\(\int \frac {x^{5/2}}{(a-b x)^{3/2}} \, dx\) [596]

Optimal result

Integrand size = 16, antiderivative size = 100 \[ \int \frac {x^{5/2}}{(a-b x)^{3/2}} \, dx=\frac {2 x^{5/2}}{b \sqrt {a-b x}}+\frac {15 a \sqrt {x} \sqrt {a-b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a-b x}}{2 b^2}-\frac {15 a^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{4 b^{7/2}} \]

[Out]

-15/4*a^2*arctan(b^(1/2)*x^(1/2)/(-b*x+a)^(1/2))/b^(7/2)+2*x^(5/2)/b/(-b*x+a)^(1/2)+5/2*x^(3/2)*(-b*x+a)^(1/2)
/b^2+15/4*a*x^(1/2)*(-b*x+a)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {49, 52, 65, 223, 209} \[ \int \frac {x^{5/2}}{(a-b x)^{3/2}} \, dx=-\frac {15 a^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{4 b^{7/2}}+\frac {15 a \sqrt {x} \sqrt {a-b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a-b x}}{2 b^2}+\frac {2 x^{5/2}}{b \sqrt {a-b x}} \]

[In]

Int[x^(5/2)/(a - b*x)^(3/2),x]

[Out]

(2*x^(5/2))/(b*Sqrt[a - b*x]) + (15*a*Sqrt[x]*Sqrt[a - b*x])/(4*b^3) + (5*x^(3/2)*Sqrt[a - b*x])/(2*b^2) - (15
*a^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/(4*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 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 {2 x^{5/2}}{b \sqrt {a-b x}}-\frac {5 \int \frac {x^{3/2}}{\sqrt {a-b x}} \, dx}{b} \\ & = \frac {2 x^{5/2}}{b \sqrt {a-b x}}+\frac {5 x^{3/2} \sqrt {a-b x}}{2 b^2}-\frac {(15 a) \int \frac {\sqrt {x}}{\sqrt {a-b x}} \, dx}{4 b^2} \\ & = \frac {2 x^{5/2}}{b \sqrt {a-b x}}+\frac {15 a \sqrt {x} \sqrt {a-b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a-b x}}{2 b^2}-\frac {\left (15 a^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx}{8 b^3} \\ & = \frac {2 x^{5/2}}{b \sqrt {a-b x}}+\frac {15 a \sqrt {x} \sqrt {a-b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a-b x}}{2 b^2}-\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^3} \\ & = \frac {2 x^{5/2}}{b \sqrt {a-b x}}+\frac {15 a \sqrt {x} \sqrt {a-b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a-b x}}{2 b^2}-\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a-b x}}\right )}{4 b^3} \\ & = \frac {2 x^{5/2}}{b \sqrt {a-b x}}+\frac {15 a \sqrt {x} \sqrt {a-b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a-b x}}{2 b^2}-\frac {15 a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{4 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {x^{5/2}}{(a-b x)^{3/2}} \, dx=-\frac {\sqrt {x} \left (-15 a^2+5 a b x+2 b^2 x^2\right )}{4 b^3 \sqrt {a-b x}}+\frac {15 a^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a-b x}}\right )}{2 b^{7/2}} \]

[In]

Integrate[x^(5/2)/(a - b*x)^(3/2),x]

[Out]

-1/4*(Sqrt[x]*(-15*a^2 + 5*a*b*x + 2*b^2*x^2))/(b^3*Sqrt[a - b*x]) + (15*a^2*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[a]
- Sqrt[a - b*x])])/(2*b^(7/2))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.27

method	result	size
risch	\(\frac {\left (2 b x +7 a \right ) \sqrt {x}\, \sqrt {-b x +a}}{4 b^{3}}+\frac {\left (-\frac {15 a^{2} \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-b \,x^{2}+a x}}\right )}{8 b^{\frac {7}{2}}}-\frac {2 a^{2} \sqrt {-b \left (-\frac {a}{b}+x \right )^{2}-\left (-\frac {a}{b}+x \right ) a}}{b^{4} \left (-\frac {a}{b}+x \right )}\right ) \sqrt {x \left (-b x +a \right )}}{\sqrt {x}\, \sqrt {-b x +a}}\)	\(127\)

[In]

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

[Out]

1/4*(2*b*x+7*a)/b^3*x^(1/2)*(-b*x+a)^(1/2)+(-15/8*a^2/b^(7/2)*arctan(b^(1/2)*(x-1/2*a/b)/(-b*x^2+a*x)^(1/2))-2
*a^2/b^4/(-a/b+x)*(-b*(-a/b+x)^2-(-a/b+x)*a)^(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 = 181, normalized size of antiderivative = 1.81 \[ \int \frac {x^{5/2}}{(a-b x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (a^{2} b x - a^{3}\right )} \sqrt {-b} \log \left (-2 \, b x - 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) - 2 \, {\left (2 \, b^{3} x^{2} + 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{8 \, {\left (b^{5} x - a b^{4}\right )}}, \frac {15 \, {\left (a^{2} b x - a^{3}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + {\left (2 \, b^{3} x^{2} + 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{4 \, {\left (b^{5} x - a b^{4}\right )}}\right ] \]

[In]

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

[Out]

[-1/8*(15*(a^2*b*x - a^3)*sqrt(-b)*log(-2*b*x - 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) - 2*(2*b^3*x^2 + 5*a*b^
2*x - 15*a^2*b)*sqrt(-b*x + a)*sqrt(x))/(b^5*x - a*b^4), 1/4*(15*(a^2*b*x - a^3)*sqrt(b)*arctan(sqrt(-b*x + a)
/(sqrt(b)*sqrt(x))) + (2*b^3*x^2 + 5*a*b^2*x - 15*a^2*b)*sqrt(-b*x + a)*sqrt(x))/(b^5*x - a*b^4)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.08 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.24 \[ \int \frac {x^{5/2}}{(a-b x)^{3/2}} \, dx=\begin {cases} - \frac {15 i a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {-1 + \frac {b x}{a}}} + \frac {5 i \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {-1 + \frac {b x}{a}}} + \frac {15 i a^{2} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} + \frac {i x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\\frac {15 a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {1 - \frac {b x}{a}}} - \frac {5 \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 - \frac {b x}{a}}} - \frac {15 a^{2} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} - \frac {x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {1 - \frac {b x}{a}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-15*I*a**(3/2)*sqrt(x)/(4*b**3*sqrt(-1 + b*x/a)) + 5*I*sqrt(a)*x**(3/2)/(4*b**2*sqrt(-1 + b*x/a)) +
15*I*a**2*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) + I*x**(5/2)/(2*sqrt(a)*b*sqrt(-1 + b*x/a)), Abs(b*x/a)
> 1), (15*a**(3/2)*sqrt(x)/(4*b**3*sqrt(1 - b*x/a)) - 5*sqrt(a)*x**(3/2)/(4*b**2*sqrt(1 - b*x/a)) - 15*a**2*a
sin(sqrt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) - x**(5/2)/(2*sqrt(a)*b*sqrt(1 - b*x/a)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int \frac {x^{5/2}}{(a-b x)^{3/2}} \, dx=\frac {8 \, a^{2} b^{2} - \frac {25 \, {\left (b x - a\right )} a^{2} b}{x} + \frac {15 \, {\left (b x - a\right )}^{2} a^{2}}{x^{2}}}{4 \, {\left (\frac {\sqrt {-b x + a} b^{5}}{\sqrt {x}} + \frac {2 \, {\left (-b x + a\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (-b x + a\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}\right )}} + \frac {15 \, a^{2} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{4 \, b^{\frac {7}{2}}} \]

[In]

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

[Out]

1/4*(8*a^2*b^2 - 25*(b*x - a)*a^2*b/x + 15*(b*x - a)^2*a^2/x^2)/(sqrt(-b*x + a)*b^5/sqrt(x) + 2*(-b*x + a)^(3/
2)*b^4/x^(3/2) + (-b*x + a)^(5/2)*b^3/x^(5/2)) + 15/4*a^2*arctan(sqrt(-b*x + a)/(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. 154 vs. \(2 (74) = 148\).

Time = 16.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.54 \[ \int \frac {x^{5/2}}{(a-b x)^{3/2}} \, dx=\frac {{\left (2 \, \sqrt {{\left (b x - a\right )} b + a b} \sqrt {-b x + a} {\left (\frac {2 \, {\left (b x - a\right )}}{b^{3}} + \frac {9 \, a}{b^{3}}\right )} + \frac {32 \, a^{3}}{{\left ({\left (\sqrt {-b x + a} \sqrt {-b} - \sqrt {{\left (b x - a\right )} b + a b}\right )}^{2} - a b\right )} \sqrt {-b} b} - \frac {15 \, a^{2} \log \left ({\left (\sqrt {-b x + a} \sqrt {-b} - \sqrt {{\left (b x - a\right )} b + a b}\right )}^{2}\right )}{\sqrt {-b} b^{2}}\right )} {\left | b \right |}}{8 \, b^{2}} \]

[In]

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

[Out]

1/8*(2*sqrt((b*x - a)*b + a*b)*sqrt(-b*x + a)*(2*(b*x - a)/b^3 + 9*a/b^3) + 32*a^3/(((sqrt(-b*x + a)*sqrt(-b)
- sqrt((b*x - a)*b + a*b))^2 - a*b)*sqrt(-b)*b) - 15*a^2*log((sqrt(-b*x + a)*sqrt(-b) - sqrt((b*x - a)*b + a*b
))^2)/(sqrt(-b)*b^2))*abs(b)/b^2

Mupad [F(-1)]

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

[In]

int(x^(5/2)/(a - b*x)^(3/2),x)

[Out]

int(x^(5/2)/(a - b*x)^(3/2), x)