3.562 \(\int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx\)
Optimal. Leaf size=81 \[ 10 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )+5 b^2 \sqrt {x} \sqrt {b x+2}-\frac {2 (b x+2)^{5/2}}{3 x^{3/2}}-\frac {10 b (b x+2)^{3/2}}{3 \sqrt {x}} \]
[Out]
-2/3*(b*x+2)^(5/2)/x^(3/2)+10*b^(3/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))-10/3*b*(b*x+2)^(3/2)/x^(1/2)+5*b^2*
x^(1/2)*(b*x+2)^(1/2)
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Rubi [A] time = 0.02, antiderivative size = 81, 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} \[ 5 b^2 \sqrt {x} \sqrt {b x+2}+10 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )-\frac {2 (b x+2)^{5/2}}{3 x^{3/2}}-\frac {10 b (b x+2)^{3/2}}{3 \sqrt {x}} \]
Antiderivative was successfully verified.
[In]
Int[(2 + b*x)^(5/2)/x^(5/2),x]
[Out]
5*b^2*Sqrt[x]*Sqrt[2 + b*x] - (10*b*(2 + b*x)^(3/2))/(3*Sqrt[x]) - (2*(2 + b*x)^(5/2))/(3*x^(3/2)) + 10*b^(3/2
)*ArcSinh[(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^{5/2}} \, dx &=-\frac {2 (2+b x)^{5/2}}{3 x^{3/2}}+\frac {1}{3} (5 b) \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx\\ &=-\frac {10 b (2+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2+b x)^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx\\ &=5 b^2 \sqrt {x} \sqrt {2+b x}-\frac {10 b (2+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2+b x)^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx\\ &=5 b^2 \sqrt {x} \sqrt {2+b x}-\frac {10 b (2+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2+b x)^{5/2}}{3 x^{3/2}}+\left (10 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )\\ &=5 b^2 \sqrt {x} \sqrt {2+b x}-\frac {10 b (2+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2+b x)^{5/2}}{3 x^{3/2}}+10 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )\\ \end {align*}
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Mathematica [C] time = 0.01, size = 30, normalized size = 0.37 \[ -\frac {8 \sqrt {2} \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};-\frac {b x}{2}\right )}{3 x^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(2 + b*x)^(5/2)/x^(5/2),x]
[Out]
(-8*Sqrt[2]*Hypergeometric2F1[-5/2, -3/2, -1/2, -1/2*(b*x)])/(3*x^(3/2))
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fricas [A] time = 0.45, size = 123, normalized size = 1.52 \[ \left [\frac {15 \, b^{\frac {3}{2}} x^{2} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (3 \, b^{2} x^{2} - 28 \, b x - 8\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, x^{2}}, -\frac {30 \, \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, b^{2} x^{2} - 28 \, b x - 8\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, x^{2}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+2)^(5/2)/x^(5/2),x, algorithm="fricas")
[Out]
[1/3*(15*b^(3/2)*x^2*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + (3*b^2*x^2 - 28*b*x - 8)*sqrt(b*x + 2)*sqr
t(x))/x^2, -1/3*(30*sqrt(-b)*b*x^2*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) - (3*b^2*x^2 - 28*b*x - 8)*sqrt(
b*x + 2)*sqrt(x))/x^2]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+2)^(5/2)/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,[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]1/abs(b)*b^2/b*(2*((9*b^4
/18/b*sqrt(b*x+2)*sqrt(b*x+2)-120*b^4/18/b)*sqrt(b*x+2)*sqrt(b*x+2)+180*b^4/18/b)*sqrt(b*x+2)*sqrt(b*(b*x+2)-2
*b)/(b*(b*x+2)-2*b)^2-10*b^2/sqrt(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 = 82, normalized size = 1.01 \[ \frac {5 \sqrt {\left (b x +2\right ) x}\, b^{\frac {3}{2}} \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{\sqrt {b x +2}\, \sqrt {x}}+\frac {3 b^{3} x^{3}-22 b^{2} x^{2}-64 b x -16}{3 \sqrt {b x +2}\, x^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+2)^(5/2)/x^(5/2),x)
[Out]
1/3*(3*b^3*x^3-22*b^2*x^2-64*b*x-16)/x^(3/2)/(b*x+2)^(1/2)+5*((b*x+2)*x)^(1/2)/(b*x+2)^(1/2)*b^(3/2)/x^(1/2)*l
n((b*x+1)/b^(1/2)+(b*x^2+2*x)^(1/2))
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maxima [A] time = 2.94, size = 96, normalized size = 1.19 \[ -5 \, b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right ) - \frac {8 \, \sqrt {b x + 2} b}{\sqrt {x}} - \frac {2 \, \sqrt {b x + 2} b^{2}}{{\left (b - \frac {b x + 2}{x}\right )} \sqrt {x}} - \frac {4 \, {\left (b x + 2\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+2)^(5/2)/x^(5/2),x, algorithm="maxima")
[Out]
-5*b^(3/2)*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x))) - 8*sqrt(b*x + 2)*b/sqrt(
x) - 2*sqrt(b*x + 2)*b^2/((b - (b*x + 2)/x)*sqrt(x)) - 4/3*(b*x + 2)^(3/2)/x^(3/2)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x+2\right )}^{5/2}}{x^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x + 2)^(5/2)/x^(5/2),x)
[Out]
int((b*x + 2)^(5/2)/x^(5/2), x)
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sympy [A] time = 5.15, size = 88, normalized size = 1.09 \[ b^{\frac {5}{2}} x \sqrt {1 + \frac {2}{b x}} - \frac {28 b^{\frac {3}{2}} \sqrt {1 + \frac {2}{b x}}}{3} - 5 b^{\frac {3}{2}} \log {\left (\frac {1}{b x} \right )} + 10 b^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {2}{b x}} + 1 \right )} - \frac {8 \sqrt {b} \sqrt {1 + \frac {2}{b x}}}{3 x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+2)**(5/2)/x**(5/2),x)
[Out]
b**(5/2)*x*sqrt(1 + 2/(b*x)) - 28*b**(3/2)*sqrt(1 + 2/(b*x))/3 - 5*b**(3/2)*log(1/(b*x)) + 10*b**(3/2)*log(sqr
t(1 + 2/(b*x)) + 1) - 8*sqrt(b)*sqrt(1 + 2/(b*x))/(3*x)
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