3.7.17 \(\int \frac {x^{3/2}}{(2+b x)^{3/2}} \, dx\) [617]

Optimal. Leaf size=63 \[ -\frac {2 x^{3/2}}{b \sqrt {2+b x}}+\frac {3 \sqrt {x} \sqrt {2+b x}}{b^2}-\frac {6 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \]

[Out]

-6*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(5/2)-2*x^(3/2)/b/(b*x+2)^(1/2)+3*x^(1/2)*(b*x+2)^(1/2)/b^2

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Rubi [A]
time = 0.01, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {49, 52, 56, 221} \begin {gather*} -\frac {6 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}+\frac {3 \sqrt {x} \sqrt {b x+2}}{b^2}-\frac {2 x^{3/2}}{b \sqrt {b x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(2 + b*x)^(3/2),x]

[Out]

(-2*x^(3/2))/(b*Sqrt[2 + b*x]) + (3*Sqrt[x]*Sqrt[2 + b*x])/b^2 - (6*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(5/2
)

Rule 49

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 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}}{(2+b x)^{3/2}} \, dx &=-\frac {2 x^{3/2}}{b \sqrt {2+b x}}+\frac {3 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{b}\\ &=-\frac {2 x^{3/2}}{b \sqrt {2+b x}}+\frac {3 \sqrt {x} \sqrt {2+b x}}{b^2}-\frac {3 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{b^2}\\ &=-\frac {2 x^{3/2}}{b \sqrt {2+b x}}+\frac {3 \sqrt {x} \sqrt {2+b x}}{b^2}-\frac {6 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{b^2}\\ &=-\frac {2 x^{3/2}}{b \sqrt {2+b x}}+\frac {3 \sqrt {x} \sqrt {2+b x}}{b^2}-\frac {6 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 54, normalized size = 0.86 \begin {gather*} \frac {\sqrt {x} (6+b x)}{b^2 \sqrt {2+b x}}+\frac {6 \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)/(2 + b*x)^(3/2),x]

[Out]

(Sqrt[x]*(6 + b*x))/(b^2*Sqrt[2 + b*x]) + (6*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]])/b^(5/2)

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Mathics [A]
time = 3.43, size = 63, normalized size = 1.00 \begin {gather*} \frac {-6 b^3 \text {ArcSinh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ] \left (2+b x\right )+6 b^{\frac {7}{2}} \sqrt {x} \sqrt {2+b x}+b^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {2+b x}}{b^{\frac {11}{2}} \left (2+b x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6 b ^ 3 ArcSinh[Sqrt[2] Sqrt[b] Sqrt[x] / 2] (2 + b x) + 6 b ^ (7 / 2) Sqrt[x] Sqrt[2 + b x] + b ^ (9 / 2) x
^ (3 / 2) Sqrt[2 + b x]) / (b ^ (11 / 2) (2 + b x))

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Maple [A]
time = 0.13, size = 55, normalized size = 0.87

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (\frac {5 b x}{2}+15\right )}{5 \sqrt {\frac {b x}{2}+1}}-6 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {5}{2}} \sqrt {\pi }}\)	\(55\)
risch	\(\frac {\sqrt {x}\, \sqrt {b x +2}}{b^{2}}+\frac {\left (-\frac {3 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{b^{\frac {5}{2}}}+\frac {4 \sqrt {\left (x +\frac {2}{b}\right )^{2} b -2 x -\frac {4}{b}}}{b^{3} \left (x +\frac {2}{b}\right )}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/b^(5/2)/Pi^(1/2)*(1/20*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(5/2*b*x+15)/(1/2*b*x+1)^(1/2)-3/2*Pi^(1/2)*arcsinh(
1/2*b^(1/2)*x^(1/2)*2^(1/2)))

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Maxima [A]
time = 0.34, size = 90, normalized size = 1.43 \begin {gather*} \frac {2 \, {\left (2 \, b - \frac {3 \, {\left (b x + 2\right )}}{x}\right )}}{\frac {\sqrt {b x + 2} b^{3}}{\sqrt {x}} - \frac {{\left (b x + 2\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{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 )}{b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*b - 3*(b*x + 2)/x)/(sqrt(b*x + 2)*b^3/sqrt(x) - (b*x + 2)^(3/2)*b^2/x^(3/2)) + 3*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 = 0.32, size = 134, normalized size = 2.13 \begin {gather*} \left [\frac {3 \, {\left (b x + 2\right )} \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (b^{2} x + 6 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{b^{4} x + 2 \, b^{3}}, \frac {6 \, {\left (b x + 2\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (b^{2} x + 6 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{b^{4} x + 2 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(3*(b*x + 2)*sqrt(b)*log(b*x - sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + (b^2*x + 6*b)*sqrt(b*x + 2)*sqrt(x))/(b^4
*x + 2*b^3), (6*(b*x + 2)*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) + (b^2*x + 6*b)*sqrt(b*x + 2)*sq
rt(x))/(b^4*x + 2*b^3)]

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Sympy [A]
time = 1.82, size = 58, normalized size = 0.92 \begin {gather*} \frac {x^{\frac {3}{2}}}{b \sqrt {b x + 2}} + \frac {6 \sqrt {x}}{b^{2} \sqrt {b x + 2}} - \frac {6 \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)**(3/2),x)

[Out]

x**(3/2)/(b*sqrt(b*x + 2)) + 6*sqrt(x)/(b**2*sqrt(b*x + 2)) - 6*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(5/2)

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Giac [A]
time = 0.01, size = 92, normalized size = 1.46 \begin {gather*} 2 \left (\frac {2 \left (\frac {\frac {1}{4} b^{2} \sqrt {x} \sqrt {x}}{b^{3}}+\frac {\frac {1}{4}\cdot 6 b}{b^{3}}\right ) \sqrt {x} \sqrt {b x+2}}{b x+2}+\frac {3 \ln \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{b^{2} \sqrt {b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x)*(x/b + 6/b^2)/sqrt(b*x + 2) + 6*log(-sqrt(b)*sqrt(x) + sqrt(b*x + 2))/b^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(b*x + 2)^(3/2),x)

[Out]

int(x^(3/2)/(b*x + 2)^(3/2), x)

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