3.7.8 \(\int \frac {x^{5/2}}{\sqrt {2+b x}} \, dx\) [608]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {52, 56, 221} \begin {gather*} -\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} \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 52

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 56

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 221

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 \text {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*}

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Mathematica [A]
time = 0.07, size = 66, normalized size = 0.75 \begin {gather*} \frac {\sqrt {x} \sqrt {2+b x} \left (15-5 b x+2 b^2 x^2\right )}{6 b^3}+\frac {5 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\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*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]])/b^(7/2)

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Mathics [A]
time = 10.12, size = 75, normalized size = 0.85 \begin {gather*} \frac {-5 \text {ArcSinh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ]}{b^{\frac {7}{2}}}+\frac {5 \sqrt {x}}{b^3 \sqrt {2+b x}}+\frac {5 x^{\frac {3}{2}}}{6 b^2 \sqrt {2+b x}}-\frac {x^{\frac {5}{2}}}{6 b \sqrt {2+b x}}+\frac {x^{\frac {7}{2}}}{3 \sqrt {2+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^(5/2)/Sqrt[2 + b*x],x]')

[Out]

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

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Maple [A]
time = 0.13, size = 104, normalized size = 1.18

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (14 x^{2} b^{2}-35 b x +105\right ) \sqrt {\frac {b x}{2}+1}}{42}-5 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\)	\(63\)
risch	\(\frac {\left (2 x^{2} b^{2}-5 b x +15\right ) \sqrt {x}\, \sqrt {b x +2}}{6 b^{3}}-\frac {5 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right ) \sqrt {x \left (b x +2\right )}}{2 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(77\)
default	\(\frac {x^{\frac {5}{2}} \sqrt {b x +2}}{3 b}-\frac {5 \left (\frac {x^{\frac {3}{2}} \sqrt {b x +2}}{2 b}-\frac {3 \left (\frac {\sqrt {x}\, \sqrt {b x +2}}{b}-\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{b^{\frac {3}{2}} \sqrt {b x +2}\, \sqrt {x}}\right )}{2 b}\right )}{3 b}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^(5/2)*(b*x+2)^(1/2)/b-5/3/b*(1/2*x^(3/2)*(b*x+2)^(1/2)/b-3/2/b*(x^(1/2)*(b*x+2)^(1/2)/b-1/b^(3/2)*(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (63) = 126\).
time = 0.34, 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)

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Fricas [A]
time = 0.32, 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]

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Sympy [A]
time = 8.85, 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)

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Giac [A]
time = 0.00, size = 116, normalized size = 1.32 \begin {gather*} 2 \left (2 \left (\left (\frac {\frac {1}{72}\cdot 6 b^{4} \sqrt {x} \sqrt {x}}{b^{5}}-\frac {\frac {1}{72}\cdot 15 b^{3}}{b^{5}}\right ) \sqrt {x} \sqrt {x}+\frac {\frac {1}{72}\cdot 45 b^{2}}{b^{5}}\right ) \sqrt {x} \sqrt {b x+2}+\frac {5 \ln \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{2 b^{3} \sqrt {b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)

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