\(\int \frac {(a-b x)^{3/2}}{x^{5/2}} \, dx\) [532]

Optimal result

Integrand size = 16, antiderivative size = 67 \[ \int \frac {(a-b x)^{3/2}}{x^{5/2}} \, dx=\frac {2 b \sqrt {a-b x}}{\sqrt {x}}-\frac {2 (a-b x)^{3/2}}{3 x^{3/2}}+2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right ) \]

[Out]

-2/3*(-b*x+a)^(3/2)/x^(3/2)+2*b^(3/2)*arctan(b^(1/2)*x^(1/2)/(-b*x+a)^(1/2))+2*b*(-b*x+a)^(1/2)/x^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 67, 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, 65, 223, 209} \[ \int \frac {(a-b x)^{3/2}}{x^{5/2}} \, dx=2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )-\frac {2 (a-b x)^{3/2}}{3 x^{3/2}}+\frac {2 b \sqrt {a-b x}}{\sqrt {x}} \]

[In]

Int[(a - b*x)^(3/2)/x^(5/2),x]

[Out]

(2*b*Sqrt[a - b*x])/Sqrt[x] - (2*(a - b*x)^(3/2))/(3*x^(3/2)) + 2*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*
x]]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {(a-b x)^{3/2}}{x^{5/2}} \, dx=\frac {2 \sqrt {a-b x} (-a+4 b x)}{3 x^{3/2}}+4 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a-b x}}\right ) \]

[In]

Integrate[(a - b*x)^(3/2)/x^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06

method	result	size
risch	\(-\frac {2 \sqrt {-b x +a}\, \left (-4 b x +a \right )}{3 x^{\frac {3}{2}}}+\frac {b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-b \,x^{2}+a x}}\right ) \sqrt {x \left (-b x +a \right )}}{\sqrt {x}\, \sqrt {-b x +a}}\)	\(71\)

[In]

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

[Out]

-2/3*(-b*x+a)^(1/2)*(-4*b*x+a)/x^(3/2)+b^(3/2)*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 = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.72 \[ \int \frac {(a-b x)^{3/2}}{x^{5/2}} \, dx=\left [\frac {3 \, \sqrt {-b} b x^{2} \log \left (-2 \, b x - 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) + 2 \, {\left (4 \, b x - a\right )} \sqrt {-b x + a} \sqrt {x}}{3 \, x^{2}}, -\frac {2 \, {\left (3 \, b^{\frac {3}{2}} x^{2} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - {\left (4 \, b x - a\right )} \sqrt {-b x + a} \sqrt {x}\right )}}{3 \, x^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.44 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.79 \[ \int \frac {(a-b x)^{3/2}}{x^{5/2}} \, dx=\begin {cases} - \frac {2 a \sqrt {b} \sqrt {\frac {a}{b x} - 1}}{3 x} + \frac {8 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}}{3} - 2 i b^{\frac {3}{2}} \log {\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} + i b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )} + 2 b^{\frac {3}{2}} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\- \frac {2 i a \sqrt {b} \sqrt {- \frac {a}{b x} + 1}}{3 x} + \frac {8 i b^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}}{3} + i b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )} - 2 i b^{\frac {3}{2}} \log {\left (\sqrt {- \frac {a}{b x} + 1} + 1 \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*a*sqrt(b)*sqrt(a/(b*x) - 1)/(3*x) + 8*b**(3/2)*sqrt(a/(b*x) - 1)/3 - 2*I*b**(3/2)*log(sqrt(a)/(s
qrt(b)*sqrt(x))) + I*b**(3/2)*log(a/(b*x)) + 2*b**(3/2)*asin(sqrt(b)*sqrt(x)/sqrt(a)), Abs(a/(b*x)) > 1), (-2*
I*a*sqrt(b)*sqrt(-a/(b*x) + 1)/(3*x) + 8*I*b**(3/2)*sqrt(-a/(b*x) + 1)/3 + I*b**(3/2)*log(a/(b*x)) - 2*I*b**(3
/2)*log(sqrt(-a/(b*x) + 1) + 1), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int \frac {(a-b x)^{3/2}}{x^{5/2}} \, dx=-2 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + \frac {2 \, \sqrt {-b x + a} b}{\sqrt {x}} - \frac {2 \, {\left (-b x + a\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

-2*b^(3/2)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) + 2*sqrt(-b*x + a)*b/sqrt(x) - 2/3*(-b*x + a)^(3/2)/x^(3/2
)

Giac [A] (verification not implemented)

none

Time = 76.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.40 \[ \int \frac {(a-b x)^{3/2}}{x^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, b^{2} \log \left ({\left | -\sqrt {-b x + a} \sqrt {-b} + \sqrt {{\left (b x - a\right )} b + a b} \right |}\right )}{\sqrt {-b}} + \frac {{\left (4 \, {\left (b x - a\right )} b^{3} + 3 \, a b^{3}\right )} \sqrt {-b x + a}}{{\left ({\left (b x - a\right )} b + a b\right )}^{\frac {3}{2}}}\right )} b}{3 \, {\left | b \right |}} \]

[In]

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

[Out]

2/3*(3*b^2*log(abs(-sqrt(-b*x + a)*sqrt(-b) + sqrt((b*x - a)*b + a*b)))/sqrt(-b) + (4*(b*x - a)*b^3 + 3*a*b^3)
*sqrt(-b*x + a)/((b*x - a)*b + a*b)^(3/2))*b/abs(b)

Mupad [F(-1)]

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

[In]

int((a - b*x)^(3/2)/x^(5/2),x)

[Out]

int((a - b*x)^(3/2)/x^(5/2), x)