3.4.32 \(\int \frac {(-a+b x)^{5/2}}{x^2} \, dx\) [332]

Optimal. Leaf size=74 \[ -5 a b \sqrt {-a+b x}+\frac {5}{3} b (-a+b x)^{3/2}-\frac {(-a+b x)^{5/2}}{x}+5 a^{3/2} b \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {43, 52, 65, 211} \begin {gather*} 5 a^{3/2} b \text {ArcTan}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )-\frac {(b x-a)^{5/2}}{x}+\frac {5}{3} b (b x-a)^{3/2}-5 a b \sqrt {b x-a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-a + b*x)^(5/2)/x^2,x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 211

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

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 64, normalized size = 0.86 \begin {gather*} -\frac {\sqrt {-a+b x} \left (3 a^2+14 a b x-2 b^2 x^2\right )}{3 x}+5 a^{3/2} b \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-a + b*x)^(5/2)/x^2,x]

[Out]

-1/3*(Sqrt[-a + b*x]*(3*a^2 + 14*a*b*x - 2*b^2*x^2))/x + 5*a^(3/2)*b*ArcTan[Sqrt[-a + b*x]/Sqrt[a]]

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Maple [A]
time = 0.11, size = 69, normalized size = 0.93

method	result	size
derivativedivides	\(2 b \left (\frac {\left (b x -a \right )^{\frac {3}{2}}}{3}-2 a \sqrt {b x -a}+a^{2} \left (-\frac {\sqrt {b x -a}}{2 b x}+\frac {5 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\)	\(69\)
default	\(2 b \left (\frac {\left (b x -a \right )^{\frac {3}{2}}}{3}-2 a \sqrt {b x -a}+a^{2} \left (-\frac {\sqrt {b x -a}}{2 b x}+\frac {5 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\)	\(69\)
risch	\(\frac {a^{2} \left (-b x +a \right )}{x \sqrt {b x -a}}+\frac {2 b \left (b x -a \right )^{\frac {3}{2}}}{3}-4 a b \sqrt {b x -a}+5 a^{\frac {3}{2}} b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 63, normalized size = 0.85 \begin {gather*} 5 \, a^{\frac {3}{2}} b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + \frac {2}{3} \, {\left (b x - a\right )}^{\frac {3}{2}} b - 4 \, \sqrt {b x - a} a b - \frac {\sqrt {b x - a} a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a^(3/2)*b*arctan(sqrt(b*x - a)/sqrt(a)) + 2/3*(b*x - a)^(3/2)*b - 4*sqrt(b*x - a)*a*b - sqrt(b*x - a)*a^2/x

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Fricas [A]
time = 0.49, size = 131, normalized size = 1.77 \begin {gather*} \left [\frac {15 \, \sqrt {-a} a b x \log \left (\frac {b x + 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (2 \, b^{2} x^{2} - 14 \, a b x - 3 \, a^{2}\right )} \sqrt {b x - a}}{6 \, x}, \frac {15 \, a^{\frac {3}{2}} b x \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (2 \, b^{2} x^{2} - 14 \, a b x - 3 \, a^{2}\right )} \sqrt {b x - a}}{3 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*sqrt(-a)*a*b*x*log((b*x + 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) + 2*(2*b^2*x^2 - 14*a*b*x - 3*a^2)*sqrt(
b*x - a))/x, 1/3*(15*a^(3/2)*b*x*arctan(sqrt(b*x - a)/sqrt(a)) + (2*b^2*x^2 - 14*a*b*x - 3*a^2)*sqrt(b*x - a))
/x]

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Sympy [C] Result contains complex when optimal does not.
time = 2.20, size = 245, normalized size = 3.31 \begin {gather*} \begin {cases} - \frac {a^{\frac {5}{2}} \sqrt {-1 + \frac {b x}{a}}}{x} - \frac {14 a^{\frac {3}{2}} b \sqrt {-1 + \frac {b x}{a}}}{3} - \frac {5 i a^{\frac {3}{2}} b \log {\left (\frac {b x}{a} \right )}}{2} + 5 i a^{\frac {3}{2}} b \log {\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - 5 a^{\frac {3}{2}} b \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} + \frac {2 \sqrt {a} b^{2} x \sqrt {-1 + \frac {b x}{a}}}{3} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {i a^{\frac {5}{2}} \sqrt {1 - \frac {b x}{a}}}{x} - \frac {14 i a^{\frac {3}{2}} b \sqrt {1 - \frac {b x}{a}}}{3} - \frac {5 i a^{\frac {3}{2}} b \log {\left (\frac {b x}{a} \right )}}{2} + 5 i a^{\frac {3}{2}} b \log {\left (\sqrt {1 - \frac {b x}{a}} + 1 \right )} + \frac {2 i \sqrt {a} b^{2} x \sqrt {1 - \frac {b x}{a}}}{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**(5/2)*sqrt(-1 + b*x/a)/x - 14*a**(3/2)*b*sqrt(-1 + b*x/a)/3 - 5*I*a**(3/2)*b*log(b*x/a)/2 + 5*I
*a**(3/2)*b*log(sqrt(b)*sqrt(x)/sqrt(a)) - 5*a**(3/2)*b*asin(sqrt(a)/(sqrt(b)*sqrt(x))) + 2*sqrt(a)*b**2*x*sqr
t(-1 + b*x/a)/3, Abs(b*x/a) > 1), (-I*a**(5/2)*sqrt(1 - b*x/a)/x - 14*I*a**(3/2)*b*sqrt(1 - b*x/a)/3 - 5*I*a**
(3/2)*b*log(b*x/a)/2 + 5*I*a**(3/2)*b*log(sqrt(1 - b*x/a) + 1) + 2*I*sqrt(a)*b**2*x*sqrt(1 - b*x/a)/3, True))

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Giac [A]
time = 0.91, size = 75, normalized size = 1.01 \begin {gather*} \frac {15 \, a^{\frac {3}{2}} b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + 2 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{2} - 12 \, \sqrt {b x - a} a b^{2} - \frac {3 \, \sqrt {b x - a} a^{2} b}{x}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(15*a^(3/2)*b^2*arctan(sqrt(b*x - a)/sqrt(a)) + 2*(b*x - a)^(3/2)*b^2 - 12*sqrt(b*x - a)*a*b^2 - 3*sqrt(b*
x - a)*a^2*b/x)/b

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Mupad [B]
time = 0.10, size = 63, normalized size = 0.85 \begin {gather*} \frac {2\,b\,{\left (b\,x-a\right )}^{3/2}}{3}-\frac {a^2\,\sqrt {b\,x-a}}{x}+5\,a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )-4\,a\,b\,\sqrt {b\,x-a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x - a)^(5/2)/x^2,x)

[Out]

(2*b*(b*x - a)^(3/2))/3 - (a^2*(b*x - a)^(1/2))/x + 5*a^(3/2)*b*atan((b*x - a)^(1/2)/a^(1/2)) - 4*a*b*(b*x - a
)^(1/2)

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