\(\int (a+b x) \sqrt {c+d x} \, dx\) [1379]

Optimal result

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

[Out]

-2/3*(-a*d+b*c)*(d*x+c)^(3/2)/d^2+2/5*b*(d*x+c)^(5/2)/d^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} \[ \int (a+b x) \sqrt {c+d x} \, dx=\frac {2 b (c+d x)^{5/2}}{5 d^2}-\frac {2 (c+d x)^{3/2} (b c-a d)}{3 d^2} \]

[In]

Int[(a + b*x)*Sqrt[c + d*x],x]

[Out]

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

Mathematica [A] (verified)

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

[In]

Integrate[(a + b*x)*Sqrt[c + d*x],x]

[Out]

(2*(c + d*x)^(3/2)*(-2*b*c + 5*a*d + 3*b*d*x))/(15*d^2)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64

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

[In]

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

[Out]

2/15*(d*x+c)^(3/2)*(3*b*d*x+5*a*d-2*b*c)/d^2

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.21 \[ \int (a+b x) \sqrt {c+d x} \, dx=\begin {cases} \frac {2 \left (\frac {b \left (c + d x\right )^{\frac {5}{2}}}{5 d} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (a d - b c\right )}{3 d}\right )}{d} & \text {for}\: d \neq 0 \\\sqrt {c} \left (a x + \frac {b x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79 \[ \int (a+b x) \sqrt {c+d x} \, dx=\frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b - 5 \, {\left (b c - a d\right )} {\left (d x + c\right )}^{\frac {3}{2}}\right )}}{15 \, d^{2}} \]

[In]

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

[Out]

2/15*(3*(d*x + c)^(5/2)*b - 5*(b*c - a*d)*(d*x + c)^(3/2))/d^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 = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.38 \[ \int (a+b x) \sqrt {c+d x} \, dx=\frac {2 \, {\left (15 \, \sqrt {d x + c} a c + 5 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a + \frac {5 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} b c}{d} + \frac {{\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} b}{d}\right )}}{15 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.69 \[ \int (a+b x) \sqrt {c+d x} \, dx=\frac {2\,{\left (c+d\,x\right )}^{3/2}\,\left (5\,a\,d-5\,b\,c+3\,b\,\left (c+d\,x\right )\right )}{15\,d^2} \]

[In]

int((a + b*x)*(c + d*x)^(1/2),x)

[Out]

(2*(c + d*x)^(3/2)*(5*a*d - 5*b*c + 3*b*(c + d*x)))/(15*d^2)