3.487 \(\int \frac {1}{x^{3/2} (-a+b x)^3} \, dx\)

Optimal. Leaf size=84 \[ -\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}}+\frac {15}{4 a^3 \sqrt {x}}-\frac {5}{4 a^2 \sqrt {x} (a-b x)}-\frac {1}{2 a \sqrt {x} (a-b x)^2} \]

[Out]

-15/4*arctanh(b^(1/2)*x^(1/2)/a^(1/2))*b^(1/2)/a^(7/2)+15/4/a^3/x^(1/2)-1/2/a/(-b*x+a)^2/x^(1/2)-5/4/a^2/(-b*x
+a)/x^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 84, 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} \[ -\frac {5}{4 a^2 \sqrt {x} (a-b x)}-\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}}+\frac {15}{4 a^3 \sqrt {x}}-\frac {1}{2 a \sqrt {x} (a-b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(-a + b*x)^3),x]

[Out]

15/(4*a^3*Sqrt[x]) - 1/(2*a*Sqrt[x]*(a - b*x)^2) - 5/(4*a^2*Sqrt[x]*(a - b*x)) - (15*Sqrt[b]*ArcTanh[(Sqrt[b]*
Sqrt[x])/Sqrt[a]])/(4*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^{3/2} (-a+b x)^3} \, dx &=-\frac {1}{2 a \sqrt {x} (a-b x)^2}-\frac {5 \int \frac {1}{x^{3/2} (-a+b x)^2} \, dx}{4 a}\\ &=-\frac {1}{2 a \sqrt {x} (a-b x)^2}-\frac {5}{4 a^2 \sqrt {x} (a-b x)}+\frac {15 \int \frac {1}{x^{3/2} (-a+b x)} \, dx}{8 a^2}\\ &=\frac {15}{4 a^3 \sqrt {x}}-\frac {1}{2 a \sqrt {x} (a-b x)^2}-\frac {5}{4 a^2 \sqrt {x} (a-b x)}+\frac {(15 b) \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{8 a^3}\\ &=\frac {15}{4 a^3 \sqrt {x}}-\frac {1}{2 a \sqrt {x} (a-b x)^2}-\frac {5}{4 a^2 \sqrt {x} (a-b x)}+\frac {(15 b) \operatorname {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{4 a^3}\\ &=\frac {15}{4 a^3 \sqrt {x}}-\frac {1}{2 a \sqrt {x} (a-b x)^2}-\frac {5}{4 a^2 \sqrt {x} (a-b x)}-\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 24, normalized size = 0.29 \[ \frac {2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {b x}{a}\right )}{a^3 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(-a + b*x)^3),x]

[Out]

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

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fricas [A]  time = 0.45, size = 213, normalized size = 2.54 \[ \left [\frac {15 \, {\left (b^{2} x^{3} - 2 \, a b x^{2} + a^{2} x\right )} \sqrt {\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {\frac {b}{a}} + a}{b x - a}\right ) + 2 \, {\left (15 \, b^{2} x^{2} - 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{2} x^{3} - 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac {15 \, {\left (b^{2} x^{3} - 2 \, a b x^{2} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \arctan \left (\frac {a \sqrt {-\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (15 \, b^{2} x^{2} - 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{2} x^{3} - 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x-a)^3,x, algorithm="fricas")

[Out]

[1/8*(15*(b^2*x^3 - 2*a*b*x^2 + a^2*x)*sqrt(b/a)*log((b*x - 2*a*sqrt(x)*sqrt(b/a) + a)/(b*x - a)) + 2*(15*b^2*
x^2 - 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 - 2*a^4*b*x^2 + a^5*x), 1/4*(15*(b^2*x^3 - 2*a*b*x^2 + a^2*x)*sq
rt(-b/a)*arctan(a*sqrt(-b/a)/(b*sqrt(x))) + (15*b^2*x^2 - 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 - 2*a^4*b*x^
2 + a^5*x)]

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giac [A]  time = 1.05, size = 63, normalized size = 0.75 \[ \frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{4 \, \sqrt {-a b} a^{3}} + \frac {2}{a^{3} \sqrt {x}} + \frac {7 \, b^{2} x^{\frac {3}{2}} - 9 \, a b \sqrt {x}}{4 \, {\left (b x - a\right )}^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x-a)^3,x, algorithm="giac")

[Out]

15/4*b*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*a^3) + 2/(a^3*sqrt(x)) + 1/4*(7*b^2*x^(3/2) - 9*a*b*sqrt(x))/(
(b*x - a)^2*a^3)

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maple [A]  time = 0.02, size = 58, normalized size = 0.69 \[ \frac {2 \left (-\frac {15 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}+\frac {\frac {7 b \,x^{\frac {3}{2}}}{8}-\frac {9 a \sqrt {x}}{8}}{\left (b x -a \right )^{2}}\right ) b}{a^{3}}+\frac {2}{a^{3} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(b*x-a)^3,x)

[Out]

2/a^3*b*((7/8*b*x^(3/2)-9/8*a*x^(1/2))/(b*x-a)^2-15/8/(a*b)^(1/2)*arctanh(1/(a*b)^(1/2)*b*x^(1/2)))+2/a^3/x^(1
/2)

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maxima [A]  time = 2.93, size = 90, normalized size = 1.07 \[ \frac {15 \, b^{2} x^{2} - 25 \, a b x + 8 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{\frac {5}{2}} - 2 \, a^{4} b x^{\frac {3}{2}} + a^{5} \sqrt {x}\right )}} + \frac {15 \, b \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x-a)^3,x, algorithm="maxima")

[Out]

1/4*(15*b^2*x^2 - 25*a*b*x + 8*a^2)/(a^3*b^2*x^(5/2) - 2*a^4*b*x^(3/2) + a^5*sqrt(x)) + 15/8*b*log((b*sqrt(x)
- sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/(sqrt(a*b)*a^3)

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mupad [B]  time = 0.15, size = 69, normalized size = 0.82 \[ \frac {\frac {2}{a}+\frac {15\,b^2\,x^2}{4\,a^3}-\frac {25\,b\,x}{4\,a^2}}{a^2\,\sqrt {x}+b^2\,x^{5/2}-2\,a\,b\,x^{3/2}}-\frac {15\,\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^(3/2)*(a - b*x)^3),x)

[Out]

(2/a + (15*b^2*x^2)/(4*a^3) - (25*b*x)/(4*a^2))/(a^2*x^(1/2) + b^2*x^(5/2) - 2*a*b*x^(3/2)) - (15*b^(1/2)*atan
h((b^(1/2)*x^(1/2))/a^(1/2)))/(4*a^(7/2))

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sympy [A]  time = 54.56, size = 802, normalized size = 9.55 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {7}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2}{a^{3} \sqrt {x}} & \text {for}\: b = 0 \\- \frac {2}{7 b^{3} x^{\frac {7}{2}}} & \text {for}\: a = 0 \\\frac {16 a^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {50 a^{\frac {3}{2}} b x \sqrt {\frac {1}{b}}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {30 \sqrt {a} b^{2} x^{2} \sqrt {\frac {1}{b}}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 a^{2} \sqrt {x} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 a^{2} \sqrt {x} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {30 a b x^{\frac {3}{2}} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {30 a b x^{\frac {3}{2}} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} 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 )}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} 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 )}}{8 a^{\frac {11}{2}} \sqrt {x} \sqrt {\frac {1}{b}} - 16 a^{\frac {9}{2}} b x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 8 a^{\frac {7}{2}} b^{2} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(b*x-a)**3,x)

[Out]

Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (2/(a**3*sqrt(x)), Eq(b, 0)), (-2/(7*b**3*x**(7/2)), Eq(a, 0)),
(16*a**(5/2)*sqrt(1/b)/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x*
*(5/2)*sqrt(1/b)) - 50*a**(3/2)*b*x*sqrt(1/b)/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b
) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 30*sqrt(a)*b**2*x**2*sqrt(1/b)/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a
**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 15*a**2*sqrt(x)*log(-sqrt(a)*sqrt(1/b) +
sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b
)) - 15*a**2*sqrt(x)*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*
sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 30*a*b*x**(3/2)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(11/
2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 30*a*b*x**(3/2
)*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7
/2)*b**2*x**(5/2)*sqrt(1/b)) + 15*b**2*x**(5/2)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/
b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 15*b**2*x**(5/2)*log(sqrt(a)*sqr
t(1/b) + sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)
*sqrt(1/b)), True))

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