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

Optimal. Leaf size=72 \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{5/2}}+\frac {3 \sqrt {x}}{4 b^2 (a-b x)}-\frac {x^{3/2}}{2 b (a-b x)^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 72, 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 = {47, 63, 208} \begin {gather*} \frac {3 \sqrt {x}}{4 b^2 (a-b x)}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{5/2}}-\frac {x^{3/2}}{2 b (a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(-a + b*x)^3,x]

[Out]

-x^(3/2)/(2*b*(a - b*x)^2) + (3*Sqrt[x])/(4*b^2*(a - b*x)) - (3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*Sqrt[a]
*b^(5/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 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 {x^{3/2}}{(-a+b x)^3} \, dx &=-\frac {x^{3/2}}{2 b (a-b x)^2}+\frac {3 \int \frac {\sqrt {x}}{(-a+b x)^2} \, dx}{4 b}\\ &=-\frac {x^{3/2}}{2 b (a-b x)^2}+\frac {3 \sqrt {x}}{4 b^2 (a-b x)}+\frac {3 \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{8 b^2}\\ &=-\frac {x^{3/2}}{2 b (a-b x)^2}+\frac {3 \sqrt {x}}{4 b^2 (a-b x)}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{4 b^2}\\ &=-\frac {x^{3/2}}{2 b (a-b x)^2}+\frac {3 \sqrt {x}}{4 b^2 (a-b x)}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 60, normalized size = 0.83 \begin {gather*} \frac {\sqrt {x} (3 a-5 b x)}{4 b^2 (a-b x)^2}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(-a + b*x)^3,x]

[Out]

(Sqrt[x]*(3*a - 5*b*x))/(4*b^2*(a - b*x)^2) - (3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*Sqrt[a]*b^(5/2))

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IntegrateAlgebraic [A]  time = 0.12, size = 65, normalized size = 0.90 \begin {gather*} \frac {3 a \sqrt {x}-5 b x^{3/2}}{4 b^2 (b x-a)^2}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*Sqrt[x] - 5*b*x^(3/2))/(4*b^2*(-a + b*x)^2) - (3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*Sqrt[a]*b^(5/2))

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fricas [A]  time = 0.71, size = 186, normalized size = 2.58 \begin {gather*} \left [\frac {3 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {a b} \log \left (\frac {b x + a - 2 \, \sqrt {a b} \sqrt {x}}{b x - a}\right ) - 2 \, {\left (5 \, a b^{2} x - 3 \, a^{2} b\right )} \sqrt {x}}{8 \, {\left (a b^{5} x^{2} - 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}, \frac {3 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {-a b} \arctan \left (\frac {\sqrt {-a b}}{b \sqrt {x}}\right ) - {\left (5 \, a b^{2} x - 3 \, a^{2} b\right )} \sqrt {x}}{4 \, {\left (a b^{5} x^{2} - 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.92, size = 51, normalized size = 0.71 \begin {gather*} \frac {3 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{4 \, \sqrt {-a b} b^{2}} - \frac {5 \, b x^{\frac {3}{2}} - 3 \, a \sqrt {x}}{4 \, {\left (b x - a\right )}^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 52, normalized size = 0.72 \begin {gather*} -\frac {3 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{2}}+\frac {-\frac {5 x^{\frac {3}{2}}}{4 b}+\frac {3 a \sqrt {x}}{4 b^{2}}}{\left (b x -a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-5/8/b*x^(3/2)+3/8*a/b^2*x^(1/2))/(b*x-a)^2-3/4/b^2/(a*b)^(1/2)*arctanh(1/(a*b)^(1/2)*b*x^(1/2))

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maxima [A]  time = 2.99, size = 78, normalized size = 1.08 \begin {gather*} -\frac {5 \, b x^{\frac {3}{2}} - 3 \, a \sqrt {x}}{4 \, {\left (b^{4} x^{2} - 2 \, a b^{3} x + a^{2} b^{2}\right )}} + \frac {3 \, \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.14, size = 58, normalized size = 0.81 \begin {gather*} -\frac {\frac {5\,x^{3/2}}{4\,b}-\frac {3\,a\,\sqrt {x}}{4\,b^2}}{a^2-2\,a\,b\,x+b^2\,x^2}-\frac {3\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,\sqrt {a}\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^(3/2)/(a - b*x)^3,x)

[Out]

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

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sympy [A]  time = 29.25, size = 673, normalized size = 9.35 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2 x^{\frac {5}{2}}}{5 a^{3}} & \text {for}\: b = 0 \\- \frac {2}{b^{3} \sqrt {x}} & \text {for}\: a = 0 \\\frac {6 a^{\frac {3}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{8 a^{\frac {5}{2}} b^{3} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{4} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{5} x^{2} \sqrt {\frac {1}{b}}} - \frac {10 \sqrt {a} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{8 a^{\frac {5}{2}} b^{3} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{4} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{5} x^{2} \sqrt {\frac {1}{b}}} + \frac {3 a^{2} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{3} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{4} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{5} x^{2} \sqrt {\frac {1}{b}}} - \frac {3 a^{2} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{3} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{4} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{5} x^{2} \sqrt {\frac {1}{b}}} - \frac {6 a b x \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{3} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{4} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{5} x^{2} \sqrt {\frac {1}{b}}} + \frac {6 a b x \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{3} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{4} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{5} x^{2} \sqrt {\frac {1}{b}}} + \frac {3 b^{2} x^{2} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{3} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{4} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{5} x^{2} \sqrt {\frac {1}{b}}} - \frac {3 b^{2} x^{2} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{3} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{4} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{5} x^{2} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/sqrt(x), Eq(a, 0) & Eq(b, 0)), (-2*x**(5/2)/(5*a**3), Eq(b, 0)), (-2/(b**3*sqrt(x)), Eq(a, 0)),
(6*a**(3/2)*b*sqrt(x)*sqrt(1/b)/(8*a**(5/2)*b**3*sqrt(1/b) - 16*a**(3/2)*b**4*x*sqrt(1/b) + 8*sqrt(a)*b**5*x*
*2*sqrt(1/b)) - 10*sqrt(a)*b**2*x**(3/2)*sqrt(1/b)/(8*a**(5/2)*b**3*sqrt(1/b) - 16*a**(3/2)*b**4*x*sqrt(1/b) +
8*sqrt(a)*b**5*x**2*sqrt(1/b)) + 3*a**2*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**3*sqrt(1/b) - 16*a**
(3/2)*b**4*x*sqrt(1/b) + 8*sqrt(a)*b**5*x**2*sqrt(1/b)) - 3*a**2*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*
b**3*sqrt(1/b) - 16*a**(3/2)*b**4*x*sqrt(1/b) + 8*sqrt(a)*b**5*x**2*sqrt(1/b)) - 6*a*b*x*log(-sqrt(a)*sqrt(1/b
) + sqrt(x))/(8*a**(5/2)*b**3*sqrt(1/b) - 16*a**(3/2)*b**4*x*sqrt(1/b) + 8*sqrt(a)*b**5*x**2*sqrt(1/b)) + 6*a*
b*x*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**3*sqrt(1/b) - 16*a**(3/2)*b**4*x*sqrt(1/b) + 8*sqrt(a)*b**
5*x**2*sqrt(1/b)) + 3*b**2*x**2*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**3*sqrt(1/b) - 16*a**(3/2)*b**
4*x*sqrt(1/b) + 8*sqrt(a)*b**5*x**2*sqrt(1/b)) - 3*b**2*x**2*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**3
*sqrt(1/b) - 16*a**(3/2)*b**4*x*sqrt(1/b) + 8*sqrt(a)*b**5*x**2*sqrt(1/b)), True))

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