3.4.31 \(\int \frac {(-a+b x)^{5/2}}{x} \, dx\)

Optimal. Leaf size=73 \[ -2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )+2 a^2 \sqrt {b x-a}-\frac {2}{3} a (b x-a)^{3/2}+\frac {2}{5} (b x-a)^{5/2} \]

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Rubi [A]  time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {50, 63, 205} \begin {gather*} 2 a^2 \sqrt {b x-a}-2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )-\frac {2}{3} a (b x-a)^{3/2}+\frac {2}{5} (b x-a)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-a + b*x)^(5/2)/x,x]

[Out]

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

Rule 50

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 63

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 205

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 60, normalized size = 0.82 \begin {gather*} \frac {2}{15} \sqrt {b x-a} \left (23 a^2-11 a b x+3 b^2 x^2\right )-2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-a + b*x)^(5/2)/x,x]

[Out]

(2*Sqrt[-a + b*x]*(23*a^2 - 11*a*b*x + 3*b^2*x^2))/15 - 2*a^(5/2)*ArcTan[Sqrt[-a + b*x]/Sqrt[a]]

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IntegrateAlgebraic [A]  time = 0.03, size = 74, normalized size = 1.01 \begin {gather*} \frac {2}{15} \left (15 a^2 \sqrt {b x-a}+3 (b x-a)^{5/2}-5 a (b x-a)^{3/2}\right )-2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-a + b*x)^(5/2)/x,x]

[Out]

(2*(15*a^2*Sqrt[-a + b*x] - 5*a*(-a + b*x)^(3/2) + 3*(-a + b*x)^(5/2)))/15 - 2*a^(5/2)*ArcTan[Sqrt[-a + b*x]/S
qrt[a]]

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fricas [A]  time = 1.16, size = 119, normalized size = 1.63 \begin {gather*} \left [\sqrt {-a} a^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + \frac {2}{15} \, {\left (3 \, b^{2} x^{2} - 11 \, a b x + 23 \, a^{2}\right )} \sqrt {b x - a}, -2 \, a^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + \frac {2}{15} \, {\left (3 \, b^{2} x^{2} - 11 \, a b x + 23 \, a^{2}\right )} \sqrt {b x - a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[sqrt(-a)*a^2*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) + 2/15*(3*b^2*x^2 - 11*a*b*x + 23*a^2)*sqrt(b*x -
a), -2*a^(5/2)*arctan(sqrt(b*x - a)/sqrt(a)) + 2/15*(3*b^2*x^2 - 11*a*b*x + 23*a^2)*sqrt(b*x - a)]

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giac [A]  time = 1.22, size = 57, normalized size = 0.78 \begin {gather*} -2 \, a^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + \frac {2}{5} \, {\left (b x - a\right )}^{\frac {5}{2}} - \frac {2}{3} \, {\left (b x - a\right )}^{\frac {3}{2}} a + 2 \, \sqrt {b x - a} a^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 58, normalized size = 0.79 \begin {gather*} -2 a^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )+2 \sqrt {b x -a}\, a^{2}-\frac {2 \left (b x -a \right )^{\frac {3}{2}} a}{3}+\frac {2 \left (b x -a \right )^{\frac {5}{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.92, size = 57, normalized size = 0.78 \begin {gather*} -2 \, a^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + \frac {2}{5} \, {\left (b x - a\right )}^{\frac {5}{2}} - \frac {2}{3} \, {\left (b x - a\right )}^{\frac {3}{2}} a + 2 \, \sqrt {b x - a} a^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 57, normalized size = 0.78 \begin {gather*} \frac {2\,{\left (b\,x-a\right )}^{5/2}}{5}-\frac {2\,a\,{\left (b\,x-a\right )}^{3/2}}{3}-2\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )+2\,a^2\,\sqrt {b\,x-a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x - a)^(5/2)/x,x)

[Out]

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

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sympy [C]  time = 4.24, size = 240, normalized size = 3.29 \begin {gather*} \begin {cases} \frac {46 a^{\frac {5}{2}} \sqrt {-1 + \frac {b x}{a}}}{15} + i a^{\frac {5}{2}} \log {\left (\frac {b x}{a} \right )} - 2 i a^{\frac {5}{2}} \log {\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + 2 a^{\frac {5}{2}} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {22 a^{\frac {3}{2}} b x \sqrt {-1 + \frac {b x}{a}}}{15} + \frac {2 \sqrt {a} b^{2} x^{2} \sqrt {-1 + \frac {b x}{a}}}{5} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\\frac {46 i a^{\frac {5}{2}} \sqrt {1 - \frac {b x}{a}}}{15} + i a^{\frac {5}{2}} \log {\left (\frac {b x}{a} \right )} - 2 i a^{\frac {5}{2}} \log {\left (\sqrt {1 - \frac {b x}{a}} + 1 \right )} - \frac {22 i a^{\frac {3}{2}} b x \sqrt {1 - \frac {b x}{a}}}{15} + \frac {2 i \sqrt {a} b^{2} x^{2} \sqrt {1 - \frac {b x}{a}}}{5} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((46*a**(5/2)*sqrt(-1 + b*x/a)/15 + I*a**(5/2)*log(b*x/a) - 2*I*a**(5/2)*log(sqrt(b)*sqrt(x)/sqrt(a))
+ 2*a**(5/2)*asin(sqrt(a)/(sqrt(b)*sqrt(x))) - 22*a**(3/2)*b*x*sqrt(-1 + b*x/a)/15 + 2*sqrt(a)*b**2*x**2*sqrt
(-1 + b*x/a)/5, Abs(b*x/a) > 1), (46*I*a**(5/2)*sqrt(1 - b*x/a)/15 + I*a**(5/2)*log(b*x/a) - 2*I*a**(5/2)*log(
sqrt(1 - b*x/a) + 1) - 22*I*a**(3/2)*b*x*sqrt(1 - b*x/a)/15 + 2*I*sqrt(a)*b**2*x**2*sqrt(1 - b*x/a)/5, True))

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