\(\int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx\) [537]

Optimal result

Integrand size = 15, antiderivative size = 58 \[ \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx=3 b \sqrt {x} \sqrt {2+b x}-\frac {2 (2+b x)^{3/2}}{\sqrt {x}}+6 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {49, 52, 56, 221} \[ \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx=6 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )-\frac {2 (b x+2)^{3/2}}{\sqrt {x}}+3 b \sqrt {x} \sqrt {b x+2} \]

[In]

Int[(2 + b*x)^(3/2)/x^(3/2),x]

[Out]

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

Rule 49

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 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*} \text {integral}& = -\frac {2 (2+b x)^{3/2}}{\sqrt {x}}+(3 b) \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx \\ & = 3 b \sqrt {x} \sqrt {2+b x}-\frac {2 (2+b x)^{3/2}}{\sqrt {x}}+(3 b) \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx \\ & = 3 b \sqrt {x} \sqrt {2+b x}-\frac {2 (2+b x)^{3/2}}{\sqrt {x}}+(6 b) \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = 3 b \sqrt {x} \sqrt {2+b x}-\frac {2 (2+b x)^{3/2}}{\sqrt {x}}+6 \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02 \[ \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx=\frac {(-4+b x) \sqrt {2+b x}}{\sqrt {x}}-12 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right ) \]

[In]

Integrate[(2 + b*x)^(3/2)/x^(3/2),x]

[Out]

((-4 + b*x)*Sqrt[2 + b*x])/Sqrt[x] - 12*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 + b*x])]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95

method	result	size
meijerg	\(\frac {3 \sqrt {b}\, \left (-\frac {8 \sqrt {\pi }\, \sqrt {2}\, \left (-\frac {b x}{4}+1\right ) \sqrt {\frac {b x}{2}+1}}{3 \sqrt {x}\, \sqrt {b}}+4 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )\right )}{2 \sqrt {\pi }}\)	\(55\)
risch	\(\frac {b^{2} x^{2}-2 b x -8}{\sqrt {x}\, \sqrt {b x +2}}+\frac {3 \sqrt {b}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\)	\(72\)

[In]

int((b*x+2)^(3/2)/x^(3/2),x,method=_RETURNVERBOSE)

[Out]

3/2*b^(1/2)/Pi^(1/2)*(-8/3*Pi^(1/2)/x^(1/2)*2^(1/2)/b^(1/2)*(-1/4*b*x+1)*(1/2*b*x+1)^(1/2)+4*Pi^(1/2)*arcsinh(
1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.71 \[ \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx=\left [\frac {3 \, \sqrt {b} x \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + \sqrt {b x + 2} {\left (b x - 4\right )} \sqrt {x}}{x}, -\frac {6 \, \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - \sqrt {b x + 2} {\left (b x - 4\right )} \sqrt {x}}{x}\right ] \]

[In]

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

[Out]

[(3*sqrt(b)*x*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + sqrt(b*x + 2)*(b*x - 4)*sqrt(x))/x, -(6*sqrt(-b)*
x*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) - sqrt(b*x + 2)*(b*x - 4)*sqrt(x))/x]

Sympy [A] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.26 \[ \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx=6 \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )} + \frac {b^{2} x^{\frac {3}{2}}}{\sqrt {b x + 2}} - \frac {2 b \sqrt {x}}{\sqrt {b x + 2}} - \frac {8}{\sqrt {x} \sqrt {b x + 2}} \]

[In]

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

[Out]

6*sqrt(b)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2) + b**2*x**(3/2)/sqrt(b*x + 2) - 2*b*sqrt(x)/sqrt(b*x + 2) - 8/(sqrt
(x)*sqrt(b*x + 2))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.40 \[ \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx=-3 \, \sqrt {b} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right ) - \frac {4 \, \sqrt {b x + 2}}{\sqrt {x}} - \frac {2 \, \sqrt {b x + 2} b}{{\left (b - \frac {b x + 2}{x}\right )} \sqrt {x}} \]

[In]

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

[Out]

-3*sqrt(b)*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x))) - 4*sqrt(b*x + 2)/sqrt(x)
- 2*sqrt(b*x + 2)*b/((b - (b*x + 2)/x)*sqrt(x))

Giac [A] (verification not implemented)

none

Time = 6.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.17 \[ \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx=\frac {{\left (\frac {\sqrt {b x + 2} {\left (b x - 4\right )}}{\sqrt {{\left (b x + 2\right )} b - 2 \, b}} - \frac {6 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{\sqrt {b}}\right )} b^{2}}{{\left | b \right |}} \]

[In]

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

[Out]

(sqrt(b*x + 2)*(b*x - 4)/sqrt((b*x + 2)*b - 2*b) - 6*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 2)*b - 2*b))
)/sqrt(b))*b^2/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(2+b x)^{3/2}}{x^{3/2}} \, dx=\int \frac {{\left (b\,x+2\right )}^{3/2}}{x^{3/2}} \,d x \]

[In]

int((b*x + 2)^(3/2)/x^(3/2),x)

[Out]

int((b*x + 2)^(3/2)/x^(3/2), x)