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

Optimal. Leaf size=88 \[ -\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}+\frac {5 \sqrt {x} \sqrt {b x+2}}{2 b^3}-\frac {5 x^{3/2} \sqrt {b x+2}}{6 b^2}+\frac {x^{5/2} \sqrt {b x+2}}{3 b} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 88, 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 {5 x^{3/2} \sqrt {b x+2}}{6 b^2}+\frac {5 \sqrt {x} \sqrt {b x+2}}{2 b^3}-\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}+\frac {x^{5/2} \sqrt {b x+2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/Sqrt[2 + b*x],x]

[Out]

(5*Sqrt[x]*Sqrt[2 + b*x])/(2*b^3) - (5*x^(3/2)*Sqrt[2 + b*x])/(6*b^2) + (x^(5/2)*Sqrt[2 + b*x])/(3*b) - (5*Arc
Sinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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 {x^{5/2}}{\sqrt {2+b x}} \, dx &=\frac {x^{5/2} \sqrt {2+b x}}{3 b}-\frac {5 \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx}{3 b}\\ &=-\frac {5 x^{3/2} \sqrt {2+b x}}{6 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{3 b}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{2 b^2}\\ &=\frac {5 \sqrt {x} \sqrt {2+b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{6 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{3 b}-\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{2 b^3}\\ &=\frac {5 \sqrt {x} \sqrt {2+b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{6 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{3 b}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=\frac {5 \sqrt {x} \sqrt {2+b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{6 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{3 b}-\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 60, normalized size = 0.68 \begin {gather*} \frac {\sqrt {x} \sqrt {b x+2} \left (2 b^2 x^2-5 b x+15\right )}{6 b^3}-\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/Sqrt[2 + b*x],x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(15 - 5*b*x + 2*b^2*x^2))/(6*b^3) - (5*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.09, size = 73, normalized size = 0.83 \begin {gather*} \frac {5 \log \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{b^{7/2}}+\frac {\sqrt {b x+2} \left (2 b^2 x^{5/2}-5 b x^{3/2}+15 \sqrt {x}\right )}{6 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2 + b*x]*(15*Sqrt[x] - 5*b*x^(3/2) + 2*b^2*x^(5/2)))/(6*b^3) + (5*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]
])/b^(7/2)

________________________________________________________________________________________

fricas [A]  time = 1.15, size = 124, normalized size = 1.41 \begin {gather*} \left [\frac {{\left (2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, 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^{4}}, \frac {{\left (2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, 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^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*((2*b^3*x^2 - 5*b^2*x + 15*b)*sqrt(b*x + 2)*sqrt(x) + 15*sqrt(b)*log(b*x - sqrt(b*x + 2)*sqrt(b)*sqrt(x)
+ 1))/b^4, 1/6*((2*b^3*x^2 - 5*b^2*x + 15*b)*sqrt(b*x + 2)*sqrt(x) + 30*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)
/(b*sqrt(x))))/b^4]

________________________________________________________________________________________

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(x^(5/2)/(b*x+2)^(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]2*abs(b)/b^2/b*(2*((12*b^
5/144/b^7*sqrt(b*x+2)*sqrt(b*x+2)-78*b^5/144/b^7)*sqrt(b*x+2)*sqrt(b*x+2)+198*b^5/144/b^7)*sqrt(b*x+2)*sqrt(b*
(b*x+2)-2*b)+5/2/b/sqrt(b)*ln(abs(sqrt(b*(b*x+2)-2*b)-sqrt(b)*sqrt(b*x+2))))

________________________________________________________________________________________

maple [A]  time = 0.00, size = 93, normalized size = 1.06 \begin {gather*} \frac {\sqrt {b x +2}\, x^{\frac {5}{2}}}{3 b}-\frac {5 \sqrt {b x +2}\, x^{\frac {3}{2}}}{6 b^{2}}+\frac {5 \sqrt {b x +2}\, \sqrt {x}}{2 b^{3}}-\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}\, b^{\frac {7}{2}} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^(5/2)*(b*x+2)^(1/2)/b-5/6*x^(3/2)*(b*x+2)^(1/2)/b^2+5/2*(b*x+2)^(1/2)/b^3*x^(1/2)-5/2*((b*x+2)*x)^(1/2)/
(b*x+2)^(1/2)/b^(7/2)/x^(1/2)*ln((b*x+1)/b^(1/2)+(b*x^2+2*x)^(1/2))

________________________________________________________________________________________

maxima [B]  time = 2.97, size = 134, normalized size = 1.52 \begin {gather*} -\frac {\frac {33 \, \sqrt {b x + 2} b^{2}}{\sqrt {x}} - \frac {40 \, {\left (b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} + \frac {15 \, {\left (b x + 2\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}}{3 \, {\left (b^{6} - \frac {3 \, {\left (b x + 2\right )} b^{5}}{x} + \frac {3 \, {\left (b x + 2\right )}^{2} b^{4}}{x^{2}} - \frac {{\left (b x + 2\right )}^{3} b^{3}}{x^{3}}\right )}} + \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 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(33*sqrt(b*x + 2)*b^2/sqrt(x) - 40*(b*x + 2)^(3/2)*b/x^(3/2) + 15*(b*x + 2)^(5/2)/x^(5/2))/(b^6 - 3*(b*x
+ 2)*b^5/x + 3*(b*x + 2)^2*b^4/x^2 - (b*x + 2)^3*b^3/x^3) + 5/2*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b
) + sqrt(b*x + 2)/sqrt(x)))/b^(7/2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}}{\sqrt {b\,x+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(b*x + 2)^(1/2),x)

[Out]

int(x^(5/2)/(b*x + 2)^(1/2), x)

________________________________________________________________________________________

sympy [A]  time = 7.40, size = 95, normalized size = 1.08 \begin {gather*} \frac {x^{\frac {7}{2}}}{3 \sqrt {b x + 2}} - \frac {x^{\frac {5}{2}}}{6 b \sqrt {b x + 2}} + \frac {5 x^{\frac {3}{2}}}{6 b^{2} \sqrt {b x + 2}} + \frac {5 \sqrt {x}}{b^{3} \sqrt {b x + 2}} - \frac {5 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**(7/2)/(3*sqrt(b*x + 2)) - x**(5/2)/(6*b*sqrt(b*x + 2)) + 5*x**(3/2)/(6*b**2*sqrt(b*x + 2)) + 5*sqrt(x)/(b**
3*sqrt(b*x + 2)) - 5*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2)

________________________________________________________________________________________