3.5.81 \(\int \frac {1}{x^{5/2} (-a+b x)^2} \, dx\) [481]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 70, 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 = {44, 53, 65, 214} \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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], 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 ) \text {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 [A]
time = 0.08, size = 69, normalized size = 0.99 \begin {gather*} \frac {-2 a^2-10 a b x+15 b^2 x^2}{3 a^3 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 {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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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Maple [A]
time = 0.11, size = 59, normalized size = 0.84

method	result	size
risch	\(-\frac {2 \left (6 b x +a \right )}{3 a^{3} x^{\frac {3}{2}}}-\frac {b^{2} \left (\frac {\sqrt {x}}{b x -a}-\frac {5 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{a^{3}}\)	\(56\)
derivativedivides	\(\frac {2 b^{2} \left (\frac {\sqrt {x}}{-2 b x +2 a}+\frac {5 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}-\frac {2}{3 a^{2} x^{\frac {3}{2}}}-\frac {4 b}{a^{3} \sqrt {x}}\)	\(59\)
default	\(\frac {2 b^{2} \left (\frac {\sqrt {x}}{-2 b x +2 a}+\frac {5 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}-\frac {2}{3 a^{2} x^{\frac {3}{2}}}-\frac {4 b}{a^{3} \sqrt {x}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.56, 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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Fricas [A]
time = 0.40, 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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 416 vs. \(2 (65) = 130\).
time = 30.59, size = 416, normalized size = 5.94 \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^{2} \sqrt {\frac {a}{b}}}{6 a^{4} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 6 a^{3} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} - \frac {15 a b x^{\frac {3}{2}} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{6 a^{4} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 6 a^{3} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} + \frac {15 a b x^{\frac {3}{2}} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{6 a^{4} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 6 a^{3} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} - \frac {20 a b x \sqrt {\frac {a}{b}}}{6 a^{4} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 6 a^{3} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} + \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{6 a^{4} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 6 a^{3} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} - \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{6 a^{4} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 6 a^{3} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} + \frac {30 b^{2} x^{2} \sqrt {\frac {a}{b}}}{6 a^{4} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 6 a^{3} b x^{\frac {5}{2}} \sqrt {\frac {a}{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**2*sqrt(a/b)/(6*a**4*x**(3/2)*sqrt(a/b) - 6*a**3*b*x**(5/2)*sqrt(a/b)) - 15*a*b*x**(3/2)*log(sqrt(x
) - sqrt(a/b))/(6*a**4*x**(3/2)*sqrt(a/b) - 6*a**3*b*x**(5/2)*sqrt(a/b)) + 15*a*b*x**(3/2)*log(sqrt(x) + sqrt(
a/b))/(6*a**4*x**(3/2)*sqrt(a/b) - 6*a**3*b*x**(5/2)*sqrt(a/b)) - 20*a*b*x*sqrt(a/b)/(6*a**4*x**(3/2)*sqrt(a/b
) - 6*a**3*b*x**(5/2)*sqrt(a/b)) + 15*b**2*x**(5/2)*log(sqrt(x) - sqrt(a/b))/(6*a**4*x**(3/2)*sqrt(a/b) - 6*a*
*3*b*x**(5/2)*sqrt(a/b)) - 15*b**2*x**(5/2)*log(sqrt(x) + sqrt(a/b))/(6*a**4*x**(3/2)*sqrt(a/b) - 6*a**3*b*x**
(5/2)*sqrt(a/b)) + 30*b**2*x**2*sqrt(a/b)/(6*a**4*x**(3/2)*sqrt(a/b) - 6*a**3*b*x**(5/2)*sqrt(a/b)), True))

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Giac [A]
time = 1.37, 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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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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