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

Optimal. Leaf size=86 \[ \frac {15 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {15 \sqrt {x} \sqrt {b x+2}}{2 b^3}+\frac {5 x^{3/2} \sqrt {b x+2}}{2 b^2}-\frac {2 x^{5/2}}{b \sqrt {b x+2}} \]

[Out]

15*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-2*x^(5/2)/b/(b*x+2)^(1/2)+5/2*x^(3/2)*(b*x+2)^(1/2)/b^2-15/2*x
^(1/2)*(b*x+2)^(1/2)/b^3

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Rubi [A]  time = 0.02, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {47, 50, 54, 215} \[ \frac {5 x^{3/2} \sqrt {b x+2}}{2 b^2}-\frac {15 \sqrt {x} \sqrt {b x+2}}{2 b^3}+\frac {15 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {2 x^{5/2}}{b \sqrt {b x+2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(2 + b*x)^(3/2),x]

[Out]

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

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 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}}{(2+b x)^{3/2}} \, dx &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx}{b}\\ &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}-\frac {15 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{2 b^2}\\ &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{2 b^3}\\ &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 30, normalized size = 0.35 \[ \frac {x^{7/2} \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};-\frac {b x}{2}\right )}{7 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(2 + b*x)^(3/2),x]

[Out]

(x^(7/2)*Hypergeometric2F1[3/2, 7/2, 9/2, -1/2*(b*x)])/(7*Sqrt[2])

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fricas [A]  time = 0.46, size = 152, normalized size = 1.77 \[ \left [\frac {15 \, {\left (b x + 2\right )} \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (b^{3} x^{2} - 5 \, b^{2} x - 30 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x + 2 \, b^{4}\right )}}, -\frac {30 \, {\left (b x + 2\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (b^{3} x^{2} - 5 \, b^{2} x - 30 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x + 2 \, b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 11.12, size = 119, normalized size = 1.38 \[ \frac {{\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left (\frac {b x + 2}{b^{3}} - \frac {9}{b^{3}}\right )} - \frac {15 \, \log \left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2}\right )}{b^{\frac {5}{2}}} - \frac {64}{{\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )} b^{\frac {3}{2}}}\right )} {\left | b \right |}}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2)*((b*x + 2)/b^3 - 9/b^3) - 15*log((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x
+ 2)*b - 2*b))^2)/b^(5/2) - 64/(((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)*b^(3/2)))*abs(b)/
b^2

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maple [A]  time = 0.03, size = 106, normalized size = 1.23 \[ \frac {\left (\frac {15 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 b^{\frac {7}{2}}}-\frac {8 \sqrt {\left (x +\frac {2}{b}\right )^{2} b -2 x -\frac {4}{b}}}{\left (x +\frac {2}{b}\right ) b^{4}}\right ) \sqrt {\left (b x +2\right ) x}}{\sqrt {b x +2}\, \sqrt {x}}+\frac {\left (b x -7\right ) \sqrt {b x +2}\, \sqrt {x}}{2 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x-7)*(b*x+2)^(1/2)*x^(1/2)/b^3+(15/2/b^(7/2)*ln((b*x+1)/b^(1/2)+(b*x^2+2*x)^(1/2))-8/b^4/(x+2/b)*(b*(x+
2/b)^2-2*x-4/b)^(1/2))*((b*x+2)*x)^(1/2)/(b*x+2)^(1/2)/x^(1/2)

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maxima [A]  time = 3.06, size = 119, normalized size = 1.38 \[ -\frac {8 \, b^{2} - \frac {25 \, {\left (b x + 2\right )} b}{x} + \frac {15 \, {\left (b x + 2\right )}^{2}}{x^{2}}}{\frac {\sqrt {b x + 2} b^{5}}{\sqrt {x}} - \frac {2 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}} - \frac {15 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(8*b^2 - 25*(b*x + 2)*b/x + 15*(b*x + 2)^2/x^2)/(sqrt(b*x + 2)*b^5/sqrt(x) - 2*(b*x + 2)^(3/2)*b^4/x^(3/2) +
(b*x + 2)^(5/2)*b^3/x^(5/2)) - 15/2*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/
b^(7/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{5/2}}{{\left (b\,x+2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(b*x + 2)^(3/2),x)

[Out]

int(x^(5/2)/(b*x + 2)^(3/2), x)

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sympy [A]  time = 7.10, size = 80, normalized size = 0.93 \[ \frac {x^{\frac {5}{2}}}{2 b \sqrt {b x + 2}} - \frac {5 x^{\frac {3}{2}}}{2 b^{2} \sqrt {b x + 2}} - \frac {15 \sqrt {x}}{b^{3} \sqrt {b x + 2}} + \frac {15 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**(5/2)/(2*b*sqrt(b*x + 2)) - 5*x**(3/2)/(2*b**2*sqrt(b*x + 2)) - 15*sqrt(x)/(b**3*sqrt(b*x + 2)) + 15*asinh(
sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2)

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