\(\int (3-x)^{3/2} (-2+x)^{3/2} \, dx\) [1160]

Optimal result

Integrand size = 17, antiderivative size = 91 \[ \int (3-x)^{3/2} (-2+x)^{3/2} \, dx=\frac {3}{64} \sqrt {3-x} \sqrt {-2+x}+\frac {1}{32} (3-x)^{3/2} \sqrt {-2+x}-\frac {1}{8} (3-x)^{5/2} \sqrt {-2+x}-\frac {1}{4} (3-x)^{5/2} (-2+x)^{3/2}-\frac {3}{128} \arcsin (5-2 x) \]

[Out]

-1/4*(3-x)^(5/2)*(-2+x)^(3/2)+3/128*arcsin(-5+2*x)+1/32*(3-x)^(3/2)*(-2+x)^(1/2)-1/8*(3-x)^(5/2)*(-2+x)^(1/2)+
3/64*(3-x)^(1/2)*(-2+x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {52, 55, 633, 222} \[ \int (3-x)^{3/2} (-2+x)^{3/2} \, dx=-\frac {3}{128} \arcsin (5-2 x)-\frac {1}{4} (x-2)^{3/2} (3-x)^{5/2}-\frac {1}{8} \sqrt {x-2} (3-x)^{5/2}+\frac {1}{32} \sqrt {x-2} (3-x)^{3/2}+\frac {3}{64} \sqrt {x-2} \sqrt {3-x} \]

[In]

Int[(3 - x)^(3/2)*(-2 + x)^(3/2),x]

[Out]

(3*Sqrt[3 - x]*Sqrt[-2 + x])/64 + ((3 - x)^(3/2)*Sqrt[-2 + x])/32 - ((3 - x)^(5/2)*Sqrt[-2 + x])/8 - ((3 - x)^
(5/2)*(-2 + x)^(3/2))/4 - (3*ArcSin[5 - 2*x])/128

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 55

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

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{4} (3-x)^{5/2} (-2+x)^{3/2}+\frac {3}{8} \int (3-x)^{3/2} \sqrt {-2+x} \, dx \\ & = -\frac {1}{8} (3-x)^{5/2} \sqrt {-2+x}-\frac {1}{4} (3-x)^{5/2} (-2+x)^{3/2}+\frac {1}{16} \int \frac {(3-x)^{3/2}}{\sqrt {-2+x}} \, dx \\ & = \frac {1}{32} (3-x)^{3/2} \sqrt {-2+x}-\frac {1}{8} (3-x)^{5/2} \sqrt {-2+x}-\frac {1}{4} (3-x)^{5/2} (-2+x)^{3/2}+\frac {3}{64} \int \frac {\sqrt {3-x}}{\sqrt {-2+x}} \, dx \\ & = \frac {3}{64} \sqrt {3-x} \sqrt {-2+x}+\frac {1}{32} (3-x)^{3/2} \sqrt {-2+x}-\frac {1}{8} (3-x)^{5/2} \sqrt {-2+x}-\frac {1}{4} (3-x)^{5/2} (-2+x)^{3/2}+\frac {3}{128} \int \frac {1}{\sqrt {3-x} \sqrt {-2+x}} \, dx \\ & = \frac {3}{64} \sqrt {3-x} \sqrt {-2+x}+\frac {1}{32} (3-x)^{3/2} \sqrt {-2+x}-\frac {1}{8} (3-x)^{5/2} \sqrt {-2+x}-\frac {1}{4} (3-x)^{5/2} (-2+x)^{3/2}+\frac {3}{128} \int \frac {1}{\sqrt {-6+5 x-x^2}} \, dx \\ & = \frac {3}{64} \sqrt {3-x} \sqrt {-2+x}+\frac {1}{32} (3-x)^{3/2} \sqrt {-2+x}-\frac {1}{8} (3-x)^{5/2} \sqrt {-2+x}-\frac {1}{4} (3-x)^{5/2} (-2+x)^{3/2}-\frac {3}{128} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,5-2 x\right ) \\ & = \frac {3}{64} \sqrt {3-x} \sqrt {-2+x}+\frac {1}{32} (3-x)^{3/2} \sqrt {-2+x}-\frac {1}{8} (3-x)^{5/2} \sqrt {-2+x}-\frac {1}{4} (3-x)^{5/2} (-2+x)^{3/2}-\frac {3}{128} \sin ^{-1}(5-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.87 \[ \int (3-x)^{3/2} (-2+x)^{3/2} \, dx=-\frac {\sqrt {-6+5 x-x^2} \left (\sqrt {-3+x} \sqrt {-2+x} \left (-225+290 x-120 x^2+16 x^3\right )+3 \text {arctanh}\left (\frac {1}{\sqrt {\frac {-3+x}{-2+x}}}\right )\right )}{64 \sqrt {-3+x} \sqrt {-2+x}} \]

[In]

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

[Out]

-1/64*(Sqrt[-6 + 5*x - x^2]*(Sqrt[-3 + x]*Sqrt[-2 + x]*(-225 + 290*x - 120*x^2 + 16*x^3) + 3*ArcTanh[1/Sqrt[(-
3 + x)/(-2 + x)]]))/(Sqrt[-3 + x]*Sqrt[-2 + x])

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {\left (16 x^{3}-120 x^{2}+290 x -225\right ) \left (-3+x \right ) \sqrt {-2+x}\, \sqrt {\left (-2+x \right ) \left (3-x \right )}}{64 \sqrt {-\left (-3+x \right ) \left (-2+x \right )}\, \sqrt {3-x}}+\frac {3 \sqrt {\left (-2+x \right ) \left (3-x \right )}\, \arcsin \left (-5+2 x \right )}{128 \sqrt {-2+x}\, \sqrt {3-x}}\)	\(86\)
default	\(\frac {\left (3-x \right )^{\frac {3}{2}} \left (-2+x \right )^{\frac {5}{2}}}{4}+\frac {\sqrt {3-x}\, \left (-2+x \right )^{\frac {5}{2}}}{8}-\frac {\sqrt {3-x}\, \left (-2+x \right )^{\frac {3}{2}}}{32}-\frac {3 \sqrt {3-x}\, \sqrt {-2+x}}{64}+\frac {3 \sqrt {\left (-2+x \right ) \left (3-x \right )}\, \arcsin \left (-5+2 x \right )}{128 \sqrt {-2+x}\, \sqrt {3-x}}\)	\(89\)

[In]

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

[Out]

1/64*(16*x^3-120*x^2+290*x-225)*(-3+x)*(-2+x)^(1/2)/(-(-3+x)*(-2+x))^(1/2)*((-2+x)*(3-x))^(1/2)/(3-x)^(1/2)+3/
128*((-2+x)*(3-x))^(1/2)/(-2+x)^(1/2)/(3-x)^(1/2)*arcsin(-5+2*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.68 \[ \int (3-x)^{3/2} (-2+x)^{3/2} \, dx=-\frac {1}{64} \, {\left (16 \, x^{3} - 120 \, x^{2} + 290 \, x - 225\right )} \sqrt {x - 2} \sqrt {-x + 3} - \frac {3}{128} \, \arctan \left (\frac {{\left (2 \, x - 5\right )} \sqrt {x - 2} \sqrt {-x + 3}}{2 \, {\left (x^{2} - 5 \, x + 6\right )}}\right ) \]

[In]

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

[Out]

-1/64*(16*x^3 - 120*x^2 + 290*x - 225)*sqrt(x - 2)*sqrt(-x + 3) - 3/128*arctan(1/2*(2*x - 5)*sqrt(x - 2)*sqrt(
-x + 3)/(x^2 - 5*x + 6))

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 15.43 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.19 \[ \int (3-x)^{3/2} (-2+x)^{3/2} \, dx=\begin {cases} - \frac {3 i \operatorname {acosh}{\left (\sqrt {x - 2} \right )}}{64} - \frac {i \left (x - 2\right )^{\frac {9}{2}}}{4 \sqrt {x - 3}} + \frac {5 i \left (x - 2\right )^{\frac {7}{2}}}{8 \sqrt {x - 3}} - \frac {13 i \left (x - 2\right )^{\frac {5}{2}}}{32 \sqrt {x - 3}} - \frac {i \left (x - 2\right )^{\frac {3}{2}}}{64 \sqrt {x - 3}} + \frac {3 i \sqrt {x - 2}}{64 \sqrt {x - 3}} & \text {for}\: \left |{x - 2}\right | > 1 \\\frac {3 \operatorname {asin}{\left (\sqrt {x - 2} \right )}}{64} + \frac {\left (x - 2\right )^{\frac {9}{2}}}{4 \sqrt {3 - x}} - \frac {5 \left (x - 2\right )^{\frac {7}{2}}}{8 \sqrt {3 - x}} + \frac {13 \left (x - 2\right )^{\frac {5}{2}}}{32 \sqrt {3 - x}} + \frac {\left (x - 2\right )^{\frac {3}{2}}}{64 \sqrt {3 - x}} - \frac {3 \sqrt {x - 2}}{64 \sqrt {3 - x}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-3*I*acosh(sqrt(x - 2))/64 - I*(x - 2)**(9/2)/(4*sqrt(x - 3)) + 5*I*(x - 2)**(7/2)/(8*sqrt(x - 3))
- 13*I*(x - 2)**(5/2)/(32*sqrt(x - 3)) - I*(x - 2)**(3/2)/(64*sqrt(x - 3)) + 3*I*sqrt(x - 2)/(64*sqrt(x - 3)),
Abs(x - 2) > 1), (3*asin(sqrt(x - 2))/64 + (x - 2)**(9/2)/(4*sqrt(3 - x)) - 5*(x - 2)**(7/2)/(8*sqrt(3 - x))
+ 13*(x - 2)**(5/2)/(32*sqrt(3 - x)) + (x - 2)**(3/2)/(64*sqrt(3 - x)) - 3*sqrt(x - 2)/(64*sqrt(3 - x)), True)
)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int (3-x)^{3/2} (-2+x)^{3/2} \, dx=\frac {1}{4} \, {\left (-x^{2} + 5 \, x - 6\right )}^{\frac {3}{2}} x - \frac {5}{8} \, {\left (-x^{2} + 5 \, x - 6\right )}^{\frac {3}{2}} + \frac {3}{32} \, \sqrt {-x^{2} + 5 \, x - 6} x - \frac {15}{64} \, \sqrt {-x^{2} + 5 \, x - 6} + \frac {3}{128} \, \arcsin \left (2 \, x - 5\right ) \]

[In]

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

[Out]

1/4*(-x^2 + 5*x - 6)^(3/2)*x - 5/8*(-x^2 + 5*x - 6)^(3/2) + 3/32*sqrt(-x^2 + 5*x - 6)*x - 15/64*sqrt(-x^2 + 5*
x - 6) + 3/128*arcsin(2*x - 5)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11 \[ \int (3-x)^{3/2} (-2+x)^{3/2} \, dx=-\frac {1}{192} \, {\left (2 \, {\left (4 \, {\left (6 \, x + 35\right )} {\left (x - 2\right )} + 523\right )} {\left (x - 2\right )} + 801\right )} \sqrt {x - 2} \sqrt {-x + 3} + \frac {7}{24} \, {\left (2 \, {\left (4 \, x + 15\right )} {\left (x - 2\right )} + 69\right )} \sqrt {x - 2} \sqrt {-x + 3} - 4 \, {\left (2 \, x + 3\right )} \sqrt {x - 2} \sqrt {-x + 3} + 12 \, \sqrt {x - 2} \sqrt {-x + 3} + \frac {3}{64} \, \arcsin \left (\sqrt {x - 2}\right ) \]

[In]

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

[Out]

-1/192*(2*(4*(6*x + 35)*(x - 2) + 523)*(x - 2) + 801)*sqrt(x - 2)*sqrt(-x + 3) + 7/24*(2*(4*x + 15)*(x - 2) +
69)*sqrt(x - 2)*sqrt(-x + 3) - 4*(2*x + 3)*sqrt(x - 2)*sqrt(-x + 3) + 12*sqrt(x - 2)*sqrt(-x + 3) + 3/64*arcsi
n(sqrt(x - 2))

Mupad [F(-1)]

Timed out. \[ \int (3-x)^{3/2} (-2+x)^{3/2} \, dx=\int {\left (x-2\right )}^{3/2}\,{\left (3-x\right )}^{3/2} \,d x \]

[In]

int((x - 2)^(3/2)*(3 - x)^(3/2),x)

[Out]

int((x - 2)^(3/2)*(3 - x)^(3/2), x)