3.6.7 \(\int \sqrt {x} \sqrt {2+b x} \, dx\) [507]

Optimal. Leaf size=64 \[ \frac {\sqrt {x} \sqrt {2+b x}}{2 b}+\frac {1}{2} x^{3/2} \sqrt {2+b x}-\frac {\sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 64, 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 = {52, 56, 221} \begin {gather*} -\frac {\sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}+\frac {1}{2} x^{3/2} \sqrt {b x+2}+\frac {\sqrt {x} \sqrt {b x+2}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Sqrt[2 + b*x],x]

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 56, normalized size = 0.88 \begin {gather*} \frac {\sqrt {x} (1+b x) \sqrt {2+b x}}{2 b}+\frac {\log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*Sqrt[2 + b*x],x]

[Out]

(Sqrt[x]*(1 + b*x)*Sqrt[2 + b*x])/(2*b) + Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]]/b^(3/2)

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Mathics [A]
time = 3.57, size = 66, normalized size = 1.03 \begin {gather*} \frac {-b \text {ArcSinh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ] \left (2+b x\right )+b^{\frac {3}{2}} \sqrt {x} \sqrt {2+b x}+\frac {b^{\frac {5}{2}} x^{\frac {3}{2}} \left (3+b x\right ) \sqrt {2+b x}}{2}}{b^{\frac {5}{2}} \left (2+b x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 79, normalized size = 1.23

method	result	size
meijerg	\(-\frac {2 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (3 b x +3\right ) \sqrt {\frac {b x}{2}+1}}{12}+\frac {\sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{2}\right )}{b^{\frac {3}{2}} \sqrt {\pi }}\)	\(55\)
risch	\(\frac {\left (b x +1\right ) \sqrt {x}\, \sqrt {b x +2}}{2 b}-\frac {\ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right ) \sqrt {x \left (b x +2\right )}}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(68\)
default	\(\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {3}{2}}}{2 b}-\frac {\sqrt {x}\, \sqrt {b x +2}+\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{\sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{2 b}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (45) = 90\).
time = 0.36, size = 98, normalized size = 1.53 \begin {gather*} \frac {\frac {\sqrt {b x + 2} b}{\sqrt {x}} + \frac {{\left (b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{3} - \frac {2 \, {\left (b x + 2\right )} b^{2}}{x} + \frac {{\left (b x + 2\right )}^{2} b}{x^{2}}} + \frac {\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 {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(b*x + 2)*b/sqrt(x) + (b*x + 2)^(3/2)/x^(3/2))/(b^3 - 2*(b*x + 2)*b^2/x + (b*x + 2)^2*b/x^2) + 1/2*log(-(
sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(3/2)

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Fricas [A]
time = 0.32, size = 101, normalized size = 1.58 \begin {gather*} \left [\frac {{\left (b^{2} x + b\right )} \sqrt {b x + 2} \sqrt {x} + \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{2 \, b^{2}}, \frac {{\left (b^{2} x + b\right )} \sqrt {b x + 2} \sqrt {x} + 2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{2 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.79, size = 71, normalized size = 1.11 \begin {gather*} \frac {b x^{\frac {5}{2}}}{2 \sqrt {b x + 2}} + \frac {3 x^{\frac {3}{2}}}{2 \sqrt {b x + 2}} + \frac {\sqrt {x}}{b \sqrt {b x + 2}} - \frac {\operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 86, normalized size = 1.34 \begin {gather*} 2 \left (2 \left (\frac {\frac {1}{16}\cdot 2 b^{2} \sqrt {x} \sqrt {x}}{b^{2}}+\frac {\frac {1}{16}\cdot 2 b}{b^{2}}\right ) \sqrt {x} \sqrt {b x+2}+\frac {\ln \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{2 b \sqrt {b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b*x + 2)*(x + 1/b)*sqrt(x) + log(-sqrt(b)*sqrt(x) + sqrt(b*x + 2))/b^(3/2)

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Mupad [B]
time = 0.10, size = 46, normalized size = 0.72 \begin {gather*} \sqrt {x}\,\left (\frac {x}{2}+\frac {1}{2\,b}\right )\,\sqrt {b\,x+2}-\frac {\ln \left (b\,x+\sqrt {b}\,\sqrt {x}\,\sqrt {b\,x+2}+1\right )}{2\,b^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(b*x + 2)^(1/2),x)

[Out]

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

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