3.6.36 \(\int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx\) [536]

Optimal. Leaf size=61 \[ \frac {3}{2} \sqrt {x} \sqrt {2+b x}+\frac {1}{2} \sqrt {x} (2+b x)^{3/2}+\frac {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[Out]

3*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(1/2)+1/2*(b*x+2)^(3/2)*x^(1/2)+3/2*x^(1/2)*(b*x+2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 61, 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, 56, 221} \begin {gather*} \frac {1}{2} \sqrt {x} (b x+2)^{3/2}+\frac {3}{2} \sqrt {x} \sqrt {b x+2}+\frac {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + b*x)^(3/2)/Sqrt[x],x]

[Out]

(3*Sqrt[x]*Sqrt[2 + b*x])/2 + (Sqrt[x]*(2 + b*x)^(3/2))/2 + (3*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/Sqrt[b]

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 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
 a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
 x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 54, normalized size = 0.89 \begin {gather*} \frac {1}{2} \sqrt {x} \sqrt {2+b x} (5+b x)-\frac {3 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(3/2)/Sqrt[x],x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(5 + b*x))/2 - (3*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]])/Sqrt[b]

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Mathics [A]
time = 3.32, size = 54, normalized size = 0.89 \begin {gather*} \frac {\sqrt {b} \sqrt {x} \left (10+7 b x+b^2 x^2\right )+6 \text {ArcSinh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ] \sqrt {2+b x}}{2 \sqrt {b} \sqrt {2+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b] Sqrt[x] (10 + 7 b x + b ^ 2 x ^ 2) + 6 ArcSinh[Sqrt[2] Sqrt[b] Sqrt[x] / 2] Sqrt[2 + b x]) / (2 Sqrt[
b] Sqrt[2 + b x])

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Maple [A]
time = 0.12, size = 72, normalized size = 1.18

method result size
meijerg \(\frac {4 \sqrt {\pi }\, \sqrt {b}\, \sqrt {x}\, \sqrt {2}\, \left (\frac {b x}{8}+\frac {5}{8}\right ) \sqrt {\frac {b x}{2}+1}+3 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{\sqrt {b}\, \sqrt {\pi }}\) \(54\)
risch \(\frac {\left (b x +5\right ) \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {3 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}\) \(65\)
default \(\frac {\left (b x +2\right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {3 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}\) \(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+2)^(3/2)/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(b*x+2)^(3/2)*x^(1/2)+3/2*x^(1/2)*(b*x+2)^(1/2)+3/2*(x*(b*x+2))^(1/2)/(b*x+2)^(1/2)/x^(1/2)*ln((b*x+1)/b^(
1/2)+(b*x^2+2*x)^(1/2))/b^(1/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (42) = 84\).
time = 0.35, size = 98, normalized size = 1.61 \begin {gather*} -\frac {3 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, \sqrt {b}} - \frac {\frac {3 \, \sqrt {b x + 2} b}{\sqrt {x}} - \frac {5 \, {\left (b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{2} - \frac {2 \, {\left (b x + 2\right )} b}{x} + \frac {{\left (b x + 2\right )}^{2}}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)^(3/2)/x^(1/2),x, algorithm="maxima")

[Out]

-3/2*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/sqrt(b) - (3*sqrt(b*x + 2)*b/sq
rt(x) - 5*(b*x + 2)^(3/2)/x^(3/2))/(b^2 - 2*(b*x + 2)*b/x + (b*x + 2)^2/x^2)

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Fricas [A]
time = 0.33, size = 105, normalized size = 1.72 \begin {gather*} \left [\frac {{\left (b^{2} x + 5 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 3 \, \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{2 \, b}, \frac {{\left (b^{2} x + 5 \, b\right )} \sqrt {b x + 2} \sqrt {x} - 6 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{2 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)^(3/2)/x^(1/2),x, algorithm="fricas")

[Out]

[1/2*((b^2*x + 5*b)*sqrt(b*x + 2)*sqrt(x) + 3*sqrt(b)*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1))/b, 1/2*((b
^2*x + 5*b)*sqrt(b*x + 2)*sqrt(x) - 6*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))))/b]

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Sympy [A]
time = 1.59, size = 76, normalized size = 1.25 \begin {gather*} \frac {b^{2} x^{\frac {5}{2}}}{2 \sqrt {b x + 2}} + \frac {7 b x^{\frac {3}{2}}}{2 \sqrt {b x + 2}} + \frac {5 \sqrt {x}}{\sqrt {b x + 2}} + \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)**(3/2)/x**(1/2),x)

[Out]

b**2*x**(5/2)/(2*sqrt(b*x + 2)) + 7*b*x**(3/2)/(2*sqrt(b*x + 2)) + 5*sqrt(x)/sqrt(b*x + 2) + 3*asinh(sqrt(2)*s
qrt(b)*sqrt(x)/2)/sqrt(b)

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Giac [A]
time = 1.17, size = 113, normalized size = 1.85 \begin {gather*} \frac {b^{2} \left (2 \left (\frac {\frac {1}{4} \sqrt {b x+2} \sqrt {b x+2}}{b}+\frac {\frac {1}{4}\cdot 3}{b}\right ) \sqrt {b x+2} \sqrt {b \left (b x+2\right )-2 b}-\frac {3 \ln \left |\sqrt {b \left (b x+2\right )-2 b}-\sqrt {b} \sqrt {b x+2}\right |}{\sqrt {b}}\right )}{\left |b\right | b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)^(3/2)/x^(1/2),x)

[Out]

1/2*(sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2)*((b*x + 2)/b + 3/b) - 6*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x
+ 2)*b - 2*b)))/sqrt(b))*b/abs(b)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (b\,x+2\right )}^{3/2}}{\sqrt {x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + 2)^(3/2)/x^(1/2),x)

[Out]

int((b*x + 2)^(3/2)/x^(1/2), x)

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