\(\int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx\) [635]

Optimal result

Integrand size = 16, antiderivative size = 89 \[ \int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx=\frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \]

[Out]

-15*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)+2*x^(5/2)/b/(-b*x+2)^(1/2)+5/2*x^(3/2)*(-b*x+2)^(1/2)/b^2+15/2
*x^(1/2)*(-b*x+2)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {49, 52, 56, 222} \[ \int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx=-\frac {15 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}+\frac {2 x^{5/2}}{b \sqrt {2-b x}} \]

[In]

Int[x^(5/2)/(2 - b*x)^(3/2),x]

[Out]

(2*x^(5/2))/(b*Sqrt[2 - b*x]) + (15*Sqrt[x]*Sqrt[2 - b*x])/(2*b^3) + (5*x^(3/2)*Sqrt[2 - b*x])/(2*b^2) - (15*A
rcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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 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*} \text {integral}& = \frac {2 x^{5/2}}{b \sqrt {2-b x}}-\frac {5 \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{b} \\ & = \frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{2 b^2} \\ & = \frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{2 b^3} \\ & = \frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b^3} \\ & = \frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx=-\frac {\sqrt {x} \left (-30+5 b x+b^2 x^2\right )}{2 b^3 \sqrt {2-b x}}+\frac {30 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{b^{7/2}} \]

[In]

Integrate[x^(5/2)/(2 - b*x)^(3/2),x]

[Out]

-1/2*(Sqrt[x]*(-30 + 5*b*x + b^2*x^2))/(b^3*Sqrt[2 - b*x]) + (30*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 -
b*x])])/b^(7/2)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91

method	result	size
meijerg	\(-\frac {8 \left (\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {7}{2}} \left (-\frac {7}{2} b^{2} x^{2}-\frac {35}{2} b x +105\right )}{112 b^{3} \sqrt {-\frac {b x}{2}+1}}-\frac {15 \sqrt {\pi }\, \left (-b \right )^{\frac {7}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{8 b^{\frac {7}{2}}}\right )}{\left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, b}\)	\(81\)
risch	\(-\frac {\left (b x +7\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{3} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}-\frac {\left (\frac {15 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 b^{\frac {7}{2}}}+\frac {8 \sqrt {-b \left (x -\frac {2}{b}\right )^{2}-2 x +\frac {4}{b}}}{b^{4} \left (x -\frac {2}{b}\right )}\right ) \sqrt {\left (-b x +2\right ) x}}{\sqrt {x}\, \sqrt {-b x +2}}\)	\(138\)

[In]

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

[Out]

-8/(-b)^(5/2)/Pi^(1/2)/b*(1/112*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(7/2)*(-7/2*b^2*x^2-35/2*b*x+105)/b^3/(-1/2*b*x+
1)^(1/2)-15/8*Pi^(1/2)*(-b)^(7/2)/b^(7/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.74 \[ \int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (b x - 2\right )} \sqrt {-b} \log \left (-b x - \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) - {\left (b^{3} x^{2} + 5 \, b^{2} x - 30 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x - 2 \, b^{4}\right )}}, \frac {30 \, {\left (b x - 2\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) + {\left (b^{3} x^{2} + 5 \, b^{2} x - 30 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x - 2 \, b^{4}\right )}}\right ] \]

[In]

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

[Out]

[-1/2*(15*(b*x - 2)*sqrt(-b)*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) - (b^3*x^2 + 5*b^2*x - 30*b)*sqrt
(-b*x + 2)*sqrt(x))/(b^5*x - 2*b^4), 1/2*(30*(b*x - 2)*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + (b^3
*x^2 + 5*b^2*x - 30*b)*sqrt(-b*x + 2)*sqrt(x))/(b^5*x - 2*b^4)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.98 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.93 \[ \int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx=\begin {cases} \frac {i x^{\frac {5}{2}}}{2 b \sqrt {b x - 2}} + \frac {5 i x^{\frac {3}{2}}}{2 b^{2} \sqrt {b x - 2}} - \frac {15 i \sqrt {x}}{b^{3} \sqrt {b x - 2}} + \frac {15 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {x^{\frac {5}{2}}}{2 b \sqrt {- b x + 2}} - \frac {5 x^{\frac {3}{2}}}{2 b^{2} \sqrt {- b x + 2}} + \frac {15 \sqrt {x}}{b^{3} \sqrt {- b x + 2}} - \frac {15 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((I*x**(5/2)/(2*b*sqrt(b*x - 2)) + 5*I*x**(3/2)/(2*b**2*sqrt(b*x - 2)) - 15*I*sqrt(x)/(b**3*sqrt(b*x
- 2)) + 15*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2), Abs(b*x) > 2), (-x**(5/2)/(2*b*sqrt(-b*x + 2)) - 5*x**
(3/2)/(2*b**2*sqrt(-b*x + 2)) + 15*sqrt(x)/(b**3*sqrt(-b*x + 2)) - 15*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2)
, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13 \[ \int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx=\frac {8 \, b^{2} - \frac {25 \, {\left (b x - 2\right )} b}{x} + \frac {15 \, {\left (b x - 2\right )}^{2}}{x^{2}}}{\frac {\sqrt {-b x + 2} b^{5}}{\sqrt {x}} + \frac {2 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (-b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}} + \frac {15 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {7}{2}}} \]

[In]

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

[Out]

(8*b^2 - 25*(b*x - 2)*b/x + 15*(b*x - 2)^2/x^2)/(sqrt(-b*x + 2)*b^5/sqrt(x) + 2*(-b*x + 2)^(3/2)*b^4/x^(3/2) +
(-b*x + 2)^(5/2)*b^3/x^(5/2)) + 15*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(7/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (66) = 132\).

Time = 1.56 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.53 \[ \int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx=\frac {{\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left (\frac {b x - 2}{b^{3}} + \frac {9}{b^{3}}\right )} - \frac {15 \, \log \left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt {-b} b^{2}} + \frac {64}{{\left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )} \sqrt {-b} b}\right )} {\left | b \right |}}{2 \, b^{2}} \]

[In]

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

[Out]

1/2*(sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2)*((b*x - 2)/b^3 + 9/b^3) - 15*log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((
b*x - 2)*b + 2*b))^2)/(sqrt(-b)*b^2) + 64/(((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)*sqrt(
-b)*b))*abs(b)/b^2

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx=\int \frac {x^{5/2}}{{\left (2-b\,x\right )}^{3/2}} \,d x \]

[In]

int(x^(5/2)/(2 - b*x)^(3/2),x)

[Out]

int(x^(5/2)/(2 - b*x)^(3/2), x)