3.4.65 \(\int \frac {1}{x (-a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=60 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2}{a^2 \sqrt {b x-a}}-\frac {2}{3 a (b x-a)^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {51, 63, 205} \begin {gather*} \frac {2}{a^2 \sqrt {b x-a}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {2}{3 a (b x-a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(-a + b*x)^(5/2)),x]

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {1}{x (-a+b x)^{5/2}} \, dx &=-\frac {2}{3 a (-a+b x)^{3/2}}-\frac {\int \frac {1}{x (-a+b x)^{3/2}} \, dx}{a}\\ &=-\frac {2}{3 a (-a+b x)^{3/2}}+\frac {2}{a^2 \sqrt {-a+b x}}+\frac {\int \frac {1}{x \sqrt {-a+b x}} \, dx}{a^2}\\ &=-\frac {2}{3 a (-a+b x)^{3/2}}+\frac {2}{a^2 \sqrt {-a+b x}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right )}{a^2 b}\\ &=-\frac {2}{3 a (-a+b x)^{3/2}}+\frac {2}{a^2 \sqrt {-a+b x}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 35, normalized size = 0.58 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};1-\frac {b x}{a}\right )}{3 a (b x-a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(-a + b*x)^(5/2)),x]

[Out]

(-2*Hypergeometric2F1[-3/2, 1, -1/2, 1 - (b*x)/a])/(3*a*(-a + b*x)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.04, size = 55, normalized size = 0.92 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {2 (a-3 (b x-a))}{3 a^2 (b x-a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*(-a + b*x)^(5/2)),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 182, normalized size = 3.03 \begin {gather*} \left [-\frac {3 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {-a} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b x - 4 \, a^{2}\right )} \sqrt {b x - a}}{3 \, {\left (a^{3} b^{2} x^{2} - 2 \, a^{4} b x + a^{5}\right )}}, \frac {2 \, {\left (3 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (3 \, a b x - 4 \, a^{2}\right )} \sqrt {b x - a}\right )}}{3 \, {\left (a^{3} b^{2} x^{2} - 2 \, a^{4} b x + a^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.06, size = 42, normalized size = 0.70 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, b x - 4 \, a\right )}}{3 \, {\left (b x - a\right )}^{\frac {3}{2}} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 49, normalized size = 0.82 \begin {gather*} -\frac {2}{3 \left (b x -a \right )^{\frac {3}{2}} a}+\frac {2 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {2}{\sqrt {b x -a}\, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x-a)^(5/2),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.94, size = 42, normalized size = 0.70 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, b x - 4 \, a\right )}}{3 \, {\left (b x - a\right )}^{\frac {3}{2}} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.09, size = 48, normalized size = 0.80 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\frac {2\,\left (a-b\,x\right )}{a^2}+\frac {2}{3\,a}}{{\left (b\,x-a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*x - a)^(5/2)),x)

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________