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

Optimal. Leaf size=70 \[ \frac {5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {5 b}{a^3 \sqrt {x}}-\frac {5}{3 a^2 x^{3/2}}+\frac {1}{a x^{3/2} (a-b x)} \]

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Rubi [A]  time = 0.02, antiderivative size = 70, 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 = {51, 63, 208} \begin {gather*} \frac {5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {5 b}{a^3 \sqrt {x}}-\frac {5}{3 a^2 x^{3/2}}+\frac {1}{a x^{3/2} (a-b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(-a + b*x)^2),x]

[Out]

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

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 208

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

Rubi steps

\begin {align*} \int \frac {1}{x^{5/2} (-a+b x)^2} \, dx &=\frac {1}{a x^{3/2} (a-b x)}-\frac {5 \int \frac {1}{x^{5/2} (-a+b x)} \, dx}{2 a}\\ &=-\frac {5}{3 a^2 x^{3/2}}+\frac {1}{a x^{3/2} (a-b x)}-\frac {(5 b) \int \frac {1}{x^{3/2} (-a+b x)} \, dx}{2 a^2}\\ &=-\frac {5}{3 a^2 x^{3/2}}-\frac {5 b}{a^3 \sqrt {x}}+\frac {1}{a x^{3/2} (a-b x)}-\frac {\left (5 b^2\right ) \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{2 a^3}\\ &=-\frac {5}{3 a^2 x^{3/2}}-\frac {5 b}{a^3 \sqrt {x}}+\frac {1}{a x^{3/2} (a-b x)}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{a^3}\\ &=-\frac {5}{3 a^2 x^{3/2}}-\frac {5 b}{a^3 \sqrt {x}}+\frac {1}{a x^{3/2} (a-b x)}+\frac {5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 26, normalized size = 0.37 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {b x}{a}\right )}{3 a^2 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(-a + b*x)^2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.08, size = 69, normalized size = 0.99 \begin {gather*} \frac {5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {-2 a^2-10 a b x+15 b^2 x^2}{3 a^3 x^{3/2} (a-b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2 - 10*a*b*x + 15*b^2*x^2)/(3*a^3*x^(3/2)*(a - b*x)) + (5*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^
(7/2)

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fricas [A]  time = 1.00, size = 187, normalized size = 2.67 \begin {gather*} \left [\frac {15 \, {\left (b^{2} x^{3} - a b x^{2}\right )} \sqrt {\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {\frac {b}{a}} + a}{b x - a}\right ) - 2 \, {\left (15 \, b^{2} x^{2} - 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (a^{3} b x^{3} - a^{4} x^{2}\right )}}, -\frac {15 \, {\left (b^{2} x^{3} - a b x^{2}\right )} \sqrt {-\frac {b}{a}} \arctan \left (\frac {a \sqrt {-\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (15 \, b^{2} x^{2} - 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} - a^{4} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(b^2*x^3 - a*b*x^2)*sqrt(b/a)*log((b*x + 2*a*sqrt(x)*sqrt(b/a) + a)/(b*x - a)) - 2*(15*b^2*x^2 - 10*a
*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 - a^4*x^2), -1/3*(15*(b^2*x^3 - a*b*x^2)*sqrt(-b/a)*arctan(a*sqrt(-b/a)/(b*s
qrt(x))) + (15*b^2*x^2 - 10*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 - a^4*x^2)]

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giac [A]  time = 0.93, size = 61, normalized size = 0.87 \begin {gather*} -\frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} a^{3}} - \frac {b^{2} \sqrt {x}}{{\left (b x - a\right )} a^{3}} - \frac {2 \, {\left (6 \, b x + a\right )}}{3 \, a^{3} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*b^2*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*a^3) - b^2*sqrt(x)/((b*x - a)*a^3) - 2/3*(6*b*x + a)/(a^3*x^(3
/2))

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maple [A]  time = 0.02, size = 60, normalized size = 0.86 \begin {gather*} -\frac {2 \left (-\frac {5 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}+\frac {\sqrt {x}}{2 b x -2 a}\right ) b^{2}}{a^{3}}-\frac {4 b}{a^{3} \sqrt {x}}-\frac {2}{3 a^{2} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.00, size = 82, normalized size = 1.17 \begin {gather*} -\frac {15 \, b^{2} x^{2} - 10 \, a b x - 2 \, a^{2}}{3 \, {\left (a^{3} b x^{\frac {5}{2}} - a^{4} x^{\frac {3}{2}}\right )}} - \frac {5 \, b^{2} \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(15*b^2*x^2 - 10*a*b*x - 2*a^2)/(a^3*b*x^(5/2) - a^4*x^(3/2)) - 5/2*b^2*log((b*sqrt(x) - sqrt(a*b))/(b*sq
rt(x) + sqrt(a*b)))/(sqrt(a*b)*a^3)

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mupad [B]  time = 0.14, size = 60, normalized size = 0.86 \begin {gather*} \frac {5\,b^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {\frac {2}{3\,a}-\frac {5\,b^2\,x^2}{a^3}+\frac {10\,b\,x}{3\,a^2}}{a\,x^{3/2}-b\,x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a - b*x)^2),x)

[Out]

(5*b^(3/2)*atanh((b^(1/2)*x^(1/2))/a^(1/2)))/a^(7/2) - (2/(3*a) - (5*b^2*x^2)/a^3 + (10*b*x)/(3*a^2))/(a*x^(3/
2) - b*x^(5/2))

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sympy [A]  time = 50.25, size = 471, normalized size = 6.73 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {7}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{7 b^{2} x^{\frac {7}{2}}} & \text {for}\: a = 0 \\- \frac {2}{3 a^{2} x^{\frac {3}{2}}} & \text {for}\: b = 0 \\- \frac {4 a^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{6 a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 6 a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {20 a^{\frac {3}{2}} b x \sqrt {\frac {1}{b}}}{6 a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 6 a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {30 \sqrt {a} b^{2} x^{2} \sqrt {\frac {1}{b}}}{6 a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 6 a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 a b x^{\frac {3}{2}} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 6 a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 a b x^{\frac {3}{2}} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 6 a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 6 a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 6 a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (-2/(7*b**2*x**(7/2)), Eq(a, 0)), (-2/(3*a**2*x**(3/2)), Eq(b,
0)), (-4*a**(5/2)*sqrt(1/b)/(6*a**(9/2)*x**(3/2)*sqrt(1/b) - 6*a**(7/2)*b*x**(5/2)*sqrt(1/b)) - 20*a**(3/2)*b*
x*sqrt(1/b)/(6*a**(9/2)*x**(3/2)*sqrt(1/b) - 6*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 30*sqrt(a)*b**2*x**2*sqrt(1/b)
/(6*a**(9/2)*x**(3/2)*sqrt(1/b) - 6*a**(7/2)*b*x**(5/2)*sqrt(1/b)) - 15*a*b*x**(3/2)*log(-sqrt(a)*sqrt(1/b) +
sqrt(x))/(6*a**(9/2)*x**(3/2)*sqrt(1/b) - 6*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 15*a*b*x**(3/2)*log(sqrt(a)*sqrt(
1/b) + sqrt(x))/(6*a**(9/2)*x**(3/2)*sqrt(1/b) - 6*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 15*b**2*x**(5/2)*log(-sqrt
(a)*sqrt(1/b) + sqrt(x))/(6*a**(9/2)*x**(3/2)*sqrt(1/b) - 6*a**(7/2)*b*x**(5/2)*sqrt(1/b)) - 15*b**2*x**(5/2)*
log(sqrt(a)*sqrt(1/b) + sqrt(x))/(6*a**(9/2)*x**(3/2)*sqrt(1/b) - 6*a**(7/2)*b*x**(5/2)*sqrt(1/b)), True))

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