3.6.61 \(\int \frac {(2+b x)^{5/2}}{x^{3/2}} \, dx\) [561]

Optimal. Leaf size=79 \[ \frac {15}{2} b \sqrt {x} \sqrt {2+b x}+\frac {5}{2} b \sqrt {x} (2+b x)^{3/2}-\frac {2 (2+b x)^{5/2}}{\sqrt {x}}+15 \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {2 (b x+2)^{5/2}}{\sqrt {x}}+\frac {5}{2} b \sqrt {x} (b x+2)^{3/2}+\frac {15}{2} b \sqrt {x} \sqrt {b x+2}+15 \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + b*x)^(5/2)/x^(3/2),x]

[Out]

(15*b*Sqrt[x]*Sqrt[2 + b*x])/2 + (5*b*Sqrt[x]*(2 + b*x)^(3/2))/2 - (2*(2 + b*x)^(5/2))/Sqrt[x] + 15*Sqrt[b]*Ar
cSinh[(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 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)^{5/2}}{x^{3/2}} \, dx &=-\frac {2 (2+b x)^{5/2}}{\sqrt {x}}+(5 b) \int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx\\ &=\frac {5}{2} b \sqrt {x} (2+b x)^{3/2}-\frac {2 (2+b x)^{5/2}}{\sqrt {x}}+\frac {1}{2} (15 b) \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx\\ &=\frac {15}{2} b \sqrt {x} \sqrt {2+b x}+\frac {5}{2} b \sqrt {x} (2+b x)^{3/2}-\frac {2 (2+b x)^{5/2}}{\sqrt {x}}+\frac {1}{2} (15 b) \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx\\ &=\frac {15}{2} b \sqrt {x} \sqrt {2+b x}+\frac {5}{2} b \sqrt {x} (2+b x)^{3/2}-\frac {2 (2+b x)^{5/2}}{\sqrt {x}}+(15 b) \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {15}{2} b \sqrt {x} \sqrt {2+b x}+\frac {5}{2} b \sqrt {x} (2+b x)^{3/2}-\frac {2 (2+b x)^{5/2}}{\sqrt {x}}+15 \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 62, normalized size = 0.78 \begin {gather*} \frac {\sqrt {2+b x} \left (-16+9 b x+b^2 x^2\right )}{2 \sqrt {x}}-15 \sqrt {b} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(5/2)/x^(3/2),x]

[Out]

(Sqrt[2 + b*x]*(-16 + 9*b*x + b^2*x^2))/(2*Sqrt[x]) - 15*Sqrt[b]*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]]

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Maple [A]
time = 0.11, size = 63, normalized size = 0.80

method	result	size
meijerg	\(-\frac {15 \sqrt {b}\, \left (\frac {16 \sqrt {\pi }\, \sqrt {2}\, \left (-\frac {1}{16} x^{2} b^{2}-\frac {9}{16} b x +1\right ) \sqrt {\frac {b x}{2}+1}}{15 \sqrt {x}\, \sqrt {b}}-2 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )\right )}{2 \sqrt {\pi }}\)	\(63\)
risch	\(\frac {b^{3} x^{3}+11 x^{2} b^{2}+2 b x -32}{2 \sqrt {x}\, \sqrt {b x +2}}+\frac {15 \sqrt {b}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right ) \sqrt {x \left (b x +2\right )}}{2 \sqrt {x}\, \sqrt {b x +2}}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/2*b^(1/2)/Pi^(1/2)*(16/15*Pi^(1/2)/x^(1/2)*2^(1/2)/b^(1/2)*(-1/16*x^2*b^2-9/16*b*x+1)*(1/2*b*x+1)^(1/2)-2*
Pi^(1/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (56) = 112\).
time = 0.52, size = 113, normalized size = 1.43 \begin {gather*} -\frac {15}{2} \, \sqrt {b} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right ) - \frac {\frac {7 \, \sqrt {b x + 2} b^{2}}{\sqrt {x}} - \frac {9 \, {\left (b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}}}{b^{2} - \frac {2 \, {\left (b x + 2\right )} b}{x} + \frac {{\left (b x + 2\right )}^{2}}{x^{2}}} - \frac {8 \, \sqrt {b x + 2}}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/2*sqrt(b)*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x))) - (7*sqrt(b*x + 2)*b^2
/sqrt(x) - 9*(b*x + 2)^(3/2)*b/x^(3/2))/(b^2 - 2*(b*x + 2)*b/x + (b*x + 2)^2/x^2) - 8*sqrt(b*x + 2)/sqrt(x)

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Fricas [A]
time = 1.13, size = 116, normalized size = 1.47 \begin {gather*} \left [\frac {15 \, \sqrt {b} x \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (b^{2} x^{2} + 9 \, b x - 16\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, x}, -\frac {30 \, \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (b^{2} x^{2} + 9 \, b x - 16\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(15*sqrt(b)*x*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + (b^2*x^2 + 9*b*x - 16)*sqrt(b*x + 2)*sqrt(x)
)/x, -1/2*(30*sqrt(-b)*x*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) - (b^2*x^2 + 9*b*x - 16)*sqrt(b*x + 2)*sqr
t(x))/x]

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Sympy [A]
time = 3.59, size = 94, normalized size = 1.19 \begin {gather*} 15 \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )} + \frac {b^{3} x^{\frac {5}{2}}}{2 \sqrt {b x + 2}} + \frac {11 b^{2} x^{\frac {3}{2}}}{2 \sqrt {b x + 2}} + \frac {b \sqrt {x}}{\sqrt {b x + 2}} - \frac {16}{\sqrt {x} \sqrt {b x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*sqrt(b)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2) + b**3*x**(5/2)/(2*sqrt(b*x + 2)) + 11*b**2*x**(3/2)/(2*sqrt(b*x +
2)) + b*sqrt(x)/sqrt(b*x + 2) - 16/(sqrt(x)*sqrt(b*x + 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((b*x+2)^(5/2)/x^(3/2),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 {{\left (b\,x+2\right )}^{5/2}}{x^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + 2)^(5/2)/x^(3/2),x)

[Out]

int((b*x + 2)^(5/2)/x^(3/2), x)

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