ) \]

[Out]

1/5*arcsin(1/21*42^(1/2)*(3+5*x)^(1/2))*10^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {56, 222} \[ \int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx=\sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{21}} \sqrt {5 x+3}\right ) \]

[In]

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

[Out]

Sqrt[2/5]*ArcSin[Sqrt[2/21]*Sqrt[3 + 5*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*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {21-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{\sqrt {5}} \\ & = \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{21}} \sqrt {3+5 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx=\sqrt {\frac {2}{5}} \arctan \left (\frac {\sqrt {\frac {6}{5}+2 x}}{\sqrt {3-2 x}}\right ) \]

[In]

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

[Out]

Sqrt[2/5]*ArcTan[Sqrt[6/5 + 2*x]/Sqrt[3 - 2*x]]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(38\) vs. \(2(18)=36\).

Time = 0.55 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {\sqrt {\left (3-2 x \right ) \left (3+5 x \right )}\, \sqrt {10}\, \arcsin \left (\frac {20 x}{21}-\frac {3}{7}\right )}{10 \sqrt {3-2 x}\, \sqrt {3+5 x}}\)	\(39\)

[In]

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

[Out]

1/10*((3-2*x)*(3+5*x))^(1/2)/(3-2*x)^(1/2)/(3+5*x)^(1/2)*10^(1/2)*arcsin(20/21*x-3/7)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (18) = 36\).

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx=-\frac {1}{5} \, \sqrt {5} \sqrt {2} \arctan \left (\frac {\sqrt {5} \sqrt {2} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 3} - 3 \, \sqrt {5} \sqrt {2}}{10 \, x}\right ) \]

[In]

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

[Out]

-1/5*sqrt(5)*sqrt(2)*arctan(1/10*(sqrt(5)*sqrt(2)*sqrt(5*x + 3)*sqrt(-2*x + 3) - 3*sqrt(5)*sqrt(2))/x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.57 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx=\begin {cases} - \frac {\sqrt {10} i \operatorname {acosh}{\left (\frac {\sqrt {210} \sqrt {x + \frac {3}{5}}}{21} \right )}}{5} & \text {for}\: \left |{x + \frac {3}{5}}\right | > \frac {21}{10} \\\frac {\sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {210} \sqrt {x + \frac {3}{5}}}{21} \right )}}{5} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-sqrt(10)*I*acosh(sqrt(210)*sqrt(x + 3/5)/21)/5, Abs(x + 3/5) > 21/10), (sqrt(10)*asin(sqrt(210)*sq
rt(x + 3/5)/21)/5, True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx=-\frac {1}{10} \, \sqrt {10} \arcsin \left (-\frac {20}{21} \, x + \frac {3}{7}\right ) \]

[In]

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

[Out]

-1/10*sqrt(10)*arcsin(-20/21*x + 3/7)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx=\frac {1}{5} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {1}{21} \, \sqrt {42} \sqrt {5 \, x + 3}\right ) \]

[In]

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

[Out]

1/5*sqrt(5)*sqrt(2)*arcsin(1/21*sqrt(42)*sqrt(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {1}{\sqrt {3-2 x} \sqrt {3+5 x}} \, dx=-\frac {2\,\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,\left (\sqrt {3}-\sqrt {3-2\,x}\right )}{2\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}\right )}{5} \]

[In]

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

[Out]

-(2*10^(1/2)*atan((10^(1/2)*(3^(1/2) - (3 - 2*x)^(1/2)))/(2*(3^(1/2) - (5*x + 3)^(1/2)))))/5