3.5.70 \(\int \frac {x^{3/2}}{-a+b x} \, dx\) [470]

Optimal. Leaf size=53 \[ \frac {2 a \sqrt {x}}{b^2}+\frac {2 x^{3/2}}{3 b}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 53, 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 = {52, 65, 214} \begin {gather*} -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}}+\frac {2 a \sqrt {x}}{b^2}+\frac {2 x^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(-a + b*x),x]

[Out]

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

Rule 52

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

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Mathematica [A]
time = 0.04, size = 49, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {x} (3 a+b x)}{3 b^2}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(-a + b*x),x]

[Out]

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

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 2.23, size = 111, normalized size = 2.09 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\text {DirectedInfinity}\left [x^{\frac {3}{2}}\right ],a\text {==}0\text {\&\&}b\text {==}0\right \},\left \{\frac {-2 x^{\frac {5}{2}}}{5 a},b\text {==}0\right \},\left \{\frac {2 x^{\frac {3}{2}}}{3 b},a\text {==}0\right \}\right \},-\frac {a^2 \text {Log}\left [\sqrt {x}+\sqrt {\frac {a}{b}}\right ]}{b^3 \sqrt {\frac {a}{b}}}+\frac {a^2 \text {Log}\left [\sqrt {x}-\sqrt {\frac {a}{b}}\right ]}{b^3 \sqrt {\frac {a}{b}}}+\frac {2 a \sqrt {x}}{b^2}+\frac {2 x^{\frac {3}{2}}}{3 b}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x^(3/2)/(-a + b*x),x]')

[Out]

Piecewise[{{DirectedInfinity[x ^ (3 / 2)], a == 0 && b == 0}, {-2 x ^ (5 / 2) / (5 a), b == 0}, {2 x ^ (3 / 2)
/ (3 b), a == 0}}, -a ^ 2 Log[Sqrt[x] + Sqrt[a / b]] / (b ^ 3 Sqrt[a / b]) + a ^ 2 Log[Sqrt[x] - Sqrt[a / b]]
/ (b ^ 3 Sqrt[a / b]) + 2 a Sqrt[x] / b ^ 2 + 2 x ^ (3 / 2) / (3 b)]

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Maple [A]
time = 0.12, size = 43, normalized size = 0.81

method	result	size
risch	\(\frac {2 \left (b x +3 a \right ) \sqrt {x}}{3 b^{2}}-\frac {2 a^{2} \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(41\)
derivativedivides	\(\frac {\frac {2 b \,x^{\frac {3}{2}}}{3}+2 a \sqrt {x}}{b^{2}}-\frac {2 a^{2} \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(43\)
default	\(\frac {\frac {2 b \,x^{\frac {3}{2}}}{3}+2 a \sqrt {x}}{b^{2}}-\frac {2 a^{2} \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 58, normalized size = 1.09 \begin {gather*} \frac {a^{2} \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (b x^{\frac {3}{2}} + 3 \, a \sqrt {x}\right )}}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.62, size = 103, normalized size = 1.94 \begin {gather*} \left [\frac {3 \, a \sqrt {\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {\frac {a}{b}} + a}{b x - a}\right ) + 2 \, {\left (b x + 3 \, a\right )} \sqrt {x}}{3 \, b^{2}}, \frac {2 \, {\left (3 \, a \sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {-\frac {a}{b}}}{a}\right ) + {\left (b x + 3 \, a\right )} \sqrt {x}\right )}}{3 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.74, size = 102, normalized size = 1.92 \begin {gather*} \begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2 x^{\frac {5}{2}}}{5 a} & \text {for}\: b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 b} & \text {for}\: a = 0 \\\frac {a^{2} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{b^{3} \sqrt {\frac {a}{b}}} - \frac {a^{2} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{b^{3} \sqrt {\frac {a}{b}}} + \frac {2 a \sqrt {x}}{b^{2}} + \frac {2 x^{\frac {3}{2}}}{3 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),x)

[Out]

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

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Giac [A]
time = 0.00, size = 70, normalized size = 1.32 \begin {gather*} -2 \left (\frac {-\frac {1}{3} \sqrt {x} x b^{2}-\sqrt {x} b a}{b^{3}}-\frac {2 a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{2 b^{2} \sqrt {-a b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.11, size = 37, normalized size = 0.70 \begin {gather*} \frac {2\,x^{3/2}}{3\,b}+\frac {2\,a\,\sqrt {x}}{b^2}-\frac {2\,a^{3/2}\,\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),x)

[Out]

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

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