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

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

[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)

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Rubi [A]  time = 0.021807, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {50, 54, 215} \[ -\frac{5 x^{3/2} \sqrt{b x+2}}{6 b^2}+\frac{5 \sqrt{x} \sqrt{b x+2}}{2 b^3}-\frac{5 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}+\frac{x^{5/2} \sqrt{b x+2}}{3 b} \]

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 = 0.0408326, size = 60, normalized size = 0.68 \[ \frac{\sqrt{x} \sqrt{b x+2} \left (2 b^2 x^2-5 b x+15\right )}{6 b^3}-\frac{5 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}} \]

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)

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Maple [A]  time = 0.005, size = 93, normalized size = 1.1 \begin{align*}{\frac{1}{3\,b}{x}^{{\frac{5}{2}}}\sqrt{bx+2}}-{\frac{5}{6\,{b}^{2}}{x}^{{\frac{3}{2}}}\sqrt{bx+2}}+{\frac{5}{2\,{b}^{3}}\sqrt{x}\sqrt{bx+2}}-{\frac{5}{2}\sqrt{x \left ( bx+2 \right ) }\ln \left ({(bx+1){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+2\,x} \right ){b}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{bx+2}}}{\frac{1}{\sqrt{x}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82923, size = 328, normalized size = 3.73 \begin{align*} \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{align*}

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]

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Sympy [A]  time = 12.5306, size = 95, normalized size = 1.08 \begin{align*} \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{align*}

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)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

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