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

Optimal. Leaf size=123 \[ -\frac{3 \sqrt{x} \sqrt{b x+2}}{8 b^2}+\frac{3 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{5/2}}+\frac{1}{5} x^{5/2} (b x+2)^{5/2}+\frac{1}{4} x^{5/2} (b x+2)^{3/2}+\frac{1}{4} x^{5/2} \sqrt{b x+2}+\frac{x^{3/2} \sqrt{b x+2}}{8 b} \]

[Out]

(-3*Sqrt[x]*Sqrt[2 + b*x])/(8*b^2) + (x^(3/2)*Sqrt[2 + b*x])/(8*b) + (x^(5/2)*Sqrt[2 + b*x])/4 + (x^(5/2)*(2 +
b*x)^(3/2))/4 + (x^(5/2)*(2 + b*x)^(5/2))/5 + (3*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*b^(5/2))

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Rubi [A]  time = 0.0282731, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {50, 54, 215} \[ -\frac{3 \sqrt{x} \sqrt{b x+2}}{8 b^2}+\frac{3 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{5/2}}+\frac{1}{5} x^{5/2} (b x+2)^{5/2}+\frac{1}{4} x^{5/2} (b x+2)^{3/2}+\frac{1}{4} x^{5/2} \sqrt{b x+2}+\frac{x^{3/2} \sqrt{b x+2}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(2 + b*x)^(5/2),x]

[Out]

(-3*Sqrt[x]*Sqrt[2 + b*x])/(8*b^2) + (x^(3/2)*Sqrt[2 + b*x])/(8*b) + (x^(5/2)*Sqrt[2 + b*x])/4 + (x^(5/2)*(2 +
b*x)^(3/2))/4 + (x^(5/2)*(2 + b*x)^(5/2))/5 + (3*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*b^(5/2))

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 x^{3/2} (2+b x)^{5/2} \, dx &=\frac{1}{5} x^{5/2} (2+b x)^{5/2}+\int x^{3/2} (2+b x)^{3/2} \, dx\\ &=\frac{1}{4} x^{5/2} (2+b x)^{3/2}+\frac{1}{5} x^{5/2} (2+b x)^{5/2}+\frac{3}{4} \int x^{3/2} \sqrt{2+b x} \, dx\\ &=\frac{1}{4} x^{5/2} \sqrt{2+b x}+\frac{1}{4} x^{5/2} (2+b x)^{3/2}+\frac{1}{5} x^{5/2} (2+b x)^{5/2}+\frac{1}{4} \int \frac{x^{3/2}}{\sqrt{2+b x}} \, dx\\ &=\frac{x^{3/2} \sqrt{2+b x}}{8 b}+\frac{1}{4} x^{5/2} \sqrt{2+b x}+\frac{1}{4} x^{5/2} (2+b x)^{3/2}+\frac{1}{5} x^{5/2} (2+b x)^{5/2}-\frac{3 \int \frac{\sqrt{x}}{\sqrt{2+b x}} \, dx}{8 b}\\ &=-\frac{3 \sqrt{x} \sqrt{2+b x}}{8 b^2}+\frac{x^{3/2} \sqrt{2+b x}}{8 b}+\frac{1}{4} x^{5/2} \sqrt{2+b x}+\frac{1}{4} x^{5/2} (2+b x)^{3/2}+\frac{1}{5} x^{5/2} (2+b x)^{5/2}+\frac{3 \int \frac{1}{\sqrt{x} \sqrt{2+b x}} \, dx}{8 b^2}\\ &=-\frac{3 \sqrt{x} \sqrt{2+b x}}{8 b^2}+\frac{x^{3/2} \sqrt{2+b x}}{8 b}+\frac{1}{4} x^{5/2} \sqrt{2+b x}+\frac{1}{4} x^{5/2} (2+b x)^{3/2}+\frac{1}{5} x^{5/2} (2+b x)^{5/2}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+b x^2}} \, dx,x,\sqrt{x}\right )}{4 b^2}\\ &=-\frac{3 \sqrt{x} \sqrt{2+b x}}{8 b^2}+\frac{x^{3/2} \sqrt{2+b x}}{8 b}+\frac{1}{4} x^{5/2} \sqrt{2+b x}+\frac{1}{4} x^{5/2} (2+b x)^{3/2}+\frac{1}{5} x^{5/2} (2+b x)^{5/2}+\frac{3 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0506403, size = 78, normalized size = 0.63 \[ \frac{\sqrt{x} \sqrt{b x+2} \left (8 b^4 x^4+42 b^3 x^3+62 b^2 x^2+5 b x-15\right )}{40 b^2}+\frac{3 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(2 + b*x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(-15 + 5*b*x + 62*b^2*x^2 + 42*b^3*x^3 + 8*b^4*x^4))/(40*b^2) + (3*ArcSinh[(Sqrt[b]*Sqr
t[x])/Sqrt[2]])/(4*b^(5/2))

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Maple [A]  time = 0.004, size = 123, normalized size = 1. \begin{align*}{\frac{1}{5\,b}{x}^{{\frac{3}{2}}} \left ( bx+2 \right ) ^{{\frac{7}{2}}}}-{\frac{3}{20\,{b}^{2}}\sqrt{x} \left ( bx+2 \right ) ^{{\frac{7}{2}}}}+{\frac{1}{20\,{b}^{2}} \left ( bx+2 \right ) ^{{\frac{5}{2}}}\sqrt{x}}+{\frac{1}{8\,{b}^{2}} \left ( bx+2 \right ) ^{{\frac{3}{2}}}\sqrt{x}}+{\frac{3}{8\,{b}^{2}}\sqrt{x}\sqrt{bx+2}}+{\frac{3}{8}\sqrt{x \left ( bx+2 \right ) }\ln \left ({(bx+1){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+2\,x} \right ){b}^{-{\frac{5}{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^(3/2)*(b*x+2)^(5/2),x)

[Out]

1/5/b*x^(3/2)*(b*x+2)^(7/2)-3/20/b^2*x^(1/2)*(b*x+2)^(7/2)+1/20/b^2*x^(1/2)*(b*x+2)^(5/2)+1/8/b^2*x^(1/2)*(b*x
+2)^(3/2)+3/8*x^(1/2)*(b*x+2)^(1/2)/b^2+3/8/b^(5/2)*(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^(3/2)*(b*x+2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.93715, size = 401, normalized size = 3.26 \begin{align*} \left [\frac{{\left (8 \, b^{5} x^{4} + 42 \, b^{4} x^{3} + 62 \, 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 )}{40 \, b^{3}}, \frac{{\left (8 \, b^{5} x^{4} + 42 \, b^{4} x^{3} + 62 \, 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 )}{40 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+2)^(5/2),x, algorithm="fricas")

[Out]

[1/40*((8*b^5*x^4 + 42*b^4*x^3 + 62*b^3*x^2 + 5*b^2*x - 15*b)*sqrt(b*x + 2)*sqrt(x) + 15*sqrt(b)*log(b*x + sqr
t(b*x + 2)*sqrt(b)*sqrt(x) + 1))/b^3, 1/40*((8*b^5*x^4 + 42*b^4*x^3 + 62*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^3]

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Sympy [A]  time = 34.2153, size = 138, normalized size = 1.12 \begin{align*} \frac{b^{3} x^{\frac{11}{2}}}{5 \sqrt{b x + 2}} + \frac{29 b^{2} x^{\frac{9}{2}}}{20 \sqrt{b x + 2}} + \frac{73 b x^{\frac{7}{2}}}{20 \sqrt{b x + 2}} + \frac{129 x^{\frac{5}{2}}}{40 \sqrt{b x + 2}} - \frac{x^{\frac{3}{2}}}{8 b \sqrt{b x + 2}} - \frac{3 \sqrt{x}}{4 b^{2} \sqrt{b x + 2}} + \frac{3 \operatorname{asinh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{4 b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(b*x+2)**(5/2),x)

[Out]

b**3*x**(11/2)/(5*sqrt(b*x + 2)) + 29*b**2*x**(9/2)/(20*sqrt(b*x + 2)) + 73*b*x**(7/2)/(20*sqrt(b*x + 2)) + 12
9*x**(5/2)/(40*sqrt(b*x + 2)) - x**(3/2)/(8*b*sqrt(b*x + 2)) - 3*sqrt(x)/(4*b**2*sqrt(b*x + 2)) + 3*asinh(sqrt
(2)*sqrt(b)*sqrt(x)/2)/(4*b**(5/2))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+2)^(5/2),x, algorithm="giac")

[Out]

Timed out