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

Optimal. Leaf size=79 \[ \frac {1}{3} \sqrt {x} (b x+2)^{5/2}+\frac {5}{6} \sqrt {x} (b x+2)^{3/2}+\frac {5}{2} \sqrt {x} \sqrt {b x+2}+\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

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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 = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {50, 54, 215} \begin {gather*} \frac {1}{3} \sqrt {x} (b x+2)^{5/2}+\frac {5}{6} \sqrt {x} (b x+2)^{3/2}+\frac {5}{2} \sqrt {x} \sqrt {b x+2}+\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + b*x)^(5/2)/Sqrt[x],x]

[Out]

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

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

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Mathematica [A]  time = 0.03, size = 57, normalized size = 0.72 \begin {gather*} \frac {1}{6} \sqrt {x} \sqrt {b x+2} \left (2 b^2 x^2+13 b x+33\right )+\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(5/2)/Sqrt[x],x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(33 + 13*b*x + 2*b^2*x^2))/6 + (5*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/Sqrt[b]

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IntegrateAlgebraic [A]  time = 0.10, size = 70, normalized size = 0.89 \begin {gather*} \frac {1}{6} \sqrt {b x+2} \left (2 b^2 x^{5/2}+13 b x^{3/2}+33 \sqrt {x}\right )-\frac {5 \log \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2 + b*x)^(5/2)/Sqrt[x],x]

[Out]

(Sqrt[2 + b*x]*(33*Sqrt[x] + 13*b*x^(3/2) + 2*b^2*x^(5/2)))/6 - (5*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]])/Sq
rt[b]

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fricas [A]  time = 1.24, size = 123, normalized size = 1.56 \begin {gather*} \left [\frac {{\left (2 \, b^{3} x^{2} + 13 \, b^{2} x + 33 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 15 \, \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{6 \, b}, \frac {{\left (2 \, b^{3} x^{2} + 13 \, b^{2} x + 33 \, b\right )} \sqrt {b x + 2} \sqrt {x} - 30 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{6 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)^(5/2)/x^(1/2),x, algorithm="fricas")

[Out]

[1/6*((2*b^3*x^2 + 13*b^2*x + 33*b)*sqrt(b*x + 2)*sqrt(x) + 15*sqrt(b)*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x)
+ 1))/b, 1/6*((2*b^3*x^2 + 13*b^2*x + 33*b)*sqrt(b*x + 2)*sqrt(x) - 30*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)
/(b*sqrt(x))))/b]

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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^(1/2),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*((1/6/b
*sqrt(b*x+2)*sqrt(b*x+2)+5/12/b)*sqrt(b*x+2)*sqrt(b*x+2)+5/4/b)*sqrt(b*x+2)*sqrt(b*(b*x+2)-2*b)-5/sqrt(b)*ln(a
bs(sqrt(b*(b*x+2)-2*b)-sqrt(b)*sqrt(b*x+2))))

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maple [A]  time = 0.00, size = 84, normalized size = 1.06 \begin {gather*} \frac {\left (b x +2\right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 \left (b x +2\right )^{\frac {3}{2}} \sqrt {x}}{6}+\frac {5 \sqrt {b x +2}\, \sqrt {x}}{2}+\frac {5 \sqrt {\left (b x +2\right ) x}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {b}\, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+2)^(5/2)/x^(1/2),x)

[Out]

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

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maxima [B]  time = 3.03, size = 129, normalized size = 1.63 \begin {gather*} -\frac {5 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, \sqrt {b}} - \frac {\frac {15 \, \sqrt {b x + 2} b^{2}}{\sqrt {x}} - \frac {40 \, {\left (b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} + \frac {33 \, {\left (b x + 2\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}}{3 \, {\left (b^{3} - \frac {3 \, {\left (b x + 2\right )} b^{2}}{x} + \frac {3 \, {\left (b x + 2\right )}^{2} b}{x^{2}} - \frac {{\left (b x + 2\right )}^{3}}{x^{3}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)^(5/2)/x^(1/2),x, algorithm="maxima")

[Out]

-5/2*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/sqrt(b) - 1/3*(15*sqrt(b*x + 2)
*b^2/sqrt(x) - 40*(b*x + 2)^(3/2)*b/x^(3/2) + 33*(b*x + 2)^(5/2)/x^(5/2))/(b^3 - 3*(b*x + 2)*b^2/x + 3*(b*x +
2)^2*b/x^2 - (b*x + 2)^3/x^3)

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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}}{\sqrt {x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + 2)^(5/2)/x^(1/2),x)

[Out]

int((b*x + 2)^(5/2)/x^(1/2), x)

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sympy [A]  time = 5.46, size = 97, normalized size = 1.23 \begin {gather*} \frac {b^{3} x^{\frac {7}{2}}}{3 \sqrt {b x + 2}} + \frac {17 b^{2} x^{\frac {5}{2}}}{6 \sqrt {b x + 2}} + \frac {59 b x^{\frac {3}{2}}}{6 \sqrt {b x + 2}} + \frac {11 \sqrt {x}}{\sqrt {b x + 2}} + \frac {5 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+2)**(5/2)/x**(1/2),x)

[Out]

b**3*x**(7/2)/(3*sqrt(b*x + 2)) + 17*b**2*x**(5/2)/(6*sqrt(b*x + 2)) + 59*b*x**(3/2)/(6*sqrt(b*x + 2)) + 11*sq
rt(x)/sqrt(b*x + 2) + 5*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/sqrt(b)

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