\(\int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx\) [570]

Optimal result

Integrand size = 15, antiderivative size = 77 \[ \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx=-\frac {3 a \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {x^{3/2} \sqrt {a+b x}}{2 b}+\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{5/2}} \]

[Out]

3/4*a^2*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)+1/2*x^(3/2)*(b*x+a)^(1/2)/b-3/4*a*x^(1/2)*(b*x+a)^(1/2)
/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {52, 65, 223, 212} \[ \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{5/2}}-\frac {3 a \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {x^{3/2} \sqrt {a+b x}}{2 b} \]

[In]

Int[x^(3/2)/Sqrt[a + b*x],x]

[Out]

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

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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {(3 a) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{4 b} \\ & = -\frac {3 a \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {x^{3/2} \sqrt {a+b x}}{2 b}+\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{8 b^2} \\ & = -\frac {3 a \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {x^{3/2} \sqrt {a+b x}}{2 b}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^2} \\ & = -\frac {3 a \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {x^{3/2} \sqrt {a+b x}}{2 b}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^2} \\ & = -\frac {3 a \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {x^{3/2} \sqrt {a+b x}}{2 b}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} (-3 a+2 b x)+6 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{4 b^{5/2}} \]

[In]

Integrate[x^(3/2)/Sqrt[a + b*x],x]

[Out]

(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-3*a + 2*b*x) + 6*a^2*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(
4*b^(5/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99

method	result	size
risch	\(-\frac {\left (-2 b x +3 a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b^{2}}+\frac {3 a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{8 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(76\)
default	\(\frac {x^{\frac {3}{2}} \sqrt {b x +a}}{2 b}-\frac {3 a \left (\frac {\sqrt {x}\, \sqrt {b x +a}}{b}-\frac {a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +a}}\right )}{4 b}\)	\(87\)

[In]

int(x^(3/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(-2*b*x+3*a)*x^(1/2)*(b*x+a)^(1/2)/b^2+3/8*a^2/b^(5/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))*(x*(b*x+
a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.55 \[ \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{2} x - 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, b^{3}}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, b^{2} x - 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, b^{3}}\right ] \]

[In]

integrate(x^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(3*a^2*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^2*x - 3*a*b)*sqrt(b*x + a)*sqrt(
x))/b^3, -1/4*(3*a^2*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (2*b^2*x - 3*a*b)*sqrt(b*x + a)*sqr
t(x))/b^3]

Sympy [A] (verification not implemented)

Time = 3.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.30 \[ \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx=- \frac {3 a^{\frac {3}{2}} \sqrt {x}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {\sqrt {a} x^{\frac {3}{2}}}{4 b \sqrt {1 + \frac {b x}{a}}} + \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

[In]

integrate(x**(3/2)/(b*x+a)**(1/2),x)

[Out]

-3*a**(3/2)*sqrt(x)/(4*b**2*sqrt(1 + b*x/a)) - sqrt(a)*x**(3/2)/(4*b*sqrt(1 + b*x/a)) + 3*a**2*asinh(sqrt(b)*s
qrt(x)/sqrt(a))/(4*b**(5/2)) + x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/a))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (55) = 110\).

Time = 0.41 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.45 \[ \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx=-\frac {3 \, a^{2} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{8 \, b^{\frac {5}{2}}} + \frac {\frac {5 \, \sqrt {b x + a} a^{2} b}{\sqrt {x}} - \frac {3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2}}{x^{\frac {3}{2}}}}{4 \, {\left (b^{4} - \frac {2 \, {\left (b x + a\right )} b^{3}}{x} + \frac {{\left (b x + a\right )}^{2} b^{2}}{x^{2}}\right )}} \]

[In]

integrate(x^(3/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 76.58 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx=-\frac {{\left (3 \, a^{2} \sqrt {b} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) - \sqrt {{\left (b x + a\right )} b - a b} {\left (2 \, b x - 3 \, a\right )} \sqrt {b x + a}\right )} {\left | b \right |}}{4 \, b^{4}} \]

[In]

integrate(x^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/4*(3*a^2*sqrt(b)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b))) - sqrt((b*x + a)*b - a*b)*(2*b*
x - 3*a)*sqrt(b*x + a))*abs(b)/b^4

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx=\int \frac {x^{3/2}}{\sqrt {a+b\,x}} \,d x \]

[In]

int(x^(3/2)/(a + b*x)^(1/2),x)

[Out]

int(x^(3/2)/(a + b*x)^(1/2), x)