\(\int \sqrt {a+b x} (A+B x) \, dx\) [391]

Optimal result

Integrand size = 15, antiderivative size = 42 \[ \int \sqrt {a+b x} (A+B x) \, dx=\frac {2 (A b-a B) (a+b x)^{3/2}}{3 b^2}+\frac {2 B (a+b x)^{5/2}}{5 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \sqrt {a+b x} (A+B x) \, dx=\frac {2 (a+b x)^{3/2} (A b-a B)}{3 b^2}+\frac {2 B (a+b x)^{5/2}}{5 b^2} \]

[In]

Int[Sqrt[a + b*x]*(A + B*x),x]

[Out]

(2*(A*b - a*B)*(a + b*x)^(3/2))/(3*b^2) + (2*B*(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 b-a B) \sqrt {a+b x}}{b}+\frac {B (a+b x)^{3/2}}{b}\right ) \, dx \\ & = \frac {2 (A b-a B) (a+b x)^{3/2}}{3 b^2}+\frac {2 B (a+b x)^{5/2}}{5 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \sqrt {a+b x} (A+B x) \, dx=\frac {2 (a+b x)^{3/2} (5 A b-2 a B+3 b B x)}{15 b^2} \]

[In]

Integrate[Sqrt[a + b*x]*(A + B*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64

method	result	size
gosper	\(\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (3 b B x +5 A b -2 B a \right )}{15 b^{2}}\)	\(27\)
pseudoelliptic	\(\frac {2 \left (\left (3 B x +5 A \right ) b -2 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{15 b^{2}}\)	\(28\)
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (A b -B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{3}}{b^{2}}\)	\(34\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (A b -B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{3}}{b^{2}}\)	\(34\)
trager	\(\frac {2 \left (3 b^{2} B \,x^{2}+5 A \,b^{2} x +B a b x +5 a b A -2 a^{2} B \right ) \sqrt {b x +a}}{15 b^{2}}\)	\(46\)
risch	\(\frac {2 \left (3 b^{2} B \,x^{2}+5 A \,b^{2} x +B a b x +5 a b A -2 a^{2} B \right ) \sqrt {b x +a}}{15 b^{2}}\)	\(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10 \[ \int \sqrt {a+b x} (A+B x) \, dx=\frac {2 \, {\left (3 \, B b^{2} x^{2} - 2 \, B a^{2} + 5 \, A a b + {\left (B a b + 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}}{15 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.21 \[ \int \sqrt {a+b x} (A+B x) \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {5}{2}}}{5 b} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (A b - B a\right )}{3 b}\right )}{b} & \text {for}\: b \neq 0 \\\sqrt {a} \left (A x + \frac {B x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79 \[ \int \sqrt {a+b x} (A+B x) \, dx=\frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B - 5 \, {\left (B a - A b\right )} {\left (b x + a\right )}^{\frac {3}{2}}\right )}}{15 \, b^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (34) = 68\).

Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.38 \[ \int \sqrt {a+b x} (A+B x) \, dx=\frac {2 \, {\left (15 \, \sqrt {b x + a} A a + 5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A + \frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} B a}{b} + \frac {{\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} B}{b}\right )}}{15 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+b x} (A+B x) \, dx=\frac {2\,{\left (a+b\,x\right )}^{3/2}\,\left (5\,A\,b-5\,B\,a+3\,B\,\left (a+b\,x\right )\right )}{15\,b^2} \]

[In]

int((A + B*x)*(a + b*x)^(1/2),x)

[Out]

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