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

Optimal. Leaf size=65 \[ \frac {2 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}-\frac {2 \sqrt {x}}{b^2 \sqrt {b x+2}}-\frac {2 x^{3/2}}{3 b (b x+2)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {47, 54, 215} \[ -\frac {2 \sqrt {x}}{b^2 \sqrt {b x+2}}+\frac {2 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}-\frac {2 x^{3/2}}{3 b (b x+2)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(2 + b*x)^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.07, size = 52, normalized size = 0.80 \[ \frac {2 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}-\frac {4 \sqrt {x} (2 b x+3)}{3 b^2 (b x+2)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(2 + b*x)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 171, normalized size = 2.63 \[ \left [\frac {3 \, {\left (b^{2} x^{2} + 4 \, b x + 4\right )} \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) - 4 \, {\left (2 \, b^{2} x + 3 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b^{5} x^{2} + 4 \, b^{4} x + 4 \, b^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{2} x^{2} + 4 \, b x + 4\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) + 2 \, {\left (2 \, b^{2} x + 3 \, b\right )} \sqrt {b x + 2} \sqrt {x}\right )}}{3 \, {\left (b^{5} x^{2} + 4 \, b^{4} x + 4 \, b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 10.77, size = 154, normalized size = 2.37 \[ -\frac {{\left (\frac {3 \, \log \left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2}\right )}{\sqrt {b}} + \frac {16 \, {\left (3 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{4} \sqrt {b} + 6 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} b^{\frac {3}{2}} + 8 \, b^{\frac {5}{2}}\right )}}{{\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )}^{3}}\right )} {\left | b \right |}}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*log((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2)/sqrt(b) + 16*(3*(sqrt(b*x + 2)*sqrt(b) - sqrt
((b*x + 2)*b - 2*b))^4*sqrt(b) + 6*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2*b^(3/2) + 8*b^(5/2))/((
sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)^3)*abs(b)/b^3

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maple [A]  time = 0.04, size = 55, normalized size = 0.85 \[ \frac {-\frac {\sqrt {\pi }\, \sqrt {2}\, \left (10 b x +15\right ) \sqrt {b}\, \sqrt {x}}{15 \left (\frac {b x}{2}+1\right )^{\frac {3}{2}}}+2 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {2}\, \sqrt {b}\, \sqrt {x}}{2}\right )}{\sqrt {\pi }\, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(b*x+2)^(5/2),x)

[Out]

4/3/b^(5/2)/Pi^(1/2)*(-1/20*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(10*b*x+15)/(1/2*b*x+1)^(3/2)+3/2*Pi^(1/2)*arcsin
h(1/2*2^(1/2)*b^(1/2)*x^(1/2)))

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maxima [A]  time = 2.97, size = 69, normalized size = 1.06 \[ -\frac {2 \, {\left (b + \frac {3 \, {\left (b x + 2\right )}}{x}\right )} x^{\frac {3}{2}}}{3 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{2}} - \frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(b + 3*(b*x + 2)/x)*x^(3/2)/((b*x + 2)^(3/2)*b^2) - log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqr
t(b*x + 2)/sqrt(x)))/b^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^{3/2}}{{\left (b\,x+2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(b*x + 2)^(5/2),x)

[Out]

int(x^(3/2)/(b*x + 2)^(5/2), x)

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sympy [B]  time = 3.58, size = 257, normalized size = 3.95 \[ - \frac {8 b^{\frac {11}{2}} x^{8}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x + 2} + 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x + 2}} - \frac {12 b^{\frac {9}{2}} x^{7}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x + 2} + 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x + 2}} + \frac {6 b^{5} x^{\frac {15}{2}} \sqrt {b x + 2} \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x + 2} + 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x + 2}} + \frac {12 b^{4} x^{\frac {13}{2}} \sqrt {b x + 2} \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x + 2} + 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x + 2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*b**(11/2)*x**8/(3*b**(15/2)*x**(15/2)*sqrt(b*x + 2) + 6*b**(13/2)*x**(13/2)*sqrt(b*x + 2)) - 12*b**(9/2)*x*
*7/(3*b**(15/2)*x**(15/2)*sqrt(b*x + 2) + 6*b**(13/2)*x**(13/2)*sqrt(b*x + 2)) + 6*b**5*x**(15/2)*sqrt(b*x + 2
)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(15/2)*x**(15/2)*sqrt(b*x + 2) + 6*b**(13/2)*x**(13/2)*sqrt(b*x + 2))
+ 12*b**4*x**(13/2)*sqrt(b*x + 2)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(15/2)*x**(15/2)*sqrt(b*x + 2) + 6*b
**(13/2)*x**(13/2)*sqrt(b*x + 2))

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