\(\int x \sqrt {a+b x} \, dx\) [286]

Optimal result

Integrand size = 11, antiderivative size = 34 \[ \int x \sqrt {a+b x} \, dx=-\frac {2 a (a+b x)^{3/2}}{3 b^2}+\frac {2 (a+b x)^{5/2}}{5 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int x \sqrt {a+b x} \, dx=\frac {2 (a+b x)^{5/2}}{5 b^2}-\frac {2 a (a+b x)^{3/2}}{3 b^2} \]

[In]

Int[x*Sqrt[a + b*x],x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int x \sqrt {a+b x} \, dx=\frac {2 \sqrt {a+b x} \left (-2 a^2+a b x+3 b^2 x^2\right )}{15 b^2} \]

[In]

Integrate[x*Sqrt[a + b*x],x]

[Out]

(2*Sqrt[a + b*x]*(-2*a^2 + a*b*x + 3*b^2*x^2))/(15*b^2)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62

method	result	size
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-3 b x +2 a \right )}{15 b^{2}}\)	\(21\)
pseudoelliptic	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-3 b x +2 a \right )}{15 b^{2}}\)	\(21\)
derivativedivides	\(\frac {\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5}-\frac {2 a \left (b x +a \right )^{\frac {3}{2}}}{3}}{b^{2}}\)	\(26\)
default	\(\frac {\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5}-\frac {2 a \left (b x +a \right )^{\frac {3}{2}}}{3}}{b^{2}}\)	\(26\)
trager	\(-\frac {2 \left (-3 b^{2} x^{2}-a b x +2 a^{2}\right ) \sqrt {b x +a}}{15 b^{2}}\)	\(32\)
risch	\(-\frac {2 \left (-3 b^{2} x^{2}-a b x +2 a^{2}\right ) \sqrt {b x +a}}{15 b^{2}}\)	\(32\)

[In]

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

[Out]

-2/15*(b*x+a)^(3/2)*(-3*b*x+2*a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int x \sqrt {a+b x} \, dx=\frac {2 \, {\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{15 \, b^{2}} \]

[In]

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

[Out]

2/15*(3*b^2*x^2 + a*b*x - 2*a^2)*sqrt(b*x + a)/b^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (31) = 62\).

Time = 0.74 (sec) , antiderivative size = 202, normalized size of antiderivative = 5.94 \[ \int x \sqrt {a+b x} \, dx=- \frac {4 a^{\frac {9}{2}} \sqrt {1 + \frac {b x}{a}}}{15 a^{2} b^{2} + 15 a b^{3} x} + \frac {4 a^{\frac {9}{2}}}{15 a^{2} b^{2} + 15 a b^{3} x} - \frac {2 a^{\frac {7}{2}} b x \sqrt {1 + \frac {b x}{a}}}{15 a^{2} b^{2} + 15 a b^{3} x} + \frac {4 a^{\frac {7}{2}} b x}{15 a^{2} b^{2} + 15 a b^{3} x} + \frac {8 a^{\frac {5}{2}} b^{2} x^{2} \sqrt {1 + \frac {b x}{a}}}{15 a^{2} b^{2} + 15 a b^{3} x} + \frac {6 a^{\frac {3}{2}} b^{3} x^{3} \sqrt {1 + \frac {b x}{a}}}{15 a^{2} b^{2} + 15 a b^{3} x} \]

[In]

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

[Out]

-4*a**(9/2)*sqrt(1 + b*x/a)/(15*a**2*b**2 + 15*a*b**3*x) + 4*a**(9/2)/(15*a**2*b**2 + 15*a*b**3*x) - 2*a**(7/2
)*b*x*sqrt(1 + b*x/a)/(15*a**2*b**2 + 15*a*b**3*x) + 4*a**(7/2)*b*x/(15*a**2*b**2 + 15*a*b**3*x) + 8*a**(5/2)*
b**2*x**2*sqrt(1 + b*x/a)/(15*a**2*b**2 + 15*a*b**3*x) + 6*a**(3/2)*b**3*x**3*sqrt(1 + b*x/a)/(15*a**2*b**2 +
15*a*b**3*x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int x \sqrt {a+b x} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}}}{5 \, b^{2}} - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} a}{3 \, b^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (26) = 52\).

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.94 \[ \int x \sqrt {a+b x} \, dx=\frac {2 \, {\left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a}{b} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}}{b}\right )}}{15 \, b} \]

[In]

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

[Out]

2/15*(5*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a/b + (3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x +
a)*a^2)/b)/b

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int x \sqrt {a+b x} \, dx=-\frac {10\,a\,{\left (a+b\,x\right )}^{3/2}-6\,{\left (a+b\,x\right )}^{5/2}}{15\,b^2} \]

[In]

int(x*(a + b*x)^(1/2),x)

[Out]

-(10*a*(a + b*x)^(3/2) - 6*(a + b*x)^(5/2))/(15*b^2)