3.7.9 \(\int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx\) [609]

Optimal. Leaf size=67 \[ -\frac {3 \sqrt {x} \sqrt {2+b x}}{2 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{2 b}+\frac {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 67, 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 {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}-\frac {3 \sqrt {x} \sqrt {b x+2}}{2 b^2}+\frac {x^{3/2} \sqrt {b x+2}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[2 + b*x],x]

[Out]

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

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Mathematica [A]
time = 0.07, size = 57, normalized size = 0.85 \begin {gather*} \frac {\sqrt {x} (-3+b x) \sqrt {2+b x}}{2 b^2}-\frac {3 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/Sqrt[2 + b*x],x]

[Out]

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

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Maple [A]
time = 0.11, size = 83, normalized size = 1.24

method	result	size
meijerg	\(\frac {-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (-5 b x +15\right ) \sqrt {\frac {b x}{2}+1}}{10}+3 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {5}{2}} \sqrt {\pi }}\)	\(55\)
risch	\(\frac {\left (b x -3\right ) \sqrt {x}\, \sqrt {b x +2}}{2 b^{2}}+\frac {3 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right ) \sqrt {x \left (b x +2\right )}}{2 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(68\)
default	\(\frac {x^{\frac {3}{2}} \sqrt {b x +2}}{2 b}-\frac {3 \left (\frac {\sqrt {x}\, \sqrt {b x +2}}{b}-\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{b^{\frac {3}{2}} \sqrt {b x +2}\, \sqrt {x}}\right )}{2 b}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(3/2)*(b*x+2)^(1/2)/b-3/2/b*(x^(1/2)*(b*x+2)^(1/2)/b-1/b^(3/2)*(x*(b*x+2))^(1/2)/(b*x+2)^(1/2)/x^(1/2)*l
n((b*x+1)/b^(1/2)+(b*x^2+2*x)^(1/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (48) = 96\).
time = 0.51, size = 102, normalized size = 1.52 \begin {gather*} \frac {\frac {5 \, \sqrt {b x + 2} b}{\sqrt {x}} - \frac {3 \, {\left (b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{4} - \frac {2 \, {\left (b x + 2\right )} b^{3}}{x} + \frac {{\left (b x + 2\right )}^{2} b^{2}}{x^{2}}} - \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 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*sqrt(b*x + 2)*b/sqrt(x) - 3*(b*x + 2)^(3/2)/x^(3/2))/(b^4 - 2*(b*x + 2)*b^3/x + (b*x + 2)^2*b^2/x^2) - 3/2*
log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(5/2)

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Fricas [A]
time = 0.48, size = 105, normalized size = 1.57 \begin {gather*} \left [\frac {{\left (b^{2} x - 3 \, 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^{3}}, \frac {{\left (b^{2} x - 3 \, 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^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 2.49, size = 75, normalized size = 1.12 \begin {gather*} \frac {x^{\frac {5}{2}}}{2 \sqrt {b x + 2}} - \frac {x^{\frac {3}{2}}}{2 b \sqrt {b x + 2}} - \frac {3 \sqrt {x}}{b^{2} \sqrt {b x + 2}} + \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,1]%%%}+%%%{-4,[1,0]%%%}+%%%{-4,[0,1]%%%}+%%%{-8,[0,0]%%%},0,%%%{6,[2,2]%%%}+%%%{4,[2,1]%%%}+%%%{6,[2,0]%%%}+%
%%{4,[1,2]%%%}+%%%{28

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{3/2}}{\sqrt {b\,x+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(b*x + 2)^(1/2),x)

[Out]

int(x^(3/2)/(b*x + 2)^(1/2), x)

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