3.3.97 \(\int \frac {(a+b x)^{3/2}}{x^2} \, dx\) [297]

Optimal. Leaf size=51 \[ 3 b \sqrt {a+b x}-\frac {(a+b x)^{3/2}}{x}-3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 52, 65, 214} \begin {gather*} -\frac {(a+b x)^{3/2}}{x}+3 b \sqrt {a+b x}-3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/x^2,x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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

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Mathematica [A]
time = 0.06, size = 45, normalized size = 0.88 \begin {gather*} -\frac {(a-2 b x) \sqrt {a+b x}}{x}-3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/x^2,x]

[Out]

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

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Maple [A]
time = 0.10, size = 48, normalized size = 0.94

method	result	size
risch	\(-\frac {a \sqrt {b x +a}}{x}+\frac {b \left (4 \sqrt {b x +a}-6 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}\right )}{2}\)	\(45\)
derivativedivides	\(2 b \left (\sqrt {b x +a}-a \left (\frac {\sqrt {b x +a}}{2 b x}+\frac {3 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\)	\(48\)
default	\(2 b \left (\sqrt {b x +a}-a \left (\frac {\sqrt {b x +a}}{2 b x}+\frac {3 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 58, normalized size = 1.14 \begin {gather*} \frac {3}{2} \, \sqrt {a} b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + 2 \, \sqrt {b x + a} b - \frac {\sqrt {b x + a} a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.40, size = 102, normalized size = 2.00 \begin {gather*} \left [\frac {3 \, \sqrt {a} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, b x - a\right )} \sqrt {b x + a}}{2 \, x}, \frac {3 \, \sqrt {-a} b x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, b x - a\right )} \sqrt {b x + a}}{x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (44) = 88\).
time = 1.26, size = 92, normalized size = 1.80 \begin {gather*} - 3 \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {a^{2}}{\sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {a \sqrt {b}}{\sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {2 b^{\frac {3}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x))) - a**2/(sqrt(b)*x**(3/2)*sqrt(a/(b*x) + 1)) + a*sqrt(b)/(sqrt(x)
*sqrt(a/(b*x) + 1)) + 2*b**(3/2)*sqrt(x)/sqrt(a/(b*x) + 1)

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Giac [A]
time = 0.95, size = 56, normalized size = 1.10 \begin {gather*} \frac {\frac {3 \, a b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {b x + a} b^{2} - \frac {\sqrt {b x + a} a b}{x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.10, size = 42, normalized size = 0.82 \begin {gather*} 2\,b\,\sqrt {a+b\,x}-3\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )-\frac {a\,\sqrt {a+b\,x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/x^2,x)

[Out]

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

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