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

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

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Rubi [A]  time = 0.02, 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 = {47, 50, 54, 215} \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 47

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 50

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 54

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 215

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) \operatorname {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 [C]  time = 0.01, size = 28, normalized size = 0.35 \begin {gather*} -\frac {8 \sqrt {2} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x}{2}\right )}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(5/2)/x^(3/2),x]

[Out]

(-8*Sqrt[2]*Hypergeometric2F1[-5/2, -1/2, 1/2, -1/2*(b*x)])/Sqrt[x]

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IntegrateAlgebraic [A]  time = 0.12, size = 62, normalized size = 0.78 \begin {gather*} \frac {\sqrt {b x+2} \left (b^2 x^2+9 b x-16\right )}{2 \sqrt {x}}-15 \sqrt {b} \log \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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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fricas [A]  time = 1.24, 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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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,[1,1]%%%}+%%%{8,[1,0]%%%}+%%%{6,[0,2]%%%}+%%%{8,[0,1]%%%}+%%%{24,[0,0]%%%},0,%%%{-4,[3,3
]%%%}+%%%{4,[3,2]%%%}+%%%{4,[3,1]%%%}+%%%{-4,[3,0]%%%}+%%%{4,[2,3]%%%}+%%%{-64,[2,2]%%%}+%%%{20,[2,1]%%%}+%%%{
8,[2,0]%%%}+%%%{4,[1,3]%%%}+%%%{20,[1,2]%%%}+%%%{-128,[1,1]%%%}+%%%{16,[1,0]%%%}+%%%{-4,[0,3]%%%}+%%%{8,[0,2]%
%%}+%%%{16,[0,1]%%%}+%%%{-32,[0,0]%%%},0,%%%{1,[4,4]%%%}+%%%{-4,[4,3]%%%}+%%%{6,[4,2]%%%}+%%%{-4,[4,1]%%%}+%%%
{1,[4,0]%%%}+%%%{-4,[3,4]%%%}+%%%{12,[3,3]%%%}+%%%{-20,[3,2]%%%}+%%%{20,[3,1]%%%}+%%%{-8,[3,0]%%%}+%%%{6,[2,4]
%%%}+%%%{-20,[2,3]%%%}+%%%{46,[2,2]%%%}+%%%{-40,[2,1]%%%}+%%%{24,[2,0]%%%}+%%%{-4,[1,4]%%%}+%%%{20,[1,3]%%%}+%
%%{-40,[1,2]%%%}+%%%{48,[1,1]%%%}+%%%{-32,[1,0]%%%}+%%%{1,[0,4]%%%}+%%%{-8,[0,3]%%%}+%%%{24,[0,2]%%%}+%%%{-32,
[0,1]%%%}+%%%{16,[0,0]%%%}] at parameters values [85.3561567818,61.7937478349]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,[1,1]%%%}+%%%{8,[1,0]%%%}+%%%{6,[0,2]%%%}+%%%{8,[0,1]%%%}+%%%{24,[0,0]%%%},0,%%%
{-4,[3,3]%%%}+%%%{4,[3,2]%%%}+%%%{4,[3,1]%%%}+%%%{-4,[3,0]%%%}+%%%{4,[2,3]%%%}+%%%{-64,[2,2]%%%}+%%%{20,[2,1]%
%%}+%%%{8,[2,0]%%%}+%%%{4,[1,3]%%%}+%%%{20,[1,2]%%%}+%%%{-128,[1,1]%%%}+%%%{16,[1,0]%%%}+%%%{-4,[0,3]%%%}+%%%{
8,[0,2]%%%}+%%%{16,[0,1]%%%}+%%%{-32,[0,0]%%%},0,%%%{1,[4,4]%%%}+%%%{-4,[4,3]%%%}+%%%{6,[4,2]%%%}+%%%{-4,[4,1]
%%%}+%%%{1,[4,0]%%%}+%%%{-4,[3,4]%%%}+%%%{12,[3,3]%%%}+%%%{-20,[3,2]%%%}+%%%{20,[3,1]%%%}+%%%{-8,[3,0]%%%}+%%%
{6,[2,4]%%%}+%%%{-20,[2,3]%%%}+%%%{46,[2,2]%%%}+%%%{-40,[2,1]%%%}+%%%{24,[2,0]%%%}+%%%{-4,[1,4]%%%}+%%%{20,[1,
3]%%%}+%%%{-40,[1,2]%%%}+%%%{48,[1,1]%%%}+%%%{-32,[1,0]%%%}+%%%{1,[0,4]%%%}+%%%{-8,[0,3]%%%}+%%%{24,[0,2]%%%}+
%%%{-32,[0,1]%%%}+%%%{16,[0,0]%%%}] at parameters values [71.707969239,78.6493344628]b/abs(b)*b^2/b*(2*((1/4*s
qrt(b*x+2)*sqrt(b*x+2)+5/4)*sqrt(b*x+2)*sqrt(b*x+2)-15/2)*sqrt(b*x+2)*sqrt(b*(b*x+2)-2*b)/(b*(b*x+2)-2*b)-15/s
qrt(b)*ln(abs(sqrt(b*(b*x+2)-2*b)-sqrt(b)*sqrt(b*x+2))))

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maple [A]  time = 0.02, size = 81, normalized size = 1.03 \begin {gather*} \frac {15 \sqrt {\left (b x +2\right ) x}\, \sqrt {b}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}}+\frac {b^{3} x^{3}+11 b^{2} x^{2}+2 b x -32}{2 \sqrt {b x +2}\, \sqrt {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]

1/2*(b^3*x^3+11*b^2*x^2+2*b*x-32)/(b*x+2)^(1/2)/x^(1/2)+15/2*((b*x+2)*x)^(1/2)/(b*x+2)^(1/2)*b^(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]  time = 2.94, 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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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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sympy [A]  time = 5.60, 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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