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

Optimal. Leaf size=68 \[ \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {(b x-a)^{3/2}}{2 x^2}-\frac {3 b \sqrt {b x-a}}{4 x} \]

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Rubi [A]  time = 0.01, antiderivative size = 68, 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, 205} \begin {gather*} \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {(b x-a)^{3/2}}{2 x^2}-\frac {3 b \sqrt {b x-a}}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-a + b*x)^(3/2)/x^3,x]

[Out]

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

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin {align*} \int \frac {(-a+b x)^{3/2}}{x^3} \, dx &=-\frac {(-a+b x)^{3/2}}{2 x^2}+\frac {1}{4} (3 b) \int \frac {\sqrt {-a+b x}}{x^2} \, dx\\ &=-\frac {3 b \sqrt {-a+b x}}{4 x}-\frac {(-a+b x)^{3/2}}{2 x^2}+\frac {1}{8} \left (3 b^2\right ) \int \frac {1}{x \sqrt {-a+b x}} \, dx\\ &=-\frac {3 b \sqrt {-a+b x}}{4 x}-\frac {(-a+b x)^{3/2}}{2 x^2}+\frac {1}{4} (3 b) \operatorname {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right )\\ &=-\frac {3 b \sqrt {-a+b x}}{4 x}-\frac {(-a+b x)^{3/2}}{2 x^2}+\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 \sqrt {a}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 72, normalized size = 1.06 \begin {gather*} -\frac {2 a^2+3 b^2 x^2 \sqrt {1-\frac {b x}{a}} \tanh ^{-1}\left (\sqrt {1-\frac {b x}{a}}\right )-7 a b x+5 b^2 x^2}{4 x^2 \sqrt {b x-a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-a + b*x)^(3/2)/x^3,x]

[Out]

-1/4*(2*a^2 - 7*a*b*x + 5*b^2*x^2 + 3*b^2*x^2*Sqrt[1 - (b*x)/a]*ArcTanh[Sqrt[1 - (b*x)/a]])/(x^2*Sqrt[-a + b*x
])

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IntegrateAlgebraic [A]  time = 0.09, size = 62, normalized size = 0.91 \begin {gather*} \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {\sqrt {b x-a} (5 (b x-a)+3 a)}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[-a + b*x]*(3*a + 5*(-a + b*x)))/x^2 + (3*b^2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/(4*Sqrt[a])

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fricas [A]  time = 1.04, size = 129, normalized size = 1.90 \begin {gather*} \left [-\frac {3 \, \sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (5 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, a x^{2}}, \frac {3 \, \sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - {\left (5 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, a x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.95, size = 66, normalized size = 0.97 \begin {gather*} \frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{\sqrt {a}} - \frac {5 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{3} + 3 \, \sqrt {b x - a} a b^{3}}{b^{2} x^{2}}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 53, normalized size = 0.78 \begin {gather*} \frac {3 b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {3 \sqrt {b x -a}\, a}{4 x^{2}}-\frac {5 \left (b x -a \right )^{\frac {3}{2}}}{4 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.02, size = 80, normalized size = 1.18 \begin {gather*} \frac {3 \, b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, \sqrt {a}} - \frac {5 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {b x - a} a b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} + 2 \, {\left (b x - a\right )} a + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.10, size = 52, normalized size = 0.76 \begin {gather*} \frac {3\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4\,\sqrt {a}}-\frac {5\,{\left (b\,x-a\right )}^{3/2}}{4\,x^2}-\frac {3\,a\,\sqrt {b\,x-a}}{4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x - a)^(3/2)/x^3,x)

[Out]

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

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sympy [A]  time = 3.30, size = 190, normalized size = 2.79 \begin {gather*} \begin {cases} \frac {i a^{2}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {7 i a \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {5 i b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} - 1}} + \frac {3 i b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 \sqrt {a}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {a \sqrt {b} \sqrt {- \frac {a}{b x} + 1}}{2 x^{\frac {3}{2}}} - \frac {5 b^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}}{4 \sqrt {x}} - \frac {3 b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 \sqrt {a}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*a**2/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) - 1)) - 7*I*a*sqrt(b)/(4*x**(3/2)*sqrt(a/(b*x) - 1)) + 5*I*
b**(3/2)/(4*sqrt(x)*sqrt(a/(b*x) - 1)) + 3*I*b**2*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*sqrt(a)), Abs(a/(b*x)) >
1), (a*sqrt(b)*sqrt(-a/(b*x) + 1)/(2*x**(3/2)) - 5*b**(3/2)*sqrt(-a/(b*x) + 1)/(4*sqrt(x)) - 3*b**2*asin(sqrt
(a)/(sqrt(b)*sqrt(x)))/(4*sqrt(a)), True))

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