\(\int x^{3/2} \sqrt {a+b x} \, dx\) [490]

Optimal result

Integrand size = 15, antiderivative size = 98 \[ \int x^{3/2} \sqrt {a+b x} \, dx=-\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b^2}+\frac {a x^{3/2} \sqrt {a+b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a+b x}+\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{5/2}} \]

[Out]

1/8*a^3*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)+1/12*a*x^(3/2)*(b*x+a)^(1/2)/b+1/3*x^(5/2)*(b*x+a)^(1/2
)-1/8*a^2*x^(1/2)*(b*x+a)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {52, 65, 223, 212} \[ \int x^{3/2} \sqrt {a+b x} \, dx=\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{5/2}}-\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b^2}+\frac {a x^{3/2} \sqrt {a+b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a+b x} \]

[In]

Int[x^(3/2)*Sqrt[a + b*x],x]

[Out]

-1/8*(a^2*Sqrt[x]*Sqrt[a + b*x])/b^2 + (a*x^(3/2)*Sqrt[a + b*x])/(12*b) + (x^(5/2)*Sqrt[a + b*x])/3 + (a^3*Arc
Tanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*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 {1}{3} x^{5/2} \sqrt {a+b x}+\frac {1}{6} a \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx \\ & = \frac {a x^{3/2} \sqrt {a+b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a+b x}-\frac {a^2 \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{8 b} \\ & = -\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b^2}+\frac {a x^{3/2} \sqrt {a+b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a+b x}+\frac {a^3 \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{16 b^2} \\ & = -\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b^2}+\frac {a x^{3/2} \sqrt {a+b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a+b x}+\frac {a^3 \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^2} \\ & = -\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b^2}+\frac {a x^{3/2} \sqrt {a+b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a+b x}+\frac {a^3 \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^2} \\ & = -\frac {a^2 \sqrt {x} \sqrt {a+b x}}{8 b^2}+\frac {a x^{3/2} \sqrt {a+b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a+b x}+\frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.87 \[ \int x^{3/2} \sqrt {a+b x} \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (-3 a^2+2 a b x+8 b^2 x^2\right )+6 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{24 b^{5/2}} \]

[In]

Integrate[x^(3/2)*Sqrt[a + b*x],x]

[Out]

(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-3*a^2 + 2*a*b*x + 8*b^2*x^2) + 6*a^3*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sq
rt[a + b*x])])/(24*b^(5/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {\left (-8 b^{2} x^{2}-2 a b x +3 a^{2}\right ) \sqrt {x}\, \sqrt {b x +a}}{24 b^{2}}+\frac {a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{16 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(87\)
default	\(\frac {x^{\frac {3}{2}} \left (b x +a \right )^{\frac {3}{2}}}{3 b}-\frac {a \left (\frac {\sqrt {x}\, \left (b x +a \right )^{\frac {3}{2}}}{2 b}-\frac {a \left (\sqrt {x}\, \sqrt {b x +a}+\frac {a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b x +a}\, \sqrt {x}\, \sqrt {b}}\right )}{4 b}\right )}{2 b}\)	\(106\)

[In]

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

[Out]

-1/24*(-8*b^2*x^2-2*a*b*x+3*a^2)*x^(1/2)*(b*x+a)^(1/2)/b^2+1/16*a^3/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 = 0.23 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.44 \[ \int x^{3/2} \sqrt {a+b x} \, dx=\left [\frac {3 \, a^{3} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 2 \, a b^{2} x - 3 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{3}}, -\frac {3 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, b^{3} x^{2} + 2 \, a b^{2} x - 3 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{3}}\right ] \]

[In]

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

[Out]

[1/48*(3*a^3*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*b^3*x^2 + 2*a*b^2*x - 3*a^2*b)*sq
rt(b*x + a)*sqrt(x))/b^3, -1/24*(3*a^3*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*b^3*x^2 + 2*a*
b^2*x - 3*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^3]

Sympy [A] (verification not implemented)

Time = 8.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.24 \[ \int x^{3/2} \sqrt {a+b x} \, dx=- \frac {a^{\frac {5}{2}} \sqrt {x}}{8 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b \sqrt {1 + \frac {b x}{a}}} + \frac {5 \sqrt {a} x^{\frac {5}{2}}}{12 \sqrt {1 + \frac {b x}{a}}} + \frac {a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {5}{2}}} + \frac {b x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

[In]

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

[Out]

-a**(5/2)*sqrt(x)/(8*b**2*sqrt(1 + b*x/a)) - a**(3/2)*x**(3/2)/(24*b*sqrt(1 + b*x/a)) + 5*sqrt(a)*x**(5/2)/(12
*sqrt(1 + b*x/a)) + a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(5/2)) + b*x**(7/2)/(3*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. 146 vs. \(2 (70) = 140\).

Time = 0.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.49 \[ \int x^{3/2} \sqrt {a+b x} \, dx=-\frac {a^{3} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {\frac {3 \, \sqrt {b x + a} a^{3} b^{2}}{\sqrt {x}} + \frac {8 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b}{x^{\frac {3}{2}}} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3}}{x^{\frac {5}{2}}}}{24 \, {\left (b^{5} - \frac {3 \, {\left (b x + a\right )} b^{4}}{x} + \frac {3 \, {\left (b x + a\right )}^{2} b^{3}}{x^{2}} - \frac {{\left (b x + a\right )}^{3} b^{2}}{x^{3}}\right )}} \]

[In]

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

[Out]

-1/16*a^3*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/b^(5/2) - 1/24*(3*sqrt(b*x
+ a)*a^3*b^2/sqrt(x) + 8*(b*x + a)^(3/2)*a^3*b/x^(3/2) - 3*(b*x + a)^(5/2)*a^3/x^(5/2))/(b^5 - 3*(b*x + a)*b^
4/x + 3*(b*x + a)^2*b^3/x^2 - (b*x + a)^3*b^2/x^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (70) = 140\).

Time = 158.64 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.86 \[ \int x^{3/2} \sqrt {a+b x} \, dx=\frac {\frac {{\left (\frac {15 \, a^{3} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}} + \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} - \frac {13 \, a}{b^{2}}\right )} + \frac {33 \, a^{2}}{b^{2}}\right )}\right )} {\left | b \right |}}{b} - \frac {6 \, {\left (3 \, a^{2} \sqrt {b} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) - \sqrt {{\left (b x + a\right )} b - a b} {\left (2 \, b x - 3 \, a\right )} \sqrt {b x + a}\right )} a {\left | b \right |}}{b^{3}}}{24 \, b} \]

[In]

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

[Out]

1/24*((15*a^3*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^(3/2) + sqrt((b*x + a)*b - a*b)*sqr
t(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 - 13*a/b^2) + 33*a^2/b^2))*abs(b)/b - 6*(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))*a*abs(b)/b
^3)/b

Mupad [F(-1)]

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

[In]

int(x^(3/2)*(a + b*x)^(1/2),x)

[Out]

int(x^(3/2)*(a + b*x)^(1/2), x)