3.6.38 \(\int \frac {(2+b x)^{3/2}}{x^{5/2}} \, dx\) [538]

Optimal. Leaf size=60 \[ -\frac {2 b \sqrt {2+b x}}{\sqrt {x}}-\frac {2 (2+b x)^{3/2}}{3 x^{3/2}}+2 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 60, 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 = {49, 56, 221} \begin {gather*} 2 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )-\frac {2 (b x+2)^{3/2}}{3 x^{3/2}}-\frac {2 b \sqrt {b x+2}}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + b*x)^(3/2)/x^(5/2),x]

[Out]

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

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Mathematica [A]
time = 0.09, size = 55, normalized size = 0.92 \begin {gather*} -\frac {4 \sqrt {2+b x} (1+2 b x)}{3 x^{3/2}}-2 b^{3/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(3/2)/x^(5/2),x]

[Out]

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

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Maple [A]
time = 0.13, size = 55, normalized size = 0.92

method	result	size
meijerg	\(\frac {3 b^{\frac {3}{2}} \left (-\frac {16 \sqrt {\pi }\, \sqrt {2}\, \left (2 b x +1\right ) \sqrt {\frac {b x}{2}+1}}{9 x^{\frac {3}{2}} b^{\frac {3}{2}}}+\frac {8 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{3}\right )}{4 \sqrt {\pi }}\)	\(55\)
risch	\(-\frac {4 \left (2 x^{2} b^{2}+5 b x +2\right )}{3 x^{\frac {3}{2}} \sqrt {b x +2}}+\frac {b^{\frac {3}{2}} \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+2)^(3/2)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

3/4*b^(3/2)/Pi^(1/2)*(-16/9*Pi^(1/2)/x^(3/2)*2^(1/2)/b^(3/2)*(2*b*x+1)*(1/2*b*x+1)^(1/2)+8/3*Pi^(1/2)*arcsinh(
1/2*b^(1/2)*x^(1/2)*2^(1/2)))

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Maxima [A]
time = 0.50, size = 67, normalized size = 1.12 \begin {gather*} -b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right ) - \frac {2 \, \sqrt {b x + 2} b}{\sqrt {x}} - \frac {2 \, {\left (b x + 2\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)^(3/2)/x^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.49, size = 108, normalized size = 1.80 \begin {gather*} \left [\frac {3 \, b^{\frac {3}{2}} x^{2} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) - 4 \, {\left (2 \, b x + 1\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, x^{2}}, -\frac {2 \, {\left (3 \, \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) + 2 \, {\left (2 \, b x + 1\right )} \sqrt {b x + 2} \sqrt {x}\right )}}{3 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)^(3/2)/x^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 1.68, size = 70, normalized size = 1.17 \begin {gather*} - \frac {8 b^{\frac {3}{2}} \sqrt {1 + \frac {2}{b x}}}{3} - b^{\frac {3}{2}} \log {\left (\frac {1}{b x} \right )} + 2 b^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {2}{b x}} + 1 \right )} - \frac {4 \sqrt {b} \sqrt {1 + \frac {2}{b x}}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)**(3/2)/x**(5/2),x)

[Out]

-8*b**(3/2)*sqrt(1 + 2/(b*x))/3 - b**(3/2)*log(1/(b*x)) + 2*b**(3/2)*log(sqrt(1 + 2/(b*x)) + 1) - 4*sqrt(b)*sq
rt(1 + 2/(b*x))/(3*x)

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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((b*x+2)^(3/2)/x^(5/2),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 = -0.02 \begin {gather*} \int \frac {{\left (b\,x+2\right )}^{3/2}}{x^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + 2)^(3/2)/x^(5/2),x)

[Out]

int((b*x + 2)^(3/2)/x^(5/2), x)

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