\(\int x^{5/2} \sqrt {a-b x} \, dx\) [497]

Optimal result

Integrand size = 16, antiderivative size = 127 \[ \int x^{5/2} \sqrt {a-b x} \, dx=-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a-b x}}{96 b^2}-\frac {a x^{5/2} \sqrt {a-b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a-b x}+\frac {5 a^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{64 b^{7/2}} \]

[Out]

5/64*a^4*arctan(b^(1/2)*x^(1/2)/(-b*x+a)^(1/2))/b^(7/2)-5/96*a^2*x^(3/2)*(-b*x+a)^(1/2)/b^2-1/24*a*x^(5/2)*(-b
*x+a)^(1/2)/b+1/4*x^(7/2)*(-b*x+a)^(1/2)-5/64*a^3*x^(1/2)*(-b*x+a)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {52, 65, 223, 209} \[ \int x^{5/2} \sqrt {a-b x} \, dx=\frac {5 a^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{64 b^{7/2}}-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a-b x}}{96 b^2}-\frac {a x^{5/2} \sqrt {a-b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a-b x} \]

[In]

Int[x^(5/2)*Sqrt[a - b*x],x]

[Out]

(-5*a^3*Sqrt[x]*Sqrt[a - b*x])/(64*b^3) - (5*a^2*x^(3/2)*Sqrt[a - b*x])/(96*b^2) - (a*x^(5/2)*Sqrt[a - b*x])/(
24*b) + (x^(7/2)*Sqrt[a - b*x])/4 + (5*a^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/(64*b^(7/2))

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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^{7/2} \sqrt {a-b x}+\frac {1}{8} a \int \frac {x^{5/2}}{\sqrt {a-b x}} \, dx \\ & = -\frac {a x^{5/2} \sqrt {a-b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a-b x}+\frac {\left (5 a^2\right ) \int \frac {x^{3/2}}{\sqrt {a-b x}} \, dx}{48 b} \\ & = -\frac {5 a^2 x^{3/2} \sqrt {a-b x}}{96 b^2}-\frac {a x^{5/2} \sqrt {a-b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a-b x}+\frac {\left (5 a^3\right ) \int \frac {\sqrt {x}}{\sqrt {a-b x}} \, dx}{64 b^2} \\ & = -\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a-b x}}{96 b^2}-\frac {a x^{5/2} \sqrt {a-b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a-b x}+\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx}{128 b^3} \\ & = -\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a-b x}}{96 b^2}-\frac {a x^{5/2} \sqrt {a-b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a-b x}+\frac {\left (5 a^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )}{64 b^3} \\ & = -\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a-b x}}{96 b^2}-\frac {a x^{5/2} \sqrt {a-b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a-b x}+\frac {\left (5 a^4\right ) \text {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a-b x}}\right )}{64 b^3} \\ & = -\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a-b x}}{96 b^2}-\frac {a x^{5/2} \sqrt {a-b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a-b x}+\frac {5 a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{64 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.77 \[ \int x^{5/2} \sqrt {a-b x} \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a-b x} \left (-15 a^3-10 a^2 b x-8 a b^2 x^2+48 b^3 x^3\right )+30 a^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a-b x}}\right )}{192 b^{7/2}} \]

[In]

Integrate[x^(5/2)*Sqrt[a - b*x],x]

[Out]

(Sqrt[b]*Sqrt[x]*Sqrt[a - b*x]*(-15*a^3 - 10*a^2*b*x - 8*a*b^2*x^2 + 48*b^3*x^3) + 30*a^4*ArcTan[(Sqrt[b]*Sqrt
[x])/(-Sqrt[a] + Sqrt[a - b*x])])/(192*b^(7/2))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {\left (-48 b^{3} x^{3}+8 a \,b^{2} x^{2}+10 a^{2} b x +15 a^{3}\right ) \sqrt {x}\, \sqrt {-b x +a}}{192 b^{3}}+\frac {5 a^{4} \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-b \,x^{2}+a x}}\right ) \sqrt {x \left (-b x +a \right )}}{128 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {-b x +a}}\)	\(102\)
default	\(-\frac {x^{\frac {5}{2}} \left (-b x +a \right )^{\frac {3}{2}}}{4 b}+\frac {5 a \left (-\frac {x^{\frac {3}{2}} \left (-b x +a \right )^{\frac {3}{2}}}{3 b}+\frac {a \left (-\frac {\sqrt {x}\, \left (-b x +a \right )^{\frac {3}{2}}}{2 b}+\frac {a \left (\sqrt {x}\, \sqrt {-b x +a}+\frac {a \sqrt {x \left (-b x +a \right )}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-b \,x^{2}+a x}}\right )}{2 \sqrt {-b x +a}\, \sqrt {x}\, \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )}{8 b}\)	\(135\)

[In]

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

[Out]

-1/192*(-48*b^3*x^3+8*a*b^2*x^2+10*a^2*b*x+15*a^3)/b^3*x^(1/2)*(-b*x+a)^(1/2)+5/128*a^4/b^(7/2)*arctan(b^(1/2)
*(x-1/2*a/b)/(-b*x^2+a*x)^(1/2))*(x*(-b*x+a))^(1/2)/x^(1/2)/(-b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.29 \[ \int x^{5/2} \sqrt {a-b x} \, dx=\left [-\frac {15 \, a^{4} \sqrt {-b} \log \left (-2 \, b x + 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) - 2 \, {\left (48 \, b^{4} x^{3} - 8 \, a b^{3} x^{2} - 10 \, a^{2} b^{2} x - 15 \, a^{3} b\right )} \sqrt {-b x + a} \sqrt {x}}{384 \, b^{4}}, -\frac {15 \, a^{4} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - {\left (48 \, b^{4} x^{3} - 8 \, a b^{3} x^{2} - 10 \, a^{2} b^{2} x - 15 \, a^{3} b\right )} \sqrt {-b x + a} \sqrt {x}}{192 \, b^{4}}\right ] \]

[In]

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

[Out]

[-1/384*(15*a^4*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) - 2*(48*b^4*x^3 - 8*a*b^3*x^2 - 1
0*a^2*b^2*x - 15*a^3*b)*sqrt(-b*x + a)*sqrt(x))/b^4, -1/192*(15*a^4*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqr
t(x))) - (48*b^4*x^3 - 8*a*b^3*x^2 - 10*a^2*b^2*x - 15*a^3*b)*sqrt(-b*x + a)*sqrt(x))/b^4]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 30.40 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.54 \[ \int x^{5/2} \sqrt {a-b x} \, dx=\begin {cases} \frac {5 i a^{\frac {7}{2}} \sqrt {x}}{64 b^{3} \sqrt {-1 + \frac {b x}{a}}} - \frac {5 i a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 b^{2} \sqrt {-1 + \frac {b x}{a}}} - \frac {i a^{\frac {3}{2}} x^{\frac {5}{2}}}{96 b \sqrt {-1 + \frac {b x}{a}}} - \frac {7 i \sqrt {a} x^{\frac {7}{2}}}{24 \sqrt {-1 + \frac {b x}{a}}} - \frac {5 i a^{4} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {7}{2}}} + \frac {i b x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {5 a^{\frac {7}{2}} \sqrt {x}}{64 b^{3} \sqrt {1 - \frac {b x}{a}}} + \frac {5 a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 b^{2} \sqrt {1 - \frac {b x}{a}}} + \frac {a^{\frac {3}{2}} x^{\frac {5}{2}}}{96 b \sqrt {1 - \frac {b x}{a}}} + \frac {7 \sqrt {a} x^{\frac {7}{2}}}{24 \sqrt {1 - \frac {b x}{a}}} + \frac {5 a^{4} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {7}{2}}} - \frac {b x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 - \frac {b x}{a}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((5*I*a**(7/2)*sqrt(x)/(64*b**3*sqrt(-1 + b*x/a)) - 5*I*a**(5/2)*x**(3/2)/(192*b**2*sqrt(-1 + b*x/a))
- I*a**(3/2)*x**(5/2)/(96*b*sqrt(-1 + b*x/a)) - 7*I*sqrt(a)*x**(7/2)/(24*sqrt(-1 + b*x/a)) - 5*I*a**4*acosh(s
qrt(b)*sqrt(x)/sqrt(a))/(64*b**(7/2)) + I*b*x**(9/2)/(4*sqrt(a)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (-5*a**(7/
2)*sqrt(x)/(64*b**3*sqrt(1 - b*x/a)) + 5*a**(5/2)*x**(3/2)/(192*b**2*sqrt(1 - b*x/a)) + a**(3/2)*x**(5/2)/(96*
b*sqrt(1 - b*x/a)) + 7*sqrt(a)*x**(7/2)/(24*sqrt(1 - b*x/a)) + 5*a**4*asin(sqrt(b)*sqrt(x)/sqrt(a))/(64*b**(7/
2)) - b*x**(9/2)/(4*sqrt(a)*sqrt(1 - b*x/a)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.34 \[ \int x^{5/2} \sqrt {a-b x} \, dx=-\frac {5 \, a^{4} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{64 \, b^{\frac {7}{2}}} + \frac {\frac {15 \, \sqrt {-b x + a} a^{4} b^{3}}{\sqrt {x}} - \frac {73 \, {\left (-b x + a\right )}^{\frac {3}{2}} a^{4} b^{2}}{x^{\frac {3}{2}}} - \frac {55 \, {\left (-b x + a\right )}^{\frac {5}{2}} a^{4} b}{x^{\frac {5}{2}}} - \frac {15 \, {\left (-b x + a\right )}^{\frac {7}{2}} a^{4}}{x^{\frac {7}{2}}}}{192 \, {\left (b^{7} - \frac {4 \, {\left (b x - a\right )} b^{6}}{x} + \frac {6 \, {\left (b x - a\right )}^{2} b^{5}}{x^{2}} - \frac {4 \, {\left (b x - a\right )}^{3} b^{4}}{x^{3}} + \frac {{\left (b x - a\right )}^{4} b^{3}}{x^{4}}\right )}} \]

[In]

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

[Out]

-5/64*a^4*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x)))/b^(7/2) + 1/192*(15*sqrt(-b*x + a)*a^4*b^3/sqrt(x) - 73*(-b
*x + a)^(3/2)*a^4*b^2/x^(3/2) - 55*(-b*x + a)^(5/2)*a^4*b/x^(5/2) - 15*(-b*x + a)^(7/2)*a^4/x^(7/2))/(b^7 - 4*
(b*x - a)*b^6/x + 6*(b*x - a)^2*b^5/x^2 - 4*(b*x - a)^3*b^4/x^3 + (b*x - a)^4*b^3/x^4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (93) = 186\).

Time = 157.01 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.02 \[ \int x^{5/2} \sqrt {a-b x} \, dx=\frac {\frac {8 \, {\left (\frac {15 \, a^{3} \log \left ({\left | -\sqrt {-b x + a} \sqrt {-b} + \sqrt {{\left (b x - a\right )} b + a b} \right |}\right )}{\sqrt {-b} b} - \sqrt {{\left (b x - a\right )} b + a b} \sqrt {-b x + a} {\left (2 \, {\left (b x - a\right )} {\left (\frac {4 \, {\left (b x - a\right )}}{b^{2}} + \frac {13 \, a}{b^{2}}\right )} + \frac {33 \, a^{2}}{b^{2}}\right )}\right )} a {\left | b \right |}}{b^{2}} - \frac {{\left (\frac {105 \, a^{4} \log \left ({\left | -\sqrt {-b x + a} \sqrt {-b} + \sqrt {{\left (b x - a\right )} b + a b} \right |}\right )}{\sqrt {-b} b^{2}} - {\left (2 \, {\left (b x - a\right )} {\left (4 \, {\left (b x - a\right )} {\left (\frac {6 \, {\left (b x - a\right )}}{b^{3}} + \frac {25 \, a}{b^{3}}\right )} + \frac {163 \, a^{2}}{b^{3}}\right )} + \frac {279 \, a^{3}}{b^{3}}\right )} \sqrt {{\left (b x - a\right )} b + a b} \sqrt {-b x + a}\right )} {\left | b \right |}}{b}}{192 \, b} \]

[In]

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

[Out]

1/192*(8*(15*a^3*log(abs(-sqrt(-b*x + a)*sqrt(-b) + sqrt((b*x - a)*b + a*b)))/(sqrt(-b)*b) - sqrt((b*x - a)*b
+ a*b)*sqrt(-b*x + a)*(2*(b*x - a)*(4*(b*x - a)/b^2 + 13*a/b^2) + 33*a^2/b^2))*a*abs(b)/b^2 - (105*a^4*log(abs
(-sqrt(-b*x + a)*sqrt(-b) + sqrt((b*x - a)*b + a*b)))/(sqrt(-b)*b^2) - (2*(b*x - a)*(4*(b*x - a)*(6*(b*x - a)/
b^3 + 25*a/b^3) + 163*a^2/b^3) + 279*a^3/b^3)*sqrt((b*x - a)*b + a*b)*sqrt(-b*x + a))*abs(b)/b)/b

Mupad [F(-1)]

Timed out. \[ \int x^{5/2} \sqrt {a-b x} \, dx=\int x^{5/2}\,\sqrt {a-b\,x} \,d x \]

[In]

int(x^(5/2)*(a - b*x)^(1/2),x)

[Out]

int(x^(5/2)*(a - b*x)^(1/2), x)