\(\int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx\) [426]

Optimal result

Integrand size = 16, antiderivative size = 65 \[ \int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx=-\frac {2 a (A b-a B) \sqrt {a+b x}}{b^3}+\frac {2 (A b-2 a B) (a+b x)^{3/2}}{3 b^3}+\frac {2 B (a+b x)^{5/2}}{5 b^3} \]

[Out]

2/3*(A*b-2*B*a)*(b*x+a)^(3/2)/b^3+2/5*B*(b*x+a)^(5/2)/b^3-2*a*(A*b-B*a)*(b*x+a)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 (a+b x)^{3/2} (A b-2 a B)}{3 b^3}-\frac {2 a \sqrt {a+b x} (A b-a B)}{b^3}+\frac {2 B (a+b x)^{5/2}}{5 b^3} \]

[In]

Int[(x*(A + B*x))/Sqrt[a + b*x],x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} \left (8 a^2 B-2 a b (5 A+2 B x)+b^2 x (5 A+3 B x)\right )}{15 b^3} \]

[In]

Integrate[(x*(A + B*x))/Sqrt[a + b*x],x]

[Out]

(2*Sqrt[a + b*x]*(8*a^2*B - 2*a*b*(5*A + 2*B*x) + b^2*x*(5*A + 3*B*x)))/(15*b^3)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.63

method	result	size
pseudoelliptic	\(-\frac {4 \left (-\frac {\left (\frac {3 B x}{5}+A \right ) x \,b^{2}}{2}+a \left (\frac {2 B x}{5}+A \right ) b -\frac {4 a^{2} B}{5}\right ) \sqrt {b x +a}}{3 b^{3}}\)	\(41\)
gosper	\(-\frac {2 \sqrt {b x +a}\, \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +4 B a b x +10 a b A -8 a^{2} B \right )}{15 b^{3}}\)	\(47\)
trager	\(-\frac {2 \sqrt {b x +a}\, \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +4 B a b x +10 a b A -8 a^{2} B \right )}{15 b^{3}}\)	\(47\)
risch	\(-\frac {2 \sqrt {b x +a}\, \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +4 B a b x +10 a b A -8 a^{2} B \right )}{15 b^{3}}\)	\(47\)
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{3}-2 a \left (A b -B a \right ) \sqrt {b x +a}}{b^{3}}\)	\(52\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{3}-2 a \left (A b -B a \right ) \sqrt {b x +a}}{b^{3}}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (3 \, B b^{2} x^{2} + 8 \, B a^{2} - 10 \, A a b - {\left (4 \, B a b - 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}}{15 \, b^{3}} \]

[In]

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

[Out]

2/15*(3*B*b^2*x^2 + 8*B*a^2 - 10*A*a*b - (4*B*a*b - 5*A*b^2)*x)*sqrt(b*x + a)/b^3

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.23 \[ \int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {5}{2}}}{5 b} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (A b - 2 B a\right )}{3 b} + \frac {\sqrt {a + b x} \left (- A a b + B a^{2}\right )}{b}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{2}}{2} + \frac {B x^{3}}{3}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(B*(a + b*x)**(5/2)/(5*b) + (a + b*x)**(3/2)*(A*b - 2*B*a)/(3*b) + sqrt(a + b*x)*(-A*a*b + B*a**2
)/b)/b**2, Ne(b, 0)), ((A*x**2/2 + B*x**3/3)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B - 5 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 15 \, {\left (B a^{2} - A a b\right )} \sqrt {b x + a}\right )}}{15 \, b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A}{b} + \frac {{\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} B}{b^{2}}\right )}}{15 \, b} \]

[In]

integrate(x*(B*x+A)/(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^2)/b

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80 \[ \int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2\,\sqrt {a+b\,x}\,\left (15\,B\,a^2+3\,B\,{\left (a+b\,x\right )}^2-15\,A\,a\,b+5\,A\,b\,\left (a+b\,x\right )-10\,B\,a\,\left (a+b\,x\right )\right )}{15\,b^3} \]

[In]

int((x*(A + B*x))/(a + b*x)^(1/2),x)

[Out]

(2*(a + b*x)^(1/2)*(15*B*a^2 + 3*B*(a + b*x)^2 - 15*A*a*b + 5*A*b*(a + b*x) - 10*B*a*(a + b*x)))/(15*b^3)