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

Optimal. Leaf size=57 \[ -\frac {3 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}}+\frac {x^{3/2}}{b (a-b x)}+\frac {3 \sqrt {x}}{b^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {47, 50, 63, 208} \begin {gather*} -\frac {3 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}}+\frac {x^{3/2}}{b (a-b x)}+\frac {3 \sqrt {x}}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(-a + b*x)^2,x]

[Out]

(3*Sqrt[x])/b^2 + x^(3/2)/(b*(a - b*x)) - (3*Sqrt[a]*ArcTanh[(Sqrt[b]*Sqrt[x])/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 50

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

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Mathematica [C]  time = 0.00, size = 26, normalized size = 0.46 \begin {gather*} \frac {2 x^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {b x}{a}\right )}{5 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(-a + b*x)^2,x]

[Out]

(2*x^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, (b*x)/a])/(5*a^2)

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IntegrateAlgebraic [A]  time = 0.06, size = 56, normalized size = 0.98 \begin {gather*} \frac {\sqrt {x} (2 b x-3 a)}{b^2 (b x-a)}-\frac {3 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 138, normalized size = 2.42 \begin {gather*} \left [\frac {3 \, {\left (b x - a\right )} \sqrt {\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {\frac {a}{b}} + a}{b x - a}\right ) + 2 \, {\left (2 \, b x - 3 \, a\right )} \sqrt {x}}{2 \, {\left (b^{3} x - a b^{2}\right )}}, \frac {3 \, {\left (b x - a\right )} \sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {-\frac {a}{b}}}{a}\right ) + {\left (2 \, b x - 3 \, a\right )} \sqrt {x}}{b^{3} x - a b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.98, size = 51, normalized size = 0.89 \begin {gather*} \frac {3 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} b^{2}} - \frac {a \sqrt {x}}{{\left (b x - a\right )} b^{2}} + \frac {2 \, \sqrt {x}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 49, normalized size = 0.86 \begin {gather*} \frac {2 \left (-\frac {3 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}-\frac {\sqrt {x}}{2 \left (b x -a \right )}\right ) a}{b^{2}}+\frac {2 \sqrt {x}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.92, size = 68, normalized size = 1.19 \begin {gather*} -\frac {a \sqrt {x}}{b^{3} x - a b^{2}} + \frac {3 \, a \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{2}} + \frac {2 \, \sqrt {x}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*sqrt(x)/(b^3*x - a*b^2) + 3/2*a*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/(sqrt(a*b)*b^2) + 2*sq
rt(x)/b^2

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mupad [B]  time = 0.11, size = 47, normalized size = 0.82 \begin {gather*} \frac {2\,\sqrt {x}}{b^2}+\frac {a\,\sqrt {x}}{a\,b^2-b^3\,x}-\frac {3\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a - b*x)^2,x)

[Out]

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

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sympy [A]  time = 9.13, size = 381, normalized size = 6.68 \begin {gather*} \begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {5}{2}}}{5 a^{2}} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{b^{2}} & \text {for}\: a = 0 \\- \frac {6 a^{\frac {3}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{- 2 a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {4 \sqrt {a} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{- 2 a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} - \frac {3 a^{2} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{- 2 a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {3 a^{2} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{- 2 a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {3 a b x \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{- 2 a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} - \frac {3 a b x \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{- 2 a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 \sqrt {a} b^{4} x \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)**2,x)

[Out]

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

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